quadratics inequalities geometry conics calculus Support rotation

1. If $a+b+c=-1$ and $ab+bc+ca=-1$, compute $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}$ ($a\ne -1$, $b\ne -1$, $c\ne -1$)

2. If $2x+y-3z=3$ and $3x+2y-z=2$ for all real numbers $x$, $y$ and $z$, then we must have $px^2+qy^2+rz^2=12$. Find the ordered triple $(p,q,r)$.

3. Find all values of $k$ such that quadratic equation $-2x^2+kx+k^2+5=0$ has two distinct roots and only one of the roots satisfies $0<x<2$.

[hide="3"]Solving for x, we get,
$\frac{-k\pm\sqrt{k^2-4\times-2\times(k^2+5)}}{-4}$
Letting $0<x<2$,
$0<\frac{-k\pm\sqrt{k^2-4\times-2\times(k^2+5)}}{-4}<2$
$0>-k\pm\sqrt{9k^2+40}>-8$
$k>\pm\sqrt{9k^2+40}>-8+k$
We can eliminate the $\pm$ and make it $-$ because $\left|\sqrt{9k^2+40}\right|>\left|k\right|$
so we are left with $-\sqrt{9k^2+40}>-8+k$. Equate the inequality and get $-3<k<1$.[/hide]

2.  By solving 2 systems of linear equations via TI, I got some really ugly answers for (p,q,r).

[hide](189/227, -12/227, -4137/227)[/hide]

But the geometry of the question doesn't make sense to me.  The intersection of the two planes is a line.  But px^2 + qy^2 + rz^2 = 12, with p positive and q and r negative, represents a 'hyperboloid of two sheets' (sort of two paraboloids in 3D).  Any line should hit this surface at most twice. No line can reside on this curved shape.   (Even if my ugly answer is wrong, an analagous contradiciton exists as long as p, q, and r are all non-zero.)

What's happening?

Well, nice work on catching yourself.  You made an error solving somewhere, although the first point I checked happened to be on the intersection of your hyperboloid and the line, which through me for a little loop.  The correct answers is [hide]$(5, \frac{20}{11}, -\frac3{11})$[/hide]

Hmm... It seems that no one got the correct ordered triple for #2 :?

I agree that this problem is a bit strange, but this was a problem from an entrance exam for a Japanese university, so I don't think there's a flaw in the problem.

[hide="Intended solution"]
Solving for $x$ and $y$ in terms of $z$, we get $x=5z+4$ and $y=-7z-5$.
Substituting these into $px^2+qy^2+rz^2=12$ gives $p(5z+4)^2+q(-7z-5)^2+rz^2=12\Longleftrightarrow (25p+49q+r)z^2+2(20p+35q)z+(16p+25q)=12$

Since the equation has to hold for any possible value for $z$, we must have $25p+49q+r=0$, $20p+35q=0$ and $16p+25q=12$.

Solving the system, we obtain $(p,q,r)=(7,-4,21)$
[/hide]

However, if you think there's any flaw in the problem/solution, feel free to post it :)

No, that's right ... I made some sort of error solving the initial two equations for x and y in terms of z.

#2.  I found my calculation errors and agree with (7,-4,21), but I still don't 'see' the geometry.  In 3D, the graph of 7x^2 - 4y^2 + 21z^2 = 12 is a 'hyperboloid of one sheet', with the y-axis being its axis of symmetry.  BTW, a nuclear cooling tower is pretty much in the shape of one-half of a 'hyperboloid of one sheet'.

The intersection of the two planes is a (straight) line.  It seems that we have proved that every point on this line is on the hyperboloid we calculated.  I don't see how any orientation of a line could lie on any hyperboloid.  It seems that this is equivalent to saying that the intersection of a plane and the hyperboloid could be a straight line.   Any help?  (These equations of lines, planes, and 3D conics usually appear in Calculus III, but of course require no calculus.)

#2.  I found my calculation errors and agree with (7,-4,21), but I still don't 'see' the geometry.  I'm stubborn  :roll: .  The algebra and the geometry usually support each other.

In 3D, the graph of 7x^2 - 4y^2 + 21z^2 = 12 is a 'hyperboloid of one sheet', with the y-axis being its axis of symmetry.  BTW, a nuclear cooling tower is pretty much in the shape of one-half of a 'hyperboloid of one sheet'.

The intersection of the two planes is a (straight) line.  It seems that we have proved that every point on this line is on the hyperboloid we calculated.  I don't see how any orientation of a line could lie on any hyperboloid.  It seems that this is equivalent to saying that the intersection of a plane and the hyperboloid could be a straight line.   Any help?  (These equations of lines, planes, and 3D conics usually appear in Calculus III, but of course require no calculus.)

Take a line which is skew to the z-axis and rotate it around the z-axis.  What you get is a hyperboloid of one sheet.  (As a degenerate case, a line which actually intersects the z-axis gives us a cone.)

[hide="1)"]let $f(x)$ be a polynomial with roots $a$, $b$, and $c$, by vieta, we get

$x^3+x^2-x+k$, we want a polynomial with roots $y=\frac{1}{1+\frac{1}{x}}$, so $\frac{y}{1-y}=x$
making this substitution, and clearing the denominators,

$y^3+y^2(1-y)-y(1-y)^2+k(1-y)^3=0$

we want the sum of the roots of $f(y)$, or

$(-1)*\frac{1+2+3k}{1-1-1-k}=3$  (by vieta's sums)[/hide]

3.  Joel, yes I can see it (at last) :D

Using Maple to plot the two planes and the hyperboloid helped me see what you described.

Then an artistic friend suggested "string art":
Take two congruent elliptical hoops; place them parallel and aligned; then  connect them with a large number of parallel pieces of yarn or wire.  If you twist one hoop, the lines (which are still straight) form a hyperboloid first (then a cone if you twist it far enough).  Each of the pieces of wire IS one of the lines that I was seeking and the set of lines was the set you described accurately.  It is a bit weird that this set of parallel lines can form a curved surface.

Thanks.  :lol:

[quote="Yooper"]Then an artistic friend suggested "string art":
Take two congruent elliptical hoops; place them parallel and aligned; then  connect them with a large number of parallel pieces of yarn or wire.  If you twist one hoop, the lines (which are still straight) form a hyperboloid first (then a cone if you twist it far enough).  Each of the pieces of wire IS one of the lines that I was seeking and the set of lines was the set you described accurately.  It is a bit weird that this set of parallel lines can form a curved surface.[/quote]

Yes, it is.  The Boston Museum of Science has a wonderful mathematics exhibit.  Among the items there (and in addition to my all-time favorite, a simple proof-without-words that a plane intersecting a cylinder gives an ellipse) is a rotating arm which repeatedly passes a straight plastic rod through a hyperbolic hole cut in a plane, which is exactly the same thing that's going on here.


calculus
integration
function
floor function
algebra
calculus computations

Evaluate the intergral :

$ I\equal{}\int_{1}^{3}[x\plus{}1]dx$

P/S : I am not sure but i think $ I\equal{}5$ :)

I assume that's meant to be the floor function? $ \lfloor \cdot\rfloor$ (\lfloor and \rfloor, which can be enlarged with \right and \left if necessary) is better notation.

Yup. Just draw the graph. Two boxes.

[quote="buddhu"]yes the evaluation of the definite integral appears to be correct.
we can evaluate this integral by evaluating the integral piece wise. :)[/quote]

I created it from an exercise in the book of James Stewart ( maybe " Calculus Early Transcendentals " ) .

[quote]

Evaluate the intergral :

$ J\equal{}\int  _{R} \int [x\plus{}y] dA$

where $ R\equal{}\{   (x,y) \ | \  1 \leq x \leq 3 \ ; 2 \leq y \leq 5 \}$

[/quote]

I think $ J\equal{}63$ :)

[quote="Materazzi"]Evaluate the intergral :

$ I \equal{} \int_{1}^{3}[x \plus{} 1]dx$

P/S : I am not sure but i think $ I \equal{} 5$ :)[/quote]

$ I \equal{} \int_{1}^{3}\lfloor x \plus{} 1\rfloor \  dx \equal{} \int_{1}^{2}\lfloor x \plus{} 1\rfloor \plus{} \lfloor 5 \minus{} x \rfloor \  dx \equal{} \int_{1}^{2} 6 \plus{} \lfloor x \rfloor \plus{} \lfloor \minus{} x \rfloor \  dx \equal{} \int_{1}^{2} 5 \  dx \equal{} 5$

Weird solution.

$ I \equal{} \int_1^3 \lfloor{x \plus{} 1}\rfloor\,dx \equal{} \int_1^2 \lfloor{x \plus{} 1}\rfloor\,dx \plus{} \int_2^3 \lfloor{x \plus{} 1}\rfloor\,dx \equal{} \int_1^2 2 dx \plus{} \int_2^3 3 dx \equal{} 5$.

[quote="makar"]$ I \equal{} \int_{1}^{3}\lfloor x \plus{} 1\rfloor \ dx \equal{} \int_{1}^{2}\lfloor x \plus{} 1\rfloor \plus{} \lfloor 5 \minus{} x \rfloor \ dx $[/quote]

plz can you explain how you change the limits ??

Makar, that's not necessarily true that those two floor functions cancel out. For example, consider 1.5. You would have -2 + 1 = 0, which is clearly false.

Well, on $[1,2]$ we have $\lfloor -x \rfloor = -2$, so that $\lfloor x \rfloor + \lfloor -x \rfloor = -1$, which he used.

$\{m,n,p\}\subset \mathbb N^*\ \wedge\ m\le n\ \implies \ I\equal{}\int_{m}^{n}[x+p]\ \mathrm{dx}=\frac 12\cdot (n-m)(n+m+2p-1)$ .


Show that if M is a metric space in which evrey infinite open covering has an own subcovering then M is compact

[quote="parrot"]an own subcovering[/quote]
What's that supposed to mean¿

By the way, it's not a good idea to post all your homework problems here!


geometry
trapezoid
ratio
Pythagorean Theorem

1. In trapezoid $ABCD$, $AD\parallel BC$, $AD=1.8$, $BC=5$, and $BD$ (diagonal) $=5$. If $AB=CD$, find the length of $AB$.

2. In quadlateral $ABCD$, diagonals $AC$ and $BD$ intersect at point $P$. The area of $\triangle ACD$ is $21$, the area of $\triangle BCD$ is $32$, and the area of $\triangle ABD$ is $24$. Find the area of $\triangle ABP$.

[hide="1"]
Draw an altitude from A and D down to the base. Call these points E from point A, and F from point D. Since the trapeziod is isosceles, BE = FC. That means that BE is 1/2(5 - 1.8) or 1.6. Now we know BF = 1.6+1.8 = 3.4. DFB is a right angle, so we can identify a right triangle with legs BF and DF, and hypotenuse BD. That means $3.4^2 + DF^2 = 5^2$. DF = $\sqrt{13.44}$. DF = AE since it's an isosceles trapezoid. Using the pythagorean theorem again, we see that $AB = \sqrt{1.6^2 + \sqrt{13.44}^2} = \sqrt{16}=4$. [/hide]

Number 1 is simple:

[hide="mod edit - I promise I will always hide all my solutions from now on. right iversonfan??"]Draw altitudes from $AD$ to $BC$, you see that there is a right triangle with the diagonal $BD$ as the hypotenuse which is $5$. Now, you know that the area between $BC$ is $1.8$ because it is the same length as $AD$ plus $3.2$. Now since it's iscoceles trapezoid, you know that each one of the sidlengths left over is $1.6$, now you have two lengths of the right triangle, so to find the leg which is the altitude of the trapezoid, you do:
$(1.6+1.8)^2+x^2=5^2-------\to x=\sqrt{13.44}$. Now you know that there is another right triangle with $1.6$ as the base, $\sqrt{13.44}$ as the height, and $AB$ as the hypotenuse, so do the pythagorean theorem and get $AB=4$.

Number 2 is not very simple:

You know the area of the quadrilateral is $ABD+BCD=24+32=56$, Now $21$ is the common region they share, I don't know exactly how to solve this, but since quadrilateral $ABCP=35$, then maybe $ABP$ is half of that? :?  :?  :([/hide]

the $2$nd one is weird,
maybe to high for me,
but does it require extra knowledge????
Some theorem????
Ive tried it by linear equations but I cannot get it??? :?  :(

You guys got #1 right :)

For #2, you don't need to know any formula, but there's a small trick. Once you find it out, the rest is very simple ;)
[hide="hint"]Can you find the ratio $AP:PC$ ?[/hide]

oh! I used your hint and I got....
$AP:PC::3:4$
So, $\text{Area of ABC=35}$ Now using ratios
we get that taking x as the common factor,
$7x=35$
$x=5$
So,
$\boxed{\text{Area of ABP=15}}$
I didnt know you can solve problems that way cooool.

That is pretty nice, I wouldn't have thought of that. Just do 3/7 * 35 and get 15. Wow.

Yeah, #2 is my favorite problem :D

[hide="Another approach to number 1"]Draw a line from $D$ that is parallel to $AB$. The line and $BC$ intersect at $E$. $EC=5-1.8=3.2$. Let $x$ be the length of $AB=DC=DE$. Since $\triangle CDE \sim \triangle DBC$, we see $\frac{EC}{DE}=\frac{DC}{BC}$, which gives us $\frac{3.2}{x}=\frac{x}{5}$. Solving for $x$, we obtain $x^2=16$, or $x=4$.[/hide]

Yeah the second problem is great. :)  I was wondering riddler, how you came up with the ratio $AP: PD=3: 4$ because I had to express each area in terms of x and then find x using $AP: PD$

my solution for number 2
let $ABP$be a,$BPC$ be b, $CPD$ be c and $APD$ be d
so u get $c+d = 21 , b+c =32, a+d =24$
it took me soooo long to solve this, anyone have a good way to solve equation like that??

[quote="PenguinIntegral"][quote="rrusczyk"]The tactic most of you used on #2 can be enormously useful.  Particularly look for it in area problems in which triangles have bases along a common line (as in this problem).[/quote]
This technique is covered in AoPS Vol 1, on the ratios of bases of triangles with the same height or ratios of height of triangles with the same bases.[/quote]

I know, it is  :lol:  but what a shame that came 3 months after reading this problem  :blush:  :(


inequalities
linear algebra
linear algebra unsolved

Suppose  A,B,C are matrices that we have AB ,BC .prove :
   rank(BC)+rank(AB)<=rank(ABC)+rank(B)
(AB means multiplication of two matrices A,B). :maybe:

Ummm... isn't that the same as proving $\min(b, c)+\min(a, b) \le \min(a, b, c)+b$? That follows from $\min(a, b, c) = \min(\min(b, c), \min(a, b))$ and $b \ge \max(\min(b, c), \min(a, b))$.

It's not the same, because the rank of a product may be less than the minimum of ranks of the factors.

Rewrite the inequality as $\mathrm{rank}(BC)-\mathrm{rank}(ABC)\le \mathrm{rank}(B)-\mathrm{rank}(AB)$. On the left we have the dimension of $\ker(A)\cap \mathrm{range}(BC)$, on the right the dimension of $\ker(A)\cap \mathrm{range}(B)$. Since $\mathrm{range}(BC)\subseteq \mathrm{range}(B)$, the inequality follows.


AMC
AMC 12
geometry
3D geometry
factorial
AIME
ARML

Does anyone have any tips on how to get faster at problem-solving in general? I am mainly concerned with my speed on some of the problems on the AMC12, as I often take more than the average 3 min to finish problems.

Fierytycoon

You can only get faster by solving more problems. There's no trick to it. It's just about how many problems you have solved.

That's not exactly true. I think there are some tricks to getting faster. Like memorizing commonly used numbers - powers of two, squares and cubes of numbers, primes up to 2 or 3 hundred,  factorials, quick ways to multiply, divide and factor numbers, common divisibility rules. All of things can greatly increase your speed when it comes to computational questions.

There are also techniques of speed reading that cut down on the time you spend reading each question. But you always have to be careful not to read too quickly that you miss something significant.

On multiple choice tests there are certain kinds of questions where testing the answer choices is more quick than solving the problem.

The best way to work on your timing is to time yourself and work on full length AMC tests (or tests of similar difficulty and length). Then you will become intuitively accustomed to the amount of time to spend on each problem.

[quote]primes up to 2 or 3 hundred[/quote]

Seriously? Do you really think that this would help on some problems?

Fierytycoon

Not on contests like the AMC12, in which calculators are allowed. On others, known the prime factorization of the current year, the previous year, and the next year.

oh. they allow calculators now? I'm so behind the times. Well forget it then. Some of those computational tricks are for dinosaurs like me then - back before they allowed calculators.

On the AMC12 they do. On many other contests, such as the AIME, they do not. But the non-calculator contests tend not to be particularly computation-intensive unless there is a better way to do it (for example, question 7 on ARML 2003: find the largest factor < 10000 of 1001001001, which can be done by finding the factor 1001 and then factoring the sum of cubes, so calculators are not necessary).


How many nine-digit numbers can be made using each of the
digits 1 through 9 exactly once with the digits alternating
between odd and even?

using the numbers 1-9, there are 5 odd numbers and 4 even numbers
therefore, there are 5 odd numbers that can go first, 4 even numbers that can go after that, 4 odd numbers after that, 3 even after that, etc...
therefore the answer to the question will be

$ 5 \times 4 \times 4 \times 3 \times 3 \times 2 \times 2 \times 1 \times 1 \equal{} 2880$

Well, you have to start with either 1,3,5,7,9- $ 5$
then, 2,4,6,8- $ 4$
then only four of the odd digits, bcuz digits cant repeat- $ 4$
three even digits- $ 3$
three odd digits- $ 3$
two even digits- $ 2$

see the pattern?

so it would be $ 5 \cdot 4 \cdot 4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 \cdot 1 \cdot 1$, or $ \boxed{2880}$

EDIT: Beaten to it, dang.

Case 1: the number is odd

Like what everyone else said, there are $ 5 \cdot 4 \cdot 4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 \cdot 1 \cdot 1\equal{}2880$ numbers, but that is only if the number is odd.

Case 2: the number is even

If the number is even, then the hundred thousands place would also be even, meaning that there are 5 even numbers. Since there are only 4 even numbers, tis case is impossible.

There are a total of 2880+0=2880 possibilities


blogs
number theory
relatively prime
Divisibility Theory

For each positive integer $n$, write the sum $\sum_{m=1}^n 1/m$ in the form $p_n/q_n$, where $p_n$ and $q_n$ are relatively prime positive integers.  Determine all $n$ such that 5 does not divide $q_n$.

$ n\equal{}5^k\minus{}l,l\equal{}1,2,3,4$. Prove by induction.

The above answer is not correct. The correct answer, which I am currently providing without full proof, is {1,2,3,4,20,21,22,23,24,100,101,102,103,104,120,121,122,123,124}.

Here is a partial proof:

[b]Theorem I[/b]: If $ \frac{a_i}{b_i}$ are fractions in their lowest terms for $ i\equal{}1, 2, 3$, $ \frac{a_1}{b_1}\plus{}\frac{a_2}{b_2}\equal{}\frac{a_3}{b_3}$ and $ b_1$ is not divisible by 5, then the highest power of 5 dividing $ b_2$ and $ b_3$ are the same.

[b]Proof[/b]: If 5 is not a factor of $ b_2$, then $ b_3$ is a factor of $ b_1b_2$ and hence not divisible by 5. If, on the other hand $ b_2$ is a multiple of 5, then $ a_2$ is not, and hence $ a_2b_1$ is not, and hence $ a_2b_1\plus{}a_1b_2$ is not a multiple of 5. Thus, since $ b_3\equal{}\frac{b_1b_2}{\gcd(b_1b_2,a_2b_1\plus{}a_1b_2)}$, $ b_3$ and $ b_2$ are divisible by the same powers of 5.

[b]Corrolary II[/b]: For $ 0\le k\le 4$ and positive integer $ n$, the highest power of 5 dividing the denominator of $ \frac{1}{1}\plus{}\frac{1}{2}\plus{}\ldots\plus{}\frac{1}{5n\plus{}k}$ is the same as that dividing the denominator of $ \frac{1}{5}\plus{}\frac{1}{10}\plus{}\ldots\plus{}\frac{1}{5n}$

[b]Proof[/b]: The integer reciprocals omitted in the latter sum give a fraction whose denominator is not divisible by 5.

[b]Corollary III[/b]: For $ 0\le k\le 4$ and positive integer $ n$, $ q_{5n\plus{}k}$ is not divisible by 5 if and only if $ p_n$ is divisible by 5.

[b]Proof[/b]: Corollary II states that the highest power of 5 dividing $ q_{5n\plus{}k}$ is the same as the highest power of five dividing the lowest-terms denominator of $ \frac{1}{5}\plus{}\frac1{10}\plus{}\ldots\plus{}\frac1{5n}\equal{}\frac{p_n}{5q_n}$. This will be a multiple of 5 if $ p_n$ is not divisible by 5. If, on the other hand, $ p_n$ is a multiple of 5, then $ q_n$ is not, so the 5 will cancel.

[b]Corollary IV[/b]: If we define $ D$ to be the set of integers such that $ q_n$ is not divisible by 5 and $ N$ to be the set of integers such that $ p_n$ is divisible by 5 then $ N$ is a subset of $ D$ and for $ 0\le k\le 4$ and $ n>0$, $ 5n\plus{}k\in D$ iff $ n\in N$.

[b]Proof[/b]: $ 5n\plus{}k\in D$ iff $ n\in N$ is the content of Corollary III. If we let $ n$ be some element of $ N$ that is not in $ D$, then the coprime integers $ p_n$ and $ q_n$ are both divisible by 5, which is a contradiction.

[b]Lemma V[/b]:$ {1,2,3}\subset (D\setminus N)$, while $ 4\in N\subset D$.

[b]Proof[/b]:Let us consider the integers 1 to 4 separately:

$ \frac11\equal{}\frac11$, so 1 is in D but not N
$ \frac11\plus{}\frac12\equal{}\frac32$, so 2 is in D but not N
$ \frac11\plus{}\frac12\plus{}\frac13\equal{}\frac{11}6$, so 3 is in D but not N
$ \frac11\plus{}\frac12\plus{}\frac13\plus{}\frac14\equal{}\frac{25}{12}$, so 4 is in D and in N.

[b]Theorem VI[/b]: $ N\equal{}\{4,20,24\}$, so $ D\equal{}\{1,2,3,4,20,21,22,23,24,100,101,102,103,104,120,121,122,123,124\}$

[b]Partial Proof[/b]:
Since 4 is in $ N$, it follows that the numbers 20 up to 24 are in $ D$. [b]By inspection[/b], 20 and 24 are the numbers in this range in $ N$. Thus the numbers in the ranges $ [100..104]$ and $ [120..124]$ are both contained in $ D$. [b]By further inspection[/b], none of these numbers are in $ N$.

Then given a minimal number $ n$ other than $ {4,20,24}$ in $ N$, $ n\ge 5$ by Lemma V. Then $ n\equal{}5k\plus{}r$ for some positive integer $ k$ and integer $ 0\le r\le 4$. Then $ n$ is in $ N$ which means that $ n$ is in $ D$ and hence $ k$ is in $ N$ by Corollary IV. By the minimality of $ N$, $ k\in\{4,20,24\}$ and hence $ n\in\{21,22,23,100,101,102,103,104,120,121,122,123,124\}$. However, we have already inspected that none of these $ n$ are in $ N$.

Luke

See my puzzle blog at [url]http://bozzball.blogspot.com[/url]


Find the ordered pair $(x,y)$, given that
\begin{eqnarray*} ax + by &=& a + b \\ a^2x + b^2y &=& 2ab \\ \end{eqnarray*}

Assume $a \ne b$

[hide]The determinants are:

$\Delta=ab^2-a^2b=ab(b-a)$
$\Delta_x=(a+b)b^2-2ab\cdot b=b^3-ab^2=b^2(b-a)$
$\Delta_y=a\cdot 2ab-a^2(a+b)=a^2b-a^3=a^2(b-a)$

Now we have

$x={\Delta_x\over\Delta}={b\over a}$
$y={\Delta_y\over\Delta}={a\over b}$[/hide]

[hide]
A "classic" solution... (using substitutions)

From the first equation, we have $ax=a+b-by$
Plug it into the second equation,
$a(a+b-by)+b^2y=2ab$
$a^2+ab-aby+b^2y=2ab$
$(b^2-ab)y=ab-a^2$

$\therefore y=\frac{a(b-a)}{b(b-a)}=\frac{a}{b}$ ($a\ne b$)

Going back to the first equation,
$ax=a+b-b\cdot\frac{a}{b}$
$x=\frac{b}{a}$[/hide]


quadratics
algebra
number theory open
number theory

[quote="sinajackson"][b]Find all positive integers [i]a[/i] and [i]b[/i] such that [i]b[/i] divides [i]a^{2}+a+1 [/i] and [i]a[/i] divides [i]b^{2}+b+1[/i].[/b]
I tryed my best Guys but I could'nt solve it completely. I think it is related to the Fibonacci sequence... :maybe:[/quote]

I think I solved it.
From $b/a^{2}+a+1$ and $a/b^{2}+b+1$ it follows that $\frac{a^{2}+b^{2}+a+b+1}{ab}\in\mathbb{N}.$ Let $\frac{a^{2}+b^{2}+a+b+1}{ab}=k.$ If we have $a=b$ then $a^{2}/2a+1$ and thus $a=b=1.$ Now let us concider the case $a>b.$ Then we have $a^{2}+(1-kb)a+b^{2}+b+1=0.$ Concider this like quadratic equation in $a.$ It has other root $a'.$ But by Viet we have that $a+a'=kb-1\in\mathbb{Z}.$ So $a'\in\mathbb{Z}.$ But $aa'=b^{2}+b+1>0$ and therefore $a'\in\mathbb{N}.$ We have that $a'=\frac{b^{2}+b+1}{a}<b+2,$ because $a>b.$ But if $a'=b+1$ then $(b+1)a=b(b+1)+1$ and therefore $b=0$ or $b=-2$ which is a contradiction. Thus we have $a'\leq b$ and so from the solution $(a,b)$ we have the solution $(b,a')$ with smaller sum of coordinates. So because this descend cannot be infinite because we work in $\mathbb{N}$ then we will have some solution with $a=b.$ Then as we showed $a=b=1$ and so $k=5.$ Similar we have by Viet that $5b-1=a+a'$ and so the solutions of the problem are two consecutive parts of the sequence $a_{n+2}=5a_{n+1}-a_{n}-1$ and $a_{1}=a_{2}=1.$

[quote="sinajackson"]that seems like a nice nice solution [b]bilarev[/b]. but i dont understand the last part: how did you reach to this consequence that we will have some solutions for [b]a=b[/b]? :maybe:[/quote]
If we don't have some $a=b$ then as I showed we will have another smaller solution and another and so on... But this is imposible in $\mathbb{N}.$
This method is called $\emph{Fermat Method of Infinite Descent}$ and is very popular...for example see Problem 6/IMO 88.
Ihis problem is not for this section and I think it's been posted in this forum.


In $\triangle ABD$, $C$ is a point on $BD$ such that $AB=CD$.
Given that $\angle CAB=30^\circ$ and $\angle ABC=40^\circ$, find $\angle CDA$.

This is from an "Olympiad" contest for kids in elementary schools in Japan, so you don't need any hard formulas. (You could brute force with trig, but it's very long.)

What do you mean by elementary kids?

I used the Law of Sine to obtain the answer, but haven't worked out the constructions yet...

[quote="shobber"]What do you mean by elementary kids?[/quote]
I meant, kids in elementary schools :oops: (Edited)

Hmm .... $AD=CD$ does fulfill the condition ... giving us $\angle CDA=40^\circ$ though i still dont know how to prove $AD=CD$ by not using trigo  :P

Suppose that $D'$ is point on line $BD$ such that $\angle BD'A=40^o$ different from
$B$. Then $\angle CAD'=180^o-40^o-70^o=70^o$ so triangle $CD'A$ is isosceles and
from that $CD'=D'A$. But $\angle ABD$ is also $40^o$ so triangle $ABD'$ is isosceles
too so $BA=AD'=CD'$ but $CD=BA$ and points $D$ and $D'$ are on the same side of $C$ on side $BD$ so $D\equiv D'$ and from that \[\angle BDA=40^o\]

Easy also by sinelaw to get sine 110 / sine 40 = sine 110-x / sine x


Let [tex]C_1 ; C_2 ; C_3 ; C_4[/tex] be four circles in the plane. Suppose that [tex]C_1[/tex] and [tex]C_2[/tex] intersect at [tex]P_1[/tex] and [tex]Q_1[/tex], [tex]C_2[/tex] and [tex]C_3[/tex] intersect at [tex]P_2[/tex] and [tex]Q_2[/tex], [tex]C_3[/tex] and [tex]C_4[/tex] intersect at [tex]P_3[/tex] and [tex]Q_3[/tex] , and [tex]C_4[/tex] and [tex]C_1[/tex] intersect at [tex]P_4[/tex] and [tex]Q_4[/tex] . Show that if [tex]P_1 ; P_2 ; P_3[/tex]  and [tex]P_4[/tex] lie on a line or circle, then [tex]Q_1 ; Q_2 ; Q_3[/tex] and [tex]Q_4[/tex] also lie on a line or circle.

This is the famous [b]Miquel six-circles theorem[/b] and the proof is easy angle chasing:

We will use directed angles modulo 180. Since the points $P_1$, $Q_1$, $P_2$, $Q_2$ lie on one circle (namely on the circle $C_2$), we have $\measuredangle Q_1Q_2P_2 = \measuredangle Q_1P_1P_2$. Since the points $P_2$, $Q_2$, $P_3$, $Q_3$ lie on one circle (namely on the circle $C_3$), we have $\measuredangle P_2Q_2Q_3 = \measuredangle P_2P_3Q_3$. Thus,

$\measuredangle Q_1Q_2Q_3 = \measuredangle Q_1Q_2P_2 + \measuredangle P_2Q_2Q_3 = \measuredangle Q_1P_1P_2 + \measuredangle P_2P_3Q_3$.

Similarly,

$\measuredangle Q_1Q_4Q_3 = \measuredangle Q_1P_1P_4 + \measuredangle P_4P_3Q_3$.

Thus,

$\measuredangle Q_1Q_2Q_3 - \measuredangle Q_1Q_4Q_3 = \left(\measuredangle Q_1P_1P_2 + \measuredangle P_2P_3Q_3\right) - \left(\measuredangle Q_1P_1P_4 + \measuredangle P_4P_3Q_3\right)$
$= \left(\measuredangle Q_1P_1P_2 - \measuredangle Q_1P_1P_4\right) + \left(\measuredangle P_2P_3Q_3 - \measuredangle P_4P_3Q_3\right) = \measuredangle P_4P_1P_2 + \measuredangle P_2P_3P_4$
$= \measuredangle P_4P_1P_2 - \measuredangle P_4P_3P_2$.

Now, since the points $P_1$, $P_2$, $P_3$, $P_4$ lie on one circle, we have $\measuredangle P_4P_1P_2 = \measuredangle P_4P_3P_2$, so that $\measuredangle P_4P_1P_2 - \measuredangle P_4P_3P_2 = 0^{\circ}$, and hence, we also have $\measuredangle Q_1Q_2Q_3 - \measuredangle Q_1Q_4Q_3 = 0^{\circ}$, and thus $\measuredangle Q_1Q_2Q_3 = \measuredangle Q_1Q_4Q_3$, what entails that the points $Q_1$, $Q_2$, $Q_3$, $Q_4$ also lie on one circle.

$\blacksquare$

One can also solve your problem using inversion, but I wouldn't call it considerably simpler.

  Darij

[quote="Myth"]Prasolov "Planimetry problems", Chapter "Using inversion"
It is a hint  :lol:  :lol:  :lol:[/quote]
How can we solve it using inversion?

Another hint: Inversion helps simplifying problems. For instance, there is an inversion which maps two of your circles $C_1$, $C_2$, $C_3$, $C_4$ to lines. Aren't lines simpler than circles?  :)

  Darij

Indeed, Prasolov used this property of inversion for reducing the problem to another one, which was solved above in his book:

Suppose $A_1$, $B_1$, $C_1$ lie on $BC$, $CA$ and $AB$ respectively. Then circumcircles of $AB_1C_1$, $A_1BC_1$ and $A_1B_1C$ intersect in the same point.

Dear Mathlinkers,

without inversion, you can see
http://perso.orange.fr/jl.ayme  vol. 2 Du théorème de Reim au théorème des six cercles, p. 9 with a rule concerning the accentuation p. 13.

Sincerely
Jean_Louis


The product of three consecutive integers is 210. What is their sum?

The three numbers are 5, 6, and 7. Their sum is 18.
Hope you find this useful!

[hide=Solution]
We guess $5\cdot5\cdot5 = 125$. This is close, so we find $5\cdot6\cdot7 = 210$. We now can just add and get $\boxed{18}$


algebra
polynomial
real analysis
real analysis unsolved

let $k>=1$ be an arbitrary real and $\epsilon>0$. Prove that there exists a real polynomial $P$ such that:
$|e^{-kx} - P(x)e^{-x}| < \epsilon$ for all $x>=0$.

Is it joke? Ha-ha-ha  :lol:  :lol:  :lol:
Sometimes I think we need beginner's corner in the College Playground Section.

Together with alekk, I don't understand Myth's post. This problem does not appear to be obvious.

my proof is not trivial I think (a kinf of recurrence on k) ! Myth, what's yours?

I have found a bug in my solution. Right estimate is
\[|e^{-kx}-P_m(x)e^{-x}|\leq \left|\frac{(k-1)^{m+1}x^{m+1}e^{-x}}{(m+1)!}\right|.\]
So my solution works only for $1\leq k\leq 2$, but I know how to make recurrence (as alekk told).


real analysis
function
limit
calculus
integration
real analysis unsolved

1. Let $f$ be a nonnegative Lebesgue integrable function on $[0,1]$. Prove that there exists nonnegative measurable function $g$ such that the pointwise product $fg$ is integrable on $[0,1]$ and $\lim_{x\to 0^{+}}g(x) = \infty$

2. Let $f$ be a nonnegative Lebesgue integrable function on $[0,1]$ with
$m(\{x: f(x)>n\}) > 0$ for all positive integer $n$. Prove that there exists nonnegative integrable function $g$ such that the pointwise product $fg$ is not integrable on $[0,1]$.

In problem 1, split the integral over $[0,1]$ into the sum of integrals over $[2^{-n},2^{1-n}]$, $n\ge 1$. Choose a function $g$ that is constant on each of these subintervals.


In the following sequence, each term after the second is the sum of the two preceding terms. Find $ a$.

\[ 7, a, \_, \_,\_, 11\]

[quote="FantasyLover"]the sequence is

7, a, a+7, 2a+7, 3a+14, 11

so 5a+21=11, 5a=-10.

answer : -2[/quote]
Hmm i don't get how you got 7,a,a+7,2a+7,3a+14,11.

[quote]each term after the second is the sum of the two preceding terms.[/quote]

very self-explanatory.


Hey,
I was trying to create an article on Wilson's theorem.
Something like:
<math>gcd((p-1)!, p) > 1</math>
Does not parse.

And I'm getting
Failed to parse (Can't write to or create math output directory): p

For something like
<math>p</math>

I fixed up that article a little bit.

The LaTeX on the wiki needs some work, it seems.  Like you said, you can't just have <math>p</math>.  To work around this, I used <math>{p}</math>.  I can imagine it might be a pain trying to change this if you have longer equations.

Also, please use \cdot instead of *.  And use commands such as \gcd instead of just putting gcd.

Edit: there will be changes made to the wiki LaTeX, right?  I have a feeling some things are going to bug me...can't just use dollar signs, no eqnarray/align, a lot of unparseable things, and so forth :(

I hope they will make the changes... wikipedia did never really grow in the math-world for this, and we're starting with the same handicap here too...

Yeah, start a wiki page (and put the link here) of a wishlist of how you'd like the wiki LaTeX to work, and I'll take a look when I've had more sleep.  I may just be able to unhook all the MW LaTeX parsing and replace it with the Forum's code for LaTeX parsing entirely, which would probably be the easiest way to go.

[quote]And I'm getting
Failed to parse (Can't write to or create math output directory): p

For something like
<math>p</math>[/quote]

I'm having the same problem with:

<math>f</math>

But <math>z</math>, for example, is fine.

I just remembered - believe it or not, this is a feature of the wiki - in your preferences, you can set the math to render as images always, never, or for the wiki to decide if it can be rendered passably well in html.  This allow people with slow connection speeds to avoid having a page of images to download, and people with fast connections to have the images.

Ah, that makes it look a lot better.

By the way,
Does anyone know how to get something like
<math>1, 2, 3, \dots, p-1</math>

To parse with spaces? (So it's not just 1,2,3,...,p-1)


Vieta
algebra
polynomial
quadratics
quadratic formula

What is it exactly? I looked up wikipedia and mathworld, but it was too confusing. Can someone explain the main idea behind it?

$ (x\minus{}a)(x\minus{}b)(x\minus{}c)\equal{}x^3\minus{}(a\plus{}b\plus{}c)x^2\plus{}(ab\plus{}ac\plus{}bc)x\minus{}abc$ <-- an example

So if you have a polynomial like x^3-6x^2+9x+3 then a+b+c=6, ab+ac+bc=9 and abc= -3 where a,b,c are the roots of the polynomial

Generalize to get vieta.

Basically they are formulas to find certain relations between the roots of a polynomial function. The most common ones are the sum and product of the roots.

Product of the roots for any even degree polynomial:
(x^n being the highest degree) the product of the roots is attained by dividing the $ x^0$ coefficient(the constant) by the coefficient of the $ x^n$ term.
ex. $ f(x)\equal{}3x^2\minus{}4x\plus{}5$ In this case we have product of the roots =$ \frac{5}{3}$

Product of the roots for any odd degree polynomial:
(x^n being the highest degree) the product of the roots is attained by dividing the NEGATIVE $ x^0$ coefficient(the constant) by the coefficient of the $ x^n$ term.
ex. $ f(x)\equal{}2x^3\plus{}3x^2\minus{}4x\plus{}5$ In this case we have product of the roots =$ \minus{}\frac{5}{2}$

Sum of the roots for any polynomial:
(x^n being the highest degree) the sum of the roots is attained by dividing the NEGATIVE coefficient of the $ x^{n\minus{}1}$ term by the coefficient of the $ x^n$ term.
ex $ f(x)\equal{}2x^3\plus{}3x^2\minus{}4x\plus{}5$  In this case we have the sum of he roots = $ \minus{}\frac{3}{2}$

Is this what you were asking?

For a quadratic:

In the form ax^2+bx+c, the sum of the roots of the quadratic is -b/a and the product is c/a

in short: Any $ n$th degree polynomial can be expressed as $ a_n \prod_{i\equal{}1}^n (x\minus{}r_i)$ where $ r_i$ are the roots of the polynomial.  The standard way to express a polynomial is $ a_nx^n \plus{} a_{n\minus{}1}x^{n\minus{}1} \plus{} \cdots \plus{} a_1x\plus{}a_0$.  So we have $ a_n \prod_{i\equal{}1}^n (x\minus{}r_i) \equal{} a_nx^n \plus{} a_{n\minus{}1}x^{n\minus{}1} \plus{} \cdots \plus{} a_1x\plus{}a_0$.  If we expand the beginning and equate coefficients, we find that $ (\minus{}1)^{n\minus{}i}\frac{a_i}{a_n} \equal{} \sum_{1 \le b_1 < \cdots < b_{n\minus{}i}\le n} \prod_{j\equal{}1}^{n\minus{}i} r_{b_j}$.

For example: if we have a quintic polynomial,
$ \minus{}\frac{a_0}{a_5} \equal{} r_1r_2r_3r_4r_5$
$ \frac{a_1}{a_5} \equal{} r_1r_2r_3r_4 \plus{} r_1r_2r_3r_5 \plus{} r_1r_2r_4r_5 \plus{} r_1r_3r_4r_5\plus{}r_2r_3r_4r_5$
$ \minus{}\frac{a_2}{a_5} \equal{} r_1r_2r_3 \plus{} r_1r_2r_4 \plus{} r_1r_2r_5 \plus{} r_1r_3r_4 \plus{} r_1r_3r_5 \plus{} r_1r_4r_5 \plus{} r_2r_3r_4 \plus{} r_2r_3r_5 \plus{} r_2r_4r_5 \plus{} r_3r_4r_5$
$ \frac{a_3}{a_5} \equal{} r_1r_2 \plus{} r_1r_3\plus{}r_1r_4 \plus{} r_1r_5 \plus{} r_2r_3\plus{}r_2r_4\plus{}r_2r_5\plus{}r_3r_4\plus{}r_3r_5\plus{}r_4r_5$
$ \minus{}\frac{a_4}{a_5} \equal{} r_1\plus{}r_2\plus{}r_3\plus{}r_4\plus{}r_5$

Viete's Formulas relate the roots of a polynomial to its coefficients. To simplify things, lets look at the polynomial $ f(x)\equal{}ax^2\plus{}bx\plus{}c$, which is a quadratic (note: this means that the polynomial has a degree of $ 2$). You might recall the quadratic formula which states that the roots of a quadratic equation are given by $ \frac{\minus{}b\pm\sqrt{b^2\minus{}4ac}}{2a}$. From this we can derive Viete's Formulas.

Lets call the roots of the quadratic equation $ ax^2\plus{}bx\plus{}c\equal{}0$, $ r_1$ and $ r_2$.

We know that the roots are given by the quadratic formula. Therefore, we can say:

$ r_1\equal{}\frac{\minus{}b\plus{}\sqrt{b^2\minus{}4ac}}{2a}$

$ r_2\equal{}\frac{\minus{}b\minus{}\sqrt{b^2\minus{}4ac}}{2a}$

By adding and multiplying $ r_1$ and $ r_2$, we can derive Viete's Formulas:

$ r_1\plus{}r_2\equal{}\minus{}\frac{b}{a}$

$ r_1r_2\equal{}\frac{c}{a}$

Vieta's generalizes the kth symmetric sum of the roots of a polynomial of power n and coefficients a_n, a_n-1, etc. The formula is $ \minus{} 1^{(n \minus{} k)}*(a_n \minus{} k)/(a_n)$

As an example, take the polynomial $ 4x^4 \plus{} 5x^3 \minus{} 2x^2 \plus{} x \plus{} 4$.
Note that as a quartic polynomial, this has four roots (though not necessarily real).
Let's call the roots x=a, b, c, and d. To find the product, you evaluate for the fourth symmetric sum, or the sum of the roots taken four at a time, which is simply abcd. Since it might be a bit hard, you use vieta's, which, plugging in 4 for k, reveals that the product of the roots is $ \minus{} 1^{(4 \minus{} 4)} *a_0/a_4 \equal{} 1*4/4 \equal{} 1.$

Similarly, you can use it to find the sum of the roots. This would be the 1st symmetric sum, or the sum of the roots taken one at a time. So, plug in 1 for k, revealing that the sum of the roots $ a \plus{} b \plus{} c \plus{} d \equal{} \minus{} 5/4.$

You can easily apply Vieta's to more complicated problems, but the above were just examples to help.


I need help with the following question.

Equal volumes of aqueous solutions (each 1.0mol/L) of two substances, W and X are mixed. The reaction below rapidly reaches equilibrium.

W(aq) + X(aq) <-> Y(aq) + Z (aq)

At equilibrium, the concentration of Y is measured as 0.40 mol/L. What is the value of Kc for the reaction above?

I calculated Kc to be 0.44 repeating but the answer key says it is 16 not 0.44. Can someone plz tell me why Kc is 16. Thanks in advance.

[b]Equal volumes[/b] of aqueous solutions (each $1.0$mol/L) of two substances, $W$ and $X$ are mixed. The reaction below rapidly reaches equilibrium.

$W(aq)+X(aq) <\to  Y(aq)+Z (aq)$

At equilibrium, the concentration of $Y$ is measured as $0.40$ mol/L. What is the value of $K_{c}$ for the reaction above?

Because equal volumes of two solutions are mixed, the concentrations of those solutions are halved.  Thus, the initial concentrations of $W$ and $X$ are $0.50$ mol/L, not $1.0$ mol/L.  This leads to a $K_{c}$ of $16$.


[hide]to answer JBL's question, we can use proportions.

let n = total number of fish in lake

so 30/n = 5/36

solving this, n = 216[/hide]

[hide]to answer the original question, we can use proportions again.

let n = number of fish in lake on may 1

the number of fish that are tagged in the lake on sept. 1 is

60 * 75% = 45

the number of fish in the lake on sept.1 is

n * 75% / 60%    which is equal to      5/4 * n

putting this in a proportion,

3/70 = 45/ (5/4*n)

we get n = 840[/hide]

yeah, this question you have to be careful

[hide]I just added 28+60+52 and got 140, although I doubt it's right.[/hide]

[hide]The ratio of (fish tagged/fish not tagged) for the first time is 60/(x-60)

The ratio of the second sampling (fish tagged/fish not tagged) is 3/67

40% of the fish not tagged (67) weren't there the first time.  That changes the seocond ratio to 3/40.2

Also, 25% of the tagged fish aren't there.  That changes the second ratio to 3.75/40.2

Now, all you have to do is solve the proportion.

60/(x-60) = 3.75/40.2

2412 =  3.75x - 225

3.75x = 2187

x = 583.2

So there were about 583.2 fish in the lake in September 1.

[/hide]

But I'm probably wrong though, I make all sorts of stupid mistakes(hence my 730 on the SAT math)

How can there be a fifth of a fish? (referring to the .2 in your answer potter)

JBL wrote:1)  A biologist wishes to measure the population of a lake.  To do so, she takes a net-load of fish and tags them, counting 30 tagged fish.  A week later, she comes back and takes a new net-load of fish, in which she finds 36 fish, 5 of whom are tagged.  Assuming that this second net-load is representative of the population of the pond and that no fish leave or die and no new fish arrive or are born, how many fish are in the pond?

[hide]5 out of every 36 fish in the pond are tagged. 30 fish have tags. 30/5=6 groups of 5 fish are tagged. For every group there are 36 fish in the pond. The answer is 6 groups*36 fish for every group=216 fish total.[/hide]

JBL wrote:Well, let's give a simpler question along with it:

1)  A biologist wishes to measure the population of a lake.  To do so, she takes a net-load of fish and tags them, counting 30 tagged fish.  A week later, she comes back and takes a new net-load of fish, in which she finds 36 fish, 5 of whom are tagged.  Assuming that this second net-load is representative of the population of the pond and that no fish leave or die and no new fish arrive or are born, how many fish are in the pond?

[hide]This one is a whole lot easier.  It's just a proportion problem.

5/36 = 30/x

5x =  1080

x = 216

216 fish are in the pond.[/hide]

somehow i got [hide]1000[/hide] but i'm sure it's wrong because of the AIME range


Chandler wants to buy a $ \$500$ mountain bike. For his birthday, his grandparents send him $ \$50$, his aunt sends him $ \$35$ and his cousin gives him $ \$15$ He earns $ \$16$ per week for his paper route. He will use all of his birthday money and all of the money he earns from his paper route. In how many weeks will he be able to buy the mountain bike?

$ \textbf{(A)}\ 24 \qquad
\textbf{(B)}\ 25 \qquad
\textbf{(C)}\ 26 \qquad
\textbf{(D)}\ 27 \qquad
\textbf{(E)}\ 28$

[hide="Solution"]$ (\text{B})$ For his birthday, Chandler gets $ 50\plus{}35\plus{}15\equal{}100$ dollars. Therefore he needs $ 500\minus{}100\equal{}400$ dollars more. It will take Chandler $ 400\div 16\equal{}25$ weeks to earn $ 400$ dollars, so he can buy his bike after 25 weeks.[/hide]

since he already has 50+35+15=100 dollars, he needs 400 left. 16x=400, where x is the number of weeks left. divide by 16 on both sides, so x=400/16. x=25, [b](B) 25[/b] is the answer.[/u]

400 over 16 is 100 over 4 which is A 25.

[quote="nikeballa96"][quote="007math"]Your solution was very similar to Dojo's, so there was no reason to post another solution that is the exact same.

Note: Do not revive posts unless your solution is different than all the others posted.  Otherwise, it is called spam, which is bad.[/quote]
same to you.[/quote]

He only posted 20 minutes after anchenyao's solution while anchenyao posted 2 months after the first solution was posted.

[quote="AIME15"]You mean, "only a month", right?[/quote]

Nope, I meant two months.
"Posted: Tue Jan 20, 2009 9:58 pm" - "Posted: Mon Mar 23, 2009 4:03 pm"
I'm pretty sure that's two months.

[quote="rd5493"]so, why did dojo ask a question, only to answer it 10 minutes later... :huh:  :huh:[/quote]

Woah, I never noticed that!

[quote="Dojo"][quote="rd5493"]so, why did dojo ask a question, only to answer it 10 minutes later... :huh:  :huh:[/quote]
Resources section: http://www.artofproblemsolving.com/Forum/resources.php[/quote]

Mind elaborating on how this explains why you posted within 10 minutes?

[quote="eli140"]I didn't even know that was allowed, i thought consecutive posts had to be 24 hours apart... :huh:[/quote]
No, it's 10 minutes.
Sometimes people want to show their prowess in the area of mathematics by posting a problem and answering it as soon as they can. It gets annoying.

<sigh>. Problems are posted in the resources section, with links to threads of the problems and the solutions. If there is no such thread, then if a user goes to find 2007 AMC 8 problem 5, they can't.

[quote="1=2"]
Sometimes people want to show their prowess in the area of mathematics by posting a problem and answering it as soon as they can. It gets annoying.[/quote]
I should have never given you that avatar. :P

Anway, the resources section asks for a solution. in a different post. I have no idea why mine wasn't posted in the resourses section though.

Did you even follow the instructions and contact an RM to put the problems in?

Yes, I followed the instructions and pmed worthawholebean.


A spider has one sock and one shoe on each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put before the shoe?

[hide]
Let A denote the first leg, B the second, etc., we can write an arrangement of socks/shoes as AABBCCDDEEFFGGHH.  This is because no matter where the A's are, we have to make the first one be the sock and the second be the shoe.  And the order of the legs doesn't matter, so we merely want the number of arrangements of that "word", which is $\frac{16!}{2!^8}$, which is a big number.
[/hide]


inequalities
real analysis
real analysis unsolved

suppose that $ X_{1}, X_{2}, .., X_{n}$ are independant random variables, with zero mean, such that $ a_{i}\leq X_{i}\leq b_{i}$. Prove that for every $ x>0$ :
$ P(\sum_{1}^{n}X_{i}\geq x)\leq\exp(\minus{}2\frac{x^{2}}{\sum_{1}^{n}(b_{i}\minus{}a_{i})^{2}})$

or more generally

[url]http://www.cs.ubc.ca/~jtillett/gtdt/Presentations/NandoTalks/Wasserman_All-Of-Stats_ch4_pg362-368.pdf[/url]


geometry
3D geometry
sphere
topology
advanced fields
advanced fields unsolved

I am trying to finish the question I posted here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=195177

However, no one has helped me.

My question is, if $ (X, \tau)$ is not connected and we have that $ a, b \in X$ belong to different connected components, is it then possible to give a disconnection $ X \equal{} A \cup B$ ($ A, B$ closed, disjoint) such that $ a \in A$ and $ b \in B$? In the given context, $ X$ is a compact metric space.

Sure, in your case, X is locally path connected, thus each connected component is open. Each component is closed in any case, and compactness guarantees that there are only finitely many of the components.

Sorry, but I am too bad at this stuff to follow.

[quote="Soarer"]Sure, in your case, X is locally path connected, thus each connected component is open.[/quote]
Why is $ X$ locally path connected? Note that $ X$ is a closed subset of the riemann sphere

[quote] Each component is closed in any case, and compactness guarantees that there are only finitely many of the components.[/quote]
Does this follow from just compactness? How?

Sorry I was wrong.  :blush: But for
[quote="Kalle"]
[quote] Each component is closed in any case, and compactness guarantees that there are only finitely many of the components.[/quote]
Does this follow from just compactness? How?[/quote]

I meant to say that if each component is open, then the components form an open cover, and compactness guarantees that there are finitely many of them.

In other words, take open sets $ G$ and $ H$ separating $ a$ and $ b$ in $ X$ on the sphere. Then $ X\cap G\equal{}X\setminus H$ is compact and the same is true for $ X\cap H$. Now you have your two sets $ A,B$ with positive distance between them.

I think the whole problem is that I don't know the existence of $ G$ and $ H$

Erm... Isn't it just the definition of being in two different connected components? :?

No... what I mean is that $ X$ splits into connected components under a suitable equivalence relation. There could be infinitely many components (AFAIK). And the condition is that $ a,b$ do not belong to the same components.

And what is that equivalence relation in your definition?

Are you asking or do you think I should see why it's true from the definition?

The connected component of a point $ x$ is the union of all connected subsets that contain $ x$.

I was asking. My memory of topology is quite rusty.
OK, Let's show that such $ G$ and $ H$ exist. Suppose not. Then consider all "disconnecting" pairs $ (G_\alpha,H_\alpha)$ of $ X$ (pairs of disjoint open subsets of the sphere whose union contains $ X$). We should always have $ a$ and $ b$ in one of them, say $ G_\alpha$. Let $ K\equal{}X\setminus\bigcup_\alpha H_\alpha$. Then $ a,b\in K$. I claim that $ K$ must be connected. Suppose not. Then there exists a disconnecting pair $ (U,V)$ of $ K$ such that $ K$ intersects both $ U$ and $ V$. Then $ U,V,H_\alpha$ give you an open cover of $ X$ and (this is the crucial step) since $ X$ is compact, we can choose a finite subcover $ U,V,H_j$. Now $ A\equal{}U\cap \bigcap_j G_j$ and $ B\equal{}V\cup\bigcup_j H_j$ is a disconnecting pair for $ X$ in which both sets have a non-empty intersection with $ K$, which clearly contradicts the way $ K$ was defined. The rest should be clear :)

Thanks!

If this was given as a problem I knew for certain to be true, I could probably solve it. But when I formulate questions like this on my own and I'm not sure, then usually I will not solve it, but instead ask someone to do it for me. Why is that? Do you think it's lack of motivation or lack of intuition or what? In any case, it was not so obvious to me that compactness is the key.

[quote="Kalle"]Why is that? Do you think it's lack of motivation or lack of intuition or what?
[/quote]
Hard to tell. I know that it helps me enormously if I know that the problem has a solution or at least i I have some evidence in favor of the statement I'm trying to prove. Probably, the psychological effect beyond this is that once you know that the problem is solvable, you start to subconciously believe that it cannot be too complicated and if you don't know that, you think it should be really hard, which is a strong discouragement, but it is only my guess.
[quote] In any case, it was not so obvious to me that compactness is the key.[/quote]
Well, you had to exclude something like that:
[asy]size(100);
D(MP("a",(0,0))); D(MP("b",(1,0)));
D((0,0)--(1,0),dashed);
int n=9;
for(int k=1;k<n;++k)
D((0,1/1.5^k)--(1,1/1.5^k),blue+linewidth(1));[/asy]


I have been taking previous AMC 8 contests and have been scoring much better on the more recent tests than the earlier ones (called AJHSMEs).

I know that the AJHSME contests are the same as the AMC 8 contests except with a different name, but is it true that they were meant to be harder than the more recent AMC 8 contests?? Just curious.

I don't think so. But a lot of the older contests were more focused on computation than today's, which might explain why you're doing better on the latest ones instead of the older ones.

Yeah, the AJHSME focused on very straightforward problems such as using multiplication and exponents. AMC 8 in recent years are all word problems and require more problem-solving strategies.

I'm pretty sure the modern AMC 8 was about the same difficulty as an average 1985-99 AJHSME...

I do agree that the earlier ones were based more on computation. Though AHSMEs were somewhat easier than the modern AMC 12s, I wonder if that suggests anything...

[quote="cheeseyicecream"]Really? When doing all the AMC 8s, I found I scored much better on the earlier ones....in addition, the more recent ones seemed a lot more inconsistent in difficulty.[/quote]

That seems to happen with Mathcounts too...

2009 was so much easier than 2008 and 2007

Yeah sometimes they're hard, but at other times they get unusually easy. I agree that the 2009 problems were very easy compared to most of the previous ones. Usually, after such an easy year, the problems get much harder.

It also has to do with the competitors' math level. Nowadays, with more resources for math problems (like AoPS), the competitions become harder in general.

Calculators are not allowed in the more recent AMC 8s, while they were in the AJHSMEs.  That's probably why.   :D

I find that I score better on the newer ones as well.

I fail at computation.

Don't you hate making stupid mistakes, ESPECIALLY with computation???

Especially in computation many people (like me) tend to rush and not think much just because it's easy. In my opinion, they're just tedious and uncreative.

In my opinion, most contests ahve been steadily increasing in difficulty throughout the years. Many of the AHJSMEs that I took for practice were extremely easy, not really focusing on the problem solving aspect of mathematics, while most of the AMC 8's that I have taken were also easy, but not as much as the AHJSMEs.

The AHJSMEs in my opinion were more a computation test than a real creative problem solving test.

Well, now people are finding ways to solve AMC 8 problems like really just for multi-choice tests, so now they must be harder to minimize the amount of strategy people use just for multi-choice

For me, it's the opposite story. I score 23+ (50% of the time a perfect score) on the AJHSME tests but only 18 - 19 on the recent AMC 8 tests.

@above IMO as $LaTeX$ said AJHSME was more computational than the AMC 8, which that if you are better at computing stuff than doing word problems(which is me) then u will do better on the AJHSME.

Yeah most older tests were mainly about computations, although there MAY have been a few creative-ish problems. Now competitions require more problem solving skills. (Except standardized tests)

I've heard older contests are typically easier because there are more resources to it. However, I feel they generally do get harder. But the 2016-2017 year mathcounts was easier than the 15-16 ones.
But the past amc 12's at about 2002-11 are about hard-end mathcounts difficulty, in my opinion. It's kind of like how it balances out standardized tests, standardized tests get dumber and competition math gets harder. :)

[quote=GrilledBrain]I am better at computation, but I can make errors. I like the AMC. They are fun competitions. Practicing for the one coming up. :weightlift:[/quote]

You know, an 8 year revive isn't necessary. How did you find this topic anyway?

You do realize we could use calculators until $2008$

If you want to practice, you might want to start with the AMC 8s. They have a lot of word problems, which will probably show up this year.


I have a question about adaptive expectations (the past is the best indicator of the future) and rational expectations (the past and past policy changes is the best indicator of the future).  Can one prove that rational expectations is a more, well, rational approach given some assumptions about the function you're studying?

Alex

Could you elaborate a bit on what you're asking about?
Perhaps give an example of each?

It's basically predicting what a random (by some distribution) function f(t) will do at t = t_0 + 1 if you know f(t) for t = 0 through t = t_0.  If the function has no dependence upon past values, clearly you cannot do this, but if it does, there is a question of whether predicting f(t_0 + 1) = f(t_0) + f'(t_0) is necessarily more accurate than predicting f(t_0 + 1) = f(t_0), based on the properties of f.  I was wondering if there is a rigorous way of evaluating this if f is, say, the real GDP per capita, or the value of the dollar in yen.

Alex


[quote="Arne"]Find eight positive integers $a_1,a_2,\dots,a_8$ such that \[\sum_{i = 1}^8\left(\sqrt{a_i} - \sqrt{a_i - 1}\right) = 2.\][/quote]

how about $\sum_{i = 1}^8\left(\sqrt[3]{a_i} - \sqrt[3]{a_i - 1}\right) = 2$ ?

that is way easy too ;)


In triangle ABC, AC = 13, altitude CD has length 12, median CE has length $6\sqrt{5}$. Find the area of CEB.

[hide]Since $CD$ is the altitude, it must be perpendicular to $AB$.
$DE=\sqrt{(6\sqrt5)^{2}-12^{2}}=6$
$DA=5$ since it is a 5-12-13 triangle.
$DA+DE=EB$ since $E$ is the midpoint.
$CEB=\frac{(5+6)(12)}{2}=66$[/hide]

[hide]Traingle CEB is a right triangle, with right angle at D. Solve for DE, getting 6. Then DA=5, because of right triangle CDA. That gives DB=11. (11)(12)(1/2)=66.[/hide]

[hide]AD=5, because ACD is a 5-12-13 right triangle. $DE=\sqrt{(6\sqrt{5}^{2}-12^{2}}=6$, from Pythag.
$AE=EB=11$, so Area of $BCE=(1/2)(12)(11)=66$.[/hide]

How do you know that it's a right triangle?

Yeah but that doesn't mean it's a side...

Because CD is the altitude of ACD, as in it makes a right angle with AD... so ACD is right.


This one is quite classical, so it may have appeared on the forum before, in which case, please, accept my apologies. It is fairly simple but I find it quite funny...

Does there exist $x>0$ such that the fractional part $\{x^n\}\in\left[\frac13,\frac23\right]$ for all $n=1,2,3,\dots$?

I think the answer is yes.
We are going to construct intervals $I_1 %Error. "superset" is a bad command.
I_2 \cdots %Error. "superset" is a bad command.
I_k %Error. "superset" is a bad command.
$ with following properties:
If $I_k=[a_k;b_k]$ then $b^k-a^k=\frac 13$, $\{a^k\}=\frac 13$, ${\{b^k} =\frac 23$. and for any $x \in I_k$ we have $\{x^i\} \in \{\frac 13; \frac 23\}$ whenever $i \leq k$.
Indeed, let $I_1=[2005+\frac 13;2005+\frac 23]$. Now let's proceed inductively.
If we constructed $I_n=[a;b]$, it's sufficient to have $b^{n+1}-a^{n+1} \geq 1$  and then we can pick up a desired inteval $I_{n+1}$.
But if $a_n=t$ then $b^n=t+\frac 13$.
We must ensure $(t+\frac 13)^{\frac {n+1}n}-t^{\frac {n+1}n} \geq 1$.
But we have $(t+\frac 13)^{\frac {n+1}n}-t^{\frac {n+1}n} \geq \frac {n+1}{3n} t^{\frac 1n}\geq 1$ because $t^{\frac 1n}=a>10>3$.
Now it's obvious the intervals $I_k$ are closed and their length converges to $0$ thus they have a common point $x$ which is the desired one.


geometry
quadratics
analytic geometry
algebra
system of equations
linear equation

For certain ordered pairs $(a,b)$ of real numbers, the system of equations \begin{eqnarray*} && ax+by =1\\ &&x^2+y^2=50\end{eqnarray*} has at least one solution, and each solution is an ordered pair $(x,y)$ of integers.  How many such ordered pairs $(a,b)$ are there?

Since $x$ and $y$ must be integers, possibilites for each (based on the second equation) are $\pm 1$, $\pm 5$, and $\pm 7$, I believe.  Then the possible points on the circle that could intersect the line are:

\[ \left(\pm 5,\pm 5\right);\quad \left(\pm 1, \pm 7\right);\quad \left(\pm 7, \pm 1\right) \]

Which gives 4 possibilities for each, or a total of 12 points.

I'm not quite sure how to finish this (there must be more solutions though).  Anyone care to finish it off?

[hide="Edit: Solution"]
Interpreted geometrically, we have a line $ax + by = 1$ intersecting a circle of radius $\sqrt{50}$ centered at the origin.  It can either intersect tangent to a lattice point or secant to two lattice points, and the lattice points are $(x, y) = (\pm 5, \pm 5), (\pm 1 \pm 7), (\pm 7, \pm 1)$ for a total of $12$ lattice points.

(Final edit:  So, I finally figured out what was wrong with my counting.)

There are $12$ tangent solutions, one for each point.  A secant solution can be found by choosing any one of the twelve and drawing the line connecting that point and any of the other $11$ points [i]except the point diametrically opposite[/i].  The line cannot pass through the origin (I didn't realize this before) because then $a(0) + b(0) = 1$, which does not have real solutions.

Hence, the number of solutions really is $12 + \frac{12(10)}{2} = \boxed{72}$.
[/hide]

I see how you did that.  But in that case, wouldn't it be $12+\frac{12\cdot 10}{2}=72$ solutions?  I'm not seeing where that 11 came from.

Is there a non-geometric way to approach this? :stink:

Good question.  The only way I can see to do it without geometry is basically the same method as with geometry, but different steps (I guess?).  I guess the 12 solutions could be found first as we did, then the linear equation could be substituted into the circle equation to find when there are two solutions and when there is one.  However, this doesn't really differ from the geometric method :(

This might be a moment of stupidity, but isn't the answer [i]infinitely many[/i]? Substituting these $12$ integer solutions into the system, we see that all pairs $(a,b)$ of [i]real[/i] numbers (check the wording of the problem) which satisfy $\pm 5a\pm 5b=1$ or $\pm 7a\pm b=1$ or $\pm a\pm 7b=1$ should be counted in. And since there are obviously infinitely many of such pairs...

And if we express $y$ from the first equation and substitute that into the second, and then discuss the sign of the discriminant of the resulting quadratic, we'll get the inequation $50b^4+51a^2b^2-a^2-b^2\geq 0$, which is obviously satisfied for infinitely many pairs of $a,b.$

Seems like a trick question to me.

I think that there should be a finite number of solutions because $a,b\in\mathbb{R}$ must also satisfy $x,y\in\mathbb{Z}$.  I can't quite explain it properly. :(

EDIT: If we consider $ax+by=1$ a line, and $x^2+y^2=50$ a circle centred at the origin, there can't be infinitely many sets of $(a,b)$ which result in integer solutions $(x,y)$; there can't be infinite intersections when $x,y\in\mathbb{Z}$ must be true.

EDIT: I just thougth about solving the problem and here it goes.

Since $(\pm 5,\pm 5),(\pm 1,\pm 7),(\pm 7,\pm 1)$ are integer coordinates on the circle, we have 12 diffrent possibilities.
Now, $\binom{12}{2}=\frac{12!}{2!(12-2)!}=\frac{12(11)}{2}=66$.  Now, tangent lines to the circle could occur at $(5,\pm 5),(1,\pm 7),(7,\pm 1)$ which gives us $6$ solutions. $66+6=\boxed{72}$.

Masoud Zargar

Oh i see, that WAS the moment of stupidity :) - ALL the solutions (be it one or two) must be integer.

Oh, thanks Farenhajt--I had the same confusion as you. :D

[quote="boxedexe"]tangent lines to the circle could occur at $ (5,\pm 5),(1,\pm 7),(7,\pm 1)$ which gives us $ 6$ solutions.
[/quote]
I don't get this, why can't tangent lines occur at $ (\minus{}5,\pm 5),(\minus{}1,\pm 7),(\minus{}7,\pm 1)$?

Don't think about it too much.  There is a flaw in the solution, even though the answer is correct.  Some of the lines pass through the origin (unwanted) when you do $ \binom{12}{2}$.

Wait, why can't the lines pass through the origin?

A line of the form $ax + by = 1$ can't pass through $(x, y) = (0, 0)$ because plugging those numbers in to the equation gives $0a + 0b = 1$, which is never true.

I had a similar issue in a proof I made in the past week or so - I used a coordinate bash to prove that three points on a circle are never collinear.  I put the circle $x^2 + y^2 = r^2$ on a coordinate plane, and let the line be $y = mx + b$.  Every line is of that form, except for vertical lines, so I made a second case for $x = k$.

In fact, the cool thing about $ax + by = 1$ is that you CAN deal with both vertical and horizontal lines.  The line $y = 2$ is given by $(a, b) = (0, \frac{1}{2})$, and the line $x=2$ is given by $(a, b) = (\frac{1}{2}, 0)$.  The problem is that you can't deal with lines through the origin - essentially because of division by zero.  If you rewrite it as $y = \frac{1 - ax}{b}$, you see that the line has interecept $\frac{1}{b}$, which is a non-zero x-intercept.

I wonder if there's a "skeleton form" that could represent ALL lines - horizontal, vertical, and passing through every point....

EDIT:  Oops.  The answer is "yes", and it's actually deceptively simple.  It's probably in every pre-algebra book....and is very close to being in this post, with one small change....

[hide="How to represent every line"]The skeleton $Ax + By = C$, which is close to the "standard form" of a line, can represent every line - horizontal, vertical, passing through the origin.

The technical standard form of a line, where $A, B, C$ have an GCD of $1$ and where $A>0$, doesn't represent every line, since $y = \sqrt{2}x$ is a perfectly good line that has no technical standard form.  But if you remove the GCD restriction, you're good.[/hide]


conics
ellipse
parabola
hyperbola
geometry
geometric transformation
reflection

Here are some nice properties I've been playing around with. They aren't too hard, but they are interesting.

(1) Let E and F be the foci of an ellipse, and P be a point on it. Prove that the tangent to the ellipse at P is the external angle bisector of EP and FP.
(2) Let l be the directrix and F the focus of a parabola, and let P be a point on it. Prove that the tangent to the parabola at P is the angle bisector of lP and FP.
(3) Let E and F be the foci of an hyperbola, and P be a point on it. Prove that the tangent to the parabola at P is the angle bisector of EP and FP.

Interesting  :D

For the first one, if you prove that the tangent's perpendicular bisects EPF, then it's done.

These are starting to scare me now :P

I'll work on the first one.

[quote="mysmartmouth"][quote="Iversonfan2007"]eh, if only these weren't conic sections :([/quote]

Um...if these weren't conic sections, then there would be no problem.  :roll:

Anyway, read AoPS Volume 2, Chapter 5.[/quote]

You know, posting random pages to read is completely pointless and does not necessarily solve the problem. It's like if someone asks you what the rules of basketball are, and you just tell them to go watch ESPN.

And yes, I've read the entire AOPS books a few months ago, and still am not great at conic sections. Have you read the chapter you just mentioned? If so, this problem should be a piece of cake, no? :wink:

Huh, well, I actually learned the rules of basketball from watching ESPN.  Bad analogy.  :wink:  :rotfl:  :rotfl:

Hint: think of the locus that defines conic sections and try to prove that no other point lies on the supposed tangents. Try to turn this into a problem that doesn't use conics.

[quote="jb05"]1. Reflect E across the proposed tangent to a point E'. Then E, E', and P are colinear. Hence if there is another point on our proposed tangent (say Q), then QE+QF=QE'+QF>FE'=PF+PE'=PF+PE, so Q lies outside the ellipse. Therefore the line is a tangent.

The next two parts are similar.[/quote]

Hm?  How are E, E' and P collinear?

Since angle E'PE is bisected by the proposed tangent (we know this since we are reflecting) which is also the angle bisector of PE and PF (by definition), then PF and PE' must be the same line.


[color=darkblue][b]Let the triangle $ ABC$ with $ M$, a point in the interior of this triangle.Let $ P,Q,R$ the projections of the point $ M$ on the sides of the triangle.Find out the minimum of the expresion $ \frac{BC}{MP}\plus{}\frac{AC}{PQ}\plus{}\frac{AB}{MR}$[/b][/color]

I think it should be $ \frac{BC}{MP}\plus{}\frac{AC}{MQ}\plus{}\frac{AB}{MR}$ and also $ P,Q,R$ should be on sides $ BC,AC,AB$ respectively.
anyway let $ a\equal{}BC,b\equal{}CA,c\equal{}AB$ and $ x\equal{}MP,y\equal{}MQ,z\equal{}MR$ so we want to find the minimum of $ \frac ax\plus{}\frac by\plus{}\frac cz$.

by Cauchy-Schwartz inequality we have:

$ (ax\plus{}by\plus{}cz)(\frac ax\plus{}\frac by\plus{}\frac cz)\geq (a\plus{}b\plus{}c)^2$ (*)

now let $ p,S$ be the semiperimeter and the area of triangle $ \triangle ABC$ respectively,we know that $ ax\plus{}by\plus{}cz\equal{}2S$ and $ a\plus{}b\plus{}c\equal{}2p$ so according to (*) e get that:

$ \frac ax\plus{}\frac by\plus{}\frac cz\geq\frac{4p^2}{2S}\equal{}\frac{2p^2}{S}\equal{}\frac{2p^2}{pr}\equal{}\frac{2p}r$

so the minimum is $ \frac{2p^2}{S}$ and equality occurs when $ x\equal{}y\equal{}z$ or equivalently $ M\equiv I$ where $ I$ is the incenter.

My mystake...sorry and thanks for observing :D ...


For the Columbia Science Honors Program test there is a challenge math section.  One of the problems was the reciprocal of the sum of the roots of a quadratic ax :^2:+bx+c.  From basic math knowledge I knew that the sum of the roots of any quadratic is -b/a (to be sure of this I did an easy equation:  x :^2:-4x+4    - there is a double root of 2, which adds to 4.  -b/a = -(-4)/1 = 4; so I made sure the sum of the roots was -b/a).

Then to get the reciprocal I did 1/-(b/a), which by multipliying by the reciprocal of the botton term on top and bottom, simplifies to -a/b.  That was an answer choice and so I selected it.  However, through discussion at another forum I go to, two people have told me that the answer I chose was incorrect.  Yet I can't seem to find the flaw in my calculations.  Am I doing it right or did I perform a mistake somewhere?  Thanks in advance.

Now you should be right, but maybe the question was the sum of the reciprocals of the roots.  That would be 1/r_1 + 1/r_2 = (r_1+r_2)/r_1r_2 = -b/a/(c/a) = -b/c.

Hmmm - I bet that could be it.  I probably read the problem wrong...  I'll post that on the other forum and see what they say.  Bah - I could have gotten that one right too...  oh well.

Yes, that would be a reading issue -- the sum of the reciprocal of the roots is different from the reciprocal of the sum of the roots.  Be careful of things like that :)  (Your reasoning, if the question did ask for the reciprocal of the sum, is correct.)

I made a mistake like this on my calculus placement test - it said to find the sum of the squares of the roots, and I did the square of the sum of the roots.

Thanks for your guys help.  I do think that I misinterpreted the wording of the problem (especially because two people at the other forum had different from what I had) - but you never know.  After some discussion my other answers to the "challenge" problems are correct, and I answered 50/50 on the first math section (a ton easier) so I'm hoping to get a good score.  (BTW, this was for the Columbia Science Honors Program entrance examination - I took it last year and didn't get in, but my performance was definately better this year - plus the only weekend I was sick last year during the test - figures...)  So hopefully I can get in this time.


How many pairs of real numbers (x,y) satisfy (x+y)^2=(x+3)(y-3)?

[hide="Solution"]Let $ a \equal{} x\plus{}3$ and $ b \equal{} y\minus{}3$. Therefore, $ (a\plus{}b)^2 \equal{} ab$. $ a \equal{} b \equal{} 0$ is the only solution since $ a^2 \plus{} ab \plus{} b^2 \equal{} (a\plus{}.5b)^2 \plus{} .75b^2$. Therefore, there is exactly $ \boxed{1}$ pair of real numbers that satisfies the given equation.[/hide]


Let $ a,b,c$ be positive numbers satisfy $ ab \plus{} bc \plus{} ca \equal{} 3$.Prove that
\[ \frac {(a^2 \plus{} ab \plus{} b^2)(a^2 \plus{} ac \plus{} c^2)}{b^3c \plus{} 105} \plus{} \frac {(a^2 \plus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)}{c^3a \plus{} 105} \plus{} \frac {(c^2 \plus{} cb \plus{} b^2)(c^2 \plus{} ac \plus{} a^2)}{a^3b \plus{} 105}\ge \frac{9}{106}
\]

It's obvious after using of Cauchy-Schwartz and Vasc's inequality.  :wink:

Could you tell me what is Vasc's inequality  :?: Thanks.

Jianxing113725, I have meant the following Vasc's inequality:
$ (x^2\plus{}y^2\plus{}z^2)^2\geq3(x^3y\plus{}y^3z\plus{}z^3x).$


$ ABCD$ is a convex quadrilateral such that the ratio of the side lengths is
\[ |AB|: |BC|: |CD|: |DA| \equal{} 1: 2: 4: 3.\]
Outside of the quadrilateral draw the lines
\[ p_1 \parallel AB, \quad p_2\parallel BC,\quad p_3 \parallel CD, \quad p_4 \parallel DA,\]
with the condition
\[ d(p_1, AB) \equal{} d(p_2, BC) \equal{} d(p_3, CD) \equal{} d(p_4, DA),\]
where $ d(p, q)$ stays for distance of the lines $ p$ and $ q$.
Define
\[ p_1\cap p_4 \equal{} A', \quad p_1\cap p_2 \equal{} B', \quad p_2\cap p_3 \equal{} C', \quad p_3\cap p_4 \equal{} D'.\]
Prove that the lines $ AA'$, $ BB'$, $ CC'$ and $ DD'$ are concurrent.

The sums of the two pairs of opposite sides are equal to $ 5$, so there is an inscrit circumference. Let $ O$ be its center. Draw $ X$ in $ A'B'$ such that $ BX\perp{A'B'}$ and $ Y$ in $ B'C'$ such that $ BY\perp{B'C'}$. Triangles $ BB'X$ and $ BB'Y$ are congruent, and so $ \angle XB'B\equal{}\angle YB'B$, and this implies that $ BB'$ is a bisector of $ \angle ABC$, and so $ O$ belongs to $ BB'$. With the same reasoning for the other lines we conclude that all of them concur at $ O$.


Using half-angle formulas to find exact values for:

Sec 15п/8

Haha yea my bad, switched the order of operation. Thanks.

[quote="10000th User"]Well, $ \sec\theta \equal{} \frac1{\cos\theta}$. Use the half angle formula for cosine: $ \cos\left(\frac {\theta}{2}\right) \equal{} \frac12\sqrt {1 \plus{} \cos\theta}$[/quote]

Actually, $ \cos\frac {\theta}{2} \equal{} \sqrt {\frac {1 \plus{} \cos\theta}{2}}$

hello, we have $ \sec\left(\frac{15\pi}{8}\right)\equal{}\sec\left(\frac{\pi}{8}\right)$
Sonnhard.

[hide="Solution"]$ \frac{15\pi}{8}=\frac{16\pi}{8}-\frac{\pi}{8}\equiv -\frac{\pi}{8}\pmod{2\pi}$.

What we want is $ \sec\left(-\frac{\pi}{8}\right)=\sec\frac{\pi}{8}$.

$ \cos\left(\frac{1}{2}\cdot\frac{\pi}{4}\right)=\sqrt{\frac{1+\cos\frac{\pi}{4}}{2}}$.

$ \cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, so we have

${ \frac{\sqrt{\sqrt{2}+2}}{2}\cdot\frac{1}{2}}=\frac{1}{2}\sqrt{\sqrt{2}+2}$.

Our answer is $ \boxed{\frac{2}{\sqrt{\sqrt{2}+2}}}$.[/hide]


Let us denote
\[S(x) : = x(1+x^{2}(1+x^{3}(1+x^{4}(1+...)).\]
Prove that $S(1/n)$ is irrational for every integer $n > 1$.

\[S(x)=\sum_{k=1}^{\infty}x^{k(k+1)/2}. \]
Therefore obviosly, that $S(1/n)$ is rational, because $S(1/n)<\frac{1}{n-1}$ and from $S(1/n)=\frac{P}{Q}$ we get contradition
$\frac{P}{Q}=\sum_{k=1}^{Q}(\frac{1}{n})^{k(k+1)/2}+R, R=\sum_{k=Q+1}^{\infty}(\frac{1}{n})^{k(k+1)/2}.$

Don't know what Rust does, but:
$S(x)=\sum_{k=1}^{\infty}x^{k(k+1)/2}$
implies that $S\left( \frac{1}{n}\right)$ is not periodic in base $n$, result follows.

[quote="t0rajir0u"]Sorry; could you explain how that's a contradiction?[/quote]
Let $\sum_{k=1}^{m}(\frac{1}{n})^{k(k+1)/2}=\frac{P_{1}}{n^{m(m+1)/2}}$, then
$\sum_{k=m+1}^{\infty}(\frac{1}{n})^{k(k+1)/2}=\frac{P}{Q}-\frac{P_{1}}{n^{m(m+1)/2}}=\frac{P_{2}}{Qn^{k(k+1)/2}}.$
It give contadition, because $n^{m(m+1)/2}\sum_{k=m+1}^{\infty}(\frac{1}{n})^{k(k+1)/2}<\frac{2}{n^{m+1}}<\frac{1}{Q}.$
Zetax solution is shorter.


If $ a,b,c$ are the sides of $ \triangle{ABC}$ and $ r_a,r_b,r_c$ are the radii of excircles, then

$ \frac {r_a}{a} \plus{} \frac {r_b}{b} \plus{} \frac {r_c}{c}\leq\frac {3\sqrt {3}}{4}\left(\frac {a^2}{r_b^2\plus{}r_c^2} \plus{}\frac {b^2}{r_c^2\plus{}r_a^2} \plus{}\frac {c^2}{r_a^2\plus{}r_b^2}\right),$

with equality if and only if $ \triangle{ABC}$ is equilateral.

"Given a triangle, extend two sides in the direction opposite their common vertex. The circle tangent to these two lines and to the other side of the triangle is called an excircle, or sometimes an escribed circle." From [url=http://mathworld.wolfram.com/Excircles.html]Math World[/url]

Every triangle has three excircles, the $ r_a,r_b,r_c$ refer to the radii of the circles where $ r_a$ is the radius of the circle opposite to the $ A$ vertex,.. and so on for $ r_b$ and $ r_c$. (and no, it's not a circumcircle)


I'm in calculus but I can't remember this rule and I can't find it in the book. I tried to check it with my calculator but the answers are the same. Is e^x * e^x = e^(x^2) or is it e^2x. I can't remember if you add then together or multiply it when you multiply e^x.

$a^{b}\cdot a^{c}= a^{b+c}$.

This is easiest to see when you're dealing with integer exponents.  For example,

$2^{2}\cdot 2^{3}= (2 \cdot 2) \cdot (2 \cdot 2 \cdot 2) = 2^{5}= 2^{2+3}$

[quote="Blueazulim06"]I tried to check it with my calculator but the answers are the same.[/quote]

You may have entered the first expression as e^x^2, which evaluates to $(e^{x})^{2}= e^{2x}$ and not $e^{ x^{2}}$.


Prove that if n is a positive integer the n^2 + 5^n is divisible by 6.

[quote="mdk"]hmm, interesting, but induction works for some reason[/quote]

That cannot be true...

How would you go about inducting this question anyways? I kinda want to see how you arrived at that conclusion.

You might go about it like...

$ a_{n+1}: =(n+1)^{2}+5^{n+1}=5(5^{n}+n^{2})+(n+1)^{2}-5n^{2}=5a_{n}+(-4n^{2}+2n+1)$

If the problem were true, then you would use inductive assumption for $ a_{n}$ and the rest of the expression would be divisible by $ 6$.

[quote="HTA"]we let $ A=n^{2}+5^{n}$
if n=3 then ....
because $ n^{2}\equiv 1$ (mod 3) for every positive interger n >3

if n is even , obviously , $ 5^{n}\equiv 1$ (mod 6) , furthermore , we have n is even so $ n^{2}$must be even so $ n^{2}\equiv 4 (mod 6)$
so $ A \equiv 5 (mod 6)$

if n is odd , similarly ,we have $ 5^{n}\equiv-1(mod 6)$ and $ n^{2}\equiv (mod 6)$
so A is divisible by 6

Conclude : the problem is true if and only if n is odd and n>3[/quote]

not quite...

$ n^{2}\equiv 1 \pmod3$ for every positive integer $ n$ that is not a multiple of 3 (not any integer greater than 3).

If you look at the quadratic residues (google or wiki if you don't know what those are) in mod 6 and add them to the values of $ 5^{n}$ in mod 6 then you find that it's divisible if $ n \equiv \pm1 \pmod6$.  So for example 27 doesn't make the expression divisible, but 29 will.  Or equivalently, the expression is divisible by 6 if $ n$ is odd and not divisible by 3.

oh , im sorry ....  :blush: , when i think 'bout those in mod 3 , i juzt thought that n is prime because i didnt read the problem clearly .....


superior algebra
superior algebra unsolved

Is there a proof that the intersection of galois extensions is galois again without Galois theory?

uhm..
galois means separable and normal: since separability is stable when passing to subfields, intersection of galois extensions is automatically separable.
for the "normal" part: take an element in the intersection, and its minimal polynomial. since both extensions are galois, it has all of its roots in both fields, so in their intersection, and so the extension is normal.


1. Which credits are colleges more likely to accept- getting a 5 on the AP exam or taking a class at a local college during high school?

2. Also, are AP test scores and classes taken at a local college both likely to be accepted?

3. Do AP test scores counts towards getting a bachelor's degree?

4. If you take a CLEP or AP test does your score affect your GPA in undergrad?

5. The summer directly after your senior year, can you take a class at a local college if you have already been accepted elsewhere and if you do, can you get credit for the course and will the course affect your gpa in undergrad?

websites checked so far:
http://www.collegeboard.com/student/testing/ap/about.html
gewgle

It varies from college to college.  Some colleges will accept local classes, some won't.  Most will accept a 5 on the AP exam, although which ones are counted varies.  You'd have to go to each individual website or contact the school and department you need.

The answers for my institution, California State University, Long Beach:

1 & 2. We accept AP credits with a 3 or better, and we accept local (most commonly community college) classes.

3. Yes, AP scores count. In some cases, the credit depends on the score - for instance, for AP Chemistry, a 4 or better counts as the first semester General Chemistry course that counts for science and engineering majors, while a 3 would count only as a general education course that does a science/engineering major no good. In all cases, the AP score counts as units; for some majors with large specific unit requirements, that doesn't help much.

4. AP or IB test credits (note: CLEP is very rare) are treated as if you took the class on a Credit-No Credit basis and are not computed into the GPA. Transferred college courses are counted as part of the overall GPA. (Although we also keep a separate "at our institution" GPA.)

5. Yes, that counts, and yes (as in #4) that's part of your GPA.

I've studied first semester freshmen taking our Calculus II course over the last several years. In what follows, "pass" means earning a C or better, and grades of W (that is, withdrawing from the class) are counted in the denominator, as a non-passing grade.

Of students with 5 on AP calculus AB, 90% pass Calc. II and 48% earn As.
Of students with 4 on AP calculus AB, 80% pass Calc. II and 25% earn As.
Of students with 3 on AP calculus AB, 63% pass Calc. II and 6% earn As.

That 63% pass rate is still better than than pass rate for continuing students (most commonly 3rd semester students) in the same course.

Obviously, I speak for no institution other than my own. It is worth noting that is is generally in the interests of state-supported universities to accept AP and transfer credits - the legislature doesn't want to spend more money on its students than it has to.

The best thing you can get is probably a survey of colleges and their policies. Here is Stanford's:

1. If you get a 5 on one of the accepted AP exams, you'll get the credit for sure. If you take a class at a local college, I believe you'll need to fill out a petition to get credit (it's a lot more work than mailing in an AP score report).

2. A small subset of AP tests are accepted, classes taken at local colleges probably have to be of comparable rigor to the Stanford course.

3. Yes.

4. AP test scores do not affect your GPA; classes taken at a local college may or may not, I'm not sure.

5. It should be the same policy regardless of when you take it.

Here's my school's AP/IB policy:

http://www.jhu.edu/admis/apply/apib.html

Almost all colleges will recognize an AP exam, however, it might not affect your course level. Some colleges will take a look at your "5" and move you onto a higher level course (for example a 5 in biology could place you in a 200-level biology course, as opposed to what most people would start in a 100-level). Few will not.

As for local college courses, these don't really show the level of understanding the top-tier colleges would like to see. An AP exam is written by CollegeBoard, which means that everyone who takes that AP course is measured. Local college courses could have different levels of grading, and the colleges you apply to won't appeal to that as much.

in most cases,

do AP exams result in being able to take harder classes earlier and less college classes in the long run?

yes i know it can depend on teh college

Usually, a 4 or 5 will allow you to place out of a lower level class and take a harder one earlier.  On your second question, it depends whether or not you receive credit for the score.  If you receive credit, then you'll need fewer credits to graduate so you can take fewer classes.  If not, then it won't affect how many classes you need to take.


Shouldn't it be $ \sum_{k\equal{}1}^{n\minus{}1}(2k\plus{}1)$?

Thnkx, really beautifull
but where does $ n\minus{}1$ come from ?

You're adding 1 each time, $ n\minus{}1$ times. You can take this out of the summation, and then proceed to simplifying it. :)

[hide="Another way"]The sum of the first n odd natural numbers is $ n^2$: $ \sum_{k \equal{} 0}^{n \minus{} 1}(2k \plus{} 1) \equal{} n^2$. But:

$ \sum_{k \equal{} 0}^{n \minus{} 1}(2k \plus{} 1) \equal{} n^2 \equal{} 1 \plus{} \sum_{k \equal{} 1}^{n \minus{} 1}(2k \plus{} 1)$

$ \equal{} > \sum_{k \equal{} 1}^{n \minus{} 1}(2k \plus{} 1) \equal{} n^2 \minus{} 1$.[/hide]


Dado dos conjuntos $A, B$ de enteros reales positivos tal que: $|A| = |B|= n; A \neq B$ y $S(A) = S(B)$, donde $|X|$ es el numero de elementos y $S(X)$ es la suma de todos los elementos en el conjunto $X$. Probar que podemos llenar en cada cuadro unitario de un tablero $n \times n$ con numeros positivos y algunos ceros tal que:\\

a) el conjunto de la suma de todos los numeros en cada fila es igual a $A$.\\
b) el conjunto de la suma de todos los numeros de cada columna es igual a $A$.\\
c) existen al menos $(n-1)^{2}+k$ numeros ceros en el arreglo $n \times n$ con $k = |A \cap B|$.\\

Es facil probar un resultado más general:

Dado dos conjuntos $ A, B$ de enteros positivos tal que: $ |A| \equal{} n; |B| \equal{} m ; A \neq B$ y $ S(A) \equal{} S(B)$. Probar que podemos llenar en cada cuadro unitario de un tablero $ n \times m$ con numeros positivos y algunos ceros tal que:\\

a) el conjunto de la suma de todos los numeros en cada fila es igual a $ A$.\\
b) el conjunto de la suma de todos los numeros de cada columna es igual a $ B$.\\
c) existen a lo más $ m \plus{} n \minus{} k \minus{} 1$ numeros no ceros en el arreglo $ n \times n$ con $ k \equal{} |A \cap B|$.

[i]Demostración:[/i]
Por inducción sobre $ m \plus{} n$:
El caso base $ n \plus{} m \equal{} 2$ es trivial.
El paso inductivo:
Si hay un entero $ x$ en $ A$ y en $ B$ se pone este en la casilla $ 1;1$ y borrando la fila y columna $ 1$ obtengo el caso $ n \plus{} m \minus{} 2$.
Si no hay tal entero, supongamos SPDG que $ min(a_1,b_1) \equal{} a_1$. Entonces se pone $ a_1$ en la casilla $ 1;1$ y se borra la primera fila obteniendo el caso $ m \plus{} n \minus{} 1$.

No pongo el conteo de como queda esactamente el caso anterior valido pero espero que lo entiendan.


The product of the positive number [tex]x_{1},x_{2},...,x_{n}[/tex] is [tex](n^{a}-1)^n[/tex] , where [tex]a\geq 1[/tex] is a real number . Prove that [tex]\sum_{i=1}^{n}\frac{1}{(x_i+1)^{1/a})}[/tex]
:)


AMC
AIME
videos
geometry
3D geometry
AIME II

I'm a freshman in high school who just took the AMC 12 and got a 96 :/

I'm going to take the 12B and see if I can qualify, but my little brother (aged 10, fifth grade) who is a fairly bright math student wants to come to my high school and take the AMC 10.

I have a few questions:
Firstly: is this allowed? I talked to the test administrator and they said that they'd be happy to allow a non-enrolled student to take it, but his results wouldn't count. I think he has a reasonably good shot of making AIME as a fifth grader, and I want to see that if he qualifies, he can take it. His school won't let him take the AMCs, because it's a a small middle school with no need for high school level tests.

Second question: If he is allowed to take the test, and his scores are graded (and presumably counted), how can I best alleviate some of the stress he's feeling about qualifying? I didn't make it as a freshman, and to have my ten year old brother come and kick my wagon at a math competition is rather sad. He doesn't seem to understand the social advantage he has, and is under a lot of duress to beat me  :P Typical sibling rivalry? I think not. I just want to be able to reassure him (successfully, I might hope) that his scores do not matter in fifth grade.

Any advice?

1. As far as I know, it's allowed and his score can be official (i.e. he can qualify for the AIME).

2. Go tell him to play some video games and stop thinking about the AMC.

[quote="DiscreetFourierTransform"]Video games aren't really his thing. Thank you for the quick response on the grading question.

I'll try to distract him somehow...[/quote]

Fair enough. Any distraction will suffice.

1. Yep. 5th graders are, in fact, allowed to take the tests as well as qualify for the AIME. And that would be pretty damn amazing.
2. If he doesn't already have one, buy him a Rubik's cube and make him solve it? I dunno. Or make him solve it within a minute. If he's into that kind of thing.

Students of any age and grade designation below the maximum eligible grade can take an AMC test officially. I know of a sixth grader who qualified for AIME, and maybe I've got another one in my testing group this year (but I haven't seen the A date scores yet this year). As far as how not to worry goes, just don't worry. There are plenty of more pleasant things to think about than how one will do on a test that hasn't happened yet.

[quote="mathgeniuse^ln(x)"][quote="randomdragoon"]I think it would be cool if your brother beat you in 5th grade (for your brother anyway...). My brother is not nearly that close...[/quote]

Yeah, same here.

He's in 2nd Grade.

AS for the AIME as 5th Grader.

A kid from MI gets 122 on 10A, and ends up getting 4 on the AIME II as 5th Grader.[/quote]
Dude that is so beast...
It's not even funny how beast it is...

My bro can probably qualify for AIME, but he'd get one or two right, maybe.

Not quite so beastly and insane as that other kid.


The radius of the circle is 10 centimeters, and the vertices
of the smaller square are the midpoints of the larger square.
What is the ratio of the area of the smaller square to the area of the circle? Express your answer as a common fraction in terms of $ \pi$.

[asy]size(75); import graph;
draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle);
draw(Circle((0,0),sqrt(2)));
draw((1,0)--(0,1)--(-1,0)--(0,-1)--cycle);[/asy]

The radius of the circle is equivalent to the sides of the smaller square (just take a 90 degree rotation). Therefore the ratio is 1/pi.


Indian team's performance was OK. Not too good. We got four silver & 2 bronze. We are 14th overall & 1st in commonwealth.
            individual scores are:

  Anand deopurkar         30
  Vipul Naik                    30
  Kshipra Bhavalkar        25
  Rohit joshi                   24           all silver

  Anupam Prakash          23
  Abhishek dang             19            bronze

                   Can anybody give detailed result of china team? :P  :P

probably Charmine has them noted somewhere, so ask in the China thread :)

Actually, I don't have the results with me.  Strangely enough, all the results I have are of teams which had rooms on the 13th floor of the hotel :D (and China wasn't on that floor ;) ).


Let $X$ be the coefficient of any power of $a$ in the expansion of

\[\frac{1}{\prod \displaystyle^i_{k=1} (1 - x^k \cdot a)},\]

where $i \in \mathbb{Z},$ and let $X$ be arranged according to the powers of $x$. Prove that $X$ is symmetric and unimodal.

We have the following identity:
\[\frac{1}{\prod \displaystyle^i_{k=1} (1 - x^k \cdot a)}=\prod_{i=1}^k\sum_{j=0}^\infty x^{kj}a^j.\]

Consider after an expansion the term
\[x^{j_1}a^{j_1}\cdot x^{2j_2}a^{j_2}\cdot ...\cdot x^{kj_k}a^{j_k},\]
then the "complementary" term will be
\[x^{j_k}a^{j_k}\cdot x^{2j_{k-1}}a^{j_{k-1}}\cdot ...\cdot x^{kj_1}a^{j_1}.\]
Indeed, these terms have the same power of $a$, namely $j=j_1+j_2+...+j_k$, and their degrees always sum up to $(k+1)j$.
That's all.


Same rules as usual Relay. Check out Relay 1 if you haven't so.

R1 - 1. Two sides of an isosceles triangle are 22 and 61. Compute the area of the triangle.

R1 - 2. Let $ T \equal{} TNYWR$ and let $ K \equal{} T \plus{} 1$.
If $ a,b,$ and $ c$ are positive integers, each less than or equal to $ K$, and if $ P * Q$ means $ \frac{P\plus{}Q}{2},$ compute the greatest possible value for the expression $ a * (b * c) \minus{} (a * b) * c$.

R1 - 3. Let $ T \equal{} TNYWR$.
(Note: In this problem, the brackets represent the Greatest Integer Function). Compute the number of values of $ n,$ with $ 1 < n < T$, for which $ \lfloor{\frac{n}{2}\rfloor} \plus{} \lfloor{\frac{n}{3}\rfloor} \equal{} \frac{n}{2} \plus{} \frac{n}{3}$.

[hide="1"]11-60-61 triangle.

so area is 660[/hide]

[hide="2"]simlifies to a/4-c/4.

let a=661, c=1 so 165 is the answer[/hide]

[hide="3"]I think it has to be a multiple of 6?165/6=27.5 so 27 is the answer[/hide]


calculus
integration
function
geometry
3D geometry
real analysis
real analysis solved

Let $B $and $B'$ two n-dimensional balls such that the radius of B is 1, the center of B' is in the boundary of B and the (n-dimensional) volume of the intersection of B and B' is half the volume of B. Let $r_n$ be the radius of B'. Find the limit of$ r_n $as$ n\rightarrow \infty$.

The asked limit is $\sqrt2$ ..

Go to : http://les-mathematiques.u-strasbg.fr/pdf/Hyperchevre.pdf

 :cool:  In France there are multi-dimensional animals  :cool:

oh excuse me could post soln in english? ;)

yeah that's the conjecture but after a waste of formulas with n-dimensional integrals and gamma functions, they still can't prove that result(yeah, i understand french a little bit, i learned something at school :) ).

Peter

sorry, at the end they proof it. anyway, i guess noone here wants to take the time to translate that proof...it is just straightforward in some sense...derive some crazy formula for $r_n$ and they try to show something about it.

Peter

We don't need to find $r_n$ for solving this problem.

It is clear $r_n<\sqrt 2$.

Let $V_F$ be volume of figure F and $S_F$ be area of figure F.

Let $1<r<\sqrt{2}$ and for some sufficiently large $n$ we have $r_n\leq r$.
Let $F$ be center of $B'$ and $AF$ is diametr of $B$. Consider ball $B''$ with center $F$ and radius $r$. Then it is clear $V_{B\cap B'}\leq V_{B\cap B''}$.

Cone $ACD$ contains spherical segment $CDE$ of ball $B''$ (because $AC$ is tangent to $B''$), so we have $V_{B\cap B''} < V_{ACD}+V_{CDF}$.

We can write $V_{CDF} = \frac{1}{2}V_B - V_{MNDC}$. On the other hand $V_{MNDC}>OG\cdot S_{CD}$. Moreover, $V_{ACD}=\frac{1}{n}(OG+1)\cdot S_{CD}$.

Thus $V_{B\cap B''}< \frac{1}{2}V_B - OG\cdot S_{CD}+\frac{1}{n}(OG+1)\cdot S_{CD}$.
We obtain that $V_{B\cap B''}< \frac{1}{2}V_B + S_{CD}\cdot (\frac{1}{n}-\frac{n-1}{n}OG)$.

It is clear that length $OG$ is independent of $n$ (actually, $OG=1-r^2/2$). So for all sufficiently large $n$ we have $\frac{1}{n}-\frac{n-1}{n}OG<0$ $\Rightarrow$ $V_{B\cap B''}< \frac{1}{2}V_B$ for this $n$.


calculus
integration
trigonometry
logarithms
calculus computations

[quote="kunny"]Calculate the following  integarls.

[1] $\int x\cos ^ 2 x dx$

[2] $\int \frac{x-1}{(3x-1)^2}dx$

[3] $\int \frac{x^3}{(2-x^2)^4}dx$

[4] $\int \left({\frac{1}{4\sqrt{x}}+\frac{1}{2x}}\right)dx$

[5] $\int (\ln x)^2 dx$[/quote]

Hello Kunny - I will leave full solutions to others - but will post the apprporaite methods I would use

1. [hide] $\cos(x)^2=\frac{1}{2}+\frac{1}{2}\cos(2x)$. The $x \cos(x)$ integral requires integration by parts[/hide]
2. [hide] $u=3x-1$ Break the resulting $\frac{u-2}{9u^2}$ up into 2 fractions[/hide]
3. [hide] $\frac{x^3}{(2-x^2)^4}=\frac{2x}{(2-x^2)^4}-\frac{x}{(2-x^2)^3}$ Simple u-substituiosn then suffice[/hide]
4. [hide]$\frac{\sqrt{x}}{2}+\frac{1}{2}\ln|x|+C$[/hide]
5. [hide] Integration by parts (twice!) [/hide]

Thank you for your reply and coorperation,dheardie. :)

kunny

[5]
Let $ I_n=\int(\ln x)^n\,dx,\quad I_1=x\ln x-x+C$
by integration by parts, we get a ricorrence formula:
\[ I_n=x(\ln x)^n-nI_{n-1}\]
Then \[I_2=x(\ln x)^2-2(x\ln x-x)+C\\
\\
I_3=x(\ln x)^3-3(x\ln^2x-2x\ln x+2x)+C\]
and so on.

[quote="amirhtlusa"]2
$\frac{1}{9}(\ln(3x-1)+\frac{2}{3x-1})$[/quote]

You forgot something,amirhtlusa.

[quote="amirhtlusa"]ok, you mean the absolout value AGAIN? :)
$\frac{1}{9}(\ln|(3x-1)|+\frac{2}{3x-1}) $[/quote]

Yes!


Prove that the number of odd binomial coeficient in any row of pascal triangle is a power of 2.

Let $n=a_0+a_1*2+...+a_k*2^k,m=b_0+b_1*2+...+b_k*2^k,a_i,b_i\in \{0,1\},$
$C_n^m$ is odd if and only if $b_i\le a_i.$ If n had l nonzero 2-adic digits, then we have $2^l$ cases for m, when $C_n^m$ is odd.

Rust you better give me some stuff to read about this "2-adic" stuff.
i see you haven't spoken of it,and as of one week now you're starting
to use this term very much

Let $n=a_0+a_1*2+...+a_k*2^k,m=b_0+b_1*2+...+b_k*2^k,a_i,b_i\in \{0,1\},$
$C_n^m$ is odd if and only if $b_i\le a_i.$ Let $l=\sum_i a_i$, then we have $2^l$ cases for m, when $C_n^m$ is odd.


1) Determine the number of ways of selecting four letters, without regard for order, from the word PARALLELOGRAM
Is there a formula for this type of questions? The Answer is 150, but why?
2) How many different sums of money can be formed from one $ 2 bill, three$5 bills, two $ 10 bills, and one$20 bills?
Please help me out.``` :oops:

Just so you know, the dollar sign represents the start/end of a $ \text{\LaTeX}$ formula. If you want to type out a dollar sign symbol, type:
[code]$\$$[/code]

My thoughts (which are in no way assumed to be correct)

[hide="1"]
There are 13 letters, but there are 3 A's, 3 L's, and 2 R's.

If the question's asking for the number of ways to pick any 4 letters (so that a set like {L, L, L, O} would be acceptable) then it would simply be $ \binom{13}{4}=715$.
Otherwise, I suppose it's $ \binom{8}{4}=70$.
[/hide]

[hide="2"]
I can't really think of a better way than listing them out...
[/hide]


1. In triangle ABC, is tanAtanBtanC (tanA times tanB times tanC) always a certain number? If yes, what is the number? If not, is there another way to express tanAtanBtanC?
2. How can we express the position of the orthocentre in triangle ABC with vectors?

Thanks...

1) $ \tan(A)\tan(B)\tan(C) \equal{} \tan(A) \plus{} \tan(B) \plus{} \tan(C)$

2) if $ O$ is the circumcenter of $ \bigtriangleup ABC$ then $ \vec{OH} \equal{} \vec{OA} \plus{} \vec{OB} \plus{} \vec{OC}$

Prove them