probability
integration
trigonometry
geometry
geometric transformation
reflection
analytic geometry

Please post all solutions here.  They will be latexed into a solutions document once we have all 50.

[hide="Problem 9"]Since the sum of the $7$-special numbers must equal $1$, and the numbers only contain $7$ and $0$, we can divide by $7$ to get $.\overline{142857}$, which is obtained by adding a series of $1$-special numbers.  Since $1$-special numbers only contain $1$ and $0$, there must be a minimum of $\boxed{8}$ numbers to obtain $.\overline{142857}$.[/hide]

[hide="Problem 20"]$30!$ factors as

\[ 30!=2^{26}\cdot 3^{14}\cdot 5^{7}\cdot 7^{4}\cdot 11^{2}\cdot 13^{2}\cdot 17\cdot 19\cdot 23\cdot 29.  \]

Therefore there are

\[ 27\cdot 15\cdot 8\cdot 5\cdot 3\cdot 3\cdot 2\cdot 2\cdot 2\cdot 2\cdot=\boxed{2332800}  \]

factors.[/hide]

[hide="Problem 22"]Since it is a multiple of $11$ and a perfect square, the square root is $11k$ for some $k$.  Therefore we have

$10^{5}<\left(11k\right)^{2}<10^{8}\implies 29\leq\left|k\right|\leq 909$

Then there are $909-29+1=\boxed{881}$ numbers.[/hide]

[hide="Problem 35"]
The usual calculation gives

$a_{1}=2-\sqrt{3}$

$a_{2}=\frac{\sqrt{3}}{3}$

$a_{3}=\sqrt{3}$

$a_{4}=-\sqrt{3}$

$a_{5}=\sqrt{3}$

...

Therefore there are $\boxed{4}$ unique terms in $\left\{a_{k}\right\}_{k=1}^{\infty}$.
[/hide]

[hide="Problem 6"]If there are n nickels and d dimes, the coins are worth 5n+10d+25(30-n-d)=500, so 20n+15d=250 and 4n+3d=50. Note that n could be 2, and then go up by 3s to get 2-5-8-11 or $\boxed{4}$ possibilities.[/hide]

[hide="Problem 7"]Star polygons can be created only when n and m are relatively prime (else they miss some points). The number of integers less than 36 and relatively prime to 36 is $\phi (2*2*3*3)=1*2*2*3=12$. Since each star polygon is double-counted (for instance, the same star polygon-the polygon itself, is created if m=1 or m=35), the answer is $\boxed{6}$.[/hide]

[hide="Problem 15"]We factor $k^{26}(k-1)(k+1)$ and notice that if k is even, this is divisible by 8. If k is odd, each of k-1 and k+1 is even and one is divisible by 4, so the whole thing is divisible by $\boxed{8}$.[/hide]

[hide="Problem 26"]Note that $\phi (a/7)=4$, $\phi (b/7)=10$, and neither is divisible by 49 (or else 24 or 60 would be divisible by 7). Therefore, $\phi (ab)=\phi (a/7*b/7*7*7)=4*10*6*7=\boxed{1680}$[/hide]

[hide="Problem 32"]We assume a<b<c. Then the probability of a given x being the minimum is $\int_x^1 \int_y^1 dz dy=\int_x^1 1-y dy=\frac{x^2}2-x+\frac 12$, and the expected value of x is $\int_0^1 \frac{x^3}2-x^2+\frac x2 dx=\frac 18-\frac 13+\frac 14=\frac 1{24}$. However, we must multiply by 6 since we assumed a certain ordering, yielding $\boxed{\frac 14}$.[/hide]

[hide="Problem 40"]Since we could use A in every name, N is at least the number of subsets of B-Z, or $2^{25}$. But 2^{25} has 8 digits, so $\boxed{8}$ is the only possible answer.[/hide]

[hide="Problem 43"]If the origin is the center of the larger circle, the center of the smaller circle is at $(\cos{\theta},\sin{\theta})$ as $\theta$ ranges from 0 to $2\pi$ radians. The line going through the hole and the center of the small circle is the reflection across a vertical line (through the center of the small circle) of the line through the centers of the circles because of the movement of the rotating circle. Since this halves the x-coordinate and adds half again to the y-coordinate, we have the general coordinate $(\frac{\cos{\theta}}2,\frac{3\sin{\theta}}2)$, which traces an ellipse with area $\frac 12*\frac 32*\pi=\boxed{\frac{3\pi}4}$[/hide]

[hide="Problem 47"]If we assume that all coordinates are positive and x<=y<=z, then $x\leq y\leq \sqrt{1-x^2}$, which also yields $x\leq \sqrt{1-x^2}$, so $x\leq \frac{\sqrt 2}2$. Also, $y\leq z\leq \sqrt{1-y^2}\leq \sqrt{1-x^2}$, which also yields $y\leq \sqrt{1-y^2}$, or $y\leq \frac{\sqrt 2}2$. Since this is necessarily a better upper bound than $\sqrt{1-x^2}$, we have area $\int_0^{\frac{\sqrt{2}}2} \int_x^{\frac{\sqrt{2}}2} \int_y^{\sqrt{1-y^2}} dz dy dx=\frac 13-\frac{\sqrt{2}}6$. Now we multiply by 6 since we assumed a certain ordering and 8 since we limited it to a certain quadrant (really octant), to yield $\boxed{16-8\sqrt{2}}$[/hide]

36: $\tan(3\cdot19x-2\cdot13x)=\tan(31x)$ solve from there

37: The longest sum of the to sides is $2$.  The longest side it could posssibly have is $\sqrt2$.  So the rectangle has these dimensions ($2$ by $\sqrt2$) and has area $2\sqrt2$.  This can fit all possible pairs of squares with sum of areas being 2 and it can't get smaller due to the conditions above.

The answe is definitely $\boxed{2\sqrt2}$

40: Lets say the first one is $A$.  Now every other one must have an $A$ to satisfy the last condition.  And now, since all have an $A$, again, the last condition is satisfied, so we just need to find the number of distinct combinations there are of the $25$ letters left becuase they cannot be permutations of eachother.  So the asnwer is $2^{25}$

[quote="jb05"][hide="Problem 32"]We assume a<b<c. Then the probability of a given x being the minimum is $\int_x^1 \int_y^1 dz dy=\int_x^1 1-y dy=\frac{x^2}2-x+\frac 12$, and the expected value of x is $\int_0^1 \frac{x^3}2-x^2+\frac x2 dx=\frac 18-\frac 13+\frac 14=\frac 1{24}$. However, we must multiply by 6 since we assumed a certain ordering, yielding $\boxed{\frac 14}$.[/hide][/quote]

I just generated like 1000 [pseudo]random numbers with my computer and it was very close to .25 :D

[quote="jb05"]Yeah, I did similar things for a lot of those problems, but I'm not sure that's what needs to go in the solutions document...  :lol:[/quote]
Same here along with process of elmination on some. :P

There is no need to go multivariate with #32. Simply note that if the lowest number is $x$, then the probability that the other two numbers are not the lowest is $(1-x)^2$. Thus the probability across all $x$ in the range is \[ \int_0^1 (1-x)^2dx \] but we wish to compute expected value, so we compute \[ \int_0^1 x(1-x)^2dx=\frac{1}{12} \] and multiplying by $3$ to account for the three variables yields $\frac{1}{4}$.

For #26, note the identity \[ \phi(ab)\phi(gcd(a,b))=\phi(a)\phi(b)gcd(a,b). \]

[hide="Problem 1"]
All terms cancel out expect for $1 + \frac{1}{2}.. + \frac{1}{5} - \frac{1}{96} - \frac{1}{97}.. - \frac{1}{100}$. Compute with a calculator to get  $\boxed{B}$.[/hide]

[hide="Problem 2"]
The total volume of the pyramid is $72$. That means one section must have a volume of $36$. Since the new pyramid will be similar to the one its contained in, the height of the smaller pyramid is $\sqrt[3]{108}$ or $3\sqrt[3]{4}$. This makes the distance from the plane to the base $\boxed{A}$.[/hide]

[hide="Problem 3"]
It took him $22\frac{1}{2}$ minutes to walk halfway to school and then $11\frac{1}{4}$ minutes to run home as he runs twice as fast as he walks. It took him $5$ minutes to find his calculator and then $22\frac{1}{2}$ minutes to run to school from his house. Sum these up and subtract from $45$ minutes, his usual time to get $\boxed{B}$.[/hide]

[hide="Problem 4"]
An octagon has 20 diagonals. 8 of those have length x, 8 of those have length y, and 4 have a length z. The probability of Joe and then David selecting length x is $\frac{2}{5}*\frac{7}{19}$ or $\frac{14}{95}$. The probability of Joe and then David selecting length y is $\frac{2}{5}*\frac{7}{19}$ or $\frac{14}{95}$. The probability of Joe and then David selecting length z is $\frac{1}{5}*\frac{3}{19}$ or $\frac{3}{95}$. Sum these up to get $\boxed{D}$.[/hide]

[hide="Problem 5"]
Let's say the radius of the largest circle is $6$. Then, the total area of the circles is $36\pi$. This means that each of the twelve sectors has an area of $3\pi$. Since the smallest circle is made up of four sectors, its area is $12\pi$ and its radius is $\sqrt{12}$ or $2\sqrt{3}$. This makes the answer $\frac{\sqrt{3}}{3}$ or $\boxed{B}$.[/hide]

[hide="Problem 14"]
Sum the first 3 terms to find out it's bigger than B, C, D, and E. By process of elimination, the answer must be $\boxed{A}$. :D[/hide]

[quote="Elemennop"]For #26, note the identity \[ \phi(ab)\phi(gcd(a,b))=\phi(a)\phi(b)gcd(a,b).  \][/quote]
Cool, I didn't know that.  I just took two numbers which satisfied the requirement and multiplied them :).

[hide="Problem 17"]
The last digit is $6$, and the last digit of $6^n$ where n>0 is always $6$.

Use a calculator to find the 1st digit, 1, and 1+6 = $\boxed{7}.$

$A$.
[/hide]

[hide="Problem 24"]
There are three ways to get $x^3$, namely
\[ x \times x \times x, \frac{x^2}{2} \times x, and \frac{x^3}{6}.  \]
There are ${128 \choose 3}$ ways to get $x^3$, $128 \times 127$ ways to get $\frac{x^3}{2}$, and $128$ ways to get $\frac{x^3}{6}$. Adding these (multiply # of ways by whatever you get), you get
\[ \frac{128 \times 127 \times 126}{6} + \frac{128 \times 127}{2} + \frac{128}{6}  \]
\[ =\frac{128(127 \times 129 +1)}{6} = \frac{2^6(2^7)^2}{3} = \frac{2^{20}}{3}  \]

Therefore, $p = 2$, $q = 20$, and $r=3$, so $p+q+r = \boxed{25}.$

$C.$

[/hide]

[hide="23"]
Let $A=s(2^{2006})$, $B=s(A)$, and $C=s(B)$. Working modulo 9, $2^3\equiv -1\Rightarrow 2^6\equiv 1\Rightarrow 2^{2004}\equiv 1\Rightarrow 2^{2006}\equiv 4$. Since $s$ preserves remainder modulo 9, $A\equiv B\equiv C\equiv 4$.

Also, if a number is less than $10^d$, then it has at most $d$ digits so its sum is at most $9d$. An absurd (but useful enough) bound for $2^{2006}$ is $2^{2006}<2^{2007}=8^{669}<10^{669}$. Then $A\leq 9\cdot 669=6021$ and $B\leq 9\cdot 4=36$.

Since $B\equiv 4\pmod{9}$ and $B\leq 36$, the possibilities for $B$ are 4, 13, 22, and 31, and each case leads to $C=\boxed{4}$.
[/hide]
[hide="31"]
By noting that the center of a regular polygon is the center of the circumcircle, the center will be in the interior of a triangle if and only if each central angle is less than 180 degrees, which occurs for acute triangles. Each side intercepts the same arc, so the triangle will be acute if and only if at most $n$ arcs are intercepted by an inscribed angle.

By symmetry, we can fix the first point. There are $\binom{2n}{2}=n(2n-1)$ ways to choose the other 2 points. If the points are labeled $P_1$, $P_2$, and $P_3$, then let $a$ be the number of arcs cut off by $P_1P_2$, $b$ by $P_2P_3$, and $c$ by $P_3P_1$. Then $a$, $b$, and $c$ are positive integers satisfying $a+b+c=2n+1$. When the triangle is acute, $\max (a,b,c)\leq n$.

For the desired outcome, each variable is less than $n+1$ so make a bijection with the substitutions $x=n+1-a$, $y=n+1-b$, $z=n+1-c$. Then $x$, $y$, and $z$ are positive integers and $x+y+z=n+2$. Any positive integer triple $(a,b,c)$ can be converted this way if and only if each value is less than or equal to $n$.

The second part to showing that this is a valid bijection is that no positive integer triple $(x,y,z)$ could be obtained by some $(a,b,c)$ where one of $a$, $b$, or $c$ is not positive. This can be shown by contradiction: suppose $a\leq 0$. Then $x\geq n+1$, but since $y,z\geq 1$, $x+y+z\geq n+3>n+2$ which is a contradiction. Therefore the bijection works.

The number of solutions in positive integers $x,y,z$ with $x+y+z=n+2$ is $\binom{n+1}{2}=\frac{n(n+1)}{2}$. Dividing the number of desired outcomes by the sample space, the probability is $\boxed{\frac{n+1}{4n-2}}$.
[/hide]
[hide="45"]
Generalizing is simple for this problem: let the radii of the two bigger circles be $a$ and $b$, with $a<b$, and that of the smaller circle be $c$. Draw in radii to the tangent line, the radii connecting the circles, and lines parallel to the tangent line through the centers of circles $A$ and $C$.

The distance between the centers of circles $A$ and $B$ is $a+b$ and the difference in height is $b-a$. Applying the Pythagorean Theorem, the component of the distance between the centers that is parallel to the tangent line is $2\sqrt{ab}$. Similarly, it is $2\sqrt{ac}$ from $A$ to $C$ and $2\sqrt{bc}$ from $B$ to $C$.

The parallel component of the distance from $A$ to $B$ must be the sum of the components from $A$ to $C$ and from $B$ to $C$, so $\sqrt{ab}=\sqrt{ac}+\sqrt{bc}\Rightarrow \frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}$. Plugging in 1 and 2 into the problem, taking reciprocal, and squaring, $c=\boxed{6-4\sqrt{2}}$.
[/hide]

[quote="ehehheehee"][quote="jb05"]
[hide="Problem 32"]We assume a<b<c. Then the probability of a given x being the minimum is $\int_x^1 \int_y^1 dz dy=\int_x^1 1-y dy=\frac{x^2}2-x+\frac 12$, and the expected value of x is $\int_0^1 \frac{x^3}2-x^2+\frac x2 dx=\frac 18-\frac 13+\frac 14=\frac 1{24}$. However, we must multiply by 6 since we assumed a certain ordering, yielding $\boxed{\frac 14}$.[/hide]
[/quote]

As a middle schooler, I do not know what those signs mean...so I just put three numbers evenly spaced apart

0 1/4 2/4 3/4 1

is this logic ok, or was I just lucky?[/quote]

I think it was just luck; for example, this clearly wouldn't work for a problem like "What is the probability that their sum is less than 1?"

[hide="13"]
Let $D=BX\cap CY$, $E=CY\cap AZ$, and $F=AZ\cap BX$; the problem asks to find the ratio of the area of $\triangle DEF$ to that of $\triangle ABC$. Since points $A$, $F$, and $Z$ are collinear, then by Menelaus's Theorem using $\triangle CXB$,
\[ \frac{CA}{XA}\cdot \frac{XF}{BF}\cdot \frac{BZ}{CZ}=1\Rightarrow \frac{XF}{BF}=\frac 16 \]
From this, $\frac{BF}{BX}=\frac 67$. If the area of $\triangle ABC$ is $K$, then the area of $\triangle AXB$ will be $\frac K3$ because it has a common height and 1/3 of the base of $\triangle ABC$. Also, $\triangle AFB$ shares a common base with $\triangle AXB$ and 6/7 of the height, so its area is $\frac 27 K$.

Similarly, the areas of $\triangle BDC$ and $\triangle CEA$ are each $\frac 27 K$. After accounting for the areas of these three triangles, only $\triangle DEF$ remains, so its area is $\frac K7\Rightarrow \boxed{\frac 17}$. (This result holds for any triangle, not just equilateral ones.)
[/hide]
[hide="42"]
Let the variable we are trying to maximize be $x$ and the sum of the $k$th powers of the other 4 variables be $s_k$. We are given $s_1=8-x$ and $s_2=16-x^2$. By RMS-AM, $4s_2\geq s_1^2\Rightarrow 64-4x^2\geq 64-16x+x^2\Rightarrow x(5x-16)\leq 0$. It follows that $x\leq \boxed{\frac{16}{5}}$, and there is equality when the other 4 variables each equal 6/5.
[/hide]
[hide="46"]
For all $\theta$, $2\cos^3 \theta -\cos \theta=\cos \theta \cos 2\theta\Rightarrow 2\cos^3 \theta -\cos^2 \theta -\cos \theta=\cos \theta (\cos 2\theta -\cos \theta)$. Then, by expressing $2\theta$ as $\frac 32 \theta +\frac 12 \theta$ and $\theta$ as $\frac 32 \theta -\frac 12 \theta$, this equals $-2\cos \theta \sin \frac 32 \theta \sin \frac 12 \theta$.

Now substitute $\frac{\pi}{7}$ in for $\theta$ ; the expression is $-2\cos \frac{\pi}{7} \sin \frac{3\pi}{14} \sin \frac{\pi}{14}=-2\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7}=2\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7}$.

Let $P=\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7}$. Then, after using double-angle formula for sine three times, $P\sin \frac{\pi}{7}=\frac 18 \sin \frac{8\pi}{7}=-\frac 18\sin \frac{\pi}{7}$. So $P=-\frac 18 \Rightarrow 2P=\boxed{-\frac 14 }$.
[/hide]

Did anyone do #28 without Lagrange Multipliers?

How do you use Lagrange Multipliers on #28?

man scorpius, you destroyed number 13...but it would've been a lot easier to just use law of cosines :|

#13 is really well-known. I've seen it so many times...

[quote="jb05"]

[hide="Problem 32"]We assume a<b<c. Then the probability of a given x being the minimum is $\int_x^1 \int_y^1 dz dy=\int_x^1 1-y dy=\frac{x^2}2-x+\frac 12$, and the expected value of x is $\int_0^1 \frac{x^3}2-x^2+\frac x2 dx=\frac 18-\frac 13+\frac 14=\frac 1{24}$. However, we must multiply by 6 since we assumed a certain ordering, yielding $\boxed{\frac 14}$.[/hide]

[hide="Problem 47"]If we assume that all coordinates are positive and x<=y<=z, then $x\leq y\leq \sqrt{1-x^2}$, which also yields $x\leq \sqrt{1-x^2}$, so $x\leq \frac{\sqrt 2}2$. Also, $y\leq z\leq \sqrt{1-y^2}\leq \sqrt{1-x^2}$, which also yields $y\leq \sqrt{1-y^2}$, or $y\leq \frac{\sqrt 2}2$. Since this is necessarily a better upper bound than $\sqrt{1-x^2}$, we have area $\int_0^{\frac{\sqrt{2}}2} \int_x^{\frac{\sqrt{2}}2} \int_y^{\sqrt{1-y^2}} dz dy dx=\frac 13-\frac{\sqrt{2}}6$. Now we multiply by 6 since we assumed a certain ordering and 8 since we limited it to a certain quadrant (really octant), to yield $\boxed{16-8\sqrt{2}}$[/hide][/quote]

Um, are there non-calculus ways of doing this?

Here are some compiled solutions:
Previously posted:

[quote="Ignite168"][hide="Problem 1"]
All terms cancel out expect for $1 + \frac{1}{2}.. + \frac{1}{5} - \frac{1}{96} - \frac{1}{97}.. - \frac{1}{100}$. Compute with a calculator to get  $\boxed{B}$.[/hide]

[hide="Problem 2"]
The total volume of the pyramid is $72$. That means one section must have a volume of $36$. Since the new pyramid will be similar to the one its contained in, the height of the smaller pyramid is $\sqrt[3]{108}$ or $3\sqrt[3]{4}$. This makes the distance from the plane to the base $\boxed{A}$.[/hide]

[hide="Problem 3"]
It took him $22\frac{1}{2}$ minutes to walk halfway to school and then $11\frac{1}{4}$ minutes to run home as he runs twice as fast as he walks. It took him $5$ minutes to find his calculator and then $22\frac{1}{2}$ minutes to run to school from his house. Sum these up and subtract from $45$ minutes, his usual time to get $\boxed{B}$.[/hide]

[hide="Problem 4"]
An octagon has 20 diagonals. 8 of those have length x, 8 of those have length y, and 4 have a length z. The probability of Joe and then David selecting length x is $\frac{2}{5}*\frac{7}{19}$ or $\frac{14}{95}$. The probability of Joe and then David selecting length y is $\frac{2}{5}*\frac{7}{19}$ or $\frac{14}{95}$. The probability of Joe and then David selecting length z is $\frac{1}{5}*\frac{3}{19}$ or $\frac{3}{95}$. Sum these up to get $\boxed{D}$.[/hide]

[hide="Problem 5"]
Let's say the radius of the largest circle is $6$. Then, the total area of the circles is $36\pi$. This means that each of the twelve sectors has an area of $3\pi$. Since the smallest circle is made up of four sectors, its area is $12\pi$ and its radius is $\sqrt{12}$ or $2\sqrt{3}$. This makes the answer $\frac{\sqrt{3}}{3}$ or $\boxed{B}$.[/hide]

[/quote]
[quote="jb05"][hide="Problem 6"]If there are n nickels and d dimes, the coins are worth 5n+10d+25(30-n-d)=500, so 20n+15d=250 and 4n+3d=50. Note that n could be 2, and then go up by 3s to get 2-5-8-11 or $\boxed{4}$ possibilities.[/hide]

[hide="Problem 7"]Star polygons can be created only when n and m are relatively prime (else they miss some points). The number of integers less than 36 and relatively prime to 36 is $\phi (2*2*3*3)=1*2*2*3=12$. Since each star polygon is double-counted (for instance, the same star polygon-the polygon itself, is created if m=1 or m=35), the answer is $\boxed{6}$.[/hide]
[/quote]
[quote="shadysaysurspammed"][hide="Problem 8"]for real solution,

$25-8c \geq 0$

$\implies c \leq 25/8 \implies largest\ value\ of\ c =\boxed{\frac{25}{8}}$[/hide]
[/quote]
[quote="amcavoy"][hide="Problem 9"]Since the sum of the $7$-special numbers must equal $1$, and the numbers only contain $7$ and $0$, we can divide by $7$ to get $.\overline{142857}$, which is obtained by adding a series of $1$-special numbers.  Since $1$-special numbers only contain $1$ and $0$, there must be a minimum of $\boxed{8}$ numbers to obtain $.\overline{142857}$.[/hide][/quote]
[hide="Problem 10"]
The four items are $a_1,a_2,a_3,a_4$ where each $a_i$ represents the number of cents in the price of the item. So the problem gives us that $(a_1/100)(a_2/100)(a_3/100)(a_4/100)=7.11$ or $a_1a_2a_3a_4=711*10^6$. Also, $a_1+a_2+a_3+a_4=711$. We see that each $a_i$ is an integer. Looking at the factorization of $719*10^6$, we see that the largest prime is $79$. Looking at the product and sum, we can determine that $316=4*79$ is the correct answer, so $\boxed{E}$.[/hide]
[quote="Xantos C. Guin"][hide="Problem 10"]
Since prices in convenience stores can usually be expressed in the form $\frac{N}{100}$, where N is an integer.

Let the four items cost: $\frac{A}{100}, \frac{B}{100}, \frac{C}{100}, \frac{D}{100}.$

Then the product: $\frac{A}{100}\cdot\frac{B}{100}\cdot\frac{C}{100}\cdot\frac{D}{100}=\frac{ABCD}{100000000}=\frac{711000000}{100000000}$

So:
$ABCD=711000000=2^{6}\cdot3^{2}\cdot5^{6}\cdot79$

The prices of the items which Patrick buys, when multiplied by $100$ should be composed of only prime factors that appears in $711000000$

Looking at our choices:

$312=2^{3}\cdot3\cdot13$
$313=313$
$314=2\cdot157$
$315=3^{2}\cdot5\cdot7$
$316=2^{2}\cdot79$

Since $312, 313, 314, 315$ all contain a prime factor that does not appear in $711000000$ these choices are impossible, so our answer is $\$3.16\Rightarrow \boxed{E}$
[/hide][/quote]
Need 11,12
[quote="scorpius119"]
[hide="Problem 13"]
Let $D=BX\cap CY$, $E=CY\cap AZ$, and $F=AZ\cap BX$; the problem asks to find the ratio of the area of $\triangle DEF$ to that of $\triangle ABC$. Since points $A$, $F$, and $Z$ are collinear, then by Menelaus's Theorem using $\triangle CXB$,
\[ \frac{CA}{XA}\cdot \frac{XF}{BF}\cdot \frac{BZ}{CZ}=1\Rightarrow \frac{XF}{BF}=\frac 16  \]
From this, $\frac{BF}{BX}=\frac 67$. If the area of $\triangle ABC$ is $K$, then the area of $\triangle AXB$ will be $\frac K3$ because it has a common height and 1/3 of the base of $\triangle ABC$. Also, $\triangle AFB$ shares a common base with $\triangle AXB$ and 6/7 of the height, so its area is $\frac 27 K$.

Similarly, the areas of $\triangle BDC$ and $\triangle CEA$ are each $\frac 27 K$. After accounting for the areas of these three triangles, only $\triangle DEF$ remains, so its area is $\frac K7\Rightarrow \boxed{\frac 17}$. (This result holds for any triangle, not just equilateral ones.)
[/hide][/quote]
[quote="shadysaysurspammed"]
[hide="Problem 14"]$\frac{1}{3}(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\cdots+\frac{1}{28\cdot29\cdot30}-\frac{1}{29\cdot30\cdot31})$

$=\boxed{\frac{749}{13485}}$[/hide][/quote]

[quote="jb05"]
[hide="Problem 15"]We factor $k^{26}(k-1)(k+1)$ and notice that if k is even, this is divisible by 8. If k is odd, each of k-1 and k+1 is even and one is divisible by 4, so the whole thing is divisible by $\boxed{8}$.[/hide][/quote]
[quote="Xantos C. Guin"][hide="Problem 16"]
By using a calculator: (Rounded to the nearest hundreth)

A: $8=8.00$
B: $e^{\pi}=23.14$
C: $\pi^{e}=22.46$
D: $(\frac{1}{\pi})^{-\frac{1}{e}}=1.52$
E: $10=10.00$

Since $23.14$ is larger that all the others, the answer is $\boxed{B}$.
[/hide]

Noone that had a graphing, or decent scientific calculator should have missed #16.[/quote]
[quote="not_trig"][hide="Problem 17"]
The last digit is $6$, and the last digit of $6^n$ where n>0 is always $6$.

Use a calculator to find the 1st digit, 1, and 1+6 = $\boxed{7}.$

$A$.
[/hide][/quote]
[hide="Problem 18"]We are given the four vertices of a quadrilateral and a point inside that minimizes the total distance from that point to the four vertices. Notice that to minimize the sum between two opposite vertices, any point on the diagonal connecting those vertices works. Therefore, the point we are looking for is at the intersection of the two diagonals. The sum of the four distances is equal to the sum of the two diagonals, which is $\boxed{C}$ , $2\sqrt{5}$.[/hide]

Need 19
[quote="amcavoy"]

[hide="Problem 20"]$30!$ factors as

\[ 30!=2^{26}\cdot 3^{14}\cdot 5^{7}\cdot 7^{4}\cdot 11^{2}\cdot 13^{2}\cdot 17\cdot 19\cdot 23\cdot 29.  \]

Therefore there are

\[ 27\cdot 15\cdot 8\cdot 5\cdot 3\cdot 3\cdot 2\cdot 2\cdot 2\cdot 2\cdot=\boxed{2332800}  \]

factors.[/hide][/quote]
[hide="Problem 21"]First, make the substitution $a=x^2$. Now, we can rewrite and factor the equation:
$a^{10}-a^9-a^8-a^7-... -2=0$
$a^{10}-1=a^9+a^8+a^7+... +1$
$a^{10}-1=\frac{a^{10}-1}{a-1}$
So when $a^{10}-1\not =0$ we can cancel it to get:
$a-1=1, a=2, x=\pm \sqrt{2}$ which are the only real solutions.
The other solutions occur when $a$ satisfies $\frac{a^{10}-1}{a-1}=0$ which are the ten roots of unity other than $1$, so all other $a$ values are nonreal (which implies that $\sqrt{a}$ is nonreal).
Thus the answer is $(\sqrt{2})^2 + (-\sqrt{2})^2=4$ , $\boxed{D}$
[/hide]
[quote="amcavoy"]
[hide="Problem 22"]Since it is a multiple of $11$ and a perfect square, the square root is $11k$ for some $k$.  Therefore we have

$10^{5}<\left(11k\right)^{2}<10^{8}\implies 29\leq\left|k\right|\leq 909$

Then there are $909-29+1=\boxed{881}$ numbers.[/hide][/quote]
[quote="scorpius119"][hide="Problem 23"]
Let $A=s(2^{2006})$, $B=s(A)$, and $C=s(B)$. Working modulo 9, $2^3\equiv -1\Rightarrow 2^6\equiv 1\Rightarrow 2^{2004}\equiv 1\Rightarrow 2^{2006}\equiv 4$. Since $s$ preserves remainder modulo 9, $A\equiv B\equiv C\equiv 4$.

Also, if a number is less than $10^d$, then it has at most $d$ digits so its sum is at most $9d$. An absurd (but useful enough) bound for $2^{2006}$ is $2^{2006}<2^{2007}=8^{669}<10^{669}$. Then $A\leq 9\cdot 669=6021$ and $B\leq 9\cdot 4=36$.

Since $B\equiv 4\pmod{9}$ and $B\leq 36$, the possibilities for $B$ are 4, 13, 22, and 31, and each case leads to $C=\boxed{4}$.
[/hide][/quote]
[quote="not_trig"]
[hide="Problem 24"]
There are three ways to get $x^3$, namely
\[ x \times x \times x, \frac{x^2}{2} \times x, and \frac{x^3}{6}.  \]
There are ${128 \choose 3}$ ways to get $x^3$, $128 \times 127$ ways to get $\frac{x^3}{2}$, and $128$ ways to get $\frac{x^3}{6}$. Adding these (multiply # of ways by whatever you get), you get
\[ \frac{128 \times 127 \times 126}{6} + \frac{128 \times 127}{2} + \frac{128}{6}  \]
\[ =\frac{128(127 \times 129 +1)}{6} = \frac{2^6(2^7)^2}{3} = \frac{2^{20}}{3}  \]

Therefore, $p = 2$, $q = 20$, and $r=3$, so $p+q+r = \boxed{25}.$

$C.$

[/hide][/quote]
The reason I am including two 24's is because the first is the typical solution expected, while the second is an out of the box simpler solution using calculus. (Very nice on both problems, Elemennop :) )
[quote="Elemennop"]
[hide="Problem 24"]\[ P(x) = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+9x^4+54x^5  \]
Observe that the coefficient of anything past $x^3$ is irrelevant;since we do not have any negative exponents, then exponents can only increase upon multiplication, and thus we can replace them with any coefficients that we want. Observing the first several terms (up to the $x^3$) of the power series for $e^x$ at the beginning of $P(x)$,
we can rewrite $P(x)$ as

\[ P(x)=\sum_{n=0}^{\infty} \frac{x^n}{n!}=e^x  \]
And the coefficient of $x^3$ here is still identical to our original problem. We raise it to the $128$th power, so we have $e^{128x}$. Expanding back along the power series, we have

\[ e^{128x}=1+128x+\frac{(128x)^2}{2}+\frac{(128x)^3}{6}+\ldots  \]
With our desired coefficient being $\frac{128^3}{6}=\frac{2^{21}}{6}=\frac{2^{20}}{3}$, so $p+q+r=2+20+3=25$.[/hide]
[hide="Problem 25"]
\begin{eqnarray*} \sum_{n=2}^{\infty} \displaystyle\left(\frac{n^5+n^4+8n^3+20n^2+20n+4}{3^n\cdot(n^5+4n)}\right)&=&\sum_{n=2}^{\infty}\frac{1}{3^n}\cdot\left(\frac{n^5+n^4+8n^3+20n^2+20n+4}{n^5+4n}\right)\\ &=&\sum_{n=2}^{\infty}\frac{1}{3^n}\cdot\left(1+\frac{n^4+8n^3+20n^2+16n+4}{n^5+4n}\right)\\ &=&\sum_{n=2}^{\infty}\frac{1}{3^n}\cdot\left(1+\frac{n^4+8n^3+20n^2+16n+4}{n(n^4+4)}\right)\\ &=&\sum_{n=2}^{\infty}\frac{1}{3^n}\cdot\left(1+\frac{n^4+8n^3+20n^2+16n+4}{n((n^2+2)^2-(2n)^2)}\right)\\ &=&\sum_{n=2}^{\infty}\frac{1}{3^n}\cdot\left(1+\frac{n^4+8n^3+20n^2+16n+4}{n(n^2+2n+2)(n^2-2n+2)}\right)\\ &=&\sum_{n=2}^{\infty}\frac{1}{3^n}\cdot\left(1+\frac{9}{n^2-2n+2}-\frac{1}{n^2+2n+2}+\frac{1}{n}\right)\end{eqnarray*}
Where the last step is a result of Partial Fraction Decomposition.
From here, we will evaluate the sums seperately. First off, we know
that

\[ \sum_{n=2}^{\infty}\frac{1}{3^n}=\frac{\frac{1}{9}}{1-\frac{1}{3}}=\frac{1}{6}  \]
Next, observe that

\begin{eqnarray*} \sum_{n=2}^{\infty}\frac{9}{3^n\cdot(n^2-2n+2)}-\frac{1}{3^n\cdot(n^2+2n+2)}&=&\sum_{n=2}^{\infty}\frac{9}{3^n\cdot(n^2-2n+2)}-\sum_{n=2}^{\infty}\frac{1}{3^n\cdot(n^2+2n+2)}\\ &=&\sum_{n=2}^{\infty}\frac{1}{3^{n-2}\cdot((n-1)^2+1)}-\sum_{n=2}^{\infty}\frac{1}{3^n\cdot((n+1)^2+1)}\\ &=&\frac{1}{2}+\frac{1}{15}=\frac{17}{30}\end{eqnarray*}
Now we can tackle our final sum. Observe that for $|x|<1$, we have \[ \frac{1}{1-x}=1+x+x^2+x^3+\ldots  \]

due to the infinite geometric series formula. Integrating with respect to $x$ on both sides, we have
\[ -\ln(1-x)=C+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\ldots  \]
Plugging in $x=0$ yields $C=0$. Now, plugging in $x=\frac{1}{3}$ yields
\begin{eqnarray*} -\ln(\frac{2}{3})&=&\frac{1}{3}+\frac{1}{2\cdot3^2}+\frac{1}{3\cdot3^3}+\ldots\\ &=&\sum_{n=1}^{\infty}\frac{1}{n3^n}\end{eqnarray*}
So our desired sum is
\[ \displaystyle\left(\sum_{n=1}^{\infty}\frac{1}{n3^n}\right)-\frac{1}{3}=\ln\left(\frac{3}{2}\right)-\frac{1}{3}  \]
Now we have all our parts, so let's put it together. Adding up everything we got, we get that our final answer is
\[ \displaystyle\frac{1}{6}+\frac{17}{30}+\ln\left(\frac{3}{2}\right)-\frac{1}{3}=\ln\left(\frac{3}{2}\right)+\frac{2}{5}.  \][/hide][/quote]
[quote="jb05"][hide="Problem 26"]Note that $\phi (a/7)=4$, $\phi (b/7)=10$, and neither is divisible by 49 (or else 24 or 60 would be divisible by 7). Therefore, $\phi (ab)=\phi (a/7*b/7*7*7)=4*10*6*7=\boxed{1680}$[/hide][/quote]
Need 27,28
[hide="Problem 29"]We know that the largest ellipse in an equilateral triangle with sides of $14$ is a perfectly centered circle (the incircle). We can find the radius by using the fact that $\frac{rp}{2}$ is the traingle's area, with $p$ as perimeter and $r$ as the inradius. The area of the triangle is $49\sqrt{3}$, and half its perimeter is $21$. Therefore, the length of the radius is $\frac{7\sqrt{3}}{3}$ so the area of this circle is $\frac{49\pi}{3}$. Now we try to transform this equilateral triangle into one with sides of 13,14, and 15. We transform by multiplying the height of the equilateral triangle by a factor of $\frac{12}{7\sqrt{3}}$, which sets this triangle's area equal to that of the 13,14,15. All we have done to the area of the incircle (which becomes an ellipse) is multiplyed by $\frac{12}{7\sqrt{3}}$, so the area is $\frac{28\pi}{3} , \boxed{A}$[/hide]
[quote="nat mc"][hide="Problem 30"]
Lets call the polygon $A_1A_2A_3\cdots A_n$ with center $O$. Let the midpoint of $A_iA_{i+1}$ be $B_i$ and $A_1A_2 \cap A_3A_4=C_1$

We can look at $\frac1n$ of the polygon and star because the ratio is still the same: $\frac{[A_2A_3O]}{[A_2C_1A_3O]}=\frac{[B_2A_2O]}{[C_1A_2O]}$  Since the height of these two triangles is the same ($B_2A_2$), then we can look at the ratio of their bases: $\frac{OB_2}{OC_1}$.  Also, since it is regular all $OB_i$ are equal. So the ratio can be $\frac{OB_1}{OC_1}$. We know $\angle{C_1B_1O}=90^\circ$, so $\frac{OB_1}{OC_1}=\sin\angle{B_1C_1O}$ Since $\angle A_3A_2C_1$ is an exterior angle and $\angle B_1C_1O$ is complementary to it, then $\angle B_1C_1O=90-\frac{360}n=\boxed{\frac{90(n-4)}n}$

[/hide][/quote]
[quote="scorpius119"]
[hide="Problem 31"]
By noting that the center of a regular polygon is the center of the circumcircle, the center will be in the interior of a triangle if and only if each central angle is less than 180 degrees, which occurs for acute triangles. Each side intercepts the same arc, so the triangle will be acute if and only if at most $n$ arcs are intercepted by an inscribed angle.

By symmetry, we can fix the first point. There are $\binom{2n}{2}=n(2n-1)$ ways to choose the other 2 points. If the points are labeled $P_1$, $P_2$, and $P_3$, then let $a$ be the number of arcs cut off by $P_1P_2$, $b$ by $P_2P_3$, and $c$ by $P_3P_1$. Then $a$, $b$, and $c$ are positive integers satisfying $a+b+c=2n+1$. When the triangle is acute, $\max (a,b,c)\leq n$.

For the desired outcome, each variable is less than $n+1$ so make a bijection with the substitutions $x=n+1-a$, $y=n+1-b$, $z=n+1-c$. Then $x$, $y$, and $z$ are positive integers and $x+y+z=n+2$. Any positive integer triple $(a,b,c)$ can be converted this way if and only if each value is less than or equal to $n$.

The second part to showing that this is a valid bijection is that no positive integer triple $(x,y,z)$ could be obtained by some $(a,b,c)$ where one of $a$, $b$, or $c$ is not positive. This can be shown by contradiction: suppose $a\leq 0$. Then $x\geq n+1$, but since $y,z\geq 1$, $x+y+z\geq n+3>n+2$ which is a contradiction. Therefore the bijection works.

The number of solutions in positive integers $x,y,z$ with $x+y+z=n+2$ is $\binom{n+1}{2}=\frac{n(n+1)}{2}$. Dividing the number of desired outcomes by the sample space, the probability is $\boxed{\frac{n+1}{4n-2}}$.
[/hide][/quote]
[quote="Elemennop"][hide="Problem 32"]There is no need to go multivariate with #32. Simply note that if the lowest number is $x$, then the probability that the other two numbers are not the lowest is $(1-x)^2$. Thus the probability across all $x$ in the range is \[ \int_0^1 (1-x)^2dx  \] but we wish to compute expected value, so we compute \[ \int_0^1 x(1-x)^2dx=\frac{1}{12}  \] and multiplying by $3$ to account for the three variables yields $\frac{1}{4}$.[/hide][/quote]
[quote="Xantos C. Guin"]
[hide="Problem 33"]
Let:
$\tan{(x)}=\frac{1}{5}$ and $\tan{(y)}=\frac{1}{239}$

We are trying to find $4x-y$

Using the double angle formula for the tangent function:

$\tan{(2x)}=\frac{2\cdot\tan{(x)}}{1-\tan^{2}{(x)}}=\frac{2\cdot\frac{1}{5}}{1-(\frac{1}{5})^{2}}=\frac{5}{12}$

$\tan{(4x)}=\frac{2\cdot\tan{(2x)}}{1-\tan^{2}{(2x)}}=\frac{2\cdot\frac{5}{12}}{1-(\frac{5}{12})^{2}}=\frac{120}{119}$

Using the diffrence of angles formula for the tangent function:

$\tan{(4x-y)}=\frac{tan{(4x)}-\tan{(y)}}{1+\tan{(4x)}\cdot\tan{(y)}}=\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119}\cdot\frac{1}{239}}=1$

$\tan^{-1}{(\tan{(4x-y))}}=\tan^{-1}{(1)}$

$4x-y=\frac{\pi}{4}\Rightarrow \boxed{B}$
[/hide][/quote]
[hide="Problem 34"]Imagine a row of 401 blocks. The problem is now, how many ways can we partition those blocks into three groups which satisfy the criterion for a,b, and c? The criterion are: a,b,c are non-zero, $a\not =b\not =c$, and $a<b<c$. We can ignore the last restriction by counting all orderings then dividing by 3!=6 at the end. So where can we put the first partition? There are 402 spots, but neither end works, so we have 400 spots. Then, we have only 396 spots for the next partition (can't be same as first makes 399, then 3 places which causes equality of variables). However, now we need to divide by two because once we have two partitions, order doesn't matter. The final step is to divide by 6, because of variable ordering. Total: $\frac{400*396}{2*6}=13200 , \boxed{B}$.[/hide]
[quote="Xantos C. Guin"][hide="Problem 35"]
Let $a_n=\tan{\theta_n}$
Then: $a_{n+1}=\frac{2\tan{\theta_n}}{1-\tan^{2}{\theta_n}}=\tan{2\theta_n}$

The first few terms are:
$a_1=\tan{15}$
$a_2=\tan{30}$
$a_3=\tan{60}$
$a_4=\tan{120}$

$a_5=\tan{240}=\tan{60}$
$a_6=\tan{120}$

Since $\tan{60}$ and $\tan{120}$ will continue to alternate in the sequence, The number of unique terms in the sequence is $4\Rightarrow \boxed{B}$

[/hide]

[hide="Solution that yeilds an answer for problem 36"]

Since the period of the tangent function is 180:

1: $13x \equiv 48\pmod{180}$
2: $19x \equiv 37\pmod{180}$

By multipling #1 by 2 and #2 by 3:

3: $26x \equiv 96\pmod{180}$
4: $57x \equiv 111\pmod{180}$

By subtracting #3 from #4:

5: $31x \equiv 15\pmod{180}$

So:
$\tan{31x}=\tan{15}=2-\sqrt{3} \Rightarrow \boxed{A}$

[/hide]

[hide="Proof that there are no values of x that fit both equations in problem 36"]

Since the period of the tangent function is 180:

$13x \equiv 48\pmod{180}$
$19x \equiv 37\pmod{180}$

Clearly $13x$ and $19x$ are integers.
Therefore $x$ must be a multiple of $\frac{1}{13}$ and $\frac{1}{19}$
Since $lcm(\frac{1}{13},\frac{1}{19})=1$: $x$ is a multiple of $1$ and is an integer.

So:
$13x \equiv 48\pmod{180}\rightarrow 13x \equiv 0\pmod{3}\rightarrow x \equiv 0\pmod{3}$
And:
$19x \equiv 37\pmod{180}\rightarrow 19x \equiv 1\pmod{3}\rightarrow x \equiv 1\pmod{3}$

This is clearly a contradiction.
Therefore there are no values of x that fit both equations in problem 36.
$\boxed{QED}$

[/hide][/quote]
[quote="nat mc"][hide="Problem 37"]The longest sum of the to sides is $2$.  The longest side it could posssibly have is $\sqrt2$.  So the rectangle has these dimensions ($2$ by $\sqrt2$) and has area $2\sqrt2$.  This can fit all possible pairs of squares with sum of areas being 2 and it can't get smaller due to the conditions above.
The answer is definitely $\boxed{2\sqrt2}$[/hide]
[/quote]
[quote="Xantos C. Guin"][hide="Problem 38"]
$2^{2^{n}}+2^{n}+1 \equiv 0 \pmod{21}$

So: $2^{2^{n}}+2^{n} \equiv 20 \pmod{21}$

Looking for a pattern in $\pmod{21}$:

\begin{tabular}[t]{|c|c|c|c|}\hline n&2^{n}&2^{2^{n}}&2^{2^{n}}+2^{n} \\\hline1&2&4&6\\\hline2&4&16&20\\\hline3&8&4&12\\\hline4&16&16&5\\\hline5&11&4&15\\\hline6&1&16&17\\\hline7&2&4&6\\\hline\end{tabular}

So:
$n \equiv 2\pmod{6}$

Since:
$9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999996$
is divisible by $6$

The largest value $n$ can have that is less than $10^{100}$ is:

$9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999998$

Then the sum of the digits is:
$99\cdot9+1\cdot8=899\Rightarrow\boxed{D}$
[/hide][/quote]
[hide="Extra piece of 38"]I remember solving by looking at it mod 3 and 7, instead of 21. I think that simplifies the solution and gets at 2 mod 6 quicker. (Simply look at powers of 2 (mod 7) which repeat as 1,4,2) [/hide]
[hide="Problem 39"]Since we are given nothing to differentiate the two other sides/altitudes, we set them equal to maximize. Now, we just draw the isosceles triangle with base length 2 and height 1, and see that the other sides are each $\frac{\sqrt{2}}{2}$. Our perimiter is thus $2+\sqrt{2} , \boxed{B}$[/hide]
[quote="jb05"]
[hide="Problem 40"]Since we could use A in every name, N is at least the number of subsets of B-Z, or $2^{25}$. But 2^{25} has 8 digits, so $\boxed{8}$ is the only possible answer.[/hide][/quote]
[quote="Xantos C. Guin"][hide="Problem 41"]
Let: $AB=AC=AD=x$

To maximize volume, let $AB, AC, AD$ be spread out evenly.

Then: $BC=CD=DB=x\sqrt{2}$

So:

$x+x+x+x\sqrt{2}+x\sqrt{2}+x\sqrt{2}=1$

$3x(1+\sqrt{2})=1$

$x=\frac{1}{3(1+\sqrt{2})}=\frac{-1+\sqrt{2}}{3}$

The volume of this pyramid will be $\frac{1}{3}\cdot (\frac{1}{2}\cdot x\cdot x) \cdot x=\frac{x^{3}}{6}$

By plugging in $\frac{-1+\sqrt{2}}{3}$ for $x$ and doing some annoying simplification, the volume is $\frac{5\sqrt{2}-7}{162}\Rightarrow \boxed{E}$
[/hide][/quote]
[quote="scorpius119"]
[hide="Problem 42"]
Let the variable we are trying to maximize be $x$ and the sum of the $k$th powers of the other 4 variables be $s_k$. We are given $s_1=8-x$ and $s_2=16-x^2$. By RMS-AM, $4s_2\geq s_1^2\Rightarrow 64-4x^2\geq 64-16x+x^2\Rightarrow x(5x-16)\leq 0$. It follows that $x\leq \boxed{\frac{16}{5}}$, and there is equality when the other 4 variables each equal 6/5.
[/hide][/quote]
[quote="jb05"]
[hide="Problem 43"]If the origin is the center of the larger circle, the center of the smaller circle is at $(\cos{\theta},\sin{\theta})$ as $\theta$ ranges from 0 to $2\pi$ radians. The line going through the hole and the center of the small circle is the reflection across a vertical line (through the center of the small circle) of the line through the centers of the circles because of the movement of the rotating circle. Since this halves the x-coordinate and adds half again to the y-coordinate, we have the general coordinate $(\frac{\cos{\theta}}2,\frac{3\sin{\theta}}2)$, which traces an ellipse with area $\frac 12*\frac 32*\pi=\boxed{\frac{3\pi}4}$[/hide][/quote]
[quote="nat mc"]
[hide="Problem 44"]See the attached picture.  You can unwind the cone and string into a part of a circle.  This circle has radius $6$ and partial circumference $2\pi$ making the angle $60^\circ$
We draw the point $M$ and point $P$ and draw a straight line between them to make the shortest path.  Then $MCP$ is right with legs $6$ and $3\sqrt3$, so $MP=\boxed{3\sqrt7}$[/hide][/quote]
[quote="scorpius119"]
[hide="Problem 45"]
Generalizing is simple for this problem: let the radii of the two bigger circles be $a$ and $b$, with $a<b$, and that of the smaller circle be $c$. Draw in radii to the tangent line, the radii connecting the circles, and lines parallel to the tangent line through the centers of circles $A$ and $C$.

The distance between the centers of circles $A$ and $B$ is $a+b$ and the difference in height is $b-a$. Applying the Pythagorean Theorem, the component of the distance between the centers that is parallel to the tangent line is $2\sqrt{ab}$. Similarly, it is $2\sqrt{ac}$ from $A$ to $C$ and $2\sqrt{bc}$ from $B$ to $C$.

The parallel component of the distance from $A$ to $B$ must be the sum of the components from $A$ to $C$ and from $B$ to $C$, so $\sqrt{ab}=\sqrt{ac}+\sqrt{bc}\Rightarrow \frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}$. Plugging in 1 and 2 into the problem, taking reciprocal, and squaring, $c=\boxed{6-4\sqrt{2}}$.
[/hide]

[hide="Problem 46"]
For all $\theta$, $2\cos^3 \theta -\cos \theta=\cos \theta \cos 2\theta\Rightarrow 2\cos^3 \theta -\cos^2 \theta -\cos \theta=\cos \theta (\cos 2\theta -\cos \theta)$. Then, by expressing $2\theta$ as $\frac 32 \theta +\frac 12 \theta$ and $\theta$ as $\frac 32 \theta -\frac 12 \theta$, this equals $-2\cos \theta \sin \frac 32 \theta \sin \frac 12 \theta$.

Now substitute $\frac{\pi}{7}$ in for $\theta$ ; the expression is $-2\cos \frac{\pi}{7} \sin \frac{3\pi}{14} \sin \frac{\pi}{14}=-2\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7}=2\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7}$.

Let $P=\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7}$. Then, after using double-angle formula for sine three times, $P\sin \frac{\pi}{7}=\frac 18 \sin \frac{8\pi}{7}=-\frac 18\sin \frac{\pi}{7}$. So $P=-\frac 18 \Rightarrow 2P=\boxed{-\frac 14 }$.
[/hide][/quote]

[quote="jb05"]
[hide="Problem 47"]If we assume that all coordinates are positive and x<=y<=z, then $x\leq y\leq \sqrt{1-x^2}$, which also yields $x\leq \sqrt{1-x^2}$, so $x\leq \frac{\sqrt 2}2$. Also, $y\leq z\leq \sqrt{1-y^2}\leq \sqrt{1-x^2}$, which also yields $y\leq \sqrt{1-y^2}$, or $y\leq \frac{\sqrt 2}2$. Since this is necessarily a better upper bound than $\sqrt{1-x^2}$, we have area $\int_0^{\frac{\sqrt{2}}2} \int_x^{\frac{\sqrt{2}}2} \int_y^{\sqrt{1-y^2}} dz dy dx=\frac 13-\frac{\sqrt{2}}6$. Now we multiply by 6 since we assumed a certain ordering and 8 since we limited it to a certain quadrant (really octant), to yield $\boxed{16-8\sqrt{2}}$[/hide][/quote]
[hide="Problem 48"]The fraction is: $\frac{1997n+1999*1998}{1998n}$. Looking at the top mod 2000, we have: $2-3n=0 mod 2000$ so $3n=2 + 2000k$ therefore $n = 334+1000a$. Now we substitute for n in the denominator to get:
$1997n+1999*1998=1997(1000a+334)+1999*1998=1000(4661 + 1997a)$. At first glance, all a values work because 1000 clearly divides the top. However, the other condition is that $gcd(4661+1997a,\mathfrak{denominator of fraction})=1$. This is tedious to check because the denominator is a mess, but with a calculator we confirm that $a=3$ is the smallest solution. Thus our $n=3334, s(n)=13 \boxed{D}$[/hide]
[quote="shadysaysurspammed"]
[hide="Problem 49"]the coefficent of $x^{39}$ is $\binom{100}{61}$

$\binom{100}{61}=\frac{100!}{61!*39!}$

 exponent of $2$ in $100!=[100/2]+[100/2^2]+...+[100/2^6]=97$

Similarly,for $61!=56$ and $39!=35$. Hence the answer is $\boxed{2^6}$ (and i gave 2^7 :ninja: )[/hide][/quote]
Need 50

[hide="Problem 16"]
By using a calculator: (Rounded to the nearest hundreth)

A: $8=8.00$
B: $e^{\pi}=23.14$
C: $\pi^{e}=22.46$
D: $(\frac{1}{\pi})^{-\frac{1}{e}}=1.52$
E: $10=10.00$

Since $23.14$ is larger that all the others, the answer is $\boxed{B}$.
[/hide]

Noone that had a graphing, or decent scientific calculator should have missed #16.

[hide="Problem 33"]
Let:
$\tan{(x)}=\frac{1}{5}$ and $\tan{(y)}=\frac{1}{239}$

We are trying to find $4x-y$

Using the double angle formula for the tangent function:

$\tan{(2x)}=\frac{2\cdot\tan{(x)}}{1-\tan^{2}{(x)}}=\frac{2\cdot\frac{1}{5}}{1-(\frac{1}{5})^{2}}=\frac{5}{12}$

$\tan{(4x)}=\frac{2\cdot\tan{(2x)}}{1-\tan^{2}{(2x)}}=\frac{2\cdot\frac{5}{12}}{1-(\frac{5}{12})^{2}}=\frac{120}{119}$

Using the diffrence of angles formula for the tangent function:

$\tan{(4x-y)}=\frac{tan{(4x)}-\tan{(y)}}{1+\tan{(4x)}\cdot\tan{(y)}}=\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119}\cdot\frac{1}{239}}=1$

$\tan^{-1}{(\tan{(4x-y))}}=\tan^{-1}{(1)}$

$4x-y=\frac{\pi}{4}\Rightarrow \boxed{B}$
[/hide]

[quote="Xantos C. Guin"]
Noone that had a graphing, or decent scientific calculator should have missed #16.
[/quote]

That's what I was about to say... :mad:

But I was occupied  ;)

#24 -

[hide]

Observe that the coefficient of anything past  is irrelevant;since we do not have any negative exponents, then exponents can only increase upon multiplication, and thus we can replace them with any coefficients that we want. Observing the first several terms (up to the ) of the power series for  at the beginning of ,

we can rewrite  as

And the coefficient of  here is still identical to our original problem. We raise it to the th power, so we have . Expanding back along the power series, we have

With our desired coefficient being , so .[/hide]

#25 -

[hide]

Where the last step is a result of Partial Fraction Decomposition.

From here, we will evaluate the sums seperately. First off, we know

that

Next, observe that

Now we can tackle our final sum. Observe that for , we have

due to the infinite geometric series formula. Integrating with respect to  on both sides, we have

Plugging in  yields . Now, plugging in  yields

So our desired sum is

Now we have all our parts, so let's put it together. Adding up everything we got, we get that our final answer is

[/hide]

[hide="Problem 10"]
Since prices in convenience stores can usually be expressed in the form $\frac{N}{100}$, where N is an integer.

Let the four items cost: $\frac{A}{100}, \frac{B}{100}, \frac{C}{100}, \frac{D}{100}.$

Then the product: $\frac{A}{100}\cdot\frac{B}{100}\cdot\frac{C}{100}\cdot\frac{D}{100}=\frac{ABCD}{100000000}=\frac{711000000}{100000000}$

So:
$ABCD=711000000=2^{6}\cdot3^{2}\cdot5^{6}\cdot79$

The prices of the items which Patrick buys, when multiplied by $100$ should be composed of only prime factors that appears in $711000000$

Looking at our choices:

$312=2^{3}\cdot3\cdot13$
$313=313$
$314=2\cdot157$
$315=3^{2}\cdot5\cdot7$
$316=2^{2}\cdot79$

Since $312, 313, 314, 315$ all contain a prime factor that does not appear in $711000000$ these choices are impossible, so our answer is $\$3.16\Rightarrow \boxed{E}$
[/hide]

[hide="Problem 41"]
Let: $AB=AC=AD=x$

To maximize volume, let $AB, AC, AD$ be spread out evenly.

Then: $BC=CD=DB=x\sqrt{2}$

So:

$x+x+x+x\sqrt{2}+x\sqrt{2}+x\sqrt{2}=1$

$3x(1+\sqrt{2})=1$

$x=\frac{1}{3(1+\sqrt{2})}=\frac{-1+\sqrt{2}}{3}$

The volume of this pyramid will be $\frac{1}{3}\cdot (\frac{1}{2}\cdot x\cdot x) \cdot x=\frac{x^{3}}{6}$

By plugging in $\frac{-1+\sqrt{2}}{3}$ for $x$ and doing some annoying simplification, the volume is $\frac{5\sqrt{2}-7}{162}\Rightarrow \boxed{E}$
[/hide]

[hide="33"]Notice that \[ (5-i)^4(i+1)=4(239-i).  \] $1+i$ has argument $\pi/4$. Also, $\arg (5-i) = -\arctan 1/5$. Recall that $\arg (xy) = \arg x + \arg y$. Hence, \[ \arg [(5-i)^4(i+1)] = -4\arctan (1/5)+\pi/4,  \] and \[ \arg [4(239-i)] = -\arg (239-i) = -\arctan 1/239.  \] Since the arguments must be the same, we have \[ 4 \arctan 1/5 - \arctan 1/239 = \boxed {\pi/4}.  \][/hide]

[quote="jb05"]
[hide="Problem 47"]If we assume that all coordinates are positive and x<=y<=z, then $x\leq y\leq \sqrt{1-x^2}$, which also yields $x\leq \sqrt{1-x^2}$, so $x\leq \frac{\sqrt 2}2$. Also, $y\leq z\leq \sqrt{1-y^2}\leq \sqrt{1-x^2}$, which also yields $y\leq \sqrt{1-y^2}$, or $y\leq \frac{\sqrt 2}2$. Since this is necessarily a better upper bound than $\sqrt{1-x^2}$, we have area $\int_0^{\frac{\sqrt{2}}2} \int_x^{\frac{\sqrt{2}}2} \int_y^{\sqrt{1-y^2}} dz dy dx=\frac 13-\frac{\sqrt{2}}6$. Now we multiply by 6 since we assumed a certain ordering and 8 since we limited it to a certain quadrant (really octant), to yield $\boxed{16-8\sqrt{2}}$[/hide][/quote]

Your avatar looks like a good diagram for this problem.  :)

[img]http://www.artofproblemsolving.com/Forum/images/avatars/195007141742aad6c60c362.jpg[/img]

[quote="nat mc"][quote="me@home"]Can someone please post a solution to #44 :([/quote]
Ok here:

[hide="Problem 44"]See the attached picture.  You can unwind the cone and string into a part of a circle.  This circle has radius $6$ and partial circumference $2\pi$ making the angle $60^\circ$
We draw the point $M$ and point $P$ and draw a straight line between them to make the shortest path.  Then $MCP$ is right with legs $6$ and $3\sqrt3$, so $MP=\boxed{3\sqrt7}$[/hide][/quote]

That was alot easier than wrapping a string around my beak which conveniently has similar dimentions.  :D

Actually, I used a 60 degree sector and solved the problem by reflecting points over an edge to find a minimum distance. Thanks for the easier solution.

Hey, the problem 44 diagram has been viewed 44 times (as of now)!

 :rotfl:

BTW, I just missed a ARML problem that was a specific version of problem 30 (1996 Individual #6) I knew there was something simpler.......

If anyone has a solution to the following problems, please post it:
11,12,19,27,28,50

:rotfl:  :rotfl: THE SC FORUM HAS BECOME UNIVERSAL.

That sounds good to me.  I have a latex template waiting to go, I just need to add in the "trivial" solutions.  So if you see a problem with no solution, please post one!

How to get Problem 47 for those of us who don't understand
$\int_0^{\frac{\sqrt{2}}2} \int_x^{\frac{\sqrt{2}}2} \int_y^{\sqrt{1-y^2}} dz dy dx=\frac 13-\frac{\sqrt{2}}6$:

[hide="Problem 47"]
The equations:

$x^{2}+y^{2}=1$
$y^{2}+z^{2}=1$
$z^{2}+x^{2}=1$

Describe 3 infinitely long cylinders of radius 1 that have the three axis as centers.

For a point $(x,y,z)$ to be a inside or on all 3 cylinders:

$x^{2}+y^{2}\le1$
$y^{2}+z^{2}\le1$
$z^{2}+x^{2}\le1$

Clearly:
$-1\le x,y,z \le1$

So the intersection can be completely enclosed in a cube whose vertices are in the form $(\pm1,\pm1,\pm1)$

Let $V$ be the volume of the intersection of all three cylinders.

Since the volume of this cube is $2^{3}=8$: $V\le8$

Since the volume of any object can't be negative: $V\ge0$

Evaluating the choices on a calculator to the nearest hundredth:

A: $16-12\sqrt{2}=-0.97$
B: $16-8\sqrt{2}=4.69$
C: $24-8\sqrt{2}=12.69$
D: $24-6\sqrt{2}=15.51$
E: $24-4\sqrt{2}=18.34$

The only answer choice that fits the inequality $0\le V \le 8$ is $16-8\sqrt{2} \Rightarrow\boxed{B}$
[/hide]

[quote="me@home"]noooooo this forum is becoming dead, I will finish up the solutions if they promise to latex up the document...[/quote]

That's a deal.  Please post solutions for:
Problem 11
Problem 12
Problem 19
Problem 27
Problem 28
Problem 50

[quote="me@home"]

[hide="Problem 29"]We know that the largest ellipse in an equilateral triangle with sides of $14$ is a perfectly centered circle (the incircle). We can find the radius by using the fact that $\frac{rp}{2}$ is the traingle's area, with $p$ as perimeter and $r$ as the inradius. The area of the triangle is $49\sqrt{3}$, and half its perimeter is $21$. Therefore, the length of the radius is $\frac{7\sqrt{3}}{3}$ so the area of this circle is $\frac{49\pi}{3}$. Now we try to transform this equilateral triangle into one with sides of 13,14, and 15. We transform by multiplying the height of the equilateral triangle by a factor of $\frac{12}{7\sqrt{3}}$, which sets this triangle's area equal to that of the 13,14,15. All we have done to the area of the incircle (which becomes an ellipse) is multiplyed by $\frac{12}{7\sqrt{3}}$, so the area is $\frac{28\pi}{3} , \boxed{A}$[/hide]

[/quote]

This solution does not work. Suppose that you do make the height larger. You still have to shift the altitude over also because we know that the altitude to the 14 side (in a 13,14,15 triangle) does not bisect the side like the altitude in the equilateral triangle

you have to pull the remaining vertex of the triangle diagonally, however, this is not a uniform stretch, ex. if you stretch vertically, then it is like doubleing the x coordinates for example, this is essentially orthogonal projection, so you can use the area ratios like that, however, this is not an orthogonal projection since it is going diagonally, therefore you cannot use the height ratio to area ratio

another way of thinking about it, suppose that you do stretch up along the heights, then you shift it over so the other sides are 13,15, this pushes the ellipse in, changing it's area

oh and of course if i am misunderstanding, sorry, but i would like some clarification, (preferably a pm since i don't visit the sc forum often)

Don't let this thread die...  Lots of work has apparently gone into getting these solutions together.  The effort shouldn't be abandoned.

I agree 100%.  So somebody please post:

Problem 11
Problem 12
Problem 19
Problem 27
Problem 28
Problem 50

Problem 28 was Mildorf's Mock AIME 4 Problem 11.

Here is the [url=http://www.tjhsst.edu/~tmildorf/math/PracticeAIME4Solutions.pdf]solution document[/url].

Well, since nobody seems to want to post 50, I will.
Let sinx=a cosx=b  tanx=a/b cot=b/a secx=1/b csc=1/a. Now you work with b/a+a/b, knowing the the numerator will become 1 since (XX)a^2+b^2=1 then you obtain a+b+1/ab+ (a+b)/ab. This is equal to (a+b+1)/ab + a+b so now you just do another substitution. Look at a+b=l(some random variable). If you square it you obtain 1+2ab=l^2 for same reason mentioned a (XX). Now, you solve for ab, which is just (l^2-1)/2. Then you substitute back into the equation, cancel some terms. You will have some quadratic looking thing. Let it be equal to a. Then you just do another quadratic equation, and look at the discrimant. It should be a^2-2a-7. If you look at this quadratic, you obtain 1+/- 2sqrt(2). Going back to the original question, it's asking for the minimum value of a. But with the quadratic equation's discriminant, this obtains a lower bound since a must be real(ie discriminant greater than or equal to o). Therefore, the answer is absolute value 1-2sqrt(2) = 2sqrt(2)-1. Excuse my lack of latex, but if you wish to blame anyone, blame all the other kids that know how to but chose not to, for atleast i tried. Might I ask what level question this is relative to AIME standards(how about for 2008 A, since they vary by year and test A and B.).


When I saw solutions of some problems in inequalities.

So many of them suppose that abc=1 ,a+b+c=1

For a+b+c=1 sometimes they let a=x/(x+y+z) b= y/(x+y+z)

c= z/(x+y+z) or a = xy/(xy+yz+xz)  b = yz/(xy+yz+xz)

c = xz/(xy+yz+xz) . But I really not understand why did they

can do that. So please help me by answer these questions.

   1) What conditions of problems we can suppose that abc=1

(and better for give me the reason.)

   2) What conditions of problems we can suppose that a+b+c=1

(and better for give me the reason.)

                   Thank you ver much for answering

                          Sorry for my English. :D

1.$k^{3}=abc$
Let $a_{1}=\frac{a}{k}$
$b_{1}=\frac{b}{k}$
$c_{1}=\frac{c}{k}$
so $a_{1}b_{1}c_{1}=1$
2.$k=a+b+c$
$a_{1}=\frac{a}{k}$
$b_{1}=\frac{b}{k}$
$c_{1}=\frac{c}{k}$
so $a_{1}+b_{1}+c_{1}=1$
and if the equality is still true if we change $a,b,c$ into $a_{1},b_{1},c_{1}$ respectively, we can standardize so. And i think it's depends on the degrees of the variables.

Continue the discussion here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135731[/url]

Topic locked.


[quote="BornForMath"]$\displaystyle X = 1 + \cfrac{1}{2 + \cfrac{1}{3 + \cfrac{1}{4 + . . .}}}$.             Find $X$.[/quote]I'm willing to bet that this has been posted before. :P

The closed form is related to Bessel functions; see [url=http://mathworld.wolfram.com/ContinuedFraction.html]this[/url] for details.

Thanks! But for the exact value of $0 + \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{3 + \cfrac{1}{4 + . . .}}}}$, it says $\frac{I_1 (2)}{I_0} (2)$. Any idea what that means?  :o

EDIT: I found that it is bessel functions, which is beyond the scope of this section.

[quote="blahblahblah"]The $I_n(x)$'s are Bessel functions.[/quote]
Is there a better way to solve it than using Bessel? I don't know what Bessel means  :?

Er, no.  Bessel functions appear to be the simplest description, which would imply that the problem can't be solved by more elementary means, so there really isn't a simpler solution.


Galois Theory
superior algebra
superior algebra unsolved

Does there exist a Galois extension $K/\mathbb{Q}$ such that $K$ is contained in $\mathbb{R}$ and with Galois group $S_{n}$ for each $n$?

Yes but i do'nt know the proof  :lol: ,actually Serre has proved that if a finite group can be realised as a galois group over rationals,then it can be done as the galoiis group of a totally reall number field and now using hilbert irreducibilty criteria we know symmetric groups can be realised as galois groups over rationals and ........  :)

Do you know where to find the proof of this theorem of Serre and Hilbert irreducibility?

hi
http://www.msri.org/communications/books/Book45/index.html
download thiis book,it has a proof of hilbert irreducibility theorem,but i don't know where to find serre's proof,if i come to know about any source i will let u know


Solve equation:
$ n\plus{}S(n)\plus{}S(S(n))\equal{}2003$
where $ S(n)$-sum of digits of $ n$

this was an AMC problem from a couple of years ago. check the resources.

[quote="Syler"]Solve equation:
$ n \plus{} S(n) \plus{} S(S(n)) \equal{} 2003$
where $ S(n)$-sum of digits of $ n$[/quote]
Very easy!
For every $ n$, $ S(S(n))\equiv S(n) \equiv n (pmod 3) \Rightarrow 3|n \plus{} S(n) \plus{} S(S(n))$.
The equation has no solution.


AMC
AIME
trigonometry
algebra
polynomial
quadratics
complex numbers

[quote="Adunakhor"]AIME 1996 #11

Let the product of the roots of $z^6+z^4+z^3+z^2+1=0$ with positive imaginary part be $r(\cos\theta+i\sin\theta)$. Find $\theta$ (in degrees).[/quote]
[hide]
if you plug in a fifth root of unity for z (which means z^5=1 and ergo z^5=z) we find out that the polynomial comes out to be z^4+z^3+z^2+z+1=0, and this is true if z is the fifth root of unity. This implies that the polynomial is divisible by every fifth root of unity, and therefore divisible by z^4+z^3+z^2+z+1. Therefore, we can divide z^6+z^4+z^3+z^2+1 by z^4+z^3+z^2+z+1, and come out with a quadratic, whos roots can be attained by quadratic formula.

the fifth roots of unity that have a positive imaginary part are cis(pi/5) and cis(2pi/5). [/hide]

I got lost in the solution you linked right after they had solved for $x$ and solved the case for $x=1$ and got $60$ as one of the answers. But how do you determine that $\sin18=\frac{-1+\sqrt{5}}{4}$ and that $\cos36=\frac{1+\sqrt{5}}{4}$? I initially thought, from what was posted, that it was simply memorized, but I don't think that that was the case after rereading.

Yes, he had that memorized. I think you can determine the exact value of $\sin (18^\circ)$ by finding the roots of \[ x^4+x^3+x^2+x+1=0\Longleftrightarrow \frac{x^5-1}{x-1}=0. \]


[quote="shobber"]$x,y,z \in R^+$, show that:
\[ \frac{x}{2x+y+z} + \frac{y}{x+2x+z} + \frac{z}{x+y+2z} \leq \frac{3}{4}  \][/quote]
Let $x+y+z=3.$
Hence $\frac{x}{2x+y+z} + \frac{y}{x+2x+z} + \frac{z}{x+y+2z} \leq \frac{3}{4}\Leftrightarrow$
$\Leftrightarrow\sum(\frac{x}{x+3}-\frac{3}{16}(x-1)-\frac{1}{4})\leq0\Leftrightarrow\sum\frac{(x-1)^2}{x+3}\geq0.$  :)

Another solution:
 $\frac{x}{2x+y+z} + \frac{y}{x+2x+z} + \frac{z}{x+y+2z} \leq \frac{3}{4}\Leftrightarrow\sum(\frac{1}{4}-\frac{x}{2x+y+z})\ge 0\Leftrightarrow\sum\frac{(b-c)^2}{(x+y+2z)(x+z+2y)}\ge 0$
 :lol:
I think this way is easier

$\frac{4}{2x+y+z} \leq \frac{1}{x+y} + \frac{1}{x+z}$
sum up and done.

Although I have posted this inequality long time ago, here is another solution. ;)
 [color=blue] [b] Problem.[/b]

   Let $x, y, z$ three positive real numbers. Prove the following inequality:

                     \[ \frac{x}{2x+y+z}+\frac{y}{x+2y+z}+\frac{z}{x+y+2z}\leq\frac{3}{4}.  \]

  [/color]

  [i] Solution. [/i]

      The inequality is equivalent with this one:

            \[ \frac{2x}{2x+y+z}+\frac{2y}{x+2y+z}+\frac{2z}{x+y+2z}\leq\frac{3}{2}\Leftrightarrow  \]
            \[ \frac{(2x+y+z)-(y+z)}{2x+y+z}+\frac{(x+2y+z)-(z+x)}{x+2y+z}+\frac{(x+y+2z)-(x+y)}{x+y+2z}\leq\frac{3}{2}\Leftrightarrow  \]
            \[ \frac{y+z}{2x+y+z}+\frac{z+x}{x+2y+z}+\frac{x+y}{x+y+2z}\geq\frac{3}{2}.  \]

      But, this last inequality is Nesbitt \[ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}  \]
      for $a=y+z, b=z+x$ and $c=x+y$. Done! :)

I have another solution
Let $x+y+z = 1$, then the problem becomes $\sum \frac{x}{x+1} \leq \frac{3}{4} \iff \sum(1-\frac{1}{x+1}) \leq \frac{3}{4} \iff \sum \frac{1}{x+1} \geq \frac{9}{4}$

But the latest inequality is just AM-HM


Suppose T is a linear operator from V to V with $ T^2 \equal{} T$.  Show it is an orthogonal projection iff it is self-adjoint.

[hide="reverse direction"]
Suppose T is self-adjoint.  Let $ \lambda$ be an eigenvalue and $ v \in V$ such that $ Tv \equal{} \lambda v$.  Since $ T \equal{} T^\ast$, $ \left<Tv, v\right> \equal{} \left<v, Tv\right> \equal{} \left< v, T^2 v\right>$ so \[ \lambda \left<v,v\right> \equal{} \lambda^2 \left<v,v\right>,\] so $ \lambda \in \{0, 1\}$.  Therefore $ T$ must be an orthogonal projection.

Unfortunately, this fails in the forward direction because $ \left<Tv, v\right> \equal{} \left<v, Tv\right>$ for all v does not imply $ T \equal{} T^\ast$.
[/hide]

($ \Leftarrow$)

For an idempotent $ T$ we know, that $ \ker{T} \oplus \mathrm{im}{T}=V$.

For a self-adjoint $ T$ we know, that $ \ker{T} \oplus \mathrm{im}{T^{\ast}} = V$ and $ \ker{T}^{\perp}=\mathrm{im}{T^{\ast}}$

We conclude $ im{T}=\mathrm{im}{T^{\ast}}$ and $ \ker{T}^{\perp}=\mathrm{im}{T}$, which means that $ T$ is an orthogonal projection.

($ \Rightarrow$)

We have to show, that $ \left<Tv,u\right>=\left<v,Tu\right>$ for all pairs $ u,v\in V$.

Since $ T$ is idempotent: $ v\in \ker{T}$, $ Tv=0$ and for $ v\in \mathrm{im} T$, $ Tv=v$.

[b]1. possibility:[/b]

$ v, u \in \ker{T}\Rightarrow \left<Tv,u\right>=\left<0,u\right>=0=\left<v,0\right>=\left<v,Tu\right>$.

[b]2. possibility:[/b]

$ v\in \ker{T}, u\in \mathrm{im}{U}\Rightarrow
\left<Tv,u\right>=\left<0,u\right>=0\\
\left<v,Tu\right>=\left<v,u\right>= 0$ (since $ T$ is orthogonal projection).

[b]3. possibility:[/b]

$ u\in \ker{T}, v\in \mathrm{im}{U}$

Same proof as above.

[b]4. possibility:[/b]

$ v\in \mathrm{im}{T}, u\in \mathrm{im}{U}:
\left<Tv,u\right>=\left<v,u\right>=\left<v,Tu\right>$.

Here is [url=http://www.mathlinks.ro/weblog_entry.php?t=276206]another solution[/url].

[quote="chess64"]so $ \lambda \in \{0, 1\}$. Therefore $ T$ must be an orthogonal projection. [/quote]

Why does $ \lambda \in \{0, 1\}$ imply, that $ T$ is an orthogonal projection?

For any idempotent we have $ \lambda \in \{0, 1\}$. Therefore we would not need $ T$ being self adjoint for being orthogonal projection.

what are sufficient conditions for T to be an orthogonal projection?  if $ T^2 \equal{} T$ and $ \text{null } T \equal{} (\text{range } T)^{\perp}$, does that mean T is an orthogonal projection?

You can read http://en.wikipedia.org/wiki/Orthogonal_projection.

A linear transformation $ T: V\rightarrow V$ is a projection, if $ T^2\equal{}T$, which means, that it leaves its image unchanged.

A projection $ T$ is orthogonal, if $ (\ker{T})^{\perp}\equal{}\mathrm{im}(T)$.

in that case you just use $ \text{im } T \equal{} \left(\text{ker } T^*\right)^\perp$.

As a follow-up to this question, how would you prove $ T$ is self-adjoint if and only if $ TT^*\equal{}T^*T$? We're assuming $ T^2\equal{}T$, obviously. "Only if" is obvious, but I'm having trouble with "if."

This is because normal operators in finite dimensions are unitarily diagonalizable.


doing this off the top of my head:

[code]
For(x,1,10)
Output(2,2,"hai guys")
 For(y,1,500)
 End
ClrHome
 For(y,1,500)
 End
End
[/code]

something like that. edit 500 as necessary.

EDIT: TI-89. Never mind.

[quote="#H34N1"]How do I create a program that flashes a message when it starts?

Thakns.[/quote]

I don't think you can creat programs on the TI-89?

[quote="funcia"]better question: #H34N1, CAN you program in ti 89?[/quote]

no i remember asking chess64 and he says you can't program on the Titanium

also, this should be moved to another forum? like Support or something?

Flashes a message?  Some variant of Treething's program (where 500 is drastically lowered and syntax is corrected for the 89) should work.
If you want a more powerful language than TI-BASIC, you can use [url=http://tigcc.ticalc.org]TIGCC[/url] if you know C.

[quote="#H34N1"]You can't program in TI-89? Then what's teh program editor for?[/quote]

Well i'm pretty sure it's not for BASIC...but i dunno...ask chess64, he has a TI-89

You can program on the TI-89. The language differs somewhat from the 83.

TI-89 has a timer, which you should use in preference to a loop. TI probably has the manual online

For x,1,10
Output 2,2,"hi guys"
etc.


Source: Games Magazine

Find a number n such that n, n+5, and n-5 all have rational square roots.

[hide]Hint: n is not an integer, but a rational number[/hide]

The answer is......

[hide](41/12)^2[/hide][/hide]

I'm stuck. I think I've reduced this to the problem of finding the arithmetic sequence (all integers) $l^2,p^2,k^2$ with a common difference of $5q^2$. Or maybe that's just complicating the problem..   :(


Find  all  polynomilas $P(x)$ with  real  coificients , such  that $P(2)=12$ and  $P(x^2)=x^2(x^2+1)P(x)$ for  every  real number  x .

You  can  find the  problems in a pdf  file  of our national  mo  here : http://www.mathlinks.ro/Forum/viewtopic.php?t=60572

Looking for degree gives degree P $= 4$.

$P(0) = 0$, $P(\pm 1) = 0$ gives quickly $P(x) = x^4 - x^2$ as only solution.

Look here. there exist a complete solution.
http://www.mathlinks.ro/Forum/viewtopic.php?t=59721


Participation in the local soccer league this year is 10% higher than last year.  The number of males increased by 5% and the number of females increased by 20%.  What fraction of the soccer league is now female?

answer: 4/11

I just can't seem to get the calculations right...

Thanx for the help! XD

Let $ f$ be the number of females now in te club and $ m$ be the number of males.

Now we have that $ \frac{11}{10}(\frac{5}{6}f\plus{}\frac{20}{21}m)\equal{}m\plus{}f$.

Multiplying out we get:

$ \frac{11f}{12}\plus{}\frac{22m}{21}\equal{}m\plus{}f$ or rather:

$ 77f\plus{}88m\equal{}84f\plus{}84m$

$ 7f\equal{}4m$

So the ratio of females to males in the club is $ 4: 7$.

Thus $ \frac{4}{11}$ are female.

Hmm quick question

This was on the 2010 National Team Round. How is it possible it was posted in 2009...?

Are you sure it was on 2010? The MN team just did 2010 and this problem was not included.

(Is this an April Fool's?)

This is interesting, but probably it was created before 2009, was relatively unknown, and so they used it in MathCounts in 2010.  Because I have also seen this problem in a different 2013 math competition

This problem was #13 on the 2002 AMC 10P, a practice AMC10 competition for select schools in Taiwan.  I've added all of the other problems into the contests page under the AMC 10 2002 section.

Back to the original problem, you can do it with only one variable...if you allow fractional people.  (Or, if you get really lucky.  Or if you get really OCD and change your numbers retroactively to fit.)

Let's say there used to be $100$ boys and $g$ girls.  After the increase, there are $105$ boys and $1.2g$ girls.

There used to be $100 + g$ kids, and now the $10\%$ overall increase gives $105 + 1.2g$ kids.  That means we get:

$(100 + g)1.1 = 105 + 1.2g$

Solving that gives $g = 50$, so we happened to get lucky and didn't have to chop any children up.  That means we used to have $100 + 50$ kids, and now we have $105 +  60$ kids.  The desired fraction is $\frac{60}{165} = \frac{4}{11}$

Take-home point:  Sometimes assuming a certain number as a "baseline" can eliminate a variable in a percentage problem.

Bonus:  I started with $100$ boys.  What numbers of boys work so that you never get a fractional child?

*FACEPALM*  I really, really need to check dates before I pound out a post....

Multiples of 20 boys work; so 20 is the lowest number of boys that can be in the league.


trigonometry
geometry
geometric sequence
arithmetic sequence
AMC

In $\triangle ABC$, $AB=BC$, and $\overline{BD}$ is an altitude. Point $E$ is on the extension of $\overline{AC}$ such that $BE=10$. The values of $\tan \angle CBE$, $\tan \angle DBE$, and $\tan \angle ABE$ form a geometric progression, and values of $\cot \angle DBE$, $\cot \angle CBE$, and $\cot DBC$ form an arithmetic progression. What is the area of $\triangle ABC$? \[ \text {(A) } 16\qquad \text {(B) } \frac{50}{3}\qquad \text {(C) } 10\sqrt{3}\qquad \text {(D) } 8\sqrt{5}\qquad \text {(E) } 18 \]

Nice problem, just takes a lot of time to fully explain :lol:

Note: During the actual test, I didn't actually write all of this out...that wouldn't have been good timewise :P

[hide]Let $\angle{DBE} = \alpha$ and $\angle{DBC} =\beta$. Then $\angle{CBE} = \alpha - \beta$ and $\angle{ABE} = \alpha + \beta$, so $\tan{(\alpha - \beta)}\tan{(\alpha+ \beta)} = \tan^2{\alpha}$. Thus, we now have:

$\frac{\tan{\alpha} - \tan{\beta}}{1 + \tan{\alpha}\tan{\beta}} \cdot \frac{\tan{\alpha} + \tan{\beta}}{1 - \tan{\alpha}\tan{\beta}} = \tan^2{\alpha}$

From this, it's clear that $\tan^2{\alpha} - \tan^2{\beta} = \tan^2{\alpha}(1 - \tan^2{\alpha}\tan^2{\beta})$

We can simplify this equation to $\tan^2{\beta}(\tan^4{\alpha} -1) = 0$, so $\tan{\alpha} = 1$, and $\alpha = \frac{\pi}{4}$. Let $DC = a$ and $BD = b$. Then $\cot{\angle{DBC}} = \tan{\angle{ABE}} = \tan{\left(\frac{\pi}{4} + \beta\right)} = \frac{1 + \frac{a}{b}}{1 - \frac{a}{b}} = \frac{b+a}{b-a}$.

Thus the numbers $1, \frac{b+a}{b-a}$ and $\frac{b}{a}$ form an arithmetic progression, so $\frac{b}{a} = \frac{b+3a}{b-a}$. Setting $b = ka$ yields $k^2-2k-3 = 0$, and the only positive solution is $k=3$. Hence $b = \frac{BE}{\sqrt{2}} = 5\sqrt{2}, a = \frac{5\sqrt{2}}{3}$, and the area of $\triangle{ABC}$ is $ab = \frac{50}{3}$, so our final answer is $\boxed{B}$.

[/hide]


function
inequalities unsolved
inequalities

[quote="Goga11"](∏a_i )/(∑a_i)^n≤(∏(1-a_i ) )/(∑(1-a_i ) )^n
a_i∈(0,1/2>
  i=1,2,…,n[/quote]
function $ f(x)\equal{}\frac{ax}{(x\plus{}b)^n}$ is increasing $ (a,b>0)$and $ 2a_i\le1$ so $ a_i\le 1\minus{}a_i$ for all i, so ineq is true


If you have a hyperbola whith one horizontal and one vertical asymtote, I undersatnd how to find the vertical one. How do you find the horizontal one?

Its y-intercept is the number to which the value of the function tends when x tends to $+\infty$ or $-\infty$.


Let a,b,c>0.Prove that
$ 2(a^6  \plus{} b^6  \plus{} c^6 ) \plus{} 3\sum\limits_{cyc} {a^2 b^2 (a^2  \plus{} b^2 } ) \ge 2(a^3 b^3  \plus{} b^3 c^3  \plus{} a^3 c^3 ) \plus{} 3\sum\limits_{cyc} {ab(a^4  \plus{} b^4 )}$  :)

Rewrite the inequality as $ \sum_{cyc} (b^3\minus{}c^3)^2\minus{}3bc(b\minus{}c)^2(b^2\plus{}bc\plus{}c^2)\equal{}\sum_{cyc} (b \minus{} c)^4 (b^2 \plus{} b c \plus{} c^2)\ge 0$


inequalities
modular arithmetic
number theory

find  all primes $p$ such that $p(p+2)$ divides $(p-1)!+1$

By Wilson's theorem, $p|(p-1)!+1$. Also, if $p$ is odd, $(p-1)!$ and $p+1$ are even, so we cannot have $(p-1)!$ being $1$ mod $p+1$. So the only case left is when $p$ is even, so $p=2$, and this one works.
So, $p=2$.

I think this is too easy for Pre-Olympiad.

sorry I made a mistake I mean find all primes p such that $p(p+2)$ divides $(p-1)!+1$

[hide]
Suppose $p(p+2)$ divides $(p-1)!+1$. Then $p+2$ divides $(p-1)!+1$. If $p+2$ were composite, then let $d$ be the smallest prime divisor of $p+2$, so that $d\leq \sqrt{p+2}$. Since $d$ divides $(p-1)!+1$, $d$ cannot divide $(p-1)!$, so $d\geq p$. In particular, $\sqrt{p+2}\geq p\Rightarrow (p+1)(p-2)\leq 0$. The only prime satisfying this inequality is $p=2$, but this fails.

Otherwise, $p+2$ is prime. Let $q=p+2$. By Wilson's Theorem, $(q-1)!\equiv-1\pmod{q}$, so $(q-2)!\equiv 1\pmod{q}$. However, the condition requires $(q-3)!\equiv-1\pmod{q}$. This is only possible when $q-2\equiv-1\pmod{q}$, or $1\equiv 0\pmod{q}$ which is impossible. Therefore $\boxed{\text{no solution}}$.
[/hide]

If $p+2$ isn't prime, then his divisors are $\leq p-1$ (except for banal cases) so $p+2\mid (p-1)!$, and $p+2\nmid (p-1)!+1$

So p is prime and p+2 is a prime too

So $p+2\mid (p+1)!+1$, so $p+2\mid (p+1)!-(p-1)!=(p-1)!*(p^{2}+p-1)$. Since $((p-1!);p+2)=1$, then $p+2\mid p^{2}+p-1= (p+2)^{2}-3(p+2)+1$, which is untrue. Neither banal cases work.


[quote="ZetaX"]http://www.mathlinks.ro/Forum/viewtopic.php?t=138329

 :mad:[/quote]

the solution is not correct, is it?

So what¿

PS: nonetheless deleted the other thread, but _not_ because the solution was not correct _but_ because your starting post was bad.

[moderator edit: rest of discussion merged into http://www.mathlinks.ro/Forum/viewtopic.php?t=74232 ]


Let $ x,y,z\geq0$ find smalest value of this expression
\[ P
\equal{}\displaystyle\frac{x^2(y\plus{}z)}{y\sqrt{y}\plus{}2z\sqrt{z}}\plus{}\displaystyle\frac{y^2(z\plus{}x)}{z\sqrt{z}\plus{}2x\sqrt{x}}\plus{}\displaystyle\frac{z^2(x\plus{}y)}{z\sqrt{x}\plus{}2y\sqrt{y}}\]

hello, i think it must be
$ P \equal{}\frac{x^{2}(y\plus{}z)}{y\sqrt{y}\plus{}2z\sqrt{z}}\plus{}\frac{y^{2}(z\plus{}x)}{z\sqrt{z}\plus{}2x\sqrt{x}}\plus{}\frac{z^{2}(x\plus{}y)}{x\sqrt{x}\plus{}2y\sqrt{y}}$
Sonnhard.

$ P$ can get arbitrarily close to zero. There should be some restriction on the variables or a homogeneous expression.  :wink:

yes. Dr Sonnhard Graubner are right. I confuse some thing. $ x\sqrt{x}$ and is not $ z\sqrt{x}$

Once again, the smallest possible value does not exist and can get arbitrarily close to zero.


[quote="hungkhtn"]It $4(a_1a_2a_3a_4)^{1/4}$ and $\frac{4}{(a_1a_2a_3a_4)^{1/4}}$
But I can't write in Latex?[/quote]

It's $4\sqrt[4]{a_1a_2a_3a_4}$, [code]4\sqrt[4]{a_1a_2a_3a_4}[/code] Similarly for the other.


inequalities LaTeX algorithm parameterization algebra polynomial calculus

I finally finished the program. :D Unzip CIB.zip and move Canmoo.jpg to C. I'm don't think you can open the maplet if you don't have Maple, but I'm really not sure. :?
Canmoo's Inequality Basher bashes symetric inequality using Muirhead. There should be tons of bugs in this program. Post it here if you find one.

[quote="TheDreamer"]Where can I get maple?  Do I have to buy it?[/quote]
You don't have to buy it. I'm quite sure that Maple can translate itself into other programming language, so I may be able to translate this maplet into java or something like that. But before I do that, let me fix the program. I found a shorter way(and hopefully faster) to do most of the steps.

Nice program... but yeah... some issues.  I entered #3 from 2005 IMO, hit solve, and the program window just closed.

Also, why is all of the output in latex?  It would be nicer if you could just get it to generate Maple notation... which can be exported to latex, mathML, etc.

And yes... you have to buy Maple.  It's quite expensive, but an extremely good program.  I know this because I WORK AT MAPLE!  W00T!

If you go to Waterloo (which you all should) then you can buy a student edition of Maple for like 20 bucks or something.  It's a rediculously good deal.

Maple is nice, but it's just too flaky at times.  There are some things you should never use Maple for (any intensive numerical computations, which I can speak about from experience in the last 24 hours; Matlab is far better for such things), and anything for which speed is important (face it, Maple is slow).  Not to mention that it crashed while I was saving an (important!) file today and destroyed the file in the process, which led to some furious coding in order to hand my assignment in time.  If I had actually paid for it, maybe it would bug me more, but I still think that it needs a lot of fine tuning as a whole.

There is actually a huge difference in the way Maple and Matlab handle data.  Matlab has simple data structures and very fast algorithms (but no symbolic features).  Maple's data structures are much more complicated to be able to handle symbolic computation, hence the numerics can suffer a bit in speed because of the amount of argument checking, parameter processing, etc that takes place.  In general, however, Maple can keep up with Matlab if you use the proper commands and optimize things.  For example, using evalhf for any floating point computations; using the fast commands seq, op, & map instead of writing loops; never using sum or product when mul or add would suffice.

If you're having trouble with slow Maple code, then are probably lots of things you could do to make things faster. I've seen examples where using the wrong command/data structure caused enormous slowdown.  An example would be using a list as an accumulator for expressions.  Each assignment to a list requires copying out the entire list, so a huge amount of time could be saved by using arrays instead.

If you type ?efficiency into Maple, it will give you lots and lots of tips about optimizing Maple code, and if you read the programming guides (which you would have received with Maple if you had paid for it), then you would get the full rundown.

I will admit that the Maple GUI is slightly slow in some aspects, but working in Classic or TTY mode fixes that.

Maple also has some of the fastest algorithms in the world for things like factoring polynomials, solving systems of equations, computing integrals, etc.

I didn't say that I didn't have a legal copy of Maple - my dad is a prof and hence has a  copy.  But I didn't pay for it.

The thing that makes Maple so unique is that it is an interpreted, interactive, functional language.  So if you're used to coding with things like loops, etc, you will find that code of that nature runs more slowly in maple than say, in C, which is a compiled, non-interactive, procedural language.  Functional programming just lends itself to mathematical uses much better than C-style code.  It's closer to something like Lisp or Scheme (which are good for writing algorithms in, but lousy for developing applications).

Okay.  I'll take your word with it, since you're an infinitely better programmer now than I ever expect to be.  Still, my criticisms of Maple are echoed by many members of the applied math faculty at SFU, who are actually very strong computationally (or maybe it's me echoing their criticisms, but whatever)

The applied math faculty at SFU, eh?

Maybe you didn't know that a lot of FAST Maple code has been developed by the Computer Algebra group at SFU... Michael Monagan, the leader of that group, has been a huge contributor of very very fast algorithms for Maple.  Talk to him... he probably knows much much more about Maple than I ever will.

[url=http://www.cecm.sfu.ca/CAG/]http://www.cecm.sfu.ca/CAG/[/url]

I keep abreast of the work done by the faculty here (and the area around my office is plastered with posters produced by said group) so yes, I was aware of what he's been doing.  I don't know Dr. Monagan personally (he's not especially friendly) but I do know a lot of students who have worked with him, mostly through NSERC USRAs.  From reputation he's particularly good to work with, because you often will get your name on a paper by the end of the summer.  I will say that most of the faculty on that list are not particularly active in the Computer Algebra group, especially the profs.

Actually I think he's coming to Waterloo sometime to spend a week at Maple sometime soon... so hopefully I'll get to meet him soon.  A large portion of my work at Maple involves code that he has developed, so I've been in contact with him quite a bit.

Maple definitely has some weaknesses, and its specialties are very different than those of, say, Matlab (which is quite fast for what it can do, but it really can't do 1/4 of what Maple can).  But yeah... Maple is great for a lotta things, and I use it often for anything math-related.

I don't know how much flexibility there is in your degree program, but if you particularly wanted to work with him, and wanted to see some of British Columbia, which is clearly a nicer place than Waterloo, you could always do an NSERC USRA (Undergraduate Student Research Award, [url=http://www.nserc.ca/sf_e.asp?nav=sfnav&lbi=1a]link[/url]) over here in the summer.  They're worth about 6000 for 16 weeks, and I think they pay your travel costs (but not housing, etc).

Right now I'm in co-op... so if that counts as a summer co-op term... maybe.  I'm doing school during this summer though.  Also, I've heard a few things about future co-op jobs in new york and even hong kong... so who knows where I'll end up.

Also, I think the party scene is much better here in Ontario (especially Toronto) than it is out west... from what I've been told at least.

[quote="Elyot"] Also, I've heard a few things about future co-op jobs in new york and even hong kong... so who knows where I'll end up.[/quote]

You got that right, this term there were probably over 40 jobs (mostly software) from US, England, Japan, France,... listed on jobmine.  A list of some companies are: Morgan Stanley, Google, Amazon.com, Microsoft, NI, ... just to name a few.  I am telling you all, Waterloo is the place to be!  :P  ... the only downside is that you don't get a summer vacation :mad:

BTW, Maple >> Matlab for undergrad students

Yes... there are lots of international jobs on jobmine... but I didn't get my Maple job on jobmine... and there's a good chance that my future jobs will not necessarily be found on jobmine.  A buddy of mine is working at Los Alamos national labs right now as a first year 4-stream co-op student.  Try getting a job like that on jobmine.

I've got a few future jobs potentially lined up... we'll see how that goes.  I have my connections.  :D  :D


Each of three, standard, six-sided dice is rolled once. What is the probability that the three numbers rolled are the lengths of the sides of an obtuse triangle? Express your answer as a common fraction.

Included one that doesn't work (334) and missed one that does (346)  Ugh.  My answer of 1/6 looked so much nicer than the right one.

I, actually, forgot to ask on forum this question.  I have:

$ a^2\plus{}b^2<c^2$, $ a\plus{}b>c$, and $ a,b,c \le 6$.  I have no idea how to solve, though.

Haha, I remember this from the state round distinctly. mewto55555 and I worked on it in the last minute, and mewto scribbled it down at the last half-second. We barely got it right. :)

The cases are

$ (2,2,3)$, which has $ 3$ permutations,

$ (2,3,4)$, which has $ 6$ permutations,

$ (2,4,5)$, which has $ 6$ permutations,

$ (2,5,6)$, which has $ 6$ permutations,

$ (3,3,5)$, which has $ 3$ permutations,

$ (3,4,6)$, which has $ 6$ permutations,

$ (3,5,6)$, which has $ 6$ permutations,

$ (4,4,6)$, which has $ 3$ permutations,

for a total of $ 3\plus{}6\plus{}6\plus{}6\plus{}3\plus{}6\plus{}6\plus{}3\equal{}39$.

Thus, our probability is $ \frac{39}{6^3}\equal{}\boxed{\frac{13}{72}}$.

How do we prove those are the only cases?

Brute-force and stuff like there are no obtuse triangles with side length $ 1$ because if $ 1\plus{}a>b$ where $ a\geq b$, we must have $ a\equal{}b$. Obviously, $ 1\plus{}a^2\not<a^2$.

Remeber in obtuse a^2+b^2<c^2 where c is side opposite obtuse angle.


calculus
probability
probability and stats

Hello there,

Could anyone please recommend a good, concise, enjoyable textbook for a general first-year statistics course? I am planning to use this textbook to self-study stats before university next year.

I took AP Calculus AB this past academic year and am comfortable with algebra and trig.

Thank you!

As a first book for self-study I recommend:

- [b]"Introduction to Mathematical Statistics and Its Applications" (Larsen, Marx)[/b]

Once you finished that I suggest to continue with:

- [b]"Probability and Random Processes" (Grimmett, Stirzaker)[/b]

The latter book also covers random processes with assuming knowledge in measure theory. Both books are good choices for self-study.


calculus
integration
trigonometry
logarithms
function
calculus computations

famous integral...

[hide="Answer"]
$\int \sqrt{\tan x}\, dx = \frac{1}{\sqrt{8}}\ln\left|\frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1}\right|+\frac{1}{\sqrt{2}}\arctan\left( \frac{\sqrt{2\tan x}}{1-\tan x}\right)+C$[/hide]

[hide="Hint "]You can use the substitution $y = \sqrt{\tan x}$. Then you will have $\int \sqrt{\tan x}\, dx = \int \frac{2y^{2}}{1+y^{4}}\, dy$. Integrationg rational functions is generally easier than dealing with transcendental functions... [/hide]

p.s. you're right, Extremal. :)

[quote="sos440"]famous integral...

$\int \sqrt{\tan x}\, dx = \frac{1}{\sqrt{8}}\ln\left|\frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1}\right|+\frac{1}{\sqrt{2}}\arctan\left( \frac{\sqrt{2\tan x}}{1-\tan x}\right)+C$[/quote]
Yes it is famouse but some don't know about this. if you only show this answer it will be not interesting I think.  :wink:

[hide="Hint"][color=darkblue]You can use the substitution $t=\phi (x)=\sqrt{\tan x}$and you 'll obtain $F(x)=2\cdot H(\sqrt{\tan x})$, where $H(t)=\int\frac{t^{2}}{1+t^{4}}\ dt$. Denote $G(t)=\int \frac{1}{1+t^{4}}\ dt$ and calculate $H+G$, $H-G$ with the substitutions $t-\frac{1}{t}$ and $t+\frac{1}{t}$ respectively  because $t^{2}+\frac{1}{t^{2}}=$ $(t-\frac{1}{t})^{2}+2=$ $(t+\frac{1}{t})^{2}-2$.[/color][/hide]

What about this?

$\int \sqrt{\tan x}\, dx = \int \sqrt{\frac{\sin x}{\cos x}}\, dx = \sqrt 2 \int \frac{\sin x}{\sqrt{2 \sin x \cos x}}\, dx =$

$= \frac{\sqrt 2}{2}\int \left( \frac{\sin x+\cos x}{\sqrt{2 \sin x \cos x}}+\frac{\sin x-\cos x}{\sqrt{2 \sin x \cos x}}\right) \, dx=$

$=\frac{1}{\sqrt 2}\int \frac{d(\sin x-\cos x)}{\sqrt{1-(\sin x-\cos x)^{2}}}\, dx-\frac{1}{\sqrt 2}\int \frac{d(\sin x+\cos x)}{\sqrt{(\sin x+\cos x)^{2}-1}}\, dx =$

$= \frac{1}{\sqrt 2}\arcsin (\sin x-\cos x)-\frac{1}{\sqrt 2}\ln \left| \sin x+\cos x+\sqrt{\sin 2x}\right|+C$

Has it some error?

oh... greengrass, your solution is very simple but still beautiful! and we have another form of antiderivative of $\sqrt{\tan x}$:

$\int \sqrt{\tan x}\, dx = \frac{1}{\sqrt{2}}\left[ \sin^{-1}(\sin x-\cos x)-\cosh^{-1}(\sin x+\cos x) \right]+C$

[quote="Extremal"][quote="mayhem"]Find $\int\sqrt{tanx}dx$[/quote]
You must do next substitute $y^{2}: =arctanx$ then it is easy.
then $d(arctany^{2})$=$\frac{1}{1+y^{2}}(2y)$ and simpl[/quote]
 :rotfl:
I mean $y^{2}: =tanx$


Να βρεθούν όλα τα ζευγάρια (k,n) θετικών ακεραίων για τα οποία το $ 7^k\minus{}3^n$ διαιρεί $ k^4\plus{}n^2$

Εφόσον το αριστερό μέλος είναι πάντα άρτιο, (έχουμε $ n,k\geq1$) παίρνουμε και ότι το δεξί μέλος είναι άρτιο, άρα $ k\equiv n (mod2)$ που συνεπάγεται ότι $ 7^k\minus{}3^n\equal{}(\minus{}1)^k\minus{}(\minus{}1)^n\equiv 0 (mod4)$ και που με τη σειρά του μας δίνει ότι $ k\equiv n\equiv0 (mod2)$. Έστω τώρα $ k\equal{}2m,n\equal{}2s$. Μ’ αυτόν τον τρόπο παίρνουμε ότι $ (7^m\minus{}3^s)(7^m\plus{}3^s)|16m^4\plus{}4s^2\equal{}4(4m^4\plus{}s^2)$. Όμως $ (7^m\minus{}3^s,7^m\plus{}3^s)\equal{}2$ και μάλιστα το $ 4$ διαιρεί έναν (αλλά όχι και τους δυο) από τους παραπάνω αριθμούς, άρα $ 8|(7^m\minus{}3^s)(7^m\plus{}3^s)| 4(4m^4\plus{}s^2) \implies 2|s$. Άρα $ s\equal{}2r$, οπότε το ζητούμενο γίνεται $ (7^m\minus{}9^r)(7^m\plus{}9^r)|16(m^4\plus{}r^2) \implies (7^m\plus{}9^r)|16(m^4\plus{}r^2)$, οπότε πρέπει να έχουμε $ 7^m\plus{}9^r\leq16(m^4\plus{}r^2)$.
Εδώ ήρθε η ώρα να δουλέψουμε ανισοτικά. Εύκολα όμως δείχνουμε ότι αν $ m\geq5$ έχουμε $ (7^m\plus{}9^r>16(m^4\plus{}r^2)$, οπότε οι τιμές του $ m$ που πρέπει να εξετάσουμε είναι οι $ m\equal{}1,2,3,4$. Για $ m\equal{}1$ έχουμε ότι $ (81^s\minus{}49)|(16\plus{}16s^2)$. Πάλι ανισοτικά αποδεικνύουμε ότι για $ s>1$ έχουμε $ 81^s\minus{}49>16\plus{}16s^2$ άρα μας μένει η περίπτωση $ s\equal{}1$ η οποία επαληθεύει τα ζητούμενα, άρα ένα ζεύγος λύσεων είναι η $ (k,n)\equal{}(2,4)$. Οι άλλες περιπτώσεις δουλεύονται πάλι ανισοτικά (στην αρχική όμως αυτή τη φορά, και προσέχοντας το πρόσημο του αριστερού μέλους) και δε δίνουν λύσεις.
Μοναδική λύση άρα η $ (k,n)\equal{}(2,4)$.

Ηλίας


geometry
conics
hyperbola
circumcircle
ratio
cyclic quadrilateral
geometry proposed

Let $ABCD$ be a quadrilateral, and let $H_{1}$, $H_{2}$, $H_{3}$, $H_{4}$ be the orthocenters of the triangles $DAB$, $ABC$, $BCD$, $CDA$, respectively.

Prove that the area of the quadrilateral $ABCD$ is equal to the area of the quadrilateral $H_{1}H_{2}H_{3}H_{4}$.

The assertion where $H_1, H_2, H_3$ and $H_4$ are the orthocenters of $ABC, BCD, CDA$ and $DAB$ respectively is true.  I have proven it before, so either my proof was bogus (which I doubt since not only was this posed in Iranian training, it is a classical fact) or your sketch was bogus.
My proof relies on the fact that $A, B, C$ and $H_1$ must lie on a conic and the only possible such conic is a rectangular hyperbola.  To observe that $H$ lies on a rectangular hyperbola when $A, B$ and $C$ do is simple.  But to prove that all 4 lying on a conic necessarily implies that they lie on a rectangular hyperbola is much more difficult (we are ignoring degenerate cases like right triangles where all 4 points lie on a circle since $H$ coincides with a vertex).  I'm embarrassed to say that it took me MUCH longer than the 15 minutes allotted.
Of course, once you establish this, the result is easy.  There is a conic passing through $A, B, C, H_1$ and $H_2$.  By the above, such a conic must be a rectangular hyperbola.  But since $D$ is the orthocenter of $BCH_2$, $D$ lies on this same conic.  Once you know that there is a rectangular hyperbola passing through all 8 points, coordinatzing everything and finding areas is very simple.  But unless Omid has a nicer method, I don't think it is reasonable to expect to complete this in 15 minutes.

In that exam the time was 15 minutes.
Well I solved it in 10 minutes!!!!!. I give you a hint.
Area of ACD and $H_1H_2H_3$ are equal, And area of $H_1H_3H_4$ and ABC are equal too.

this looks like a generalisation of BkMO 1984, problem 2, which stated that if $ABCD$ is cyclic, then $H_1H_2H_3H_4$ not only have the same area, but they are even congruent.  :)

Valentin: I consider this a huge generalization of BkMO 1984.  The cyclic case is SO much easier; since the 4 triangles share the same circumcenter, the orthocentres are easy to coordinatize, i.e. $OH_1=OA+OB+OC$, etc.  This makes the congruencies readily apparent (you'll get that $H_1H_2H_3H_4$ is $ABCD$ translated).

Omid: I proved exactly the same thing as you, namely that $[ACD]=[H_1H_2H_3]$ in proving that $[ABCD]=[H_1H_2H_3H_4]$.  But even that result required the rather involved procedure I described in my other post.  But since you consider this problem so readily soluble in 10 minutes, I'd love to see you post your solution.  It must be brilliant!

Of course the two quadrilaterals ABCD and $H_1H_2H_3H_4$ are equal and their sides are parrrel when ABCD is cyclic!!!!!!!!  :D  :D  :D

Yes, but that does not apply in the general case.  Do you mean to tell me that the problem posed during training was the cyclic case?  Or was it the general case?  If it is the general case, I'm very interested in seeing your solution, Omid.

I tried to use that well-known problem: I tried to imagine a transformation of the plane which conserves perpendicularity and the ratio of the areas, in order to turn the quadrilateral $ABCD$ into a cyclic quadrilateral, but I think the only ones are similarities, and I can't change the nature of $ABCD$ by applying a similarity, so I got nowhere.

I'd also like to see Omid's solution.

My solution is not very beautiful :D  :D  :D .You see it below.

Thank you very much Grobber :P  :P  :P  :P


What is the smallest n such that $\displaystyle 3^{2^n}-1$ is divisable by $2^{2005}$?

:lol:  :lol:  :lol:

Do you know why I am laughing?

Yup, i do. My bad! Is fixed now... :blush:

I have just notice your signature!  :lol:  :lol:  :lol: It is so funny  :lol:  :lol:  :lol:

Note that $3^{2^n}-1=(3^{2^{n-1}}+1)(3^{2^{n-2}}+1)...(3^{2^0}+1)(3^{2^0}-1)$.
You can easily check that maximal power of 2 which divides $3^{2^{m}}+1$ is 1 for $m\geq 1$.

[Edit: I wrote something stupid here. I hope it is correct now]

If you need a hint, (ok i know YOU don't :) ) check the highest power of two that divides the first few terms, then prove the obvious conjecture

[quote="seamusoboyle"]Thanks! Do you have a solution to it? I did it in the training contest and im fairly sure i got it right...[/quote]
How do you think what I wrote above? Of course, I can't say it is a complete proof, but all major steps. I always let other people to bring their touch to the proof  :)

I have just corrected a stupid error in my reply. Sorry.  :blush:


geometry
trapezoid
rectangle
Pythagorean Theorem

[hide]$72+72\sqrt{2}$ or $72(1+\sqrt{2})$. The trapezoid BCDE can be divided into 3 parts: a rectangle and 2 right isosceles triangles. The area of each trianlge is $\frac{x^{2}}{2}$. Together, they are simply $x^{2}$. The area of the rectangle is $12x$. Using the Pythagorean Theorem, I found $x = 6\sqrt{2}$.[/hide]

[quote="anirudh"]ABCDEFGH is a regular octagon of side 12 cm. Find the area in square centimeters of trapezoid BCDE. Express your answer in simplest radical form.[/quote]

[hide](12+12root2)*6root2 /2
36+36root2[/hide]


Let $p$ be a prime. Find all positive integers $k$ such that the set $\{1,2, \cdots, k\}$ can be partitioned into $p$ subsets with equal sum of elements.

[quote="alphachapmtl"]Does anyone knows the answer?
What about the case p=2 for a start?[/quote]

For p=2, we need $ k\equiv 0,3 \bmod{4}$, as noted by Peter, since we must have $ p\mid (1\plus{}2\plus{}...\plus{}k)$
If $ k\equiv 0 \bmod{4}$, we make the sets $ \{1,4,5,8,\cdots,i,i\plus{}3,\cdots,k\minus{}3,k\}$ and $ \{2,3,6,7,\cdots ,i\plus{}1,i\plus{}2,\cdots,k\minus{}2,k\minus{}1\}$, because $ i\plus{}i\plus{}3\equal{}i\plus{}1\plus{}i\plus{}2$.
If $ k\equiv 3 \bmod{4}$, we make the sets $ \{1,2\}\cup\{4,7,\cdots ,i,i\plus{}3,\cdots,k\minus{}3,k\}$ and $ \{3\}\cup\{5,6,\cdots,i\plus{}1,i\plus{}2,\cdots,k\minus{}2,k\minus{}1\}$, which again work because $ i\plus{}i\plus{}3\equal{}i\plus{}1\plus{}i\plus{}2$.

[quote="Peter"]Has this anything to do with number theory at all, other than that we need $ p|\frac {k(k + 1)}2$?[/quote]
You're right,$ p|\frac {k(k + 1)}2$.Using simple analysis,we can reduce it to $ k=2p,3p,2p-1,3p-1$ Then dealing with these four cases is not difficult :)

If $ k$ is solution, $ k\plus{}2p$ - solution too. We can add $ S_i, i\equal{}1,2,...,p$ for k new elements $ \{S_i,k\plus{}i,k\plus{}2p\minus{}i\}$.
If $ k\equal{}2p$, then $ S_i\equal{}\{i,2p\plus{}1\minus{}i\}$ - solution. If $ k\equal{}2p\minus{}1$, then $ S_i\equal{}\{i,2p\minus{}1\minus{}i\},i\equal{}1,..,p\minus{}1, S_p\equal{}\{2p\minus{}1\}$ is solution.
Therefore $ k\equal{}np,np\minus{}1$ are solutions, when n is even. If $ p\equal{}2$ $ k\equal{}np,np\minus{}1$ are solutions for any n.

If $ p \equal{} 2$ then $ k \equal{} 4n \minus{} 1,4n$,not true for any $ n$.


we are to express the number in terms of n--correct!?!?!?!?

ok well you didnt put that in your problems so i was just wondering!!!Just to make sure and avoid looking stupid!

[hide]
The total area of the table is $ 2n$. The area of the domino is $ 2$ So the maximum dominoes you can put on is $ \frac{2n}{2}\equal{}n$.
[/hide]

No,no, it's not about the number of dominoes(you always put n to cover the table) but about how you place them(vertically or horizontaly) and the number of different coverings...

[quote="sunnykim"]How many ways are there to cover a 2Xn table with 1x2 dominoes?[/quote]

[hide]
Let's put the first domino on one end of the table. There are two ways: domino facing up-down or domino facing left-right.  Also, denote the function by $ f(n)$.

Case 1: up-down.  This becomes the number of ways to cover a 2X(n-1) table, so $ f(n\minus{}1)$

Case 2: left-right.  We have to put another one next to it, or it won't work, so it becomes $ f(n\minus{}2)$.

So $ f(n)\equal{}f(n\minus{}1)\plus{}f(n\minus{}2)$.

We can find that $ f(1)\equal{}1$ and $ f(2)\equal{}2$, so $ f(n)\equal{}F_{n\plus{}1}$, which $ F_{k}$ is the $ k^{\text{th}}$ Fibonacci Number
[/hide]


trigonometry parameterization integration calculus function conics parabola

Hmm, I have been at this problem for awhile already, and iv either made a mistake somewhere along the line or im not too sure where to go next.
For the equation: $ a\sin{(\pi x/3)}$, find a formula $ L(a)$ for the arclengthin terms of the parameter $ a$.

Here is what I have done thus far:

$ \frac{dy}{dx} \equal{} \frac{\pi a}{3}\cos{(\pi x /3)}$

$ L \equal{} \int \sqrt{1\plus{}(\frac{dy}{dx})^2} \, \, dx  \equal{} \int \sqrt{1\plus{}(\frac{\pi a}{3}\cos{(\pi x /3)})^2}\,\,dx$

$ \equal{} \int \sqrt{1\plus{} (\displaystyle\frac{\pi ^2}{9}) a^2 \cos^2{(\pi x /3)}}\,\,dx$

or
$ \equal{} \int \sqrt{1\plus{} (\displaystyle\frac{\pi a}{3})^2\cos^2{(\pi x /3)}}\,\,dx$

let $ b \equal{} (\frac{\pi a}{3})$
let $ \theta \equal{} (\frac{\pi x}{3})$

then:

$ \equal{} \int \sqrt {1\plus{}b^2\cos^2{\theta}}\,\,dx$

$ \equal{} b^2 \int \sqrt {\frac{1}{b^2} \cos^2{\theta}}\,\,dx$

$ \equal{} b^2 \int \sqrt {(\frac{1}{b})^2 \cos^2{\theta}}\,\,dx$

It is here that i have to make a  trignometric subsitution that im a bit unsure where to go.

any and all help is much appreciate, thank you :)

This is an elliptic integral. It can't be done in terms of elementary functions.

The integral $ \int \sqrt{1\plus{}b^2\cos^2\theta}\,d\theta$ is the same integral one encounters in trying to find the arclength of an ellipse. In fact it's one variation of the objects known as elliptic integrals. It is rather famous that function cannot be integrated in closed form. There is nowhere to go with any further substitutions, and in particular, your last couple of steps in which you magically no longer have a binomial under the integral sign are incorrect.

Most arclengths cannot be computed in closed form; it would probably be more efficient to list those that can rather than those that cannot.

(A list of arclengths that can be computed in closed form: lines, parabolas, circles, cycloids and hypocycloids, 2/3 and 3/2 powers, $ \cosh$ and some related cases in which a binomial squares to having $ \minus{}\frac12$ as a middle term.)