Dataset Viewer
competition_id
string | problem_id
int64 | difficulty
int64 | category
string | problem_type
string | problem
string | solutions
sequence | solutions_count
int64 | source_file
string | competition
string |
|---|---|---|---|---|---|---|---|---|---|
1992_AHSME_Problems | 20 | 0 | Geometry | Multiple Choice | [asy] draw((1,0)--(2*cos(pi/8),2*sin(pi/8))--(cos(pi/4),sin(pi/4))--(2*cos(3*pi/8),2*sin(3*pi/8))--(cos(pi/2),sin(pi/2))--(2*cos(5*pi/8),2*sin(5*pi/8))--(cos(3*pi/4),sin(3*pi/4))--(2*cos(7*pi/8),2*sin(7*pi/8))--(-1,0),black+linewidth(.75)); MP("A_1",(2*cos(5*pi/8),2*sin(5*pi/8)),N);MP("A_2",(2*cos(3*pi/8),2*sin(3*pi/8)),N);MP("A_3",(2*cos(1*pi/8),2*sin(1*pi/8)),N); MP("A_n",(2*cos(7*pi/8),2*sin(7*pi/8)),N); MP("B_1",(cos(4*pi/8),sin(4*pi/8)),S);MP("B_2",(cos(2*pi/8),sin(2*pi/8)),S);MP("B_n",(cos(6*pi/8),sin(6*pi/8)),S); [/asy]
Part of an "n-pointed regular star" is shown. It is a simple closed polygon in which all $2n$ edges are congruent, angles $A_1,A_2,\cdots,A_n$ are congruent, and angles $B_1,B_2,\cdots,B_n$ are congruent. If the acute angle at $A_1$ is $10^\circ$ less than the acute angle at $B_1$, then $n=$
$\text{(A) } 12\quad \text{(B) } 18\quad \text{(C) } 24\quad \text{(D) } 36\quad \text{(E) } 60$
| [
"If we sum up the angles to obtain 360, we can see that the B angles add to the sum and the A angles subtract from the sum (an easy way of looking at this is by using the opposing angle theorem: if A[n] = B[n] than their total contribution is 0). Thus we have B[1] + B[2] + ... + B[n] - A[1] - A[2] - ... A[n] = 360. But every pair of A,B has a total 'angle contribution' of 10, thus there are 36 $\\fbox{D}$ pairs of A,B.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/20.json | AHSME |
1992_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | If
\[\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}\]
for three positive numbers $x,y$ and $z$, all different, then $\frac{x}{y}=$
$\text{(A) } \frac{1}{2}\quad \text{(B) } \frac{3}{5}\quad \text{(C) } \frac{2}{3}\quad \text{(D) } \frac{5}{3}\quad \text{(E) } 2$
| [
"$\\fbox{E}$ We have $\\frac{x+y}{z} = \\frac{x}{y} \\implies xy+y^2=xz$ and $\\frac{y}{x-z} = \\frac{x}{y} \\implies y^2=x^2-xz \\implies x^2-y^2=xz$. Equating the two expressions for $xz$ gives $xy+y^2=x^2-y^2 \\implies x^2-xy-2y^2=0 \\implies (x+y)(x-2y)=0$, so as $x+y$ cannot be $0$ for positive $x$ and $y$, we must have $x-2y=0 \\implies x=2y \\implies \\frac{x}{y}=2$.\n\n\n",
"We cross multiply the first and third fractions and the second and third fractions, respectively, for \\[x(x-z)=y^2\\] \\[y(x+y)=xz\\] Notice how the first equation can be expanded and rearranged to contain an $(x+y)$ term. \\[x^2-xz=y^2\\] \\[x^2-y^2=xz\\] \\[(x+y)(x-y)=xz\\] We can divide this by the second equation to get \\[\\frac{(x+y)(x-y)}{y(x+y)}=\\frac{xz}{xz}\\] \\[\\frac{x-y}{y}=1\\] \\[\\frac{x}{y}-1=1\\] \\[\\frac{x}{y}=2 \\rightarrow \\boxed{E}\\]\n\n\n",
"Since \\[\\frac{a_1}{b_1} = \\frac{a_2}{b_2} = ... = \\frac{a_n}{b_n} = \\frac{a_1 + a_2+ ... + a_n}{b_1 + b_2 + ... + b_n},\\] we can say that \\[\\frac{y}{x-z} = \\frac{x+y}{z} = \\frac{x}{y} = \\frac {y+(x+y)+x}{(x-z)+z+y} = \\boxed{2}.\\]\n$\\implies \\boxed{(E)}$.\n\n\n$\\textbf{Proof of the used property:}$\n\n\nLet \\[\\frac{a_1}{b_1} = \\frac{a_2}{b_2} = \\frac{a_3}{b_3} = ... = \\frac{a_n}{b_n} = r.\\]\n\n\nTherefore, \\begin{align*} a_1 &= b_1r,\\\\ a_2 &= b_2r,\\\\ ...,\\\\ a_n &= b_nr \\end{align*}\nso \\[\\frac{a_1 + a_2+ ... + a_n}{b_1 + b_2 + ... + b_n} = \\frac{b_1r + b_2r + ... + b_nr}{b_1 + b_2 + ... + b_n}\\] \\[= \\frac{(b_1 + b_2 + ... + b_n)(nr)}{(b_1 + b_2 + ... + b_n)(n)} = r\\] --- NamelyOrange\n\n\n"
] | 3 | ./CreativeMath/AHSME/1992_AHSME_Problems/16.json | AHSME |
1992_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | If $x>y>0$ , then $\frac{x^y y^x}{y^y x^x}=$
$\text{(A) } (x-y)^{y/x}\quad \text{(B) } \left(\frac{x}{y}\right)^{x-y}\quad \text{(C) } 1\quad \text{(D) } \left(\frac{x}{y}\right)^{y-x}\quad \text{(E) } (x-y)^{x/y}$
| [
"We see that this fraction can easily be factored as $\\frac{x^y}{y^y}\\times\\frac{y^x}{x^x}$. Since $\\frac{y^x}{x^x}=\\frac{x^{-x}}{y^{-x}}$, this fraction is equivalent to $\\left(\\frac{x}{y}\\right)^y\\times\\left(\\frac{x}{y}\\right)^{-x}=\\left(\\frac{x}{y}\\right)^{y-x}\\quad$, which corresponds to answer choice $\\fbox{D}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/6.json | AHSME |
1992_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | The ratio of $w$ to $x$ is $4:3$, of $y$ to $z$ is $3:2$ and of $z$ to $x$ is $1:6$. What is the ratio of $w$ to $y$?
$\text{(A) } 1:3\quad \text{(B) } 16:3\quad \text{(C) } 20:3\quad \text{(D) } 27:4\quad \text{(E) } 12:1$
| [
"$\\fbox{B}$\n\n\nConverting from ratios to fractions\n\n\n$\\frac{w}{x}=\\frac{4}{3}$,\n$\\frac{y}{z}=\\frac{3}{2}$,\n$\\frac{z}{x}=\\frac{1}{6}$\n\n\n\n\n$\\frac{y}{z} \\cdot \\frac{z}{x} = \\frac{3}{2} \\cdot \\frac{1}{6}$\n\n\n$\\frac{y}{x} = \\frac{1}{4}$\n\n\n$\\frac{(\\frac{w}{x})}{(\\frac{y}{x})}=\\frac{\\frac{4}{3}}{\\frac{1}{4}}$\n\n\n$\\frac{w}{y} = \\frac{16}{3}$\n\n\n$w$ to $y$ is $16:3$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/7.json | AHSME |
1992_AHSME_Problems | 17 | 0 | Number Theory | Multiple Choice | The 2-digit integers from 19 to 92 are written consecutively to form the integer $N=192021\cdots9192$. Suppose that $3^k$ is the highest power of 3 that is a factor of $N$. What is $k$?
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) more than } 3$
| [
"We can determine if our number is divisible by $3$ or $9$ by summing the digits. Looking at the one's place, we can start out with $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ and continue cycling though the numbers from $0$ through $9$. For each one of these cycles, we add $0 + 1 + ... + 9 = 45$. This is divisible by $9$, thus we can ignore the sum. However, this excludes $19$, $90$, $91$ and $92$. These remaining units digits sum up to $9 + 1 + 2 = 12$, which means our units sum is $3 \\pmod 9$. As for the tens digits, for $2, 3, 4, \\cdots , 8$ we have $10$ sets of those: \\[\\frac{8 \\cdot 9}{2} - 1 = 35,\\] which is congruent to $8 \\pmod 9$. We again have $19, 90, 91$ and $92$, so we must add \\[1 + 9 \\cdot 3 = 28\\] to our total. $28$ is congruent to $1 \\pmod 9$. Thus our sum is congruent to $3 \\pmod 9$, and $k = 1 \\implies \\boxed{B}$.\n\n\n",
"As our first trial with 3, We can say that $N\\equiv 1+9+2+0+2+1+2+2+...+9+1+9+2\\pmod{3}$. Since if the sum of the digits of a number is divisible by 3, then the number is also divisible by 3, we reverse that logic to say that $N\\equiv 19+20+21+22+...+91+92\\pmod{3}$, and adding that up using the rainbow strategy, we get $N\\equiv (111\\times37)\\pmod{3}$. We can see that since 3 divides 111, the original number is divisible by 3. But, since 111 only has one 3 and 37 has no 3's, it is not divisble by 9. Thus, the answer is $\\boxed{B}$\n\n\n~mathpro12345\n\n\npapa I wrote this second one\n\n\n"
] | 2 | ./CreativeMath/AHSME/1992_AHSME_Problems/17.json | AHSME |
1992_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | For a finite sequence $A=(a_1,a_2,...,a_n)$ of numbers, the \textit{Cesáro sum} of A is defined to be
$\frac{S_1+\cdots+S_n}{n}$ , where $S_k=a_1+\cdots+a_k$ and $1\leq k\leq n$. If the Cesáro sum of
the 99-term sequence $(a_1,...,a_{99})$ is 1000, what is the Cesáro sum of the 100-term sequence
$(1,a_1,...,a_{99})$?
$\text{(A) } 991\quad \text{(B) } 999\quad \text{(C) } 1000\quad \text{(D) } 1001\quad \text{(E) } 1009$
| [
"Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is \n\\[S_1 + S_2 + S_3 + ... + S_n\\]\n\\[= a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \\ldots + (a_1 + a_2 + \\ldots + a_n)\\]\n\\[= n \\cdot a_1 + (n - 1) \\cdot a_2 + \\ldots + a_n.\\]\nIf we take this to be the Cesáro total for the second sequence, we can see that all the terms but the first term make up the first sequence. Since we know that the Cesáro total of the original sequence is $1000 * 99 = 99,000,$ than the Cesáro total of the second sequence is $n \\cdot a_1 + 99,000 = 100 \\cdot 1 + 99,000 = 99,100.$ Thus the Cesáro sum of the second sequence is $\\frac{99,100}{100} = \\boxed{991, A}\\, .$\n\n\n",
"We know that the Cesáro sum of the first $99$ numbers is \n\n\n\\[\\frac{a_1+a_1+a_2+a_1+a_2+a_3+...+a_{99}}{99}=100.\\]\n\n\nMultiplying by $99$, we see that \n\n\n\\[a_1+a_1+a_2+a_1+a_2+a_3+...+a_{99}=100\\cdot{99}.\\]\n\n\nNow, we append $1$ to the list of $a_n,$ so our new sum becomes \n\n\n\\[\\frac{1+1+a_1+1+a_1+a_2+...+a_{99}}{100} \\rightarrow\\]\n\\[\\frac{100+a_1+a_1+a_2+...+a_{99}}{100}.\\]\n\n\nWe've estabished that $a_1+a_1+a_2+a_1+a_2+a_3+...+a_{99}=100\\cdot{99}$, so\n\n\n\\[\\frac{100+a_1+a_1+a_2+...+a_{99}}{100}=\\frac{100+99\\cdot{100}}{100}.\\]\n\n\nDividing everything by $100$, the sum of the sought sequence is $1+990=991 \\rightarrow \\boxed{A}$\n\n\n~Benedict T (countmath1)\n\n\n"
] | 2 | ./CreativeMath/AHSME/1992_AHSME_Problems/21.json | AHSME |
1992_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | The number of positive integers $k$ for which the equation
\[kx-12=3k\]
has an integer solution for $x$ is
$\text{(A) } 3\quad \text{(B) } 4\quad \text{(C) } 5\quad \text{(D) } 6\quad \text{(E) } 7$
| [
"$\\fbox{D}$\n\n\n$kx -12 = 3k$\n\n\n$-12=3k-kx$\n\n\n$-12=k(3-x)$\n\n\n$\\frac{-12}{k}=3-x$\n\n\nPositive factors of $-12$:\n$1,2,3,4,6,12$\n\n\n6 factors, each of which have an integer solution for $x$ in $\\frac{-12}{k}=3-x$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/10.json | AHSME |
1992_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | [asy] fill((1,0)--arc((1,0),2,180,225)--cycle,grey); fill((-1,0)--arc((-1,0),2,315,360)--cycle,grey); fill((0,-1)--arc((0,-1),2-sqrt(2),225,315)--cycle,grey); fill((0,0)--arc((0,0),1,180,360)--cycle,white); draw((1,0)--arc((1,0),2,180,225)--(1,0),black+linewidth(1)); draw((-1,0)--arc((-1,0),2,315,360)--(-1,0),black+linewidth(1)); draw((0,0)--arc((0,0),1,180,360)--(0,0),black+linewidth(1)); draw(arc((0,-1),2-sqrt(2),225,315),black+linewidth(1)); draw((0,0)--(0,-1),black+linewidth(1)); MP("C",(0,0),N);MP("A",(-1,0),N);MP("B",(1,0),N); MP("D",(0,-.8),NW);MP("E",(1-sqrt(2),-sqrt(2)),SW);MP("F",(-1+sqrt(2),-sqrt(2)),SE); [/asy]
Semicircle $\widehat{AB}$ has center $C$ and radius $1$. Point $D$ is on $\widehat{AB}$ and $\overline{CD}\perp\overline{AB}$. Extend $\overline{BD}$ and $\overline{AD}$ to $E$ and $F$, respectively, so that circular arcs $\widehat{AE}$ and $\widehat{BF}$ have $B$ and $A$ as their respective centers. Circular arc $\widehat{EF}$ has center $D$. The area of the shaded "smile" $AEFBDA$, is
$\text{(A) } (2-\sqrt{2})\pi\quad \text{(B) } 2\pi-\pi \sqrt{2}-1\quad \text{(C) } (1-\frac{\sqrt{2}}{2})\pi\quad\\ \text{(D) } \frac{5\pi}{2}-\pi\sqrt{2}-1\quad \text{(E) } (3-2\sqrt{2})\pi$
| [
"$\\fbox{B}$ The area of the entire outer shape is the area of sector $ABE$, plus the area of sector $ABF$, minus the area of triangle $ABD$ (since it is part of both sectors), plus the area of sector $DEF$. We know $AC = CD = 1$, so the sector angles for $ABE$ and $ABF$ are $45$ degrees, and the radius of both of them is $2$. The radius of $DEF$ is $DE = BE - BD = 2 - BD$, and $BD$ can be found using Pythagoras in triangle $BCD$, giving $BD = \\sqrt{2}$ and $DE = 2 - \\sqrt{2}$, so after doing all the calculations, the area of the entire outer shape is $\\pi(\\frac{5}{2} - \\sqrt{2}) - 1$. To get the area of the smile, we need to subtract the area of semicircle $ABD$, which is $\\frac{1}{2} \\pi 1^2 = \\frac{\\pi}{2}$, so the answer is $\\pi(\\frac{5}{2} - \\frac{1}{2} - \\sqrt{2}) - 1$ = $2\\pi - \\pi \\sqrt{2} - 1$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/26.json | AHSME |
1992_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | Let $ABCD$ be an isosceles trapezoid with bases $AB=92$ and $CD=19$. Suppose $AD=BC=x$ and a circle with center on $\overline{AB}$ is tangent to segments $\overline{AD}$ and $\overline{BC}$. If $m$ is the smallest possible value of $x$, then $m^2$=
$\text{(A) } 1369\quad \text{(B) } 1679\quad \text{(C) } 1748\quad \text{(D) } 2109\quad \text{(E) } 8825$
| [
"Note that the center of the circle is the midpoint of $AB$, call it $M$. When we decrease $x$, the limiting condition is that the circle will eventually be tangent to segment $AD$ at $D$ and segment $BC$ at $C$. That is, $MD\\perp AD$ and $MC\\perp BC$. \n\n\nFrom here, we drop the altitude from $D$ to $AM$; call the base $N$. Since $\\triangle DNM \\sim \\triangle ADM$, we have\n\\[\\frac{DM}{19/2}=\\frac{46}{DM}.\\]\nThus, $DM=\\sqrt{19\\cdot 23}$. Furthermore, $x^2=AM^2-DM^2=46^2-19\\cdot 23=1679, \\boxed{B}.$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/30.json | AHSME |
1992_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | A circle of radius $r$ has chords $\overline{AB}$ of length $10$ and $\overline{CD}$ of length 7. When $\overline{AB}$ and $\overline{CD}$ are extended through $B$ and $C$, respectively, they intersect at $P$, which is outside of the circle. If $\angle{APD}=60^\circ$ and $BP=8$, then $r^2=$
$\text{(A) } 70\quad \text{(B) } 71\quad \text{(C) } 72\quad \text{(D) } 73\quad \text{(E) } 74$
| [
"[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; path circ = Circle(origin, 1); pair A = dir(degrees(7pi/12)); pair D = dir(degrees(-5pi/12)); pair B = dir(degrees(2pi/12)); pair C = dir(degrees(-2pi/12)); pair P = extension(A, B, C, D); draw(circ); draw(A--P--D); label('$A$', A, N); label('$D$', D, S); label('$C$', C, SE); label('$B$', B, NE); label('$P$', P, E); label('$60^\\circ$', P, 2 * (dir(P--A) + dir(P--D))); label('$10$', A--B, S); label('$8$', B--P, NE); label('$7$', C--D, N); [/asy]\n\n\nApplying Power of a Point on $P$, we find that $PC=9$ and thus $PD=16$. Observing that $PD=2BP$ and that $\\angle BPD=60^{\\circ}$, we conclude that $BPD$ is a $30-60-90$ right triangle with right angle at $B$. Thus, $BD=8\\sqrt{3}$ and triangle $ABD$ is also right. Using that fact that the circumcircle of a right triangle has its diameter equal to the hypotenuse, we compute using the Pythagorean Theorem $AD=2r=2\\sqrt{73}$. From here we see that $r^2=73$. The answer is thus $\\fbox{D}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/27.json | AHSME |
1992_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | If $3(4x+5\pi)=P$ then $6(8x+10\pi)=$
$\text{(A) } 2P\quad \text{(B) } 4P\quad \text{(C) } 6P\quad \text{(D) } 8P\quad \text{(E) } 18P$
| [
"We can see that $8x+10\\pi$ is equal to $2(4x+5\\pi),$ and we know that $2^2 = 4,$ so the answer is $\\boxed{B}\\, .$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/1.json | AHSME |
1992_AHSME_Problems | 11 | 0 | Geometry | Multiple Choice | [asy] draw(circle((0,0),18),black+linewidth(.75)); draw(circle((0,0),6),black+linewidth(.75)); draw((-18,0)--(18,0)--(-14,8*sqrt(2))--cycle,black+linewidth(.75)); dot((-18,0));dot((18,0));dot((-14,8*sqrt(2))); MP("A",(-18,0),W);MP("C",(18,0),E);MP("B",(-14,8*sqrt(2)),W); [/asy]
The ratio of the radii of two concentric circles is $1:3$. If $\overline{AC}$ is a diameter of the larger circle, $\overline{BC}$ is a chord of the larger circle that is tangent to the smaller circle, and $AB=12$, then the radius of the larger circle is
$\text{(A) } 13\quad \text{(B) } 18\quad \text{(C) } 21\quad \text{(D) } 24\quad \text{(E) } 26$
| [
"We are given that $BC$ is tangent to the smaller circle. Using that, we know where the circle intersects $BC$, it creates a right triangle. We can also point out that since $AC$ is the diameter of the bigger circle and triangle $ABC$ is inscribed the semi-circle, that angle $B$ is a right angle. Therefore, we have $2$ similar triangles. Let's label the center of the smaller circle (which is also the center of the larger circle) as $D$. Let's also label the point where the smaller circle intersects $BC$ as $E$. So $ABC$ is similar to $DEC$. Since $DE$ is the radius of the smaller circle, call the length $x$ and since $DC$ is the radius of the bigger circle, call that length $3x$. The diameter, $AC$ is $6x$. So,\n\n\n$\\frac{AB}{AC} = \\frac {DE}{DC} \\Rightarrow \\frac{12}{6x} = \\frac{x}{3x} \\Rightarrow 36x = 6x^2 \\Rightarrow 6 = x$\n\n\nBut they are asking for the larger circle radius, so \n\n\n$3x= \\fbox{18}$\n\n\n\n\n$\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/11.json | AHSME |
1992_AHSME_Problems | 2 | 0 | Probability | Multiple Choice | An urn is filled with coins and beads, all of which are either silver or gold. Twenty percent of the objects in the urn are beads. Forty percent of the coins in the urn are silver. What percent of objects in the urn are gold coins?
$\text{(A) } 40\%\quad \text{(B) } 48\%\quad \text{(C) } 52\%\quad \text{(D) } 60\%\quad \text{(E) } 80\%$
| [
"If twenty percent of the objects in the urn are beads, then the other $80$% of the objects must be coins. Since $40$% of the coins are silver, then $60$% of the coins must be gold. Therefore, $60$% of $80$% of the objects in the urn are gold coins. Since $.6\\times.8=.48$, 48 percent of the objects in the urn must be gold coins. Thus the answer is $\\fbox{B}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/2.json | AHSME |
1992_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | Let $i=\sqrt{-1}$. The product of the real parts of the roots of $z^2-z=5-5i$ is
$\text{(A) } -25\quad \text{(B) } -6\quad \text{(C) } -5\quad \text{(D) } \frac{1}{4}\quad \text{(E) } 25$
| [
"Applying the quadratic formula gives \\[z=\\frac{-1\\pm\\sqrt{21-20i}}{2}\\]\n\n\nLet \\[\\sqrt{21-20i}=a+bi\\] where $a$ and $b$ are real. Squaring both sides and equating real and imaginary parts gives \\[a^2-b^2=21\\] \\[2ab=-20\\] Substituting $b=-\\frac{10}{a}$, letting $a^2=n$ and solving gives $n=25, -4$, from which we see that $a=5$ and $b=-2$.\n\n\nReplacing the square root in the quadratic formula with $5-2i$ and simplifying gives the two roots $2-i$ and $-3+i$. The product of their real parts is $-6$. The answer is $\\fbox{B}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/28.json | AHSME |
1992_AHSME_Problems | 12 | 0 | Geometry | Multiple Choice | Let $y=mx+b$ be the image when the line $x-3y+11=0$ is reflected across the $x$-axis. The value of $m+b$ is
$\text{(A) -6} \quad \text{(B) } -5\quad \text{(C) } -4\quad \text{(D) } -3\quad \text{(E) } -2$
| [
"$\\fbox{C}$\nFirst we want to put this is slope-intercept form, so we get $y=\\dfrac{1}{3}x+\\dfrac{11}{3}$. When we reflect a line across the x-axis, the line's x-intercept stays the same, while the y-intercept and the slope are additive inverses of the original line. Since $m+b$ is the sum of the slope and the y-intercept, we get $-\\dfrac{1}{3}-\\dfrac{11}{3}=\\boxed{-4}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/12.json | AHSME |
1992_AHSME_Problems | 24 | 0 | Geometry | Multiple Choice | Let $ABCD$ be a parallelogram of area $10$ with $AB=3$ and $BC=5$. Locate $E,F$ and $G$ on segments $\overline{AB},\overline{BC}$ and $\overline{AD}$, respectively, with $AE=BF=AG=2$. Let the line through $G$ parallel to $\overline{EF}$ intersect $\overline{CD}$ at $H$. The area of quadrilateral $EFHG$ is
$\text{(A) } 4\quad \text{(B) } 4.5\quad \text{(C) } 5\quad \text{(D) } 5.5\quad \text{(E) } 6$
| [
"$\\fbox{C}$ Use vectors. Place an origin at $A$, with $B = p, D = q, C = p + q$. We know that $\\|p \\times q\\|=10$, and also $E=\\frac{2}{3}p, F=p+\\frac{2}{5}q, G = \\frac{2}{5}q$, and now we can find the area of $EFHG$ by dividing it into two triangles and using cross-products (the expressions simplify using the fact that the cross-product distributes over addition, it is anticommutative, and a vector crossed with itself gives zero).\n\n\n",
"We note that $ABFG$ is a parallelogram because $AG = BF = 2$ and $AG \\parallel BF$. Using the same reasoning, $GFCD$ is also a parallelogram. \n\n\n\n\nAssume that the height of parallelogram $ABFG$ with respect to base $AB$ is $x$. Then, the area of parallelogram $ABFG$ is $AB * x$. The area of triangle $EFG$ is $\\frac{AB * x}{2}$, which is half of the area of parallelogram $ABFG$.\n\n\n\n\nLikewise, the area of triangle $FGH$ is half the area of parallelogram $GFCD$.\n\n\n\n\nThus, $[EFHG] = [EFG] + [FGH] = 1/2[ABFG] + 1/2[GFCD] = 1/2[ABCD] = 1/2(10) = \\boxed{5}$\n\n\n"
] | 2 | ./CreativeMath/AHSME/1992_AHSME_Problems/24.json | AHSME |
1992_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | In $\triangle{ABC}$, $\angle ABC=120^\circ,AB=3$ and $BC=4$. If perpendiculars constructed to $\overline{AB}$ at $A$ and to $\overline{BC}$ at $C$ meet at $D$, then $CD=$
$\text{(A) } 3\quad \text{(B) } \frac{8}{\sqrt{3}}\quad \text{(C) } 5\quad \text{(D) } \frac{11}{2}\quad \text{(E) } \frac{10}{\sqrt{3}}$
| [
"We begin by drawing a diagram.\n[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = A+dir(55); pair D = A+dir(0); pair B = extension(A,A+dir(90),C,C+dir(-155)); label(\"$A$\",A,S); label(\"$C$\",C,NE); label(\"$D$\",D,SE); label(\"$B$\",B,NW); label(\"$4$\",B--C,NW); label(\"$3$\",A--B,W); draw(A--C--D--cycle); draw(A--B--C); draw(rightanglemark(B,C,D,2)); draw(rightanglemark(B,A,D,2)); [/asy]\nWe extend $CB$ and $DA$ to meet at $E.$ This gives us a couple right triangles in $CED$ and $BEA.$\n[asy] import olympiad; import cse5; import geometry; size(250); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = A+dir(55); pair D = A+dir(0); pair B = extension(A,A+dir(90),C,C+dir(-155)); pair E = extension(A,A+2*dir(180),B,B+2*dir(-155)); label(\"$A$\",A,S); label(\"$C$\",C,NE); label(\"$D$\",D,SE); label(\"$B$\",B,NW); label(\"$4$\",B--C,NW); label(\"$3$\",A--B,W); label(\"$E$\",E,SW); draw(A--C--D--cycle); draw(A--B--C); draw(rightanglemark(B,C,D,2)); draw(rightanglemark(B,A,D,2)); draw(A--E--B,dashed); [/asy]\nWe see that $\\angle E = 30^\\circ$. Hence, $\\triangle BEA$ and $\\triangle DEC$ are 30-60-90 triangles.\n\n\nUsing the side ratios of 30-60-90 triangles, we have $BE=2BA=6$. This tells us that $CE=BC+BE=4+6=10$. Also, $EA=3\\sqrt{3}$.\n\n\nBecause $\\triangle DEC\\sim\\triangle BEA$, we have \\[\\frac{10}{3\\sqrt{3}}=\\frac{CD}{3}.\\]\nSolving the equation, we have\n\\begin{align*} \\frac{CD}3&=\\frac{10}{3\\sqrt{3}}\\\\ CD&=3\\cdot\\frac{10}{3\\sqrt{3}}\\\\ CD&=\\frac{10}{\\sqrt{3}}\\ \\end{align*}\nHence, $CD=\\boxed{\\textbf{E}}$.\n\n\n",
"Since $\\angle{A}+\\angle{C} = 180^{\\circ}, ABCD$ is cyclic. Using Ptolemy's Theorem gets $\\text{(1): } 4AD + 3CD = \\sqrt{37}BD.$\n\n\nRight triangles $\\Delta ABD$ and $\\Delta CBD$ obtain $\\text{(2): } 9+AD^2=BD^2$ and $\\text{(3):} 16+CD^2=BD^2,$ respectively.\n\n\nSeeing squares in $\\text{(2)}$ and $\\text{(3)}$, we square $\\text{(1)}$ and get $\\text{(4): } 16AD^2+9CD^2 + 24AD\\cdot CD = 37BD^2.$\n\n\nWe don't like that $AD\\cdot CD$ term, but fortunately LoC exists: $37 = AD^2 + CD^2 - 2*AD*CD*\\cos(60^{\\circ})$. Solving for $AD\\cdot CD$ and plugging it into $\\text{(4)}$, and using $AD^2 = 7 + CD^2$ and $BD^2 = 16 + CD^2$ from the first two equations, gets $40(CD^2+7)+33CD^2 - 24\\cdot 37 = 37(16+CD^2).$\n\n\nSolve for $CD = \\boxed{\\textbf{E}}$.\n\n\n~PureSwag\n\n\n"
] | 2 | ./CreativeMath/AHSME/1992_AHSME_Problems/25.json | AHSME |
1992_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | How many pairs of positive integers (a,b) with $a+b\le 100$ satisfy the equation
\[\frac{a+b^{-1}}{a^{-1}+b}=13?\]
$\text{(A) } 1\quad \text{(B) } 5\quad \text{(C) } 7\quad \text{(D) } 9\quad \text{(E) } 13$
| [
"$\\fbox{C}$ We can rewrite the left-hand side as $\\frac{a^2b+a}{b+ab^2}$ by multiplying the numerator and denominator by $ab$. Now we can multiply the top part of the fraction by the denominator to give $a^2b+a = 13b + 13ab^2 \\implies ab(a-13b)=13b-a = -(a-13b) \\implies (ab+1)(a-13b) = 0$. Now, as $a$ and $b$ are positive, we cannot have $ab+1=0$, so we must have $a-13b=0 \\implies a=13b$, so the solutions are $(13,1), (26,2), (39,3), (52,4), (65,5), (78,6), (91,7)$, which is 7 ordered pairs.\n\n\n",
"The equation is equivalent to $\\frac{a+\\frac{1}{b}}{\\frac{1}{a}+b}=13$ or $\\frac{\\frac{ab+1}{b}}{\\frac{ab+1}{a}}=13$. The $ab+1$s cancel out leaving us with $\\frac{a}{b}=13$. This is asking us for the positive multiples of $13$ up to $100$, and we know that $13 \\cdot 7 = 91$, so our answer is $\\fbox{C}$.\n\n\n"
] | 2 | ./CreativeMath/AHSME/1992_AHSME_Problems/13.json | AHSME |
1992_AHSME_Problems | 29 | 0 | Probability | Multiple Choice | An unfair coin has a $2/3$ probability of turning up heads. If this coin is tossed $50$ times, what is the probability that the total number of heads is even?
$\text{(A) } 25\bigg(\frac{2}{3}\bigg)^{50}\quad \text{(B) } \frac{1}{2}\bigg(1-\frac{1}{3^{50}}\bigg)\quad \text{(C) } \frac{1}{2}\quad \text{(D) } \frac{1}{2}\bigg(1+\frac{1}{3^{50}}\bigg)\quad \text{(E) } \frac{2}{3}$
| [
"Doing casework on the number of heads (0 heads, 2 heads, 4 heads...), we get the equation \\[P=\\left(\\frac{1}{3} \\right)^{50}+\\binom{50}{2}\\left(\\frac{2}{3} \\right)^{2}\\left(\\frac{1}{3} \\right)^{48}+\\dots+\\left(\\frac{2}{3} \\right)^{50}\\] This is essentially the expansion of $\\left(\\frac{2}{3}+\\frac{1}{3} \\right)^{50}$ but without the odd power terms. To get rid of the odd power terms in $\\left(\\frac{2}{3}+\\frac{1}{3} \\right)^{50}$, we add $\\left(\\frac{2}{3}-\\frac{1}{3} \\right)^{50}$ and then divide by $2$ because the even power terms that were not canceled were expressed twice. Thus, we have \\[P=\\frac{1}{2}\\cdot\\left(\\left(\\frac{1}{3}+\\frac{2}{3} \\right)^{50}+\\left(\\frac{2}{3}-\\frac{1}{3} \\right)^{50} \\right)\\] Or \\[\\frac{1}{2}\\left(1+\\left(\\frac{1}{3} \\right)^{50} \\right)\\] which is equivalent to answer choice $\\fbox{D}$.\n\n\n",
"Denote $P_n$ as the probability that the number of heads is even. Then we have two cases:\n\n\n$\\bold{Case 1}$: (There are an odd number of heads for $(n-1)$ coin flips).\nWell to solve this case, we first ask the question; \"What is the probability we get an odd number of heads for $(n-1)$ flips?\" That is precisely $1-P_{n-1}$. Now, since we want an even number of heads, what is the probability that on the $n-th$ coin flip we get a head? $\\frac{2}{3}$. So we have $\\frac{2}{3}(1-P_{n-1})$.\n\n\n$\\bold{Case 2}$: (There are an even number of heads for $(n-1)$ coin flips). Since we now already have an even number of heads, we now want to flip and get a tail, or else the number of heads for $n$ flips will be odd. :/ So what is the probability we get a tail? $1-\\frac{2}{3}=\\frac{1}{3}$. Thus we have $\\frac{1}{3}(P_{n-1})$. \n\n\nAdding both cases, we have that $P_n=\\frac{2}{3}-\\frac{1}{3}P_{n-1}$. Now doing some tedious casework because wolfram alpha cannot find a pattern, we get the same answer as in $\\boxed {D)}$. \n\n\nSorry for the very hand-wavy and non-rigorous answer! At least mine has a cool recurrence relation! :D\n\n\n(Or just come up with the general formula $P_n=\\frac{1}{2}(1+\\frac{1}{3^n})$ Simple observation really. If you really want to get mathy, prove this through induction).\n\n\n~th1nq3r\n\n\n"
] | 2 | ./CreativeMath/AHSME/1992_AHSME_Problems/29.json | AHSME |
1992_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | If $m>0$ and the points $(m,3)$ and $(1,m)$ lie on a line with slope $m$, then $m=$
$\text{(A) } 1\quad \text{(B) } \sqrt{2}\quad \text{(C) } \sqrt{3}\quad \text{(D) } 2\quad \text{(E) } \sqrt{5}$
| [
"We know that the formula for slope is $m = \\frac{y_2-y_1}{x_2-x_1}$\nWe are give the points $(m,3)$ and $(1,m)$.\nSubstituting into the slope formula, we get $\\frac{m-3}{1-m}$ \nAfter taking the cross-products and solving, we get $\\sqrt{3}$, which is answer choice $\\fbox{C}$.\n\n\n",
"Using the formula for slope as above, we know that $m=\\frac{m-3}{1-m}$. Multiplying both sides of this equation by $1-m$, we get that $m(1-m)=m-3$ which expands to $m-m^2=m-3$. This forms the quadratic equation $-m^2+3=0$. This means that $m^2=3$, so $m=\\sqrt{3}$, which corresponds to answer choice $\\fbox{C}$.\n\n\n"
] | 2 | ./CreativeMath/AHSME/1992_AHSME_Problems/3.json | AHSME |
1992_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | [asy] draw((-10,-10)--(-10,10)--(10,10)--(10,-10)--cycle,dashed+linewidth(.75)); draw((-7,-7)--(-7,7)--(7,7)--(7,-7)--cycle,dashed+linewidth(.75)); draw((-10,-10)--(10,10),dashed+linewidth(.75)); draw((-10,10)--(10,-10),dashed+linewidth(.75)); fill((10,10)--(10,9)--(9,9)--(9,10)--cycle,black); fill((9,9)--(9,8)--(8,8)--(8,9)--cycle,black); fill((8,8)--(8,7)--(7,7)--(7,8)--cycle,black); fill((-10,-10)--(-10,-9)--(-9,-9)--(-9,-10)--cycle,black); fill((-9,-9)--(-9,-8)--(-8,-8)--(-8,-9)--cycle,black); fill((-8,-8)--(-8,-7)--(-7,-7)--(-7,-8)--cycle,black); fill((10,-10)--(10,-9)--(9,-9)--(9,-10)--cycle,black); fill((9,-9)--(9,-8)--(8,-8)--(8,-9)--cycle,black); fill((8,-8)--(8,-7)--(7,-7)--(7,-8)--cycle,black); fill((-10,10)--(-10,9)--(-9,9)--(-9,10)--cycle,black); fill((-9,9)--(-9,8)--(-8,8)--(-8,9)--cycle,black); fill((-8,8)--(-8,7)--(-7,7)--(-7,8)--cycle,black); [/asy]
A square floor is tiled with congruent square tiles. The tiles on the two diagonals of the floor are black. The rest of the tiles are white. If there are 101 black tiles, then the total number of tiles is
$\text{(A) } 121\quad \text{(B) } 625\quad \text{(C) } 676\quad \text{(D) } 2500\quad \text{(E) } 2601$
| [
"$\\fbox{E}$ Let the side length of the square be $s$. If $s$ is even, there are $s$ black tiles on each of the two diagonals, so the total number is $2s$. If $s$ is odd, we double-count the middle square which is on both diagonals, so the number is instead $2s-1$. In this case, since $101$ is odd, we clearly must have the second case, so $2s-1=101 \\implies 2s=102 \\implies s=51$, and indeed $s$ is odd as we expected. Thus the overall number of tiles is $51^2 = 2601$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/8.json | AHSME |
1992_AHSME_Problems | 22 | 0 | Other | Multiple Choice | Ten points are selected on the positive $x$-axis,$X^+$, and five points are selected on the positive $y$-axis,$Y^+$. The fifty segments connecting the ten points on $X^+$ to the five points on $Y^+$ are drawn. What is the maximum possible number of points of intersection of these fifty segments that could lie in the interior of the first quadrant?
$\text{(A) } 250\quad \text{(B) } 450\quad \text{(C) } 500\quad \text{(D) } 1250\quad \text{(E) } 2500$
| [
"$\\fbox{B}$ We can pick any two points on the $x$-axis and any two points on the $y$-axis to form a quadrilateral, and the intersection of its diagonals will thus definitely be inside the first quadrant. Hence the answer is $10C2 \\times 5C2 = 450$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/22.json | AHSME |
1992_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | The increasing sequence of positive integers $a_1,a_2,a_3,\cdots$ has the property that
\[a_{n+2}=a_n+a_{n+1} \text{ for all } n\ge 1.\]
If $a_7=120$, then $a_8$ is
$\text{(A) } 128\quad \text{(B) } 168\quad \text{(C) } 193\quad \text{(D) } 194\quad \text{(E) } 210$
| [
"$\\fbox{D}$ Let $a_{1} = a, a_{2} = b$, so $5a + 8b = 120$. Now $8b$ and $120$ are divisible by $8$, so $5a$ is divisible by 8, so $a$ is divisible by 8. It's now easy to try the multiples of $8$ to get that $a = 8, b=10$ (all the other possibilities violate the condition $a < b$, which comes from the fact that the sequence is increasing). Hence $a_8 = 8a + 13b = 8 \\times 8 + 13 \\times 10 = 194$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/18.json | AHSME |
1992_AHSME_Problems | 4 | 0 | Number Theory | Multiple Choice | If $a,b$ and $c$ are positive integers and $a$ and $b$ are odd, then $3^a+(b-1)^2c$ is
$\text{(A) odd for all choices of c} \quad \text{(B) even for all choices of c} \quad\\ \text{(C) odd if c is even; even if c is odd} \quad\\ \text{(D) odd if c is odd; even if c is even} \quad\\ \text{(E) odd if c is not a multiple of 3; even if c is a multiple of 3}$
| [
"Since 3 has no factors of 2, $3^a$ will be odd for all values of $a$. Since $b$ is odd as well, $b-1$ must be even, so $(b-1)^2$ must be even. This means that for all choices of $c$, $(b-1)^2c$ must be even because any integer times an even number is still even. Since an odd number plus an even number is odd, $3^a+(b-1)^2c$ must be odd for all choices of $c$, which corresponds to answer choice $\\fbox{A}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/4.json | AHSME |
1992_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | Which of the following equations have the same graph?
$I.\quad y=x-2 \qquad II.\quad y=\frac{x^2-4}{x+2}\qquad III.\quad (x+2)y=x^2-4$
$\text{(A) I and II only} \quad \text{(B) I and III only} \quad \text{(C) II and III only} \quad \text{(D) I,II,and III} \quad \\ \text{(E) None. All of the equations have different graphs}$
| [
"The equations differ at $x=-2$. The graph of $I$ would contain the point $(-2, -4)$; $-2$ is undefined in the graph of $II$ because it gives a denominator of $0$; and the graph of $III$ contains the whole the vertical line $x = -2$ as for any $y$ value, the equation still satisfies $0 = 0$. So the answer is ${E}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/14.json | AHSME |
1992_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | Let $i=\sqrt{-1}$. Define a sequence of complex numbers by
\[z_1=0,\quad z_{n+1}=z_{n}^2+i \text{ for } n\ge1.\]
In the complex plane, how far from the origin is $z_{111}$?
$\text{(A) } 1\quad \text{(B) } \sqrt{2}\quad \text{(C) } \sqrt{3}\quad \text{(D) } \sqrt{110}\quad \text{(E) } \sqrt{2^{55}}$
| [
"$\\fbox{B}$ Write out some terms: $0, i, -1+i, -i, -1+i, -i, -1+i, -i$, etc., and it keeps alternating between $-1+i$ and $-i$, so as $111$ is odd, $z_{111}$ is $-1+i$. Thus its distance from the origin is $\\sqrt{(-1)^2+1^2} = \\sqrt{2}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/15.json | AHSME |
1992_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | $6^6+6^6+6^6+6^6+6^6+6^6=$
$\text{(A) } 6^6 \quad \text{(B) } 6^7\quad \text{(C) } 36^6\quad \text{(D) } 6^{36}\quad \text{(E) } 36^{36}$
| [
"We notice that $6^6+6^6+6^6+6^6+6^6+6^6=6(6^6)=\\boxed{\\textbf{(B) }6^7}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/5.json | AHSME |
1992_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | For each vertex of a solid cube, consider the tetrahedron determined by the vertex and the midpoints of the three edges that meet at that vertex. The portion of the cube that remains when these eight tetrahedra are cut away is called a cubeoctahedron. The ratio of the volume of the cubeoctahedron to the volume of the original cube is closest to which of these?
$\text{(A) } 75\%\quad \text{(B) } 78\%\quad \text{(C) } 81\%\quad \text{(D) } 84\%\quad \text{(E) } 87\%$
| [
"$\\fbox{D}$ Let the cube have side length 1, and place the cube in the coordinate plane. Then we can pick any vertex, get the coordinates of the midpoints, and hence find the three vectors that define the tetrahedron (the vectors from the chosen vertex to each midpoint). Now using $\\frac{1}{6}\\|a.(b\\times c)\\|$, we can find the volume of one of the tetrahedra, then multiply it by 8, and subtract from 1 to get $\\frac{5}{6}$, which is closest to $84\\%$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/19.json | AHSME |
1992_AHSME_Problems | 23 | 0 | Number Theory | Multiple Choice | Let $S$ be a subset of $\{1,2,3,...,50\}$ such that no pair of distinct elements in $S$ has a sum divisible by $7$. What is the maximum number of elements in $S$?
$\text{(A) } 6\quad \text{(B) } 7\quad \text{(C) } 14\quad \text{(D) } 22\quad \text{(E) } 23$
| [
"The fact that $x \\equiv 0 \\mod 7 \\Rightarrow 7 \\mid x$ is assumed as common knowledge in this answer.\n\n\nFirst, note that there are $8$ possible numbers that are equivalent to $1 \\mod 7$, and there are $7$ possible numbers equivalent to each of $2$-$6 \\mod 7$.\n\n\nSecond, note that there can be no pairs of numbers $a$ and $b$ such that $a \\equiv -b$ mod $7$, because then $a+b | 7$. These pairs are $(0,0)$, $(1,6)$, $(2,5)$, and $(3,4)$. Because $(0,0)$ is a pair, there can always be $1$ number equivalent to $0 \\mod 7$, and no more.\n\n\nTo maximize the amount of numbers in S, we will use $1$ number equivalent to $0 \\mod 7$, $8$ numbers equivalent to $1$, and $14$ numbers equivalent to $2$-$5$. This is obvious if you think for a moment. Therefore the answer is $1+8+14=23$ numbers.\n$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/23.json | AHSME |
1992_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | [asy] draw((-7,0)--(7,0),black+linewidth(.75)); draw((-3*sqrt(3),0)--(-2*sqrt(3),3)--(-sqrt(3),0)--(0,3)--(sqrt(3),0)--(2*sqrt(3),3)--(3*sqrt(3),0),black+linewidth(.75)); draw((-2*sqrt(3),0)--(-1*sqrt(3),3)--(0,0)--(sqrt(3),3)--(2*sqrt(3),0),black+linewidth(.75)); [/asy]
Five equilateral triangles, each with side $2\sqrt{3}$, are arranged so they are all on the same side of a line containing one side of each vertex. Along this line, the midpoint of the base of one triangle is a vertex of the next. The area of the region of the plane that is covered by the union of the five triangular regions is
$\text{(A) 10} \quad \text{(B) } 12\quad \text{(C) } 15\quad \text{(D) } 10\sqrt{3}\quad \text{(E) } 12\sqrt{3}$
| [
"$\\fbox{E}$\nFirst, we calculate the area of 1 triangle. For an equilateral triangle with side $s$, its area is $\\frac{\\sqrt{3}s^{2}}{4}$. If the side of the equilateral triangle is $2\\sqrt{3}$, the area of one such triangle is $3\\sqrt{3}$.\nThere are 5 equilateral triangles in total, overlapping by an area of 4 smaller equilateral triangles. Each smaller triangle is one fourth as big as the big equilateral triangles. Therefore, we subtract the overlapping area, which is equivalent to the area of 1 big equilateral triangle. \nHence, the total area is equal to the area of 4 equilateral triangles, which is $3\\sqrt{3} \\times 4 = 12\\sqrt{3}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1992_AHSME_Problems/9.json | AHSME |
1987_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | Evaluate
$\log_{10}(\tan 1^{\circ})+\log_{10}(\tan 2^{\circ})+\log_{10}(\tan 3^{\circ})+\cdots+\log_{10}(\tan 88^{\circ})+\log_{10}(\tan 89^{\circ}).$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2}\log_{10}(\frac{\sqrt{3}}{2}) \qquad \textbf{(C)}\ \frac{1}{2}\log_{10}2\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ \text{none of these}$
| [
"Because $\\tan x \\tan (90^\\circ - x) = \\tan x \\cot x = 1$, $\\tan 45^\\circ = 1$, and $\\log a + \\log b = \\log {ab}$, the answer is $\\log_{10} {\\tan 1^\\circ \\tan 2^\\circ \\dots \\tan 89^\\circ} = \\log_{10} 1 = 0.$ $\\boxed{\\textbf{(A)}}.$\n\n\n",
"We have $\\log{(\\frac{\\sin 1^\\circ}{\\cos 1^\\circ} \\times \\frac{\\sin 2^\\circ}{\\cos 2^\\circ} \\times ... \\times \\frac{\\sin 89^\\circ}{\\cos 89^\\circ})}$. However, $\\sin{(x)} = \\cos{(90^\\circ - x)}$. Thus each pair of $\\sin, \\cos$ (for example, $\\sin{1^\\circ}, \\cos{89^\\circ}$) multiplies to $1$. Hence we have $\\log{1} = 0$.\n\n\n"
] | 2 | ./CreativeMath/AHSME/1987_AHSME_Problems/20.json | AHSME |
1987_AHSME_Problems | 16 | 0 | Number Theory | Multiple Choice | A cryptographer devises the following method for encoding positive integers. First, the integer is expressed in base $5$.
Second, a 1-to-1 correspondence is established between the digits that appear in the expressions in base $5$ and the elements of the set
$\{V, W, X, Y, Z\}$. Using this correspondence, the cryptographer finds that three consecutive integers in increasing
order are coded as $VYZ, VYX, VVW$, respectively. What is the base-$10$ expression for the integer coded as $XYZ$?
$\textbf{(A)}\ 48 \qquad \textbf{(B)}\ 71 \qquad \textbf{(C)}\ 82 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 113$
| [
"Since $VYX + 1 = VVW$, i.e. adding $1$ causes the \"fives\" digit to change, we must have $X = 4$ and $W = 0$. Now since $VYZ + 1 = VYX$, we have $X = Z + 1 \\implies Z = 4 - 1 = 3$. Finally, note that in $VYX + 1 = VVW$, adding $1$ will cause the \"fives\" digit to change by $1$ if it changes at all, so $V = Y + 1$, and thus since $1$ and $2$ are the only digits left (we already know which letters are assigned to $0$, $3$, and $4$), we must have $V = 2$ and $Y = 1$. Thus $XYZ = 413_{5} = 4 \\cdot 5^{2} + 1 \\cdot 5 + 3 = 100 + 5 + 3 = 108$, which is answer $\\boxed{\\text{D}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/16.json | AHSME |
1987_AHSME_Problems | 6 | 0 | Geometry | Multiple Choice | In the $\triangle ABC$ shown, $D$ is some interior point, and $x, y, z, w$ are the measures of angles
in degrees. Solve for $x$ in terms of $y, z$ and $w$.
[asy] draw((0,0)--(10,0)--(2,7)--cycle); draw((0,0)--(4,3)--(10,0)); label("A", (0,0), SW); label("B", (10,0), SE); label("C", (2,7), W); label("D", (4,3), N); label("x", (2.25,6)); label("y", (1.5,2), SW); label("$z$", (7.88,1.5)); label("w", (4,2.85), S); [/asy]
$\textbf{(A)}\ w-y-z \qquad \textbf{(B)}\ w-2y-2z \qquad \textbf{(C)}\ 180-w-y-z \qquad \\ \textbf{(D)}\ 2w-y-z\qquad \textbf{(E)}\ 180-w+y+z$
| [
"By angles in a quadrilateral, $x = 360^{\\circ} - \\text{reflex } \\angle ADB - y - z$, and by angles at a point, $\\text{reflex } \\angle ADB = 360^{\\circ} - w$, so our expression becomes $360^{\\circ} - (360^{\\circ} - w) - y - z = w - y - z$, which is $\\boxed{\\text{A}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/6.json | AHSME |
1987_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | If $a-1=b+2=c-3=d+4$, which of the four quantities $a,b,c,d$ is the largest?
$\textbf{(A)}\ a \qquad \textbf{(B)}\ b \qquad \textbf{(C)}\ c \qquad \textbf{(D)}\ d \qquad \textbf{(E)}\ \text{no one is always largest}$
| [
"We have $a = a$, $b = a - 3$, $c = a + 2$, and $d = a - 5$, so now as $2 > 0 > -3 > -5$, $c$ is always largest. Hence the answer is $\\boxed{\\text{C}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/7.json | AHSME |
1987_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | In a mathematics competition, the sum of the scores of Bill and Dick equalled the sum of the scores of Ann and Carol.
If the scores of Bill and Carol had been interchanged, then the sum of the scores of Ann and Carol would have exceeded
the sum of the scores of the other two. Also, Dick's score exceeded the sum of the scores of Bill and Carol.
Determine the order in which the four contestants finished, from highest to lowest. Assume all scores were nonnegative.
$\textbf{(A)}\ \text{Dick, Ann, Carol, Bill} \qquad \textbf{(B)}\ \text{Dick, Ann, Bill, Carol} \qquad \textbf{(C)}\ \text{Dick, Carol, Bill, Ann}\\ \qquad \textbf{(D)}\ \text{Ann, Dick, Carol, Bill}\qquad \textbf{(E)}\ \text{Ann, Dick, Bill, Carol}$
| [
"Let the scores be $A$, $B$, $C$, and $D$. Then we have $A + C = B + D$, $A + B > C + D$, and $D > B + C$. Call these equations $(1)$, $(2)$, and $(3)$ respectively. Then adding $(1)$ and $(2)$ gives $2A + B + C > B + C + 2D \\implies 2A > 2D \\implies A > D$. Now subtracting $(1)$ from $(2)$ gives $A + B - A - C > C + D - B - D \\implies B - C > C - B \\implies 2B > 2C \\implies B > C$. Finally from $(3)$, since $D > B + C$ and $C$ is non-negative, we must have $D > B$, so putting our results together we get $A > D > B > C$, which is answer $\\boxed{\\text{E}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/17.json | AHSME |
1987_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | There are two natural ways to inscribe a square in a given isosceles right triangle.
If it is done as in Figure 1 below, then one finds that the area of the square is $441 \text{cm}^2$.
What is the area (in $\text{cm}^2$) of the square inscribed in the same $\triangle ABC$ as shown in Figure 2 below?
[asy] draw((0,0)--(10,0)--(0,10)--cycle); draw((-25,0)--(-15,0)--(-25,10)--cycle); draw((-20,0)--(-20,5)--(-25,5)); draw((6.5,3.25)--(3.25,0)--(0,3.25)--(3.25,6.5)); label("A", (-25,10), W); label("B", (-25,0), W); label("C", (-15,0), E); label("Figure 1", (-20, -5)); label("Figure 2", (5, -5)); label("A", (0,10), W); label("B", (0,0), W); label("C", (10,0), E); [/asy]
$\textbf{(A)}\ 378 \qquad \textbf{(B)}\ 392 \qquad \textbf{(C)}\ 400 \qquad \textbf{(D)}\ 441 \qquad \textbf{(E)}\ 484$
| [
"We are given that the area of the inscribed square is $441$, so the side length of that square is $21$. Since the square divides the $45-45-90$ larger triangle into 2 smaller congruent $45-45-90$, then the legs of the larger isosceles right triangle ($BC$ and $AB$) are equal to $42$. \n[asy] draw((0,0)--(10,0)--(0,10)--cycle); draw((6.5,3.25)--(3.25,0)--(0,3.25)--(3.25,6.5)); label(\"A\", (0,10), W); label(\"B\", (0,0), W); label(\"C\", (10,0), E); label(\"S\", (25/3,11/6), E); label(\"S\", (11/6,25/3), E); label(\"S\", (5,5), NE); [/asy]\n\n\nWe now have that $3S=42\\sqrt{2}$, so $S=14\\sqrt{2}$. But we want the area of the square which is $S^2=(14\\sqrt{2})^2= \\boxed{\\mathrm{(B)}\\ 392}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/21.json | AHSME |
1987_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | How many ordered triples $(a, b, c)$ of non-zero real numbers have the property that each number is the product of the other two?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$
| [
"We have $ab = c$, $bc = a$, and $ca = b$, so multiplying these three equations together gives $a^{2}b^{2}c^{2} = abc \\implies abc(abc-1)=0$, and as $a$, $b$, and $c$ are all non-zero, we cannot have $abc = 0$, so we must have $abc = 1$. Now substituting $bc = a$ gives $a(bc) = 1 \\implies a^2 = 1 \\implies a = \\pm 1$. If $a = 1$, then the system becomes $b = c, bc = 1, c = b$, so either $b = c = 1$ or $b = c = -1$, giving $2$ solutions. If $a = -1$, the system becomes $-b = c, bc = -1, -c = b$, so $-b = c = 1$ or $b = -c = 1$, giving another $2$ solutions. Thus the total number of solutions is $2 + 2 = 4$, which is answer $\\boxed{\\text{D}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/10.json | AHSME |
1987_AHSME_Problems | 26 | 0 | Probability | Multiple Choice | The amount $2.5$ is split into two nonnegative real numbers uniformly at random,
for instance, into $2.143$ and $.357$, or into $\sqrt{3}$ and $2.5-\sqrt{3}$.
Then each number is rounded to its nearest integer, for instance, $2$ and $0$ in the first case above,
$2$ and $1$ in the second. What is the probability that the two integers sum to $3$?
$\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{2}{5} \qquad \textbf{(C)}\ \frac{1}{2} \qquad \textbf{(D)}\ \frac{3}{5}\qquad \textbf{(E)}\ \frac{3}{4}$
| [
"The two parts may round to $0$ and $2$; $1$ and $2$; $1$ and $1$; $2$ and $1$; or $2$ and $0$. By considering the possible ranges for each case, it is easy to see that each case is equally likely (they divide the interval from $0$ to $2.5$, in which one of the parts is found, into five equal ranges of $0$ to $0.5$, $0.5$ to $1$, $1$ to $1.5$, $1.5$ to $2$, and $2$ to $2.5$). As exactly two of the five cases give a sum of $1 + 2 = 3$, the answer is $\\frac{2}{5}$, which is answer $\\boxed{B}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/26.json | AHSME |
1987_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | In the figure, $\triangle ABC$ has $\angle A =45^{\circ}$ and $\angle B =30^{\circ}$. A line $DE$, with $D$ on $AB$
and $\angle ADE =60^{\circ}$, divides $\triangle ABC$ into two pieces of equal area.
(Note: the figure may not be accurate; perhaps $E$ is on $CB$ instead of $AC.)$
The ratio $\frac{AD}{AB}$ is
[asy] size((220)); draw((0,0)--(20,0)--(7,6)--cycle); draw((6,6)--(10,-1)); label("A", (0,0), W); label("B", (20,0), E); label("C", (7,6), NE); label("D", (9.5,-1), W); label("E", (5.9, 6.1), SW); label("$45^{\circ}$", (2.5,.5)); label("$60^{\circ}$", (7.8,.5)); label("$30^{\circ}$", (16.5,.5)); [/asy]
$\textbf{(A)}\ \frac{1}{\sqrt{2}} \qquad \textbf{(B)}\ \frac{2}{2+\sqrt{2}} \qquad \textbf{(C)}\ \frac{1}{\sqrt{3}} \qquad \textbf{(D)}\ \frac{1}{\sqrt[3]{6}}\qquad \textbf{(E)}\ \frac{1}{\sqrt[4]{12}}$
| [
"First we show that $E$ is on $AC$, as in the given figure, by demonstrating that if $E = C$, then $\\triangle ADE$ has more than half the area, so $DE$ is too far to the right. Specifically, assume $E = C$ and drop an altitude from $C$ to $AB$ that meets $AB$ at $F$. Without loss of generality, assume that $CF = 1$, so that $\\frac{\\text{Area} \\ \\triangle EAD}{\\text{Area} \\ \\triangle EAB} = \\frac{AD}{AB}$ (as they have the same height, and area is $\\frac{1}{2}$ times the product of base and height), which is $\\frac{1 + \\frac{1}{\\sqrt{3}}}{1 + \\sqrt{3}} = \\frac{1}{\\sqrt{3}} > \\frac{1}{2}$, as required. Thus we must move $DE$ to the left, scaling $\\triangle EAD$ by a factor of $k$ such that $\\text{Area} \\ \\triangle EAD = \\frac{1}{2} \\ \\text{Area} \\ \\triangle CAB \\implies \\frac{1}{2}k^{2}(1+\\frac{1}{\\sqrt{3}}) = \\frac{1}{4}(1+\\sqrt{3}) \\implies k^{2} = \\frac{\\sqrt{3}}{2} \\implies k = (\\frac{3}{4})^{\\frac{1}{4}}$. Thus $\\frac{AD}{AB} = \\frac{k(1+\\frac{1}{\\sqrt{3}})}{1+\\sqrt{3}} = \\frac{k}{\\sqrt{3}} = (\\frac{3}{4})^{\\frac{1}{4}} \\times (\\frac{1}{9})^{\\frac{1}{4}} = (\\frac{1}{12})^{\\frac{1}{4}} = \\frac{1}{\\sqrt[4]{12}}$, which is answer $\\boxed{E}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/30.json | AHSME |
1987_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | A cube of cheese $C=\{(x, y, z)| 0 \le x, y, z \le 1\}$ is cut along the planes $x=y, y=z$ and $z=x$. How many pieces are there?
(No cheese is moved until all three cuts are made.)
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$
| [
"The cut $x = y$ separates the cube into points with $x < y$ and points with $x > y$, and analogous results apply for the other cuts. Thus, which piece a particular point is in depends only on the relative sizes of its coordinates $x$, $y$, and $z$ - for example, all points with the ordering $x < y < z$ are in the same piece. Thus, as there are $3! = 6$ possible orderings, there are $6$ pieces, which is answer $\\boxed{B}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/27.json | AHSME |
1987_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | $(1+x^2)(1-x^3)$ equals
$\text{(A)}\ 1 - x^5\qquad \text{(B)}\ 1 - x^6\qquad \text{(C)}\ 1+ x^2 -x^3\qquad \\ \text{(D)}\ 1+x^2-x^3-x^5\qquad \text{(E)}\ 1+x^2-x^3-x^6$
| [
"We multiply: $(1+x^2)(1-x^3) = 1 - x^3 + x^2 - x^5$. Thus the answer is $\\boxed{D}$.\n-slackroadia\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/1.json | AHSME |
1987_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | Let $c$ be a constant. The simultaneous equations
\begin{align*}x-y = &\ 2 \\cx+y = &\ 3 \\\end{align*}
have a solution $(x, y)$ inside Quadrant I if and only if
$\textbf{(A)}\ c=-1 \qquad \textbf{(B)}\ c>-1 \qquad \textbf{(C)}\ c<\frac{3}{2} \qquad \textbf{(D)}\ 0<c<\frac{3}{2}\\ \qquad \textbf{(E)}\ -1<c<\frac{3}{2}$
| [
"We can easily solve the equations algebraically to deduce $x = \\frac{5}{c+1}$ and $y = \\frac{3-2c}{c+1}$. Thus we firstly need $x > 0 \\implies c + 1 > 0 \\implies c > -1$. Now $y > 0$ implies $\\frac{3-2c}{c+1} > 0$, and since we now know that $c+1$ must be $>0$, the inequality simply becomes $3-2c > 0 \\implies 3 > 2c \\implies c < \\frac{3}{2}$. Thus we combine the inequalities $c > -1$ and $c < \\frac{3}{2}$ to get $-1 < c < \\frac{3}{2}$, which is answer $\\boxed{\\text{E}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/11.json | AHSME |
1987_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | A triangular corner with side lengths $DB=EB=1$ is cut from equilateral triangle ABC of side length $3$.
The perimeter of the remaining quadrilateral is
[asy] draw((0,0)--(2,0)--(2.5,.87)--(1.5,2.6)--cycle, linewidth(1)); draw((2,0)--(3,0)--(2.5,.87)); label("3", (0.75,1.3), NW); label("1", (2.5, 0), S); label("1", (2.75,.44), NE); label("A", (1.5,2.6), N); label("B", (3,0), S); label("C", (0,0), W); label("D", (2.5,.87), NE); label("E", (2,0), S); [/asy]
$\text{(A)} \ 6 \qquad \text{(B)} \ 6\frac{1}{2} \qquad \text{(C)} \ 7 \qquad \text{(D)} \ 7\frac{1}{2} \qquad \text{(E)} \ 8$
| [
"$\\triangle DBE$ is similar to $\\triangle ABC$ by AA, so $\\overline{DE}$ = 1 by similarity, and $\\overline{CE} = \\overline{AD} = 2$, by subtraction. Thus the perimeter is $3+2+2+1 = 8$, or $\\boxed{E}$. -slackroadia\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/2.json | AHSME |
1987_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | Let $a, b, c, d$ be real numbers. Suppose that all the roots of $z^4+az^3+bz^2+cz+d=0$ are complex numbers lying on a circle
in the complex plane centered at $0+0i$ and having radius $1$. The sum of the reciprocals of the roots is necessarily
$\textbf{(A)}\ a \qquad \textbf{(B)}\ b \qquad \textbf{(C)}\ c \qquad \textbf{(D)}\ -a \qquad \textbf{(E)}\ -b$
| [
"Let's denote the roots of the polynomial as $z_1, z_2, z_3, z_4$. We know that the magnitudes of these 4 roots are 1 as given in the problem statement. Therefore, we have $z_1 \\overline{z_1}, z_2 \\overline{z_2}, z_3 \\overline{z_3}, z_4 \\overline{z_4} = 1$. We want to find $\\frac{1}{z_1} + \\frac{1}{z_2} + \\frac{1}{z_3} + \\frac{1}{z_4}$ which is $\\overline{z_1} + \\overline{z_2} + \\overline{z_3} + \\overline{z_4}$. Remember that these are the roots of polynomials. Whenever complex numbers are the roots of a polynomial with real coefficients, we know they come in complex-conjugate pairs. Therefore, $\\overline{z_1} + \\overline{z_2} + \\overline{z_3} + \\overline{z_4} = z_1 + z_2 + z_3 + z_4$. However, by Vieta's, $z_1 + z_2 + z_3 + z_4 = -a$. Thus, the answer is $\\boxed{\\textbf{(D) } -a }$. --$\\text{lucasxia01}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/28.json | AHSME |
1987_AHSME_Problems | 12 | 0 | Other | Multiple Choice | In an office, at various times during the day the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's in-box. When there is time, the secretary takes the top letter off the pile and types it. If there are five letters in all, and the boss delivers them in the order $1\ 2\ 3\ 4\ 5$, which of the following could not be the order in which the secretary types them?
$\textbf{(A)}\ 1\ 2\ 3\ 4\ 5 \qquad \textbf{(B)}\ 2\ 4\ 3\ 5\ 1 \qquad \textbf{(C)}\ 3\ 2\ 4\ 1\ 5 \qquad \textbf{(D)}\ 4\ 5\ 2\ 3\ 1 \qquad \textbf{(E)}\ 5\ 4\ 3\ 2\ 1 \qquad$
| [
"We can show that $A$, $B$, $C$, and $E$ are possible.\n\n\n$A$: the boss delivers $1$, then the secretary does $1$, then the boss delivers $2$, then the secretary does $2$, etc. up to $5$.\n\n\n$B$: the boss delivers $1$, then delivers $2$, then the secretary does $2$, then the boss delivers $3$, then delivers $4$, then the secretary does $4$, then the secretary does $3$, then the boss delivers $5$, then the secretary does $5$, then the secretary does $1$.\n\n\n$C$: the boss delivers $1$, then delivers $2$, then delivers $3$, then the secretary does $3$, then does $2$, then the boss delivers $4$, then the secretary does $4$, then does $1$, then the boss delivers $5$, then the secretary does $5$.\n\n\n$E$: the boss delivers $1$, then $2$, then $3$, then $4$, then $5$, then the secretary does $5$, does $4$, etc. down to $1$.\n\n\nHence the answer must be $\\boxed{\\text{D}}$. To give a complete proof, we need to show $D$ is impossible. In order for the secretary to start with $4$, the boss must deliver $1$, $2$, $3$, and $4$, then the secretary does $4$. To get $5$ next, the boss next delivers $5$, then the secretary does $5$, but now we can't get $2$, $3$, $1$: the remaining letters must start with $3$, which is on the top of the pile. Hence $D$ is impossible, as required.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/12.json | AHSME |
1987_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | How many polynomial functions $f$ of degree $\ge 1$ satisfy
$f(x^2)=[f(x)]^2=f(f(x))$ ?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \text{finitely many but more than 2}\\ \qquad \textbf{(E)}\ \infty$
| [
"Let $f(x) = \\sum_{k=0}^{n} a_{k} x^{k}$ be a polynomial satisfying the condition, so substituting it in, we find that the highest powers in each of the three expressions are, respectively, $a_{n}x^{2n}$, $a_{n}^{2}x^{2n}$, and $a_{n}^{n+1}x^{n^{2}}$. If polynomials are identically equal, each term must be equal, so we get $2n = n^2$ and $a_{n} = a_{n}^{2} = a_{n}^{n+1}$, so since $n \\geq 1$, we must have $n = 2$, and since $a_{n} \\neq 0$, we have $a_{n} = 1$. The given condition now becomes $x^4 + bx^2 + c \\equiv (x^2 + bx + c)^2$, so we must have $b = 0$, or else the right-hand side would have a cubic term that the left-hand side does not. Thus we get $x^4 + c \\equiv (x^2 + c)^2$, so we must have $c = 0$, or else the right-hand side would have an $x^2$ term that the left-hand side does not. Thus the only possibility is $f(x) = x^2$, i.e. there is only $1$ solution, so the answer is $\\boxed{\\text{B}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/24.json | AHSME |
1987_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | $ABC$ is a triangle: $A=(0,0), B=(36,15)$ and both the coordinates of $C$ are integers. What is the minimum area $\triangle ABC$ can have?
$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \frac{13}{2}\qquad \textbf{(E)}\ \text{there is no minimum}$
| [
"Let $C$ have coordinates $(p, q)$. Then by the Shoelace Formula, the area of $\\triangle ABC$ is $\\frac{3}{2} \\lvert {12q-5p} \\rvert$. Since $p$ and $q$ are integers, $\\lvert {12q-5p} \\rvert$ is a positive integer, and by Bezout's Lemma, it can equal $1$ (e.g. with $q = 2, p = 5$), so the minimum area is $\\frac{3}{2} \\times 1 = \\frac{3}{2}$, which is answer $\\boxed{C}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/25.json | AHSME |
1987_AHSME_Problems | 13 | 0 | Geometry | Multiple Choice | A long piece of paper $5$ cm wide is made into a roll for cash registers by wrapping it $600$ times around a cardboard tube of diameter $2$ cm,
forming a roll $10$ cm in diameter. Approximate the length of the paper in meters.
(Pretend the paper forms $600$ concentric circles with diameters evenly spaced from $2$ cm to $10$ cm.)
$\textbf{(A)}\ 36\pi \qquad \textbf{(B)}\ 45\pi \qquad \textbf{(C)}\ 60\pi \qquad \textbf{(D)}\ 72\pi \qquad \textbf{(E)}\ 90\pi$
| [
"Notice (by imagining unfolding the roll), that the length of the paper is equal to the sum of the circumferences of the concentric circles, which is $\\pi$ times the sum of the diameters. Now the, the diameters form an arithmetic series with first term $2$, last term $10$, and $600$ terms in total, so using the formula $\\frac{1}{2}n(a+l)$, the sum is $300 \\times 12 = 3600$, so the length is $3600\\pi$ centimetres, or $36\\pi$ metres, which is answer $\\boxed{\\text{A}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/13.json | AHSME |
1987_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | Consider the sequence of numbers defined recursively by $t_1=1$ and for $n>1$ by $t_n=1+t_{(n/2)}$ when $n$ is even
and by $t_n=\frac{1}{t_{(n-1)}}$ when $n$ is odd. Given that $t_n=\frac{19}{87}$, the sum of the digits of $n$ is
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 17 \qquad \textbf{(C)}\ 19 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 23$
| [
"If $n$ is even, then $t_{(n/2)}$ would be negative, which is not possible. Therefore, $n$ is odd. With this function, backwards thinking is the key. If $t_x < 1$, then $x$ is odd, and $t_{(x-1)} = \\frac{1}{t_{x}}$. Otherwise, you keep on subtracting 1 and halving x until $t_\\frac{x}{2^{n}} < 1$.\nWe can use this logic to go backwards until we reach $t_1 = 1$, like so:\n\n\n$t_n=\\frac{19}{87}\\\\\\\\t_{n-1} = \\frac{87}{19}\\\\\\\\t_{\\frac{n-1}{2}} = \\frac{68}{19}\\\\\\\\t_{\\frac{n-1}{4}} = \\frac{49}{19}\\\\\\\\t_{\\frac{n-1}{8}} = \\frac{30}{19}\\\\\\\\t_{\\frac{n-1}{16}} = \\frac{11}{19}\\\\\\\\t_{\\frac{n-1}{16} - 1} = \\frac{19}{11}\\\\\\\\t_{\\frac{\\frac{n-1}{16} - 1}{2}} = \\frac{8}{11}\\\\\\\\t_{\\frac{\\frac{n-1}{16} - 1}{2} - 1} = \\frac{11}{8}\\\\\\\\t_{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2}} = \\frac{3}{8}\\\\\\\\t_{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1} = \\frac{8}{3}\\\\\\\\t_{\\frac{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{2}} = \\frac{5}{3}\\\\\\\\t_{\\frac{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4}} = \\frac{2}{3}\\\\\\\\t_{\\frac{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1} = \\frac{3}{2}\\\\\\\\t_{\\frac{\\frac{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2}} = \\frac{1}{2}\\\\\\\\t_{\\frac{\\frac{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1} = 2\\\\\\\\t_{\\frac{\\frac{\\frac{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1}{2}} = t_1 = 1 \\Rightarrow \\frac{\\frac{\\frac{\\frac{\\frac{\\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1}{2} = 1 \\Rightarrow n = 1905$, so the answer is $\\boxed{\\textbf{(A)} 15}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/29.json | AHSME |
1987_AHSME_Problems | 3 | 0 | Number Theory | Multiple Choice | How many primes less than $100$ have $7$ as the ones digit? (Assume the usual base ten representation)
$\text{(A)} \ 4 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 6 \qquad \text{(D)} \ 7 \qquad \text{(E)} \ 8$
| [
"List out all numbers that have 7 as the ones digit less than 100: ${7, 17, 27, 37, 47, 57, 67, 77, 87, 97}$. Only $7, 17,37, 47,67,$ and $97$ are prime. Thus, it is $\\boxed{C}$. -slackroadia\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/3.json | AHSME |
1987_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | In the figure the sum of the distances $AD$ and $BD$ is
[asy] draw((0,0)--(13,0)--(13,4)--(10,4)); draw((12.5,0)--(12.5,.5)--(13,.5)); draw((13,3.5)--(12.5,3.5)--(12.5,4)); label("A", (0,0), S); label("B", (13,0), SE); label("C", (13,4), NE); label("D", (10,4), N); label("13", (6.5,0), S); label("4", (13,2), E); label("3", (11.5,4), N); [/asy]
$\textbf{(A)}\ \text{between 10 and 11} \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ \text{between 15 and 16}\qquad\\ \textbf{(D)}\ \text{between 16 and 17}\qquad \textbf{(E)}\ 17$
| [
"Using Pythagoras' Theorem, $BD = \\sqrt{3^2 + 4^2} = 5$, and $AD = \\sqrt{(13 - 3)^2 + 4^2} = \\sqrt{116}$. Thus the sum is $5 + \\sqrt{116}$, and as $100 < 116 < 121$, $10 < \\sqrt{116} < 11$, so that the sum is between $5+10$ and $5+11$, or $15$ and $16$, which is answer $\\boxed{\\text{C}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/8.json | AHSME |
1987_AHSME_Problems | 22 | 0 | Geometry | Multiple Choice | A ball was floating in a lake when the lake froze. The ball was removed (without breaking the ice),
leaving a hole $24$ cm across as the top and $8$ cm deep. What was the radius of the ball (in centimeters)?
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 8\sqrt{3} \qquad \textbf{(E)}\ 6\sqrt{6}$
| [
"Consider a cross-section of this problem in which a circle lies with its center somewhere above a line. A line segment of $8$ cm can be drawn from the line to the bottom of the ball. Denote the distance between the center of the circle and the line as $x$. We can construct a right triangle by dragging the center of the circle to the intersection of the circle and the line. We then have the equation $x^2+(12)^2=(x+8)^2$, $x^2+144=x^2+16x+64$. Solving, the answer is ${\\textbf{(C)}\\ 13}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/22.json | AHSME |
1987_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | It takes $A$ algebra books (all the same thickness) and $H$ geometry books (all the same thickness,
which is greater than that of an algebra book) to completely fill a certain shelf.
Also, $S$ of the algebra books and $M$ of the geometry books would fill the same shelf.
Finally, $E$ of the algebra books alone would fill this shelf. Given that $A, H, S, M, E$ are distinct positive integers,
it follows that $E$ is
$\textbf{(A)}\ \frac{AM+SH}{M+H} \qquad \textbf{(B)}\ \frac{AM^2+SH^2}{M^2+H^2} \qquad \textbf{(C)}\ \frac{AH-SM}{M-H}\qquad \textbf{(D)}\ \frac{AM-SH}{M-H}\qquad \textbf{(E)}\ \frac{AM^2-SH^2}{M^2-H^2}$
| [
"Let $x$ and $y$ be the thicknesses of an algebra book and geometry book, respectively, and let $z$ be the length of the shelf. Then from the given information,\n\\[Ax + Hy = z\\]\n\\[Sx + My = z\\]\n\\[Ex = z.\\] \nFrom the third equation, $x = z/E$. Substituting into the first two equations, we get\n\\[\\frac{A}{E} z + Hy = z,\\]\n\\[\\frac{S}{E} z + My = z.\\]\n\n\nFrom the first equation,\n\\[Hy = z - \\frac{A}{E} z = \\frac{E - A}{E} z,\\]\nso\n\\[\\frac{y}{z} = \\frac{E - A}{EH}.\\]\nFrom the second equation,\n\\[My = z - \\frac{S}{E} z = \\frac{E - S}{E} z,\\]\nso\n\\[\\frac{y}{z} = \\frac{E - S}{ME}.\\]\nHence,\n\\[\\frac{E - A}{EH} = \\frac{E - S}{ME}.\\]\nMultiplying both sides by $HME$, we get $ME - AM = HE - HS$. Then $(M - H)E = AM - HS$, so\n\\[E = \\boxed{\\frac{AM - HS}{M - H}}.\\]\nThe answer is (D).\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/18.json | AHSME |
1987_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | $\frac{2^1+2^0+2^{-1}}{2^{-2}+2^{-3}+2^{-4}}$ equals
$\text{(A)} \ 6 \qquad \text{(B)} \ 8 \qquad \text{(C)} \ \frac{31}{2} \qquad \text{(D)} \ 24 \qquad \text{(E)} \ 512$
| [
"We can factorise to give $\\frac{2^1 + 2^0 + 2^{-1}}{2^{-3}(2^1 + 2^0 + 2^{-1})} = \\frac{1}{2^{-3}} = 8$, which is $\\boxed{\\text{B}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/4.json | AHSME |
1987_AHSME_Problems | 14 | 0 | Geometry | Multiple Choice | $ABCD$ is a square and $M$ and $N$ are the midpoints of $BC$ and $CD$ respectively. Then $\sin \theta=$
[asy] draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((0,0)--(2,1)); draw((0,0)--(1,2)); label("A", (0,0), SW); label("B", (0,2), NW); label("C", (2,2), NE); label("D", (2,0), SE); label("M", (1,2), N); label("N", (2,1), E); label("$\theta$", (.5,.5), SW); [/asy]
$\textbf{(A)}\ \frac{\sqrt{5}}{5} \qquad \textbf{(B)}\ \frac{3}{5} \qquad \textbf{(C)}\ \frac{\sqrt{10}}{5} \qquad \textbf{(D)}\ \frac{4}{5}\qquad \textbf{(E)}\ \text{none of these}$
| [
"Use the Sine Area Formula. We can isolate the triangle for which the angle $\\theta$ is contained in. WLOG, denote the side length of a triangle as $2$. Our midpoints are then $1$. Subtract the areas of the triangles that don't include the area of our desired triangle: $4-1-1-\\frac{1}{2} = \\frac{3}{2}.$ The Sine Area Formula tells us $\\frac{1}{2}(\\sqrt{5})^2\\sin\\theta=\\frac{3}{2}.$ Solving this equation, we get $\\sin\\theta=\\textbf{(B)}\\ \\frac{3}{5} \\qquad$\n\n\nSolution: Everyoneintexas\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/14.json | AHSME |
1987_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | If $(x, y)$ is a solution to the system
$xy=6$ and $x^2y+xy^2+x+y=63$,
find $x^2+y^2$.
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ \frac{1173}{32} \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 69 \qquad \textbf{(E)}\ 81$
| [
"First note that $x^2y+xy^2+x+y= (xy+1)(x+y)$. Substituting $6$ for $xy$ gives $7(x+y)= 63$, giving a result of $x+y=9$. Squaring this equation and subtracting by $12$, gives us $x^2+y^2= \\boxed{69}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/15.json | AHSME |
1987_AHSME_Problems | 5 | 0 | Number Theory | Multiple Choice | A student recorded the exact percentage frequency distribution for a set of measurements, as shown below.
However, the student neglected to indicate $N$, the total number of measurements. What is the smallest possible value of $N$?
\[\begin{tabular}{c c}\text{measured value}&\text{percent frequency}\\ \hline 0 & 12.5\\ 1 & 0\\ 2 & 50\\ 3 & 25\\ 4 & 12.5\\ \hline\ & 100\\ \end{tabular}\]
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 16 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 50$
| [
"Note that $12.5\\% = \\frac{1}{8}$, $25\\% = \\frac{1}{4}$, and $50\\% = \\frac{1}{2}$. Thus, since the frequencies must be integers, $N$ must be divisible by $2$, $4$, and $8$ (so that $\\frac{N}{8}$ etc. are integers), or in other words, $N$ is divisible by $8$. Thus the smallest possible value of $N$ is the smallest positive multiple of $8$, which is $8$ itself, or $\\boxed{\\text{B}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/5.json | AHSME |
1987_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | Which of the following is closest to $\sqrt{65}-\sqrt{63}$?
$\textbf{(A)}\ .12 \qquad \textbf{(B)}\ .13 \qquad \textbf{(C)}\ .14 \qquad \textbf{(D)}\ .15 \qquad \textbf{(E)}\ .16$
| [
"We have $\\sqrt{65} > 8 > 7.5$. Also $7.5^2 = (7 + 0.5)^2 = 7^2 + 2 \\cdot 7 \\cdot 0.5 + 0.5^2 = 49 + 7 + 0.25 = 56.25 < 63$, so $\\sqrt{63} > 7.5$. Thus $\\sqrt{65} + \\sqrt{63} > 7.5 + 7.5 = 15$. Now notice that $\\sqrt{65} - \\sqrt{63} = \\frac{(\\sqrt{65} - \\sqrt{63})(\\sqrt{65} + \\sqrt{63})}{\\sqrt{65} + \\sqrt{63}} = \\frac{2}{\\sqrt{65} + \\sqrt{63}}$, so $\\sqrt{65} - \\sqrt{63} < \\frac{2}{15} = 0.1333333...$, so the answer must be $A$ or $B$. To determine which, we write $\\sqrt{65} - \\sqrt{63} > 0.125 \\iff 65 - 2\\sqrt{65 \\cdot 63} + 63 > 0.015625 \\iff 128 - 0.015625 > 2\\sqrt{4095} \\iff \\sqrt{4095} < 64 - 0.0078125 \\iff 4095 < 4096 - 128 \\cdot 0.0078125 + 0.0078125^2 = 4096 - 1 + 0.0078125^2$ which is true. Hence as the expression is greater than $0.125$, and less than or equal to $0.13$ (since we showed it is certainly less than $0.1333333...$), it is closest to $0.13$, which is answer $\\boxed{\\text{B}}$.\n\n\n($\\sqrt{65} - \\sqrt{63}$ is approximately equal to $0.125003815$)\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/19.json | AHSME |
1987_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | If $p$ is a prime and both roots of $x^2+px-444p=0$ are integers, then
$\textbf{(A)}\ 1<p\le 11 \qquad \textbf{(B)}\ 11<p \le 21 \qquad \textbf{(C)}\ 21< p \le 31 \\ \qquad \textbf{(D)}\ 31< p\le 41\qquad \textbf{(E)}\ 41< p\le 51$
| [
"For integer roots, we need the discriminant, which is $p^2 - 4 \\cdot 1 \\cdot (-444p) = p^2 + 1776p = p(p+1776)$, to be a perfect square. Now, this means that $p$ must divide $p + 1776$, as if it did not, there would be a lone prime factor of $p$, and so this expression could not possibly be a perfect square. Thus $p$ divides $p + 1776$, which implies $p$ divides $1776 = 2^{4} \\cdot 3 \\cdot 37$, so we must have $p = 2$, $3$, or $37$. It is easy to verify that neither $p = 2$ nor $p = 3$ make $p(p+1776)$ a perfect square, but $p = 37$ does, so the answer is $31 < p \\le 41$, which is $\\boxed{\\text{D}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/23.json | AHSME |
1987_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | The first four terms of an arithmetic sequence are $a, x, b, 2x$. The ratio of $a$ to $b$ is
$\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{2} \qquad \textbf{(D)}\ \frac{2}{3}\qquad \textbf{(E)}\ 2$
| [
"To get from the second term to the fourth term, we add $2$ lots of the common difference, so if the common difference is $d$, we have $2d = 2x -x = x \\implies d = \\frac{x}{2}$. Thus the sequence is $\\frac{x}{2}, x, \\frac{3x}{2}, 2x$, so the ratio is $\\frac{\\frac{x}{2}}{\\frac{3x}{2}} = \\frac{1}{3}$, which is answer $\\boxed{\\text{B}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1987_AHSME_Problems/9.json | AHSME |
1978_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | If $a,b,c$ are non-zero real numbers such that \[\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a},\] and \[x=\frac{(a+b)(b+c)(c+a)}{abc},\] and $x<0,$ then $x$ equals
$\textbf{(A) }{-}1\qquad \textbf{(B) }{-}2\qquad \textbf{(C) }{-}4\qquad \textbf{(D) }{-}6\qquad \textbf{(E) }{-}8$
| [
"From the equation \\[\\frac{a+b-c}{c}=\\frac{a-b+c}{b}=\\frac{-a+b+c}{a},\\] we add $2$ to each fraction to get \\[\\frac{a+b+c}{c}=\\frac{a+b+c}{b}=\\frac{a+b+c}{a}.\\]\nWe perform casework on $a+b+c:$\n\n\n\\begin{itemize}\n\\item If $a+b+c\\neq0,$ then $a=b=c,$ from which $x=\\frac{(2a)(2a)(2a)}{a^3}=8.$ However, this contradicts the precondition $x<0.$\n\\end{itemize}\n\\begin{itemize}\n\\item If $a+b+c=0,$ then $x=\\frac{(-c)(-a)(-b)}{abc}=\\boxed{\\textbf{(A) }{-}1}.$\n\\end{itemize}\n~MRENTHUSIASM\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/20.json | AHSME |
1978_AHSME_Problems | 16 | 0 | Other | Multiple Choice | In a room containing $N$ people, $N > 3$, at least one person has not shaken hands with everyone else in the room.
What is the maximum number of people in the room that could have shaken hands with everyone else?
$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }N-1\qquad \textbf{(D) }N\qquad \textbf{(E) }\text{none of these}$
| [
"No solutions yet!\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/16.json | AHSME |
1978_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | The number of distinct pairs $(x,y)$ of real numbers satisfying both of the following equations:
\[x=x^2+y^2 \ \ y=2xy\]
is
$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }3\qquad \textbf{(E) }4$
| [
"If $x=x^2+y^2$ and $y=2xy$, then we can break this into two cases.\n\n\nCase 1: $y = 0$\n\n\nIf $y = 0$, then $x = x^2$ and $0 = 0$\n\n\nTherefore, $x = 0$ or $x = 1$\n\n\nThis yields 2 solutions\n\n\nCase 2: $x = \\frac{1}{2}$\n\n\nIf $x = \\frac{1}{2}$, this means that $y = y$, and $\\frac{1}{2} = \\frac{1}{4} + y^2$. \n\n\nBecause y can be negative or positive, this yields $y = \\frac{1}{2}$ or $y = -\\frac{1}{2}$\n\n\nThis yields another 2 solutions.\n\n\n$2+2 = \\boxed{\\textbf{(E) 4}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/6.json | AHSME |
1978_AHSME_Problems | 7 | 0 | Geometry | Multiple Choice | Opposite sides of a regular hexagon are $12$ inches apart. The length of each side, in inches, is
$\textbf{(A) }7.5\qquad \textbf{(B) }6\sqrt{2}\qquad \textbf{(C) }5\sqrt{2}\qquad \textbf{(D) }\frac{9}{2}\sqrt{3}\qquad \textbf{(E) }4\sqrt{3}$
| [
"Draw a perpendicular through the midpoint of the line of length $12$ such that it passes through a vertex. We now have created $2$ $30-60-90$ triangles. Using the ratios, we get that the hypotenuse is $6 \\times \\frac {2}{\\sqrt{3}}$ $= \\frac {12}{\\sqrt {3}}$ $= 4\\sqrt{3} \\rightarrow \\boxed {\\textbf{(E)}}$\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/7.json | AHSME |
1978_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | For all positive numbers $x$ distinct from $1$,
\[\frac{1}{\log_3(x)}+\frac{1}{\log_4(x)}+\frac{1}{\log_5(x)}\]
equals
$\text{(A) }\frac{1}{\log_{60}(x)}\qquad\\ \text{(B) }\frac{1}{\log_{x}(60)}\qquad\\ \text{(C) }\frac{1}{(\log_{3}(x))(\log_{4}(x))(\log_{5}(x))}\qquad\\ \text{(D) }\frac{12}{\log_{3}(x)+\log_{4}(x)+\log_{5}(x)}\qquad\\ \text{(E) }\frac{\log_{2}(x)}{\log_{3}(x)\log_{5}(x)}+\frac{\log_{3}(x)}{\log_{2}(x)\log_{5}(x)}+\frac{\log_{5}(x)}{\log_{2}(x)\log_{3}(x)}$
| [
"\\[\\frac{1}{\\log_3(x)}+\\frac{1}{\\log_4(x)}+\\frac{1}{\\log_5(x)}\\implies \\frac{1}{\\frac{\\log(x)}{\\log(3)}}+\\frac{1}{\\frac{\\log(x)}{\\log(4)}}+\\frac{1}{\\frac{\\log(x)}{\\log(5)}}\\implies \\frac{\\log(3)+\\log(4)+\\log(5)}{\\log(x)}=\\frac{\\log(60)}{\\log(x)}\\implies \\log_x(60)\\implies \\frac{1}{\\log_{60}(x)}\\]\nThus, the answer is $(\\textbf{A})$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/21.json | AHSME |
1978_AHSME_Problems | 10 | 0 | Geometry | Multiple Choice | If $\mathit{B}$ is a point on circle $\mathit{C}$ with center $\mathit{P}$, then the set of all points $\mathit{A}$ in the plane of circle $\mathit{C}$ such that the distance between $\mathit{A}$ and $\mathit{B}$ is less than or equal to the distance between $\mathit{A}$
and any other point on circle $\mathit{C}$ is
$\textbf{(A) }\text{the line segment from }P \text{ to }B\qquad\\ \textbf{(B) }\text{the ray beginning at }P \text{ and passing through }B\qquad\\ \textbf{(C) }\text{a ray beginning at }B\qquad\\ \textbf{(D) }\text{a circle whose center is }P\qquad\\ \textbf{(E) }\text{a circle whose center is }B$
| [
"Begin by drawing circle P and point B. To satisfy the conditions of the problem, A needs to be in a position where it is closer to B. This can only happen if A and B are on the same line, so we choose from answer choices A and B. \n\n\nWe can pick some arbitrary point A outside circle P that is collinear with B and see that the conditions still hold, so the answer is $\\boxed{\\textbf{(B)}}.$\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/10.json | AHSME |
1978_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | [asy] size(100); real a=4, b=3; // import cse5; pathpen=black; pair A=(a,0), B=(0,b), C=(0,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,SW)--cycle); pair X=IP(B--A,(0,0)--(b,a)); D(CP((X+C)/2,C)); D(MP("R",IP(CP((X+C)/2,C),B--C),NW)--MP("Q",IP(CP((X+C)/2,C),A--C+(0.1,0)))); //Credit to chezbgone2 for the diagram [/asy]
In $\triangle ABC, AB = 10~ AC = 8$ and $BC = 6$. Circle $P$ is the circle with smallest radius which passes through $C$ and is tangent to $AB$. Let $Q$ and $R$ be the points of intersection, distinct from $C$ , of circle $P$ with sides $AC$ and $BC$, respectively. The length of segment $QR$ is
$\textbf{(A) }4.75\qquad \textbf{(B) }4.8\qquad \textbf{(C) }5\qquad \textbf{(D) }4\sqrt{2}\qquad \textbf{(E) }3\sqrt{3}$
| [
"We know that triangle $RCQ$ is similar to triangle $ABC$. We draw a line to point $D$ on hypotenuse $AB$ such that $\\angle QDR$ is $90 ^\\circ$ and that $RDQC$ is a rectangle. Since triangle $RCQ$ is similar to triangle $ABC$, let $RC$ be $4x$ and $RD/CQ$ be $3x$. Now we have line segment $AQ$ = $8-3x$, and line segment $RB$ = $6-4x$. Since $BD + DA = AB$, we use simple algebra and Pythagorean Theorem to get $\\sqrt {(3x)^2 + (6-4x)^2}$ + $\\sqrt {(4x)^2 + (8-3x)^2}$ = $10$. Expanding and simplifying gives us $\\sqrt {25x^2-48x+36}$ + $\\sqrt {25x^2-48x+64}$ = $10$. \n\n\nSquaring both sides is not an option since it is both messy and time consuming, so we should proceed to subtracting both sides by $\\sqrt {25x^2-48x+36}$. Now, we can square both sides and simplify to get $0 = 72 - 20 \\sqrt{25x^2-48x+36}$. Dividing both sides by $4$, we get $18 - 5 \\sqrt {25x^2-48x+36}$ = $0$. We then add $5 \\sqrt {25x^2-48x+36}$ to both sides to get $18 = 5 \\sqrt {25x^2-48x+36}$. Since this is very messy, let $25x^2 - 48x = y$. Squaring both sides, we get $324 = 25y + 900, 25y = -576$. Solving for $y$, we have $y = -23.04$. Plugging in $y$ as $25x^2-48x$, we have $25x^2-48x+23.04 = 0$. Using the quadratic equation, we get $\\frac {48+0}{50}$. Therefore, $x = \\frac {48}{50}$. \n\n\nRemember that our desired answer is the hypotenuse of the triangle $3x - 4x - 5x$. Since $5x$ is the hypotenuse, our answer is $\\boxed {(B)4.8}$\n\n\n~Arcticturn\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/26.json | AHSME |
1978_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | In a tennis tournament, $n$ women and $2n$ men play, and each player plays exactly one match with every other player.
If there are no ties and the ratio of the number of matches won by women to the number of matches won by men is $7/5$, then $n$ equals
$\textbf{(A) }2\qquad \textbf{(B) }4\qquad \textbf{(C) }6\qquad \textbf{(D) }7\qquad \textbf{(E) }\text{none of these}$
| [
"Since there are $n$ women, the number of matches between only women is $\\frac{n(n-1)}{2},$ and similarly, there are $(2n - 1)n$ matches between only men. Since every woman plays every man exactly once, there are $2n\\cdot n = 2n^2$ matches which are between a man and a woman. Call these $2n^2$ matches co-ed matches, and let $w$ be the number of co-ed matches won by women.\n\n\nThen it follows that\n\\[\\frac 75 = \\frac{\\frac{n(n-1)}{2} + w}{(2n - 1)n + 2n^2 - w},\\]\nwhich can be simplified to\n\\[w = \\frac{17}{8}n^2 - \\frac38 n.\\]\n\n\nThe number of matches won by women must be less than the total number of matches, so we obtain the inequality\n\\[\\frac{17}{8}n^2 - \\frac38 n\\leq 2n^2.\\]\nRearranging and factoring gives\n\\[0\\leq -n(n-3),\\]\nand the only integers which satisfy this inequality are $n = 0,1,2,$ and $3.$\n\n\nClearly, there could not have been $0$ people in the tournament, so $n\\neq 0.$ If $n = 1,$ then there would have been only one woman and two men in the tournament, in which case the woman could not have won the majority of the matches.\n\n\nWe can now plug $n = 2$ back into the equation\n\\[\\frac 75 = \\frac{\\frac{n(n-1)}{2} + w}{(2n - 1)n + 2n^2 - w},\\]\nand solving for $w$ gives $w = \\frac{31}{4}.$ Since $w$ must be an integer, $n$ cannot be $2.$ It follows that $n = 3,$ so the answer is (E). (When $n = 3,$ solving gives $w = 18.$)\n\n\n",
"Since there are $n$ women and $2n$ men, there are a total of $n+2n=3n$ players. Hence, the number of total matches must be $\\binom{3n}{2}=\\frac{3n(3n-1)}{2}$. We also know that the ratio of the number of matches won by women to the number of matches won by men is $\\frac{7}{5}$ and that there were no draws, so the total number of matches must be $(7+5)x=12x$ for some value $x$. This gives\\[\\frac{3n(3n-1)}{2}=12x\\]So\\[n(3n-1)=8x\\]It follows that $n=8,4,3,2,1$ (various ways of splitting $8x$ into two factors). However, for solutions $n=4,2,1$, $x$ isn't an integer, therefore leaving solutions $n=8$ and $n=3$. \n\n\n$n=8$ is invalid. This can be shown as when $n=8$,\\[\\frac{24(23)}{2}=12x\\]\\[x=23\\]The ratio of matches won by gender is $\\frac{7x}{5x}=\\frac{161}{115}$, so the number of matches won by women must be at least $161$. The maximum number of matches won by women is equivalent to the number of matches between men (generates 1 match won by men no matter the outcome) subtracted from the total number of matches, which is $\\binom{24}{2}-\\binom{16}{2}=156$. As the maximum number of matches won is smaller than $161$, the solution is extraneous/invalid.\n\n\nHence, the answer must be $n=3$, or $\\boxed{\\textbf{(E) }\\text{none of these}}$\n\n\n~JcCC\n\n\n"
] | 2 | ./CreativeMath/AHSME/1978_AHSME_Problems/30.json | AHSME |
1978_AHSME_Problems | 27 | 0 | Number Theory | Multiple Choice | There is more than one integer greater than $1$ which, when divided by any integer $k$ such that $2 \le k \le 11$, has a remainder of $1$.
What is the difference between the two smallest such integers?
$\textbf{(A) }2310\qquad \textbf{(B) }2311\qquad \textbf{(C) }27,720\qquad \textbf{(D) }27,721\qquad \textbf{(E) }\text{none of these}$
| [
"Let this integer be $n$. We have\n$n\\equiv 1 \\mod 2$, $n \\equiv 1 \\mod 3$, $n \\equiv 1 \\mod 4$ $\\ldots$ $n \\equiv 1 \\mod 11$.\nRecall that if \n\\[a \\equiv b \\mod c\\]\nand \\[a \\equiv b \\mod d\\]\nthen \\[a \\equiv b \\mod (lcm (c,d))\\]\nWe see that since\n$n\\equiv 1 \\mod 2$, $n \\equiv 1 \\mod 3$, $n \\equiv 1 \\mod 4$ $\\ldots$ $n \\equiv 1 \\mod 11$.\nWe have \n\\[n \\equiv 1 \\mod (lcm (2, 3, \\ldots , 10, 11))\\]\n\n\nFrom $2$ to $11$, $8$ contains the largest power of $2$, $9$ contains the largest power of $3$, and $10$ contains the largest power of $5$. Thus, our lcm is equal to\n\\[2^3 \\cdot 3^2 \\cdot 5 \\cdot 7 \\cdot 11\\]\nSince $n > 1$, our $2$ smallest values of $n$ are \n\\[n = 1 + (2^3 \\cdot 3^2 \\cdot 5 \\cdot 7 \\cdot 11)\\] \nand\n\\[n = 1 + 2(2^3 \\cdot 3^2 \\cdot 5 \\cdot 7 \\cdot 11)\\]\nThe difference between these values is simply the value of \\[(2^3 \\cdot 3^2 \\cdot 5 \\cdot 7 \\cdot 11) = 40 \\cdot 99 \\cdot 7\\]\n\\[= 3960 \\cdot 7\\]\n\\[= 27720, \\boxed{C}\\]\n\n\n~JustinLee2017\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/27.json | AHSME |
1978_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | If $1-\frac{4}{x}+\frac{4}{x^2}=0$, then $\frac{2}{x}$ equals
$\textbf{(A) }-1\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }-1\text{ or }2\qquad \textbf{(E) }-1\text{ or }-2$
| [
"By guessing and checking, 2 works. \n$\\frac{2}{x} = \\boxed{\\textbf{(B) }1}$\n~awin\n\n\n",
"Multiplying each side by $x^2$, we get $x^2-4x+4 = 0$. Factoring, we get $(x-2)(x-2) = 0$. Therefore, $x = 2$. \n$\\frac{2}{x} = \\boxed{\\textbf{(B) }1}$\n~awin\n\n\n",
"Directly factoring, we get $(1-\\frac{2}{x})^2 = 0$. Thus $\\frac{2}{x}$ must equal $\\boxed{\\textbf{(B) }1}$\n\n\n"
] | 3 | ./CreativeMath/AHSME/1978_AHSME_Problems/1.json | AHSME |
1978_AHSME_Problems | 11 | 0 | Geometry | Multiple Choice | If $r$ is positive and the line whose equation is $x + y = r$ is tangent to the circle whose equation is $x^2 + y ^2 = r$, then $r$ equals
$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }\sqrt{2}\qquad \textbf{(E) }2\sqrt{2}$
| [
"The circle $x^2 + y^2 = r$ has center $(0,0)$ and radius $\\sqrt{r}$. Therefore, if the line $x + y = r$ is tangent to the circle $x^2 + y^2 = r$, then the distance between $(0,0)$ and the line $x + y = r$ is $\\sqrt{r}$.\n\n\nThe distance between $(0,0)$ and the line $x + y = r$ is\n\\[\\frac{|0 + 0 - r|}{\\sqrt{1^2 + 1^2}} = \\frac{r}{\\sqrt{2}}.\\]\nHence,\n\\[\\frac{r}{\\sqrt{2}} = \\sqrt{r}.\\]\nThen $r = \\sqrt{r} \\cdot \\sqrt{2}$, so $\\sqrt{r} = \\sqrt{2}$, which means $r = \\boxed{2}$ or (B), $2$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/11.json | AHSME |
1978_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | If four times the reciprocal of the circumference of a circle equals the diameter of the circle, then the area of the circle is
$\textbf{(A) }\frac{1}{\pi^2}\qquad \textbf{(B) }\frac{1}{\pi}\qquad \textbf{(C) }1\qquad \textbf{(D) }\pi\qquad \textbf{(E) }\pi^2$
| [
"Creating equations, we get $4\\cdot\\frac{1}{2\\pi r} = 2r$. Simplifying, we get $\\frac{1}{\\pi r} = r$. Multiplying each side by $r$, we get $\\frac{1}{\\pi} = r^2$. Because the formula of the area of a circle is $\\pi r^2$, we multiply each side by $\\pi$ to get $1 = \\pi r^2$.\nTherefore, our answer is $\\boxed{\\textbf{(C) }1}$\n\n\n~awin\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/2.json | AHSME |
1978_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | [asy] size(100); import cse5; pathpen=black; pair A1=(0,0), A2=(1,0), A3=(0.5,sqrt(3)/2); D(MP("A_1",A1)--MP("A_2",A2)--MP("A_3",A3,N)--cycle); pair A4=(A1+A2)/2, A5 = (A3+A2)/2, A6 = (A4+A3)/2; D(MP("A_4",A4,S)--MP("A_6",A6,W)--A3); D(A6--MP("A_5",A5,NE)--A4); //Credit to chezbgone2 for the diagram [/asy]
If $\triangle A_1A_2A_3$ is equilateral and $A_{n+3}$ is the midpoint of line segment $A_nA_{n+1}$ for all positive integers $n$,
then the measure of $\measuredangle A_{44}A_{45}A_{43}$ equals
$\textbf{(A) }30^\circ\qquad \textbf{(B) }45^\circ\qquad \textbf{(C) }60^\circ\qquad \textbf{(D) }90^\circ\qquad \textbf{(E) }120^\circ$
| [
"$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/28.json | AHSME |
1978_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | If the distinct non-zero numbers $x ( y - z),~ y(z - x),~ z(x - y )$ form a geometric progression with common ratio $r$, then $r$ satisfies the equation
$\textbf{(A) }r^2+r+1=0\qquad \textbf{(B) }r^2-r+1=0\qquad \textbf{(C) }r^4+r^2-1=0\qquad\\ \textbf{(D) }(r+1)^4+r=0\qquad \textbf{(E) }(r-1)^4+r=0$
| [
"Let the geometric progression be $a,$ $ar,$ $ar^2,$ so $a = x(y - z)$, $ar = y(z - x)$, and $ar^2 = z(x - y).$ Adding these equations, we get\n\\[a + ar + ar^2 = xy - xz + yz - xy + xz - yz = 0.\\]\nSince $a$ is nonzero, we can divide by $a$, to get $\\boxed{r^2 + r + 1 = 0}$. The answer is (A).\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/24.json | AHSME |
1978_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | Let $a$ be a positive number. Consider the set $S$ of all points whose rectangular coordinates $(x, y )$ satisfy all of the following conditions:
$\text{(i) }\frac{a}{2}\le x\le 2a\qquad \text{(ii) }\frac{a}{2}\le y\le 2a\qquad \text{(iii) }x+y\ge a\\ \\ \qquad \text{(iv) }x+a\ge y\qquad \text{(v) }y+a\ge x$
The boundary of set $S$ is a polygon with
$\textbf{(A) }3\text{ sides}\qquad \textbf{(B) }4\text{ sides}\qquad \textbf{(C) }5\text{ sides}\qquad \textbf{(D) }6\text{ sides}\qquad \textbf{(E) }7\text{ sides}$
| [
"Draw a picture. The answer is $\\fbox{D}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/25.json | AHSME |
1978_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | If $a,b,c$, and $d$ are non-zero numbers such that $c$ and $d$ are the solutions of $x^2+ax+b=0$ and $a$ and $b$ are
the solutions of $x^2+cx+d=0$, then $a+b+c+d$ equals
$\textbf{(A) }0\qquad \textbf{(B) }-2\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }(-1+\sqrt{5})/2$
| [
"By Vieta's formulas, $c + d = -a$, $cd = b$, $a + b = -c$, and $ab = d$. From the equation $c + d = -a$, $d = -a - c$, and from the equation $a + b = -c$, $b = -a - c$, so $b = d$.\n\n\nThen from the equation $cd = b$, $cb = b$. Since $b$ is nonzero, we can divide both sides of the equation by $b$ to get $c = 1$. Similarly, from the equation $ab = d$, $ab = b$, so $a = 1$. Then $b = d = -a - c = -2$. Therefore, $a + b + c + d = 1 + (-2) + 1 + (-2) = \\boxed{-2}$. The answer is (B).\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/13.json | AHSME |
1978_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | Sides $AB,~ BC, ~CD$ and $DA$, respectively, of convex quadrilateral $ABCD$ are extended past $B,~ C ,~ D$ and $A$ to points $B',~C',~ D'$ and $A'$. Also, $AB = BB' = 6,~ BC = CC' = 7, ~CD = DD' = 8$ and $DA = AA' = 9$; and the area of $ABCD$ is $10$. The area of $A 'B 'C'D'$ is
| [
"Notice that the area of $\\triangle$ $DAB$ is the same as that of $\\triangle$ $A'AB$ (same base, same height). Thus, the area of $\\triangle$ $A'AB$ is twice that (same height, twice the base). Similarly, [$\\triangle$ $BB'C$] = 2 $\\cdot$ [$\\triangle$ $ABC$], and so on.\n\n\nAdding all of these, we see that the area the four triangles around $ABCD$ is twice [$\\triangle$ $DAB$] + [$\\triangle$ $ABC$] + [$\\triangle$ $BCD$] + [$\\triangle$ $CDA$], which is itself twice the area of the quadrilateral $ABCD$. Finally, [$A'B'C'D'$] = [$ABCD$] + 4 $\\cdot$ [$ABCD$] = 5 $\\cdot$ [$ABCD$] = $\\fbox{50}$.\n\n\n~ Mathavi\n\n\nNote: Anyone with a diagram would be of great help (still new to LaTex).\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/29.json | AHSME |
1978_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | For all non-zero numbers $x$ and $y$ such that $x = 1/y$, $\left(x-\frac{1}{x}\right)\left(y+\frac{1}{y}\right)$ equals
$\textbf{(A) }2x^2\qquad \textbf{(B) }2y^2\qquad \textbf{(C) }x^2+y^2\qquad \textbf{(D) }x^2-y^2\qquad \textbf{(E) }y^2-x^2$
| [
"Using substitution, we can substitute y into the equation in the first parentheses. \nTherefore, we'll get \\[\\left(x-\\frac{1}{\\frac{1}{y}}\\right)\\left(y+\\frac{1}{y}\\right)\\]\n\n\nBecause $x = \\frac{1}{y}$, we can also see that $y = \\frac{1}{x}$. Using substitution again, we can substitute x into the second equation getting\n\\[\\left(x-\\frac{1}{\\frac{1}{y}}\\right)\\left(y+\\frac{1}{\\frac{1}{x}}\\right)\\]\n\n\nSimplifying, we get $(x-y)(y+x)$. Multiplying, we get $\\boxed{\\textbf{(D) }x^2-y^2}$\n\n\n~awin\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/3.json | AHSME |
1978_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | If $x\neq y$ and the sequences $x,a_1,a_2,y$ and $x,b_1,b_2,b_3,y$ each are in arithmetic progression, then $(a_2-a_1)/(b_2-b_1)$ equals
$\textbf{(A) }\frac{2}{3}\qquad \textbf{(B) }\frac{3}{4}\qquad \textbf{(C) }1\qquad \textbf{(D) }\frac{4}{3}\qquad \textbf{(E) }\frac{3}{2}$
| [
"WLOG, let $x =2$ and $y = 5$. From the first sequence, we get\n\\[2, 3, 4, 5\\] so $a_2 - a_1 = 1$\nFrom the second sequence, we get\n\\[2, 2+r, 2+2r, 2+3r, 5\\] so $2+4r = 5$ and $r = \\frac{3}{4}$\nThus, we have \\[2, 2 \\frac{3}{4}, 3 \\frac{1}{2} ...\\] and $b_2 - b_1 = \\frac{3}{4}$\nSo $\\frac{1}{\\frac{3}{4}} = \\frac{4}{3}$\n\\[\\boxed{D}\\]\n\n\n~JustinLee2017\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/8.json | AHSME |
1978_AHSME_Problems | 22 | 0 | Other | Multiple Choice | The following four statements, and only these are found on a card:
[asy] pair A,B,C,D,E,F,G; A=(0,1); B=(0,5); C=(11,5); D=(11,1); E=(0,4); F=(0,3); G=(0,2); draw(A--B--C--D--cycle); label("On this card exactly one statement is false.", B, SE); label("On this card exactly two statements are false.", E, SE); label("On this card exactly three statements are false.", F, SE); label("On this card exactly four statements are false.", G, SE); [/asy]
(Assume each statement is either true or false.) Among them the number of false statements is exactly
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
\subsection{Contents}
\begin{itemize}
| [
"There can be at most one true statement on the card, eliminating $\\textbf{(A)}, \\textbf{(B)},$ and $\\textbf{(C)}$. If there are $0$ true on the card, statement $4$ (\"On this card exactly four statements are false\") will be correct, causing a contradiction. Therefore, the answer is $\\boxed{\\textbf{(D) } 3}$, since $3$ are false and only the third statement (\"On this card exactly three statements are false\") is correct.\n\n\n",
"If all of them are false, that would mean that the $4$th one is false too. Therefore, $E$ is not the correct answer. If exactly $3$ of them are false, that would mean that only $1$ statement is true. This is correct since if only $1$ statement is true, the card that is true is the one that has $3$ of these statements are false. If we have $1$ or $2$ false statements, that would mean that there is more than $1$ true statement. Therefore, our answer is $\\boxed {(D)}$.\n\n\n~Arcticturn\n\n\n"
] | 2 | ./CreativeMath/AHSME/1978_AHSME_Problems/22.json | AHSME |
1978_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | What is the smallest positive integer $n$ such that $\sqrt{n}-\sqrt{n-1}<.01$?
$\textbf{(A) }2499\qquad \textbf{(B) }2500\qquad \textbf{(C) }2501\qquad \textbf{(D) }10,000\qquad \textbf{(E) }\text{There is no such integer}$
| [
"Adding $\\sqrt{n - 1}$ to both sides, we get\n\\[\\sqrt{n} < \\sqrt{n - 1} + 0.01.\\]\nSquaring both sides, we get\n\\[n < n - 1 + 0.02 \\sqrt{n - 1} + 0.0001,\\]\nwhich simplifies to\n\\[0.9999 < 0.02 \\sqrt{n - 1},\\]\nor\n\\[\\sqrt{n - 1} > 49.995.\\]\nSquaring both sides again, we get\n\\[n - 1 > 2499.500025,\\]\nso $n > 2500.500025$. The smallest positive integer $n$ that satisfies this inequality is $\\boxed{2501}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/18.json | AHSME |
1978_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | If $a = 1,~ b = 10, ~c = 100$, and $d = 1000$, then $(a+ b+ c-d) + (a + b- c+ d) +(a-b+ c+d)+ (-a+ b+c+d)$ is equal to
$\textbf{(A) }1111\qquad \textbf{(B) }2222\qquad \textbf{(C) }3333\qquad \textbf{(D) }1212\qquad \textbf{(E) }4242$
| [
"Adding all four of the equations up, we can see that it equals \n\\[3(a+b+c+d)\\]\nThis is equal to $3(1111) = \\boxed{\\textbf{(C) }3333}$ ~awin\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/4.json | AHSME |
1978_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | If an integer $n > 8$ is a solution of the equation $x^2 - ax+b=0$ and the representation of $a$ in the base-$n$ number system is $18$,
then the base-n representation of $b$ is
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 80 \qquad \textbf{(D)}\ 81 \qquad \textbf{(E)}\ 280$
| [
"Assuming the solutions to the equation are n and m, by Vieta's formulas, $n_n + m_n = 18_n$. \n\n\n$n_n = 10_n$, so $10_n + m_n = 18_n$. \n\n\n\\[m_n = 8_n\\].\n\n\nAlso by Vieta's formulas, $n_n \\cdot m_n = b_n$. \n\\[10_n \\cdot 8_n = \\boxed{80_n}\\].\n\n\nThe answer is (C) $80$\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/14.json | AHSME |
1978_AHSME_Problems | 15 | 0 | Other | Multiple Choice | If $\sin x+\cos x=1/5$ and $0\le x<\pi$, then $\tan x$ is
$\textbf{(A) }-\frac{4}{3}\qquad \textbf{(B) }-\frac{3}{4}\qquad \textbf{(C) }\frac{3}{4}\qquad \textbf{(D) }\frac{4}{3}\qquad\\ \textbf{(E) }\text{not completely determined by the given information}$
| [
"Squaring the equation, we get\n\\[\\sin ^2 x + \\cos ^2 x + 2 \\sin x \\cos x = \\frac{1}{25} =\\Rrightarrow 2\\sin x \\cos x = -\\frac{24}{25} \\Rrightarrow \\sin x \\cos x = -\\frac{12}{25}\\]\nRecall that $(a-b)^2 = (a+b)^2 - 4ab$, so\n\\[(\\sin x - \\cos x)^2 = (\\sin x + \\cos x)^2 - 4 \\sin x \\cos x\\]\n\\[(\\sin x - \\cos x)^2 = \\frac{1}{25} - 4 (- \\frac{12}{25})\\]\n\\[(\\sin x - \\cos x) = \\frac{7}{5}\\]\nWe can now solve for the values of $\\sin x$ and $\\cos x$\n\\[\\sin x + \\cos x = \\frac{1}{5}\\]\n\\[\\sin x - \\cos x = \\frac{7}{5}\\]\n\\[2 \\sin x = \\frac{8}{5} \\Rrightarrow \\sin x = \\frac{4}{5}, \\cos x = -\\frac{3}{5}\\]\nSince $\\tan x = \\frac{\\sin x}{\\cos x}$, we have\n\\[\\tan x = \\frac{\\frac{4}{5}}{-\\frac{3}{5}} = -\\frac{4}{3}\\]\n$\\boxed{A}$\n\n\n~JustinLee2017\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/15.json | AHSME |
1978_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | Four boys bought a boat for $\textdollar 60$. The first boy paid one half of the sum of the amounts paid by the other boys;
the second boy paid one third of the sum of the amounts paid by the other boys;
and the third boy paid one fourth of the sum of the amounts paid by the other boys. How much did the fourth boy pay?
$\textbf{(A) }\textdollar 10\qquad \textbf{(B) }\textdollar 12\qquad \textbf{(C) }\textdollar 13\qquad \textbf{(D) }\textdollar 14\qquad \textbf{(E) }\textdollar 15$
| [
"If the first boy paid one half of the sum of the amounts paid by the other boys, that means that he paid $\\frac{1}{3}$ of the total.\nIf the second boy paid one third of the sum of the amounts paid by the other boys, that means that he paid $\\frac{1}{4}$ of the total.\nIf the third boy paid one fourth of the sum of the amounts paid by the other boys, that means that he paid $\\frac{1}{5}$ of the total. \n\n\nSumming it up, we get $\\textdollar 20 + \\textdollar 15 + \\textdollar 12 = \\textdollar 47$.\nTherefore, our answer is $\\textdollar 60 - \\textdollar 47 = \\boxed{\\textbf{(C) }\\textdollar 13}$\n~awin\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/5.json | AHSME |
1978_AHSME_Problems | 19 | 0 | Probability | Multiple Choice | A positive integer $n$ not exceeding $100$ is chosen in such a way that if $n\le 50$, then the probability of choosing $n$ is $p$, and if $n > 50$, then the probability of choosing $n$ is $3p$. The probability that a perfect square is chosen is
$\textbf{(A) }.05\qquad \textbf{(B) }.065\qquad \textbf{(C) }.08\qquad \textbf{(D) }.09\qquad \textbf{(E) }.1$
| [
"Let's say that we will have $3$ slips for every number not exceeding $100$ but bigger than $50.$ This is to account for the $3p$ probability part. Let's now say that we will only have one slip for each number below or equal to $50.$ The probability(or $p$) will then be $\\frac{1}{200}.$ Now let's have all the squares under $50,$ which are $1,4,9,16,25,36,49.$ The probability for these are $\\frac{7}{200}.$ The numbers above $50$ that are squares are $64,81,100.$ We then need to multiply the probability by $3$ so the probability of these are $\\frac{9}{200}.$ The answer is $\\frac{7}{200}+\\frac{9}{200}=0.08\\implies\\boxed{\\textbf{(C).}}$\n\n\n~volkie thangy\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/19.json | AHSME |
1978_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | Vertex $E$ of equilateral $\triangle ABE$ is in the interior of square $ABCD$, and $F$ is the point of intersection of diagonal $BD$ and line segment $AE$. If length $AB$ is $\sqrt{1+\sqrt{3}}$ then the area of $\triangle ABF$ is
$\textbf{(A) }1\qquad \textbf{(B) }\frac{\sqrt{2}}{2}\qquad \textbf{(C) }\frac{\sqrt{3}}{2}\qquad \textbf{(D) }4-2\sqrt{3}\qquad \textbf{(E) }\frac{1}{2}+\frac{\sqrt{3}}{4}$
| [
"Place square ABCD on the coordinate plane with A at the origin. \nIn polar form, line BD is rsinθ=sqrt(1+sqrt(3))-rcosθ and line AF is θ=(pi)/3\nThis means that the length of the intersection (r) is\nr*(sqrt3)/2=sqrt(1+sqrt(3))-r/2\nSolving for r you get: r=2/sqrt(1+sqrt(3))\nUsing the firmly for area of a triangle (A=ab(Sinθ)/2) you get A=2/sqrt(1+sqrt(3))*sqrt(1+sqrt(3))*sin(pi/3)/2=(sqrt3)/2\nGetting C as the answer\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/23.json | AHSME |
1978_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | If $x<0$, then $\left|x-\sqrt{(x-1)^2}\right|$ equals
$\textbf{(A) }1\qquad \textbf{(B) }1-2x\qquad \textbf{(C) }-2x-1\qquad \textbf{(D) }1+2x\qquad \textbf{(E) }2x-1$
| [
"We have $\\sqrt{x^2} = |x|$, so we rewrite the expression as follows.\n\\[|x - \\sqrt{(x-1)^2}| = |x - |x-1||\\]\nWe know that $x < 0$, so $x-1 < 0$. Thus, we can rewrite $|x-1|$ as $1-x$. So\n\\[|x - |x-1|| = |x - (1-x)| = |2x - 1|\\].\nSince $x< 0, 2x-1 < 0$. Thus, we can write this as\n\\[|2x - 1| = 1- 2x\\]\n$\\boxed{B}$\n\n\n\n\n~JustinLee2017\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1978_AHSME_Problems/9.json | AHSME |
1965_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | For every $n$ the sum of n terms of an arithmetic progression is $2n + 3n^2$. The $r$th term is:
$\textbf{(A)}\ 3r^2 \qquad \textbf{(B) }\ 3r^2 + 2r \qquad \textbf{(C) }\ 6r - 1 \qquad \textbf{(D) }\ 5r + 5 \qquad \textbf{(E) }\ 6r+2\qquad$
| [
"In order to calculate the $r$th term of this arithmetic sequence, we can subtract the sum of the first $r-1$ terms from the sum of the first $r$ terms of the sequence. Plugging in $r$ and $r-1$ as values of $n$ in the given expression and subtracting yields \\[(3r^2+2r)-(3r^2-4r+1).\\] Simplifying gives us the final answer of $\\boxed{6r-1}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1965_AHSME_Problems/20.json | AHSME |
1965_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | If $10^{\log_{10}9} = 8x + 5$ then $x$ equals:
$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ \frac {1}{2} \qquad \textbf{(C) }\ \frac {5}{8} \qquad \textbf{(D) }\ \frac{9}{8}\qquad \textbf{(E) }\ \frac{2\log_{10}3-5}{8}$
| [
"Notice that $10^{\\log_{10} 9} = 9$. Therefore, the condition we are looking for is $9=8x+5$, or $x=\\frac{1}{2}$. $\\text{So the answer is } \\boxed{\\textbf{(B)}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1965_AHSME_Problems/6.json | AHSME |
1965_AHSME_Problems | 40 | 0 | Algebra | Multiple Choice | Let $n$ be the number of integer values of $x$ such that $P = x^4 + 6x^3 + 11x^2 + 3x + 31$ is the square of an integer. Then $n$ is:
$\textbf{(A)}\ 4 \qquad \textbf{(B) }\ 3 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 1 \qquad \textbf{(E) }\ 0$
| [
"First, we wish to factor $P$ into a more manageable form. \nFrom the beginning of $P$, we notice $x^4+6x^3$, which gives us the idea to use $(x^2+3x)^2=x^4+6x^3+9x^2$. \n\n\nThis gives us \n\\[P=(x^2+3x)^2+2x^2+3x+31\\]\nThis is not useful, but it gives us a place to start from. \n\n\nWe can then try $(x^2+3x+1)=x^4+6x^3+11x^2+6x+1$. \n\\[P=(x^2+3x+1)^2-3(x-10)\\]\nThis is much more useful, as it moves all non-linear terms inside of a squared expression.\n\n\nWe can then say $P=(x^2+3x+1)^2-3(x-10)=a^2$, where $a^2$ is the square of an integer mentioned on the problem. Right from here, we can set $x=10$, which cancels out the $3(x-10)$, giving $(100+30+1)^2=a^2$. This gives us one solution, $x=10$, $a=131$.\n\n\n\n\nWe can then rearrange the expression, giving us $(x^2+3x+1)^2-a^2=3(x-10)$. Factoring using difference of squares, we obtain \n\\[(x^2+3x+1+|a|)(x^2+3x+1-|a|)=3(x-10)\\]\n\n\nWe can then state that when $x$ is greater than $3$ and less than $-9$, $x^2+3x+1$ will be greater than $|3x-10|$. This is obtained by setting $x^2+3x+1>|3x-10|$ and then solving the inequality. We can then conclude that $x^2+3x+1+|a|>|3x-10|$. \n\n\nNext, we claim that $x^2+3x+1-|a| \\geq 1$ or $x^2+3x+1-|a| \\leq -1$ when $x \\neq 10$. We can prove this by first noting that since $x$ and $a$ are integers, $x^2+3x+1-|a|$ is an integer. Next, we shall assume that $x^2+3x+1-|a|=0$. Solving this and plugging back into the original equation, we obtain $(2|a|)(0)=3(x-10)$. Solving we obtain $x=10$, which is a contraction to $x \\neq 10$. Therefore, $x^2+3x+1-|a| \\neq 0$ and $x^2+3x+1-|a| \\geq 1$ or $x^2+3x+1-|a| \\leq -1$.\n\n\nFinally, we can go back to the equation \n\\[(x^2+3x+1+|a|)(x^2+3x+1-|a|)=3(x-10)\\]\nWe note that since $(x^2+3x+1+|a|)$ is larger than $3(x-10)$, in order for there to be solutions, $(x^2+3x+1-|a|)$ must be in the range $(-1,1)$. However, this contradicts what was proven earlier, so when $x \\neq 10$ and $x < -9$ or $x > 3$, there are no solutions for $x$. \n\n\n\n\nNow, all that remains to be checked are values of $x$ between $-9$ and $3$. Using brute force and checking each value individually, we can assert that there are no such solutions for $x$, leaving us with only $1$ solution, $x=10$. Therefore, the answer is $\\boxed{(D) }$\n\n\nSolution by treetor10145\n\n\n"
] | 1 | ./CreativeMath/AHSME/1965_AHSME_Problems/40.json | AHSME |
1965_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | The statement $x^2 - x - 6 < 0$ is equivalent to the statement:
$\textbf{(A)}\ - 2 < x < 3 \qquad \textbf{(B) }\ x > - 2 \qquad \textbf{(C) }\ x < 3 \\ \textbf{(D) }\ x > 3 \text{ and }x < - 2 \qquad \textbf{(E) }\ x > 3 \text{ and }x < - 2$
| [
"To solve this problem, we may begin by factoring $x^2-x-6$ as $(x-3)(x+2)$. This is an upward opening parabola, therefore the solutions to $(x-3)(x+2) < 0$ are inbetween the roots of the equation. That means our solutions are all $x$ such that $-2 < x < 3$, or simply $\\boxed{\\textbf{(A)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1965_AHSME_Problems/10.json | AHSME |
1965_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | Let $BC$ of right triangle $ABC$ be the diameter of a circle intersecting hypotenuse $AB$ in $D$.
At $D$ a tangent is drawn cutting leg $CA$ in $F$. This information is not sufficient to prove that
$\textbf{(A)}\ DF \text{ bisects }CA \qquad \textbf{(B) }\ DF \text{ bisects }\angle CDA \\ \textbf{(C) }\ DF = FA \qquad \textbf{(D) }\ \angle A = \angle BCD \qquad \textbf{(E) }\ \angle CFD = 2\angle A$
| [
"We will prove every result except for $\\fbox{B}$. \n\n\nBy Thales' Theorem, $\\angle CDB=90^\\circ$ and so $\\angle CDA= 90^\\circ$. $FC$ and $FD$ are both tangents to the same circle, and hence equal. Let $\\angle CFD=\\alpha$. Then $\\angle FDC = \\frac{180^\\circ - \\alpha}{2}$, and so $\\angle FDA = \\frac{\\alpha}{2}$. We also have $\\angle AFD = 180^\\circ - \\alpha$, which implies $\\angle FAD=\\frac{\\alpha}{2}$. This means that $CF=DF=FA$, so $DF$ indeed bisects $CA$. We also know that $\\angle BCD=90-\\frac{180^\\circ - \\alpha}{2}=\\frac{\\alpha}{2}$, hence $\\angle A = \\angle BCD$. And $\\angle CFD=2\\angle A$ as $\\alpha = \\frac{\\alpha}{2}\\times 2$. \n\n\nSince all of the results except for $B$ are true, our answer is $\\fbox{B}$.\n\n\n",
"It's easy to verify that $\\angle CDA$ always equals $90^\\circ$. Since $\\angle CDF$ changes depending on the sidelengths of the triangle, we cannot be certain that $\\angle CDF=45^\\circ$. Hence our answer is $\\fbox{B}$.\n\n\n"
] | 2 | ./CreativeMath/AHSME/1965_AHSME_Problems/30.json | AHSME |
1965_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | The number of real values of $x$ satisfying the equation $2^{2x^2 - 7x + 5} = 1$ is:
$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 3 \qquad \textbf{(E) }\ \text{more than 4}$
| [
"Solution by e_power_pi_times_i\n\n\nTake the logarithm with a base of $2$ to both sides, resulting in the equation $2x^2-7x+5 = 0$. Factoring results in $(2x-5)(x-1) = 0$, so there are $\\boxed{\\textbf{(C) } 2}$ real solutions.\n\n\n",
"Notice that $a^0=1, a>0$. So $2^0=1$. So $2x^2-7x+5=0$. Evaluating the discriminant, we see that it is equal to $7^2-4*2*5=9$. So this means that the equation has two real solutions. Therefore, select $\\boxed{B}$.\n\n\n~hastapasta\n\n\n"
] | 2 | ./CreativeMath/AHSME/1965_AHSME_Problems/1.json | AHSME |
1965_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | A regular hexagon is inscribed in a circle. The ratio of the length of a side of the hexagon to the length of the shorter of the arcs intercepted by the side, is:
$\textbf{(A)}\ 1: 1 \qquad \textbf{(B) }\ 1: 6 \qquad \textbf{(C) }\ 1: \pi \qquad \textbf{(D) }\ 3: \pi \qquad \textbf{(E) }\ 6:\pi$
| [
"Suppose that each side of the hexagon is $3$. Then the distance from each vertex of the hexagon to the center is also $3$, so that the circle has radius $3$. Since the circle has circumference $2\\pi(3) = 6\\pi$, the arc intercepted by any side (which measures $60^\\circ$) has length $\\frac{1}{6}*6\\pi = \\pi$, so that our answer is $\\boxed{\\text{(D)}}$ and we are done.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1965_AHSME_Problems/2.json | AHSME |
1965_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | The expression $(81)^{-2^{-2}}$ has the same value as:
$\textbf{(A)}\ \frac {1}{81} \qquad \textbf{(B) }\ \frac {1}{3} \qquad \textbf{(C) }\ 3 \qquad \textbf{(D) }\ 81\qquad \textbf{(E) }\ 81^4$
| [
"Let us recall $\\text{PEMDAS}$. We calculate the exponent first. $(-2)^{-2}=\\frac{1}{(-2)^2}=\\frac{1}{4}$ When we substitute, we get $81^{\\frac{1}{4}}=\\sqrt[4]{81}=\\boxed{\\textbf{(C) }3}$.\n\n\n~Mathfun1000\n\n\n"
] | 1 | ./CreativeMath/AHSME/1965_AHSME_Problems/3.json | AHSME |
1965_AHSME_Problems | 34 | 0 | Algebra | Multiple Choice | For $x \ge 0$ the smallest value of $\frac {4x^2 + 8x + 13}{6(1 + x)}$ is:
$\textbf{(A)}\ 1 \qquad \textbf{(B) }\ 2 \qquad \textbf{(C) }\ \frac {25}{12} \qquad \textbf{(D) }\ \frac{13}{6}\qquad \textbf{(E) }\ \frac{34}{5}$
| [
"To begin, lets denote the equation, $\\frac {4x^2 + 8x + 13}{6(1 + x)}$ as $f(x)$. Let's notice that:\n\n\n\\begin{align*} f(x) & = \\frac {4x^2 + 8x + 13}{6(1 + x)}\\\\\\\\ \t & = \\frac{4(x^2+2x) + 13}{6(x+1)}\\\\\\\\ & = \\frac{4(x^2+2x+1-1)+13}{6(x+1)}\\\\\\\\ & = \\frac{4(x+1)^2+9}{6(x+1)}\\\\\\\\ & = \\frac{4(x+1)^2}{6(x+1)} + \\frac{9}{6(1+x)}\\\\\\\\ & = \\frac{2(x+1)}{3} + \\frac{3}{2(x+1)}\\\\\\\\ \\end{align*}\n\n\nAfter this simplification, we may notice that we may use calculus, or the AM-GM inequality to finish this problem because $x\\ge 0$, which implies that both $\\frac{2(x+1)}{3} \\text{ and } \\frac{3}{2(x+1)}$ are greater than zero. Continuing with AM-GM:\n\n\n\\begin{align*} \\frac{\\frac{2(x+1)}{3} + \\frac{3}{2(x+1)}}{2} &\\ge {\\small \\sqrt{\\frac{2(x+1)}{3}\\cdot \\frac{3}{2(x+1)}}}\\\\\\\\ f(x) &\\ge 2 \\end{align*}\n\n\nTherefore, $f(x) = \\frac {4x^2 + 8x + 13}{6(1 + x)} \\ge 2$, $\\boxed{\\textbf{(B)}}$\n\n\n\n\n$(\\text{With equality when } \\frac{2(x+1)}{3} = \\frac{3}{2(x+1)}, \\text{ or } x=\\frac{1}{2})$\n\n\n",
"Take the derivative of f(x) and f'(x) using the quotient rule.\n\\begin{align*} f(x) & = \\frac {4x^2 + 8x + 13}{6(1 + x)}\\\\\\\\ f'(x) & = \\frac{(4x^2 + 8x + 13)'(1 + x) - (4x^2 + 8x + 13)(1 + x)'}{(1 + x)^2} & = \\frac{(8x + 8)(1 + x) - (4x^2 + 8x + 13)(1)}{(1 + x)^2} & = \\frac{4x^2 + 8x - 5}{(1 + x)^2} f''(x) & = \\frac{(4x^2 + 8x - 5)'(1 + x)^2 - (4x^2 + 8x - 5)((1 + x)^2)'}{(1 + x)^4} \\end{align*}\n\n\n",
"We go from A through E and we look to find the smallest value so that $x \\ge 0$, so we start from A:\n\n\n\\[\\frac{4x^2 + 8x + 13}{6x+6} = 1\\]\n\n\n\\[4x^2 + 8x + 13 = 6x + 6\\]\n\n\n\\[4x^2 + 2x + 7 = 0\\]\n\n\nHowever by the quadratic formula there are no real solutions of $x$, so $x$ cannot be greater than 0. We move on to B:\n\n\n\\[\\frac{4x^2 + 8x + 13}{6x + 6} = 2\\]\n\n\n\\[4x^2 + 8x + 13 = 12x + 12\\]\n\n\n\\[4x^2 - 4x + 1 = 0\\]\n\n\n\\[(2x-1)^2 = 0\\]\n\n\nThere is one solution: $x = \\frac{1}{2}$, which is greater than 0, so 2 works as a value. Since all the other options are bigger than 2 or invalid, the answer must be $\\boxed{\\textbf{B}}$\n\n\n"
] | 3 | ./CreativeMath/AHSME/1965_AHSME_Problems/34.json | AHSME |
1965_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | The sum of the numerical coefficients in the complete expansion of $(x^2 - 2xy + y^2)^7$ is:
$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 7 \qquad \textbf{(C) }\ 14 \qquad \textbf{(D) }\ 128 \qquad \textbf{(E) }\ 128^2$
| [
"Notice that the given equation, $(x^2 - 2xy + y^2)^7$ can be factored into $(x-y)^{14} = \\sum_{k=0}^{14} \\binom{14}{k}(-1)^k\\cdot x^{14-k}\\cdot y^k$. \n\n\nNotice that if we plug in $x = y$ = 1, then we are simply left with the sum of the coefficients from each term. Therefore, the sum of the coefficients is $(1-1)^2 = 0, \\boxed{\\textbf{(A)}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1965_AHSME_Problems/14.json | AHSME |
1965_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | When the repeating decimal $0.363636\ldots$ is written in simplest fractional form, the sum of the numerator and denominator is:
$\textbf{(A)}\ 15 \qquad \textbf{(B) }\ 45 \qquad \textbf{(C) }\ 114 \qquad \textbf{(D) }\ 135 \qquad \textbf{(E) }\ 150$
| [
"We let $x=0.\\overline{36}$. Thus, $100x=36.\\overline{36}$. We find that $100x-x=99x=36.\\overline{36}-0.\\overline{36}=36$, or $x=\\frac{36}{99}=\\frac{4}{11}$. Since $4+11=15$, the answer is $\\boxed{\\textbf{(A)}\\ 15}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1965_AHSME_Problems/5.json | AHSME |
1965_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | If $x^4 + 4x^3 + 6px^2 + 4qx + r$ is exactly divisible by $x^3 + 3x^2 + 9x + 3$, the value of $(p + q)r$ is:
$\textbf{(A)}\ - 18 \qquad \textbf{(B) }\ 12 \qquad \textbf{(C) }\ 15 \qquad \textbf{(D) }\ 27 \qquad \textbf{(E) }\ 45 \qquad$
| [
"Let $f(x)=x^3+3x^2+9x+3$ and $g(x)=x^4+4x^3+6px^2+4qx+r$. \n\n\nLet 3 roots of $f(x)$ be $r_1, r_2$ and $r_3$. As $f(x)|g(x)$ , 3 roots of 4 roots of $g(x)$ will be same as roots of $f(x)$. Let the 4th root of $g(x)$ be $r_4$. By vieta's formula\n\n\nIn $f(x)$\n\n\n$r_1+r_2+r_3=-3$\n\n\n$r_1r_2+r_2r_3+r_1r_3=9$\n\n\n$r_1r_2r_3=-3$\n\n\nIn $g(x)$\n\n\n$r_1+r_2+r_3+r_4=-4$\n\n\n$=>r_4=-1$\n\n\n$r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4=6p$\n\n\n$=>r_1r_2+r_1r_3+r_2r_3+r_4(r_1+r_2+r_3)=6p$\n\n\n$=>9+(-1)(-3)=6p$\n\n\n$=>p=2$\n\n\n$r_1r_2r_3+r_1r_3r_4+r_1r_2r_4+r_2r_3r_4=-4q$\n\n\n$=>-3+r_4(r_1r_3+r_1r_2+r_2r_3)=-4q$\n\n\n$=>-3+(-1)(9)=-4q$\n\n\n$=>q=3$\n\n\n$r_1r_2r_3r_4=r$\n\n\n$=>(-3)(-1)=r$\n\n\n$=>r=3$\n\n\nso $(p+q)r=\\fbox{15}$\n\n\nBy ~Ahmed_Ashhab\n\n\n\n\n\n\n",
"Notice that to obtain the $x^4$ term one must multiply $x^4+4x^3+6px^2+4qx+r$ by some linear function of the form $x-a$. Looking at the $x^3$ term, it is clear that $a$ must equal $1$. Therefore by multiplying $x^4+4x^3+6px^2+4qx+r$ by $x+1$, the product will be $x^4+4x^3+12x^2+12x+3$. Therefore $p=2$, $q=3$, $r=3$. Thus $(2+3)3=\\fbox{15}$\n\n\n"
] | 2 | ./CreativeMath/AHSME/1965_AHSME_Problems/19.json | AHSME |
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