ARML
logarithms
function
geometry
trigonometry
pigeonhole principle
inequalities

AIM Math Meet was a fun idea but it was hard to continue due to the time. I decide to create a new one.

This is called "ARML Homework Assignment." Basically, I'll post 10 ARML-level problems. Some may come from actual ARML contest but there will be some that come from other sources. You're not [b]expected[/b] to answer every one of them. If you get at least one, post here and share with us. But there is one requirement: [b]your solution must be complete and in hide text[/b]. This is to respect others and give full answer to hard questions.

Problems will be up every 4 days. Yes, I know... This is fairly small amount of days to do 10 problems. But imagine how many problems we can cover!! To respect ARML, there will be NEVER all problems of ARML posted at once (this is how ARML problems are posted in AoPS anyway, just like marathons).

Please post the solutions here. Also, if you disagree with other person's solution/answer, feel free to criticize. If you have alternate way, that's fine too.

[size=200]4/16 to 4/20[/size]

1. For a positive integer $n$, there is a squence $(a_0,a_1,a_2,...,a_n)$ of real values such that $a_0 = 11$ and $(a_k + a_{k+1})(a_k - a_{k+1}) = 5$ for every $k$ with $0 \leq k \leq n-1$. Find the maximum possible value of $n$.

2. If $\displaystyle \frac{\log_b a}{\log_c a} = \frac{19}{99},$ then $\frac{b}{c} = c^k$. Compute $k$.

3. For what real value of $N$ does the [i]range[/i] of the function

\[ y = f(x) = \frac{4x^2+Nx+N}{x+1}, \]

where $x$ is real [and $x \neq -1$], consist of all the real numbers except for a single interval of the form $-L < y < L$?

4. Circles of radii 5,5,8, and $m/n$ are mutally externally tangent, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

5. You have 121 marbles, some of which are red, some white, and the rest blue. You also have 10 jars. If all the marbles are distributed into the jars, there [i]must[/i] be a jar with at least $n$ marbles of the same color. Compute the maximum possible value for $n$.

6. Points $P,Q,$ and $R$ are the midpoints of the medians of triangle $ABC$. If the area of triangle $ABC$ is 1024, compute the area of triangle $PQR$.

7. Compute the smallest positive 3-digit multiple of 7 for which the sum of its digit is also a multiple of 7.

8. If

\[ \tan(120^\circ - x) = \frac{\sin 120^\circ - \sin x}{\cos 120^\circ - \cos x}, \]

where $0^\circ < x < 180^\circ,$ compute $x$.

9. The equation $z^6+z^3+1$ has complex root with argument (angle) $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane. Determine the degree measure of $\theta$.

10. There is a unique real value $A$ such that for all $x$ with $1 < x < 3$ and $x \neq 2,$

\[ \left|\frac{A}{x^2-x-2} + \frac{1}{x^2-6x+8}\right| < 1999. \]

Compute $A$.

[quote="Silverfalcon"]
1. For a positive integer $n$, there is a squence $(a_0,a_1,a_2,...,a_n)$ of real values such that $a_0 = 11$ and $(a_k + a_{k+1})(a_k - a_{k+1}) = 5$ for every $k$ with $0 \leq k \leq n-1$. Find the maximum possible value of $n$.
[/quote]
[hide]
we can assume all $a>0$, it yields the same result
$a_k^2-a_{k+1}^2=5$
$a_{k+1}=\sqrt{a_k^2-5}$
so $a_{k+1}$ will not be real when $a_k^2<5$
since $a_k^2$ decreases by $5$ as $k$ increases by $1$, we can rewrite
$a_k=\sqrt{a_0^2-5k}$
what is the greatest value of k for which $121-5k>0$? when $\boxed{k=24}$
[/hide]
[quote="Silverfalcon"]
4. Circles of radii 5,5,8, and $m/n$ are mutally externally tangent, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
[/quote]
[hide]
We have circles with radii 5, 5, and 8 tangent to each other normally, and the last circle of radius $r$ is between the other circles. so draw an isoceles triangle with sides of length (8+5), (8+5), and base (5+5). the points of the isoceles triangle represent the center of the appropriate circles..
draw an altitude to the base; the length of the altitude is 12 by pythagorean triple and the center of the circle lies on this altitude.
since the smallest distance from the top point to the circumference of the smallest circle is 8 (because they are tangent) the distance from the center of the smallest circle to the base of the triangle is $12-8-r$
then we have an equation with one variable, $r$:
$(r+5)^2=5^2+(12-8-x)^2$
$r=\frac{8}{9}$
$\boxed{17}$
[/hide]

Are these really ARML problems? They seem a little easy for ARML problems.  :huh:

[hide="Problem 5"]Double Pigeonhole!  :D (The wording gives it away)

We know that $r+b+w=121$, so there must be at least one color with at least 41 marbles. Dividing that into 10 jars, one jar must have $5$ marbles. Therefore, the largest is $n=5$.[/hide]

[hide="Number 6"]Notice that the triangle being described is the midpoint triangle of the midpoint triangle, so the answer is $\frac{1024}{16}=64$.[/hide]

[hide="Number 7"]Keep on adding multiples of 7 starting from $105$. The smallest number that satisfies the condition is $133$.

If it were much larger, I would've given up on this method. Good thing it wasn't.  :P [/hide]

[hide="Number 9"]Notice that the expression given is equal to $\frac{z^9-1}{z^3-1}$. THerefore, we're looking for a 9th root of unity that isn't a 3rd root of unity, which lies in the given interval.

Some quick searching yields $160^\circ$ as the answer.[/hide]

Edit: Oops. I typed the wrong button in on my calculator, and got 120 instead of 160. :wallbash:

[quote="quantum leap"]i've been wondering :huh: ...how do we hide text?
edit: i disagree with the answer above this post, and if i get my question answered i can post answers to probs 2 and 7[/quote]You hide test with this:

[code][hide]Text Goes Here[/hide][/code]

And the answer to number 2 is correct.  :D

[quote="4everwise"]Are these really ARML problems? They seem a little easy for ARML problems.  :huh:

[hide="Problem 5"]Double Pigeonhole!  :D (The wording gives it away)

We know that $r+b+w=121$, so there must be at least one color with at least 41 marbles. Dividing that into 10 jars, one jar must have $5$ marbles. Therefore, the largest is $n=5$.[/hide]

[hide="Number 6"]Notice that the triangle being described is the midpoint triangle of the midpoint triangle, so the answer is $\frac{1024}{16}=64$.[/hide]

[hide="Number 7"]Keep on adding multiples of 7 starting from $105$. The smallest number that satisfies the condition is $133$.

If it were much larger, I would've given up on this method. Good thing it wasn't.  :P [/hide]

[hide="Number 9"]Notice that the expression given is equal to $\frac{z^9-1}{z^3-1}$. THerefore, we're looking for a 9th root of unity that isn't a 3rd root of unity, which lies in the given interval.

Some quick searching yields $120^\circ$ as the answer.[/hide][/quote]

[hide="..."]120 is a 3rd root of unity  ;) [/hide]

[hide="3"]
$y=N+\frac{4x^2}{x+1}$, so let $g(x)=\frac{4x^2}{x+1}$. This problem is equivalent to finding the range of $g$. By the trivial inequality, only the $x+1$ part affects the sign, so $g$ is nonnegative for $x>1$ and is 0 at $x=0$. When $x<-1$, $g$ is strictly negative (because no roots).

For $x<-1$, let $x=-1-t$. $-g(x)=-g(-1-t)=\frac{4(1+t)^2}{t}=4\left(\sqrt{t}+\frac{1}{\sqrt{t}}\right)^2$. By AM-GM or completing the square, $\sqrt{t}+\frac{1}{\sqrt{t}}\geq 2$ for all positive $t$, and there is equality when $t=1$. So when $x<-1$, the maximum is $-4(2^2)=-16$.

Therefore the range of $g$ is all real numbers except the interval $(-16,0)\Rightarrow N=\boxed{8}$.
[/hide]

[hide="7"]
Let the number be $100a+10b+c$, where $0\leq a,b,c\leq 9$ and $a\neq 0$. Then $7|(100a+10b+c)$ and $7|(a+b+c)$. Then 7 must divide their difference, $99a+9b$, and 7 is relatively prime to 9, so $7|(11a+b)$.

To get the smallest solution, set $a=1$. Then $b$ must equal 3 to make $11a+b$ divisible by 7, and $a$ must be 3 because $130\equiv -10\equiv -3\pmod{7}$. So the number must be greater than or equal to $\boxed{133}$, which clearly works.
[/hide]

[hide="8"]
Rather than use the tangent addition formula on the left hand side, cross multiply, and try to cancel everything, let $a=60^\circ+\frac x2$ and $b=60^\circ-\frac x2$. The right hand side simplifies nicely:

\[ \frac{\sin 120^\circ -\sin x}{\cos 120^\circ -\cos x}=\frac{\sin (a+b) -\sin (a-b)}{\cos (a+b)-\cos (a-b)}=\frac{2\cos a\sin b}{-2\sin a\sin b}=-\cot a \]

Writing $-\cot a$ in terms of tangent function and $x$, $-\cot a=-\tan(90^\circ -a)=\tan(a-90^\circ)=\tan \left( \frac x2 -30^\circ \right)$.

This is equal to $\tan (120^\circ -x)$. Setting angles equal, $120^\circ -x=\frac x2 -30^\circ\Rightarrow x=\boxed{100^\circ}$.
[/hide]

[hide="10"]Number $10$ is strange, I'm getting that the inequality cannot hold unless the numerator is $0$ at $x=2$, so $A=\boxed{\frac{3}{2}}$  :? If that is correct, I will post a full solution[/hide]

No one's gotten 9 right so far? :o

[hide="9"]$z^6+z^3+1=0$ and $z^9+z^6+z^3=0$, so subtracting the two, we have $z^9=1$, and $\theta=40^\circ$, so $\boxed{160^\circ}$ is the only one that works for the given intervals. [/hide]

[hide="5."] Use pigeonhole to see at least one jar has to have at least 13, then at least one color has to have at least [u]5[/u].[/hide]

[hide="7."]
Guess and check, starting with 106, 115, etc., and eventually get [u]133[/u]
[/hide]

[hide="4."]
5+1=6?
since the quadrilateral formed by the radii is cyclic...?  :?
[/hide]

EDIT: oops to 4.  :oops:

[quote="Silverfalcon"]9. The equation $z^6+z^3+1$ has complex root with argument (angle) $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane. Determine the degree measure of $\theta$.[/quote]

[hide="9"]
Divide the equation by $z^3\ne 0$ to get $z^3+\frac{1}{z^3}=-1$. By de Moivre's formula, this is $2\cos 3\theta=-1$. Since $\theta\in (90,180)$, $3\theta \in (270,540)$. Clearly $\theta \in \{ \ldots, 120^\circ, 480^\circ, \ldots\}$, so $3\theta = 480^\circ$ works, and in this case $\theta = \boxed {160^\circ}$.
[/hide]

pieterminate: the circle with radius $m/n$ is a tiny circle tangent to the three other big circles... that's why they are mutually externally tangent, each circle is externally tangent to the three other circles, not 4 circles forming a 'quadrilateral shape', it's more like a 'triangular shape'.

People who wonder how to systematically do number 7, see divisibility by 7 here: if $abc$ is a 3-digit number, the number is divisible by 7 iff $2a+3b+c\equiv0\pmod7$

[hide="For number 7"]$2a+3b+c\equiv a+b+c\Rightarrow a+2b\equiv0\pmod 7$. For smallest number, let $a=1$ so that $b=3$ (7-1=6). Then $1+3+c=7$ we find $c=3$.

So, isn't it easier now than a wild guess? ;)[/hide]

You can use my method and you can find the next smallest such number or the largest such number (do find them so you can practice) :)

[EDIT]Iversonfan2005: you are correct for #10, please put up your solution, I want see how you did it :D

[quote="10000th User"]

[EDIT]Iversonfan2005: you are correct for #10, please put up your solution, I want see how you did it :D[/quote]
[hide="10"]
well, we can simplify it to $|\frac{(x-4)A+(x+1)}{(x-2)(x+1)(x-4)}|$, so therefore, when we examine this, we find that if $x=2$, the denominator is $0$, but the numerator is $-2A+3$, but as $x$ gets closer and closer to $2$, the absolute value reaches infinity, so the inequality doesn't make any sense. So the only way is if the the absolute value is $0$, which means that the numerator is $0$, when $x=2$, so plugging that in, we have $\boxed{A=\frac{3}{2}}$[/hide]

[hide="4"]By drawing a triangle with the radii of each of the larger circles, we form a triangle with sides of $13,13,10$ and so the altitude must be $12$. In one of the two $5,12,13$ right triangles, we can make another right triangle with sides of $5, 5+x$ and $4-x$, where $x$ is the radius of the smallest circle. Note that I got $4-x$ from the altitude, since the radius of the biggest circle plus the smaller circle, which is $8+x$, so the remaining part must be $4-x$. From Pythagorus, we have $5^2+(4-x)^2=(5+x)^2\Longrightarrow x=\frac{8}{9}$, so the answer is $\boxed{17}$[/hide]

[hide="4."]
I'm not going to draw it, but take the centers of the outside circles, call them A,B,C, with A being the large circle.  Now, draw in the circle O on the inside with radius r, and draw triangle ABC.  Now, draw perpendicular from A, and it intersects BC at D.  Now, we see that $AD=\sqrt{13^2-5^2}=12$, and then we use algebra, and get:
$(12-AO)^2+5^2=(OC)^2$
$(4-r)^2+25=(5+r)^2$
So $18r=16$, and r=[u][b][i]8/9[/i][/b][/u][/hide]

[hide="Number 2"]

I agree with above answer for this question. Here is my way of doing it.

Rewriting, we get $b^x = a$ and $c^y = a$ where $x/y = 19/99$ by the problem ($x,y$ are just random ratios). From the second equation, $b = c^{k+1}$. So, substitute this and set the first equation equal to each other:

$c^{(k+1)x} = c^y$

$kx+x = y$

$kx = y-x$

$k = \frac{y}{x} - 1$

Since $x/y = 19/99,$ we know that $y/x = 99/19$ so $k = 99/19 - 1 = 80/19$. [/hide]

[hide="Number 7"]

This problem was actually from ARML Relay Round (only Relay round in this set of problems). So, I wouldn't recommend random guessing unless you know for sure that it will work. To get Altheman's equation, let the three digit number be $abc$. This can be rewritten as $100a+10b+c$. Taking this in base 7, we get $2a+3b+c$. Let $a = 1$ and start from $b = 0$. The smallest one that works is when $a = 1, b = 3, c = 3$. So, 133 is the answer. [/hide]

I have bit different way to do this.. I wonder if it works.

[hide="Number 9"]

Let $Z = z^3$. This gives $Z^2+Z+1 = 0$ and $Z = \frac{-1}{2} \pm \frac{i \sqrt{3}}{2}$. This is either $\cos(120) + i \sin (120)$ or $\cos(240) + i \sin (240)$. If you use DeMoivre's and take cubic root for each one, you'll see that there isn't any root in given interval for the first case other than 120 (and this can't be answer.. or can it be?). For the second one, you get 80 and 80,160, and 240. Only 160 is in the interval so this is the answer. [/hide]

[hide="Number 2"]
$\displaystyle \frac{\log_b a}{\log_c a} = \frac{19}{99}$
${\frac{\frac{\log a}{\log b}}{\frac{\log a}{\log c}}}= \frac{\log c}{\log b}=\log_b c=\frac{19}{99}$
$b^{\frac{19}{99}}=c$
$b=c^{\frac{99}{19}}=c^{k+1}$
$k+1=\frac{99}{19}$
$k=\frac{80}{19}$ :huh: [hide="Taking this into consideration..."] [quote="quantum leap"] I disagree with the answer above this post[/quote]  yep, i was wrong :oops: [/hide][/hide]

[hide="Number 7"]  [i]Merely counting above 100[/i] the multiples of 7, $105, 112, 119, 126, 133$, and $133$ is the first one.[/hide]

[quote="pieterminate"][quote="Silverfalcon"]

[hide="Number 7"]
This problem was actually from ARML Relay Round (only Relay round in this set of problems). So, I wouldn't recommend random guessing unless you know for sure that it will work. To get Altheman's equation, let the three digit number be $abc$. This can be rewritten as $100a+10b+c$. Taking this in base 7, we get $2a+3b+c$. Let $a = 1$ and start from $b = 0$. The smallest one that works is when $a = 1, b = 3, c = 3$. So, 133 is the answer. [/hide]
[/quote]

Guessing numbers from 106, 115, 124, etc. would definitely work as it isn't random guessing...and they are all incongruent mod 7, thus at least one work within 7 tries.[/quote]Even if it isn't random guessing, you would spent many minutes just to solve this question and I am made awared that Relay Rounds you need to solve the problem in less than a minute or two. Solution like scorpius119 or even mine makes reasoning+calculation sound and fast... That's the point that Silverfalcon is stressing because if there is a systematic way of solving it, then it is better than guessing (no matter what type of guessing).

PS: no one tried getting the next smallest 3-digit number multiple of 7 whose digit sum is also a multiple of 7? also what is the largest 3-digit with these properties? :)

[quote="Silverfalcon"]7. Compute the smallest positive 3-digit multiple of 7 for which the sum of its digit is also a multiple of 7.[/quote]

[hide]$100a+10b+c\equiv 0 \pmod {7}$

$\iff 2a+3b+c\equiv 0 \pmod {7}$

Also $a+b+c\equiv 0 \pmod {7}$.

Put $a=1$ and $b+c\equiv 6\pmod {7}$ while $3b+c\equiv 2b+6\equiv 5\pmod {7}\implies b\equiv 3 \pmod {7}$.

This gives 133 as the smallest 3-digit number that works.

Put $a=9$ and we get $b+c\equiv 5\pmod {7}$ while $3b+c\equiv 3\iff b\equiv c\equiv 6\pmod {7}$.

This gives 966 as the largest 3-digit number that works.[/hide]

Excellent job guys! I appreciate you guys participating in this! Just like most people, I'm not capable of answering ALL the problems so it helped to get some help.  :D And it's fun.  :P

[size=200]4/21 to 4/24[/size]

1. For $0 < x < 1,$ let $f(x) = (1+x)(1+x^4)(1+x^{16})(1+x^{64})(1+x^{256})...$. Compute $f^{-1} \left(\frac{8}{5f(\frac{3}{8})}\right)$

2. If $a$ and $b$ are the roots of $11x^2-4x-2 = 0,$ then compute the product:

$(1+a+a^2+a^3 + \cdots )(1 + b + b^2 + b^3 + \cdots)$

3. A set of distinct, nonzero real numbers is placed along the circumference of a circle. Each of the numbers is equal to the product of the two numbers adjacent to it. What is the least possible number of numbers in the set?

4. In convex quadrilateral $PQRS,$ diagonals $\overline{PR}$ and $\overline{QS}$ intersect at $T$, with $PT: TR = 5: 4$ and $QT: TS = 2: 5.$ Point $X$ is chosen between $T$ and $S$ so that $QT = TX,$ and $\overline{RX}$ is extended its own length to $Y$. If point $Y$ is [i]outside[/i] the quadrilateral, find the ratio of the area of triangle $PSY$ so that of triangle $QRT$.

5. Compute the smallest positive integer $n > 100$ such that $\binom{n}{101}$ is divisible by $\binom{n}{100},$ but it is not equal to it.

6. Compute the smallest positive integer $n$ such that the product $(13)(19)(n)$ is also the product of three [i]consecutive[/i] integers.

7. The arithmetic mean of the positive numbers $a_1,a_2,...a_k$ is one-fourth of the arithmetic mean of the positive numbers $b_1,b_2,...b_{7k}$ (where $k$ is a positive integer). [i]If both of these means are integers[/i], compute the smallest possible integer value for the arithmetic mean of $a_1,a_2,...,a_k,b_1,b_2,...b_{7k}$.

8. Let $\omega$ be a primitive 49th root of unity. Compute:

$\displaystyle \prod^{49}_{{k=1, \gcd(k,49) = 1}} (1 - \omega^7)$

9. In a quadrilateral $ABCD, AB = 2 \sqrt{6}, BC = 7 - 2 \sqrt{3} , CD = 5, \measuredangle B = 135^\circ,$ and $\measuredangle C = 120^\circ$. Compute $AD$.

10. If $\sin 9^\circ + \cos 9^\circ = \frac{1}{a} \sqrt{b + \sqrt{c}}$, find $a+b+c.$

[hide="2"]
$\frac{1}{1-a} \cdot \frac{1}{1-b}$
$\Rightarrow \frac{1}{1-a-b+ab}$
Using piete's
$\Rightarrow \frac{1}{1-\frac{4}{11}-\frac{2}{11}}=\frac{11}{5}$
[/hide]

[hide="6"]
We see that in order to make $13$ and $19$ somewhat close to each other, we need some number $13a$ and $19b$ such that they are at most 2 apart. If $a=3$, $b=2$, then we have $38$ and $39$. Then since we need to make $n$ as small as possible, the least number of the three consecutive will be 37. Thus $n= 2\cdot3\cdot37=222$
[/hide]

Also 5:
[hide="solution"]
Well $\binom{n}{101} = \frac{n!}{(n-101)!101!}$, $\binom{n}{100}=\frac{n!}{(n-100)!100!}$
Divide the first by the latter, we get:
$\frac{n!}{(n-101)!101!} \cdot \frac{(n-100)!100!}{n!}$
$= \frac{(n-100)}{101}=k$, for some $k \in \mathbb{Z^+}$

Since $k=1$ gives us the same value of the combination, we set $k=2$
Then $n=202+100=302$
[/hide]

[hide="Problem 2"]

Clearly, the given quadratic does not factor so you don't want to use Quadratic Formula. Instead, note that we have two infinite series multiplied. The given product is actually same as $\frac{1}{1-a} \cdot \frac{1}{1-b} = \frac{1}{1-(a+b)+ab}$

We see that $a+b = 4/11$ and $ab = -2/11$ from the given quadratic so putting this in, we get the total product as $\frac{11}{5}$. [/hide]

[hide="Problem 5"]

This is actually very easy problem because note that the problem states $\binom{n}{100} | \binom{n}{101}$ so simplifying this gives $\frac{n-100}{101} = \frac{n}{101} - \frac{100}{101}$. For this to be an integer, it is necessary that $n \equiv 100 \mod 101$. Since $n > 100$, we know that we need to consider next number. This is $201$. But note that $\binom{201}{100} = \binom{201}{100}$ by $\binom{n}{k} = \binom{n}{n-k}$ identity. So, we go to the next number, and this is 302. This is the final answer. [/hide]

[hide="Problem 6"]

Note:

$13 \times 2 = 13$
$13 \times 3 = 39$

$19 \times 2 = 38$

So, to minimze $n$, we know that it must contain at least 2 and 3 and smallest number that will form consecutive integers with 38 and 39. This is 37. So, the answer is $2 \cdot 3 \cdot 37 = \fbox{222}$. [/hide]

[hide="Problem 7"]

The given equation gives:

$\frac{a_1 + \cdots + a_k}{k} = \frac{1}{4} \left(\frac{b_1 + \cdots + b_{7k}}{7k}\right)$

This leads to $28(a_1 + \cdots + a_k) = b_1 + \cdots + b_{7k}$

The given question asks for minimum of ${\frac{28(a_1 + \cdots + a_k) + a_1 + \cdots + a_k}{8k}} = \frac{29(a_1 + \cdots + a_k)}{8}$

Then this means that $a_1 + \cdots + a_k$ should contain 8. So, let that be $8$ and this gives our minimum ---- $\frac{29}{8k}$. Because $k$ is an integer and this average should be an integer, we know that $k = 1$. This gives the minimum of $29$. [/hide]

[hide="Problem 7"]
Take the mean of $a_{i}$ to be $k$, and the mean of $b_{i}$ to be $4k$. Then, the mean of $a_{i}$ and $b_{i}$ is $\frac{29k}8$. If this mean must be an integer (and $k$ is an integer), then it is at a minimum when $k=8$, so $\frac{29k}8 = 29$.
[/hide]

Okay.. I had to make a new post because I worked on two problems and got some progress but I'm unable to finish them. Perhaps you guys can help me out using this fact.

For #10, I got two angle formulas through binomial theorem (YES, I did all algebra for you).

$\sin(5x) = 16 \sin^5 x - 20 \sin^3 x + 5 \sin x$

$\cos(5x) = 16 \sin^4 x \sqrt{1 - \sin^2 x} - 12 \sin^2 x \sqrt{1 - \sin^2 x} + \sqrt{1 - \sin^2 x}$

If $x = 9$, this gives nice number on LHS but then, to simplify the RHS is very difficult. Unless I'm missing something obvious, I must be doing something wrong.

P.S. This is NOT an ARML problem so I DO have an answer to it. The original problem asked to prove it so if you must know the answer (or if it helps), here it is:

[hide="The original problem stated to prove.."]

$\sin 9 + \cos 9 = \frac{1}{2} \sqrt{3 + \sqrt{5}}$ [/hide]

For #1, I only got the general form down but it's not complete.

$\displaystyle \prod^{n}_{i=0} (1 + x^{4^i}) = f(x)$

I [i]so[/i] called that problem.  I'm not sure if anyone got that at our arml tryouts, but i know the one of the kids said you had to find a closed form for it, perhaps by squaring it where it becomes $1+x+x^2+x^3+\cdots$ or something like that.   Just check out [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47851&postorder=asc&start=1240]www.artofproblemsolving.com/Forum/viewtopic.php?t=47851&postorder=asc&start=1240[/url]
at post #1251, problem #245 ;they just gave up on it after finding that shorthand form for the product which IS NOT THE CLOSED FORM :mad: .  btw, i forgot the exact problem when i posted it, so i just asked for what would amount to the first part of it.
EDIT: ah, i was beaten to the post. nice solution though

To some people, $\sin 18^\circ$ is well-known. I actually didn't remember its value so I had to look it up (see AoPS volume 2, Problem 35)
\begin{eqnarray*}\sin9^\circ+\cos9^\circ&=&\sqrt{1+\sin18^\circ}\\ &=&\sqrt{1+\frac{\sqrt{5}-1}{4}}\\ &=&\sqrt{\frac{\sqrt{5}+3}{4}}\\ &=&\frac{1}{2}\sqrt{3+\sqrt{5}}\end{eqnarray*}Maybe there is some other easy way to do the problem without remembering $\sin18^\circ$...

[quote="tarquin"]3
[hide]
4
[/hide][/quote]

care to show work?

:huh:
for number 3, 4 doesn't work because then the numbers opposite each other are the same..

[hide="i think..."]
ok, so for every number of numbers n, define the first two adjacent numbers. lets call them x and y.
$n=3$: $x, y, \frac{y}{x}$ so $\frac{y}{x}=x\cdot y\rightarrow x^2=1$ but you would have some sequence of 1's or -1's, which wouldn't work by the Pigeon hole
$n=4$: $x, y, \frac{y}{x}, \frac{1}{x}$ opposite numbers come out to be the same
$n=5$: $x, y, \frac{y}{x}, \frac{1}{x}, \frac{1}{y}$ so $\frac{1}{y}=x\cdot\frac{1}{x}\rightarrow y=1$ but then all numbers result in a 1
$n=6$: $x, y, \frac{y}{x}, \frac{1}{x}, \frac{1}{y}, \frac{x}{y}$ choose $x=2$ and $y=3$: $\boxed{6}$ is the lowest number that works.
[/hide]

You're right. The correct answer is 6 (I checked with the answer key and that's the same thing I got from solving as well).

My solution is bit different but here it is.

[hide="Solution for number 3"]

Because of the problem, we know that the number of numbers in the set, say $n$, is at least 3. So, consider cases.

Let $n = 3$ and be $X,Y,Z$ for three numbers. We're given that $Y = XZ, X = YZ, Z = XY$. Multiplying them and taking square roots (the problem didn't specify which one but if it doesn't work for EITHER of positive or negative, it won't work at all because it only said "real"), we get $1 = XYZ$. Dividing $Y$ by both sides and combine with given expressions give $Y^2 = 1 \rightarrow Y = \pm 1$. Similar result is achieved for $X$ and $Z$ and clearly, this is not possible.

Let $n =4$ and be $A,B,C,D$ for four integers. We don't need to do anything in here because already, we get $B = AC = D$ and $C= BD = A$. Contradiction.

Let $n = 5$ and be $A,B,C,D,E$ for five integers. Multiplying the five equations and simplifying gives $1 = ABCDE$. Grouping them yields $(AC)(BD)(E) = B \cdot C \cdot E$. But depends on how you group them, you can get $B = 1$ and $D = 1$. But this contradicts $B$ and $D$ being distinct.

Let $n = 6$ and be $A,B,C,D,E,$ and $F$ for six integers. Do the same thing as the others and get $1 = ABCDEF$. Grouping gives $(AC)(BD)(E)(F) = B \cdot C \cdot E \cdot F = (BF)(CE) = AD = 1$. Let $\frac{1}{A} = D$. Then by the given problem, we know that $C = BD$ and $B = AC$. So, substitute on the first equation to get $C = B(1/A) = B/A \rightarrow AC = B$ which is what originally stated. So, this is possible. You can test few numbers to see that this is indeed valid number. [/hide]

Since 4,8, and 9 are unanswered, I'll post these on the other forum and hope to get answers.

Here are the links for the unanswered ones. If you know the solution, discuss them on THOSE THREADS.

Problem 4
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=495668#p495668

Problem 8
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=495665#p495665

Problem 9
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=495667#495667

Hopefully, people will help us out. Now, for next four day problems. Because most people, including myself, aren't super strong at geometry, I think it is best if we practice in that area.

[size=200]4/25 to 4/28[/size]

1. If $ABCDEFG$ is a regular heptagon and diagonals $AC$ and $BF$ intersect at $P,$ compute angle $APB$.

2. In an isosceles triangle $ABC, AB = AC,$ and $\angle A = 20^\circ.$ Point $P$ is taken on the side $AC$ such that $AP = BC$. Find $\angle PBC$.

3. $ABC$ is an isosceles triangle with $\angle ABC = \angle ACB = 80^\circ.$ $P$ is the point on $AB$ such that $\angle PCB = 70^\circ.$ $Q$ is the point on $AC$ such that $\angle QBC = 60^\circ.$ Find $\angle PQA$.

4. In triangle $ABC$, point $D$ is such that $\angle DCA = \angle DCB = \angle DBC = 10^\circ$ and $\angle DBA = 20^\circ$. Find the measure of $\angle CAD$.

5. In quadrilateral $ABCD$ with diagonals $BD$ and $AC, \angle ABD = 40^\circ, \angle CBD = 70^\circ, \angle CDB = 50^\circ, \angle ADB = 80^\circ$. Find the measure of $\angle CAD$.

6. Given triangle $ABC,$ with $AB = 15$ and $CA = 12$. The median from $B$, the altitude from $C$, and the angle bisector from $A$ are all concurrent inside the triangle. Compute the length of $BC$.

7. Three parallel lines $l1, l2$ and $l3$ are drawn through the vertex $A,B,$ and $C$ respectively of a square $ABCD$. If the distance between $l1$ and $l2$ is 7 and between $l2$ and $l3$ is 12, find the area of $ABCD$.

I'm going to have only 8 problems because #8 is very very tough problem.

8. Let $P$ be a point inside triangle $ABC$ with $AB = BC, \angle ABC = 80^\circ, \angle PAC = 40^\circ$ and $\angle ACP = 30^\circ$. Find the measure of $\angle BPC$.

[hide="8"] it is pretty easy to evaluate the angles except ABP, and CBP, let ABP=x, by trig ceva we have

$\sin 40*\sin (80-x)*\sin30=\sin20*\sin x*\sin10$
(using double angle, cos to sin, etc)
$\sin 70*\sin(80-x)=\sin x*\sin 10$
clearly $x=70$, this can be the only solution since the triangle is defined
that makes $\angle BPC=150$[/hide]

[hide="6"]
Let $D$ be the intersection of $\overline{BC}$ with the bisector of $\angle A$ and $H$ be the point on $\overline{AB}$ such that $CH\perp HA$. By the Angle Bisector Theorem, $\frac{BD}{15}=\frac{DC}{12}\Rightarrow \frac{BD}{DC}=\frac 54$.

Now use Ceva's Theorem. The two parts of $AC$ split by the median have ratio 1, so $\frac{AH}{HB}=\frac 45$. This gives $AH=\frac{20}{3}$ and $HB=\frac{25}{3}$. From the right triangles, we can now find $BC$:

\[ BC=\sqrt{12^2-(20/3)^2+(25/3)^2}=\sqrt{12^2+(45/3)(5/3)}=\sqrt{12^2+5^2}  \]

\[ =\boxed{13}  \]
[/hide]


Given that $ 140\%$ of $ (x\plus{}2)$ equals 5.6, find the value of $ x$.

I do these problems a weird way.
For this one, we do: 5.6/7 =0.8 ( I do this to make the "base" 20% )
Then, since it is 20% and we want 100%, We do 0.8X5=4 ( now it is in "base" 100%)
Now: x+2=4
x=2

your method might be a little unclear. (base $ 20%
$?)

we can set up an equation. $ (\frac{140}{100})(x + 2) = 5.6$. multiplying by $ \frac{100}{140}$ on both sides, we get $ x + 2 = (\frac{100}{140})(5.6)$. simplifying this, we get $ x + 2 = \frac{560}{140} = 4$ so $ x = 2$

Yes I am pretty bad at explaining things


HMMT geometry articles Harvard college MIT AMC

First off, many thanks to the listed coordinators:

Anna Chen, Eric Price, Angela Ying, Thomas Mildorf, Anders Kaseorg, Yi-Hsin Lin, Fan Zhang, Anders Kaseorg, Amali Vaz

as well as everyone else who helped make HMMT happen this year.

It was my first time going to the actual competition and I really enjoyed it.

I must say, it was a lot more difficult than I expected based on any of the previous tests (I took Alg and Calc), but some of the problems were very educative.

It was our first year sponsoring the competition and we certainly enjoyed it as well.  The organizers did a very good job managing such a large number of students on site and having relevant, engaging contests and side events.

Congratulations to all students and teams that won awards!

Congratulations to Anders for being listed twice in the above list! :D

I wish I could have gone to HMMT, but had orchestra stuff scheduled... :ninja:

yea, the contest was really good!

i am impressed with the quality of the problems

i have to say that i got owned on the geometry test...too many careless errors, also i was unable to solve the angle and random sides equal problem...

i wish that i had seen the factorization for that last algebra problem, i have posed similar problems, and the 2005 #10 was pretty much the same, i just did not see the simple factorization to finish it.......but 3rd...not bad

it was pretty awesome that the woot team won the competition; they did not necessarily have many winners, but they had depth...

[quote="Altheman"]yea, the contest was really good!

i am impressed with the quality of the problems

i have to say that i got owned on the geometry test...too many careless errors, also i was unable to solve the angle and random sides equal problem...
[/quote]

Same here, I also made many careless mistakes on the geometry round, even though I was able to solve every geometry problem. I have to re-read the article on avoiding careless mistakes since recently I have made a lot of careless errors on competitions.

Yeah HMMT was really fun. Heh definitely great problems, that's what you get with Harvard + MIT math students

Well I did general, that was pretty good, got 8 on the first half and 6 on the second half (which was really hard). I should have check #9 on the general, that might have made me even better, but it's nice I got 3rd place in general :D

Yeah I thought it was pretty cool that WOOT won too. I think they won the team round because of all that nice proof writing practice they're getting. Heh and I'm doing WOOT too (although I was with my school team...CTU LA...lol). It's interesting to think that some of the people in WOOT today could very well (although many of them aren't at the ultimate top yet) be the USAMO winners/IMO participants in a couple years.

It was also neat to see so many math people in one area...I was hoping to go over and talk to the WOOT team and meet various people I've only seen online, but unfortunately I didn't get to do that. It was pretty neat when I'm like "hey look over there there's K1807 or Altheman or MysticTerminator," when a lot of us live in different parts of the country (I think I saw you Altheman like sitting right in front of me 4 seats over during awards). I did get to see a couple people I haven't seen in a while and meet JSteinhardt (who's a friend of someone I know from school), so it was still pretty nice.

Who exactly was on your team? I know you, SplashD, k8107, who else?

I thought it was very well run, the problems were great, except alot were repeats (ABCD-DCBA=BCAD or whatever it was was a repeat, 2007*2006*...*1 was a repeat, I'm not sure if the function was the same, there were others). The only criticism I would have is that we should try doing it in a gym or something, so teams could do guts around a table instead of trying to talk down a row. And the auditorium was really crowded, even after separating A and B.

I thought it was pretty fun, for my first time.

Hah, it was a lot of fun, I n00bed geometry with a 25.
The questions were great, the guts round was terrific as I expected.
At first we thought the auditorium chairs would be a restraint in our progress on guts, but we did fine, communication worked out great :)

I really enjoyed the contest.
EDIT:
I really liked guts problems 10, 11, and 12 (that was the $A_{10}$, $A_{11}$, $A_{12}$ one).

[quote="13375P34K43V312"]Who exactly was on your team? I know you, SplashD, k8107, who else?
[/quote]

Following is the list:

chess64
SplashD
k81o7
pkerichang (LOL that's me and I came from California :rotfl: )
E^{\pi*i} = -1
n^4+4
Calculus
NeverOddOrEven

Oh yeah we don't know what to do with the 1st-place plate thing, so we decided to take picture of it and send the real one to Mr. Rusczyk instead (hopefully we get WOOT rebates :rotfl: )

[quote="The Zuton Force"]I must say, it was a lot more difficult than I expected based on any of the previous tests (I took Alg and Calc), but some of the problems were very educative.[/quote]

My thoughts on the difficulty - the last two on Calc were really hard, too hard in hindsight. Geometry was a little too easy, while Algebra was just right. For the heck of it, I'll post the original Geometry proposal tomorrow, which now seems as though it would have been about right.

You have picked up on one of this year's unannounced initiatives - making the problems educational. A number of the solutions are accompanied by remarks to that end, introducing theorems, ideas, or even tid bits of more sophisticated math. The A division team round brought in those number theoretic functions, and the ideas of  convolution (factor into those primes) and analysis (9 and 10 were inequalities in disguise!) In fact, it once had a middle section of quintessentially Topological problems, but we decided to scrap that plan in the interest of fairness and lightening the reading. The B division team round introduced tangential quadrilaterals, which for reasons unknown to me are generally neglected in contests compared to their cyclic counterparts. (USAMO 2004 lives on!) And I wanted to make sure A division heard about them too, hence their #12. Then Carmichael numbers made an appearance in guts 34 (won with the answer 14, btw)... etc. ad nauseum.

[quote="bpms"]I really liked guts problems 10, 11, and 12 (that was the $A_{10}$, $A_{11}$, $A_{12}$ one).[/quote]

I recall recoiling in fear whenever I saw codependent triplets in the guts rounds, so the aim there was to create the same reaction only to have it dissolve away amusingly.

There may have been another guts problem having unnecessary angle information, I forget whether I wound up including it, and it too was intended to have similar effect.

How about the limericks?

The limericks were annoying, but good. It really was like "what is this question asking?" a lot.

As for the guts round, someone on TJ A figured out that if you added up the maximum points for all the other problems and subtracted it from 400, you got 14, which was the answer ;). Apparently it's better to guess based on the system rather than the actual math.

Wait so did we get that right during the actual contest? I thought that we just randomly guessed 3.

Unfortunately, the Guts round did not have as much insane combinatorics this year as it did last year (actually, last year's Guts round played a nontrivial role in getting me interested in Combinatorics). It did seem to have a lot of Geometry, the result being that I actually had to think about Geometry during a team-based contest, which was semi-annoying. I guess that just tells me that I need to practice Geometry more. Fortunately, though, we had Haitao, who helped us own the Guts round.

The Power round was interesting...about a month ago I wanted to give a lecture on Mobius inversion at TJUSAMO, but I warned people that it would probably be entirely useless on contests. We took a vote and most people seemed to prefer that I talk about generators and techniques for proving that there are infinitely many numbers of a given form. However, I sort of refused to listen to them entirely and half the lecture was on multiplicative functions and Mobius inversion (with some of the excercises identical to problems on the Power round, like $\phi * 1 = n$). Then we open the power round, and I see $\phi$ and $\mu$ and am like "hey this is pretty cool." I think we made a slight mistake in having me and Brian do most of the number theoretical part of the contest while everyone else worked on the grabbag problems. The assumption was that since we already knew about the topic that the power round was on, it should be pretty easy, but while it was true that no individual part provided too much trouble, we still ended up running out of time at the end so didn't get to prove number 10. Still, I am pretty happy with our results on the Power Round, since I really didn't expect us to beat the WOOTlings.

The individual round was quite nice, though the Algebra round seemed HARD. I was surprised that so many people had high scores. My favorite problem from the whole test has to be the
\[\sum_{i+j+k=17}ijk \]
There is a very nice solution involving generating functions, if you haven't found it yet.

[hide="hint"]
$\binom{19}{5}= \binom{-6}{14}$
[/hide]

What exactly were the limericks asking?
I didn't get any of it...

Congratulations!  A job well done.

But hey, I am not surprised you all took first.  :-D

Best,
Dr. A

[quote="chess64"]Whoa, we were definitely not expecting to get 1st place. We were definitely hoping to place, but we were sure TJ was going to win, and getting first on the team round was also unexpected. I thought it would be 1. TJ, 2. Quagga 3. Wootlings or something. I guess it was just because instead of having just a few really good people (*cough* Quagga and TJ *cough*) we had 8 pretty good people.[/quote]

i thought this was pretty funny: [posted elsewhere]

Some quotes from the grading room:

-"Wait, is this written in Russian?"
-"Well, that's definitely Cyrillic"
-"Is there actually a rule stating the solutions need to be in English?"
-"Umm..."

-[solution reads]: "[random crap]...please give me some points...I'm dying here"

-[solution reads]: [random crap involving repeated use of the word 'owned']"
-Grader: "0. Owned."

[quote="bpms"]I really liked guts problems 10, 11, and 12 (that was the $A_{10}$, $A_{11}$, $A_{12}$ one).[/quote]

We guessed on 10, so we got an answer for both, and then we guessed for 11,  but we got all three of them right. That was really random. :?:

11 could be done without either of the others

[quote="Altheman"]

-[solution reads]: "[random crap]...please give me some points...I'm dying here"

-[solution reads]: [random crap involving repeated use of the word 'owned']"
-Grader: "0. Owned."[/quote]

LOL that's what i wrote...except i believe mine read

"PLEASE GIVE US POINTS OR WE'LL HUNT YOU DOWN"

[quote="JBL"]Heh, that conversation was half me.  I didn't know that paper ever made it to anyone who spoke Russian, though -- I did wander around trying to find someone for a few minutes, but unsuccessfully.   (Incidentally, there is no rule that requires solutions to be in English.  Obviously though, if no one in the grading room can read it, it won't get credit.)

Where is the [elsewhere] that it was [posted]?[/quote]

sukrit posted this on the gbs facebook group...

Results are up!
[url=http://web.mit.edu/hmmt/www/datafiles/results/2007.pdf]HMMT Results 2007[/url]

hmm I got the same score as 4th place but i'm officially 6th. Oh well thats pretty good for a sophomore first time at HMMT I think.

Is there any way to find how a team did on the Team A round if they didn't place top 10?

Does anyone know where the results are posted for everything? Like which team got  42nd on the power round, etc?

CRILLIC??? :rotfl:

2 things:

thing 1. on the calculus test, why is (-1, 0) considered a point of tangency of the circle and the curve thing? what kind of definition of tangency is this? this is a PRETTY WEIRD DEFINITION

yeah I was sort of pissed about that interpretation ... it broke my streak of 42s

of course then I took combo and my streak went to - but w/e

thing 2. dude what is with the solution to guts 31 can't we just say b_n = a_n - 1 and bam b_{n+1} = b_n^3 so we want to solve x^(3^2007) = x which pretty clearly has 3^2007 distinct solutions (0 and a bunch of roots of unity, to be precise)

[quote="MysticTerminator"]on the calculus test, why is (-1, 0) considered a point of tangency of the circle and the curve thing? what kind of definition of tangency is this? [/quote] It's the standard definition of tangency.  The curve is smooth at that point, and a circle of radius 5 is tangent to it there.

[quote]
My favorite problem from the whole test has to be the
\[\sum_{i+j+k=17}ijk \]
There is a very nice solution involving generating functions, if you haven't found it yet.

[hide="hint"]
$\binom{19}{5}= \binom{-6}{14}$
[/hide][/quote]

Hey, that's my problem :). I came up with it by the counting solution, but discovered the generating functions solution soon after.


calculus
integration
trigonometry
algebra
functional equation
calculus computations

prove that...

$ \int_0^{\frac{\pi}{2}}\sin^{2n}x \text{ } dx\equal{} \frac{2n\minus{}1}{2n} \cdot \frac{2n\minus{}3}{2n\minus{}2} .... \frac34 \cdot \frac12 \cdot \frac{\pi}{2}$

$ (n\in N)$

Define $ f(n)=\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta\,d\theta;\forall n\in\mathbb{N}$

So we get ${ f(n)=\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta\,d\theta=-\bigg[\cos\theta\sin^{n-1}\theta\bigg]_{0}^{\frac{\pi}{2}}+(n-1)\int_{0}^{\frac{\pi}{2}}\cos^{2}\theta\sin^{n-2}\,d\theta}$

Then write $ \cos^{2}\theta=1-\sin^{2}\theta$

The result is funtional equation $ f(n)=\frac{n-1}{n}f(n-2)$

The rest is simple :D

This integration by parts is usually called a "reduction formula".

Could someone show the intermediate steps one would take to get to that functional equation $ f(n) \equal{} \cfrac{n \minus{} 1}{n}f(n \minus{} 2)$?

Look at the third line. Apply it to the second, and simplify as far as possible. You should assume $ n\ge 2$ here, for the sake of the boundary terms.

ah oops I thought $ (1\minus{}\sin^2\theta)\sin^{n\minus{}2}\theta\equal{}1\minus{}\sin^n\theta$ :D

mk so with proper distribution, it would be

$ (n\minus{}1)\int_0^\frac{\pi}{2}\sin^{n\minus{}2}\theta\: d\theta\minus{}(n\minus{}1)\int_0^\frac{\pi}{2}\sin^n\theta\: d\theta$

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

..and I'm stuck there. :huh:

[quote="Timestopper_STG"]Define $ f(n) = \int_{0}^{\frac {\pi}{2}}\sin^{n}\theta\,d\theta;\forall n\in\mathbb{N}$

So we get ${ f(n) = \int_{0}^{\frac {\pi}{2}}\sin^{n}\theta\,d\theta = - \bigg[\cos\theta\sin^{n - 1}\theta\bigg]_{0}^{\frac {\pi}{2}} + (n - 1)\int_{0}^{\frac {\pi}{2}}\cos^{2}\theta\sin^{n - 2}\,d\theta}$

Then write $ \cos^{2}\theta = 1 - \sin^{2}\theta$

The result is funtional equation $ f(n) = \frac {n - 1}{n}f(n - 2)$

The rest is simple :D[/quote]

could you please tell more detail about $ f(n) = \frac {n - 1}{n}f(n - 2)$ and how did you get it?[img]http://l.yimg.com/us.yimg.com/i/mesg/emoticons7/106.gif[/img]
I may have the same first idea..
$ \sin^{2(n+1)}=(-\cos x)'\cdot\sin^{2n+1}x$

but,,anyone has another idea?

Did you write $ \cos^2 \theta \equal{} 1 \minus{} \sin^2 \theta$?  Did you evaluate the first term?

ohh okay so continuing from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1187081#1187081]this post[/url]

\begin{align*}
f(n)&=(n-1)(f(n-2)-f(n))\\
f(n)&=(n-1)f(n-2)-n\cdot{f(n)}+f(n)\\
n\cdot{f(n)}&=(n-1)f(n-2)\\
f(n)&=\cfrac{n-1}{n}f(n-2)
\end{align*}

That could be useful. :lol:


Putnam
function
integration
vector
calculus
derivative
analytic geometry

Let $ f: [0,1]^2\to\mathbb{R}$ be a continuous function on the closed unit square such that $ \frac{\partial f}{\partial x}$ and $ \frac{\partial f}{\partial y}$ exist and are continuous on the interior of $ (0,1)^2.$ Let $ a\equal{}\int_0^1f(0,y)\,dy,\ b\equal{}\int_0^1f(1,y)\,dy,\ c\equal{}\int_0^1f(x,0)\,dx$ and $ d\equal{}\int_0^1f(x,1)\,dx.$ Prove or disprove: There must be a point $ (x_0,y_0)$ in $ (0,1)^2$ such that

$ \frac{\partial f}{\partial x}(x_0,y_0)\equal{}b\minus{}a$ and $ \frac{\partial f}{\partial y}(x_0,y_0)\equal{}d\minus{}c.$

It's false. It's a failed attempt a a two-dimensional version of a mean value theorem, but it doesn't work. We'll get to the counterexample later.

I wondered a little about the fact that the hypotheses didn't seem strong enough to assure that $ \frac {\partial f}{\partial x}$ and $ \frac {\partial f}{\partial y}$ were integrable on the square, but that turns out to be an unneeded concern, as we can kill this off with very smooth counterexamples.

Assume $ f$ is as smooth as we need. Note that $ \int_0^1\int_0^1\frac {\partial f}{\partial x}(x,y)\,dx\,dy \equal{} b \minus{} a$ and $ \int_0^1\int_0^1\frac {\partial f}{\partial y}(x,y)\,dy\,dx \equal{} d \minus{} c.$ Using the notation $ \langle \alpha,\beta\rangle$ to mean the vector in $ \mathbb{R}^2$ with components $ \alpha$ and $ \beta,$ we write the hypotheses of the problem as
\[ \int_0^1\int_0^1\left\langle\frac {\partial f}{\partial x},\frac {\partial f}{\partial y}\right\rangle \minus{} \langle b \minus{} a,d \minus{} c\rangle \,dx\,dy \equal{} \langle 0,0\rangle.\]
We are asked to conclude that the integrand is the zero vector somewhere in the interior of the square. For a (continuous) function with codomain $ \mathbb{R},$ that would be a mean value theorem; for a function with codomain $ \mathbb{R}^2,$ it is false.

My first thought was to make the integrand $ \langle \cos(2\pi x),\sin(2\pi x)\rangle,$ but that isn't a gradient. But a little trial and error on the same basic theme led me to a counterexample:
\[ f(x,y) \equal{} y\sin(2\pi x);\ \nabla f \equal{} \langle 2\pi y\cos(2\pi x),\sin(2\pi x)\rangle.\]
Note that $ f$ satisfies the hypotheses with $ a \equal{} b \equal{} c \equal{} d \equal{} 0$ but $ \nabla f$ is never zero on the interior of the square.

[quote="Kent Merryfield"]
My first thought was to make the integrand $ \langle \cos(2\pi x),\sin(2\pi x)\rangle,$ but that isn't a gradient. But a little trial and error on the same basic theme led me to a counterexample:
\[ f(x,y) \equal{} y\sin(2\pi x);\ \nabla f \equal{} \langle y\cos(2\pi x),\sin(2\pi x)\rangle.\]
Note that $ f$ satisfies the hypotheses with $ a \equal{} b \equal{} c \equal{} d \equal{} 0$ but $ \nabla f$ is never zero on the interior of the square.[/quote]

There should be a factor $ 2\pi$ in $ \frac {\partial f}{\partial x}$ there

OK, I fixed that. Of course the constant is unimportant, but it should be there for accuracy.

I knew this problem was way more complicated than I thought, haha, the fact that it was A6 should have given that away to me.  Anyways, I knew it was false...but I made my proof based on that the assumption would only be satisfied at (1,1) which is not on the interior of the square so the partial derivatives are not defined at (1,1).  This is probably wrong but it was a last ditch effort in the final 30-45 minutes.

Of course, one could find functions for which the conclusion is true. If you try $ f(x)\equal{}\alpha x\plus{}\beta y,$ then the conclusion would be true at every point of the square. The issue, after you subtract off the appropriate linear function, is to make the zeros of one component of the gradient not intersect the zeros of the other component of the gradient.

Here's another counterexample.

Take $ f(x,y) \equal{} (x \minus{} \frac {1}{2})^2 y$. We get $ a \equal{} b \equal{} 1/6, c \equal{} 0, d\equal{} 1/12$.

Then $ \frac {df}{dx} (x_0, y_0) \equal{} 2 (x_0 \minus{} 1/2) y_0 \equal{} 0 \equal{} b \minus{} a$, so $ x_0 \equal{} 1/2$. ($ y_0 \equal{} 0$ is not in the interior of the square.)

But $ \frac {df}{dy} (x_0, y_0) \equal{} (x_0 \minus{} 1/2)^2 \equal{} 0$ when $ x_0 \equal{} 1/2$, when we would need it to be $ d \minus{} c \equal{} 1/12$. A contradiction.

Instead of defining the function f, I only defined the partial derivatives instead. Hope this also work.

For a point (x,y), let the partial derivative according to x be $ (x \plus{} y)^2$, according to y be $ (2 \minus{} x \minus{} y)^2$. I think if I also set f(0,0)=0, then the function f is already uniquely determined.

That should work as long as you checked that a function can have those partial derivatives, which I believe can be done by showing $ \frac {\partial }{\partial y} (x \plus{} y)^2 \equal{} \frac {\partial}{\partial x}(2 \minus{} x \minus{} y)^2$.

EDIT: And as jmerry showed that equality fails... yikes. :(

In general, the mixed partials $ f_{xy}$ and $ f_{yx}$ must be equal. What are they in your [greendtreeroad] example? They're $ 2(x\plus{}y)$ and $ \minus{}2(2\minus{}x\minus{}y)$. Not equal, so there is no function with those partial derivatives.

[quote="jmerry"]In general, the mixed partials $ f_{xy}$ and $ f_{yx}$ must be equal. What are they in your [greendtreeroad] example? They're $ 2(x \plus{} y)$ and $ \minus{} 2(2 \minus{} x \minus{} y)$. Not equal, so there is no function with those partial derivatives.[/quote]

Why? After defining partial derivatives according to x and y, can you just start at f(0,0) and compute(I know there are infinite points, but you get my meaning) all the rest points by just walking up or to the right?

That has to agree no matter what path you use. If the mixed partials don't match, it doesn't agree. We can't have $ f(1,1)$ depend on whether we got there through $ (1,0)$ or $ (0,1)$.

Because if this identity is not satisfied, you get a different answer if you go right and then up then you do if you go up and then right.

This should be dealt with in considerable detail in any multivariable or vector calculus course.

Here's an easy motivation: draw the 3d graph, and set a = b = c = d = 0 because they really don't matter.
Then, consider when $\frac{df}{dx}=0$: we can get vertical lines, or we can get horizontal lines.
Then, consider when $\frac{df}{dy}=0$: we can get vertical lines, or we can get horizontal lines.
If $\frac{df}{dx}$ has horizontal lines and $\frac{df}{dy}=0$ has vertical lines, or vice-versa, then we'll have intersections. So choose vertical lines for both. Picking f(x, y)=sin(2pi*x) sets $\frac{df}{dx}=0$ to vertical lines at x=1/4 and 3/4. We want $\frac{df}{dy}$ to have non-zero slope at x=1/4, 3/4. Multiplying by y gives constant nonzero slope appropriately, and since y is positive over the relevant area, the zero-ness of $\frac{df}{dx}$ is unchanged.


I just dont know why is physics so hard to me. i am good at math but i  just dont get the logic behind physics,  this semester i had to drop my physics class cause i just dont understand it. why is calculus is easy to me  but not physics.  is it really me  or  am i cursed or what ?

Oh,one more thing.If you want help from the physics forums here,check out the earlier pages of the "Advanced physics" forum,like about 13 pages back to now.We discussed lots of easy but tricky and important concepts back then.It was fun.Back then, there was just one physics forum.Also ask your doubts here,even the dumbest ones.

thanks for the answer , i will try to focus more in physics than my math classes.

Currently am a freshmen in college , and I am still undecided about  my future goals,  i am very interested in majoring in Applied math or Computer Science.
I would really appreciate some advice from you guys , please tell me whether CS or applied would be better for the future or which of those majors is the best one.  also what are the job prospects of those careers.

[quote="shadysaysurspammed"]Physics is $FUN$,not hard.[/quote]

No, $\text{Physics is PHUN}$ :P

[quote="mathcnt2005"][quote="shadysaysurspammed"]What grade are you in?

Physics is $FUN$,not hard.To get introduced to physdics,the most important things are a good text,an awesome teacher and $\text{the ability to think about things around you ON YOUR OWN}$.

Notrhing comes easy,you will have to be patient till you get used to think like you need to for physics.The more you think and ask,more you develop your concepts and the main thing is that you should be interested. :)

I think you shouldn't try to jump to higher concepts,first develop your basics.[/quote]

Actually, physics is easy because I was studying it since I was in first grade. I knew my SNF, WNF, Electomagnetic force and gravity in second grade, even though I never knew the math behind them. I'm in middle school right now.[/quote]

hmmm..if you study hard,you'll have mastered physics(not olympiad level) by 11th :)   try to learn more,not just from the book ;)   I got introduced to real physics in 9yh grade,so ur lucky to know so  much so fast.


Today evening, I was watching a Conservative Convention on C-Span, and what the guy was saying was completely unreasonable and somewhat dimwitted. He was preaching Bush, and attacking Kerry, which can only be expected from a conservative champion, but it was ludicrous - though very hilarious. The man's name was Rich Lowry, and he is an editor for a Conservative newspaper. He started talking against drug prevention, and was stating that Marijuana was harmless. He said smoking Marijuana wasn't addictive, and people only did it occasionally. He also said that it would be reasonable to attack Iraq, even if evidence pointed out that weapons of mass destruction didn't exist. He claimed the US was cleansing Iraq, and Iraq was the only Middle Eastern country that promoted the ideals of Democracy. He claimed going into other troubled countries and bombing their cities and mountainsides were wrong, but bombing Iraq(Even civilian territory) was the ideal thing to do. He said Somalia was the right thing at the moment, even though we know it has come to be a regreted event in history.

I don't understand people like him, who can't see the light when it is coming at 300,000 miles per second right into his face. He is so thick headed - and many of his articles prove what he is worth - here's the site -[url]http://www.townhall.com/columnists/richlowry/[/url]. This is so much BS, no offense.

heh funny :)

makes me think of the pope in some way :?

Actually, a lot of conservatives are rather upset by many of Bush's policies (i.e. immigration, budget deficits).

What I really, don't understand, though, is the right's position on the Iraq issue.  There is now significant evidence that there never was much of a case for war based on existing weapons of mass destruction.  Perhaps there was a distant threat.  But how is it worth it to pull forces--who are fighting people that have already attacked us--out of Afghanistan for this?  How does that make us safer?

I know this is late news, but wasn't the President supossed to have planned that war in Iraq At the very beginning of his Presidency?  That kind of makes me feel like he's only theree for oil, or to finish off his dad's job, or something stupid like that.

[quote="farmertim"][quote="white_horse_king88"]I don't understand people like him, who can't see the light when it is coming at 300,000 [b]miles [/b]per second right into his face.[/quote]

(emphasis mine)

kilometers.  although i have to admit there is something inherently amusing about the notion of light traveling faster than the speed of light..[/quote]
And the same to you, farmertim.
c = 299,792.458 km / s < 300,000 km / s  :maybe:

how ironic that in pointing out my 'mistake' Matt did essentially the same thing i did.  i suppose we could go back and forth holding each other to arbitrary higher standards of precision, but that would become tedious and take us far afield from the topic at hand.  instead i propose a truce.  perhaps the question of whether c is transcendental would best be reserved for a separate thread devoted to the subject..

in the interest of bringing this back to the intended topic, i agree with Matt's cautioning of everyone not to confuse conservative policies with those of the Republican Party.  of course, that brings up the question of just what the philosophical base of the Republican Party is in that case..

[quote="confuted"]To all posters: Please do not make the mistake of confusing conservatism with the policies of this administration or the Republican party as a whole. They are not equivalent.[/quote]

modulo election, they are equivalent. Do you really think there's any conservative dude gonna vote for Kerry? :D
No, bush clearly wants to get the votes of those conservatives, and he adjusts his behaviour to attract them. That is, by being more conservative.
Luckily those people are only a minority of the population... :maybe:

[quote="Peter VDD"]modulo election, they are equivalent. Do you really think there's any conservative dude gonna vote for Kerry? :D
No, bush clearly wants to get the votes of those conservatives, and he adjusts his behaviour to attract them. That is, by being more conservative.
Luckily those people are only a minority of the population... :maybe:[/quote]
There will be conservatives who vote for Kerry, as there will be liberals who vote for Bush. There will also be middle-of-the-roaders voting for both. But there will also be liberals voting for Nader and conservatives voting for Peroutka, and those are perhaps the liberals and conservatives who are truest to their beliefs.

[quote="white_horse_king88"]He was preaching Bush, and attacking Kerry, which can only be expected from a conservative champion, but it was ludicrous - though very hilarious...[/quote]

Well, apparently our dude is gonna vote for bush :D

[quote="farmertim"]how ironic that in pointing out my 'mistake' Matt did essentially the same thing i did.  i suppose we could go back and forth holding each other to arbitrary higher standards of precision, but that would become tedious and take us far afield from the topic at hand.  instead i propose a truce.  perhaps the question of whether c is transcendental would best be reserved for a separate thread devoted to the subject...[/quote]

c is about as far from trascendental as a number can get.  A meter is defined as the distance light travels in a vacuum in exactly 1/299,792,458 seconds, so c = 299 792 458 m/s exactly.


calculus
Asymptote
geometry
trigonometry
function
limit
calculus computations

So i am doing trigonometric limits in my calculus book and they have this weird figure which i can't draw in asymptote because i don't know how to draw sectors :lol: . So i will give you a few characteristics about the figure and then maybe someone other than me can draw it in asymptote. So $ \alpha$ (an angle) is smaller than $ \frac {\pi}{2}$. $ \alpha$ is an angle that is in both triangles $ OPQ$ and $ OP'Q$. Triangles $ OPQ$ and $ OP'Q$ are similar. $ OP'Q$ is right and is  the larger one of the two. $ OP$ and $ OQ$ are the radii of a sector whose central angle is $ \alpha$  $ OQ \equal{} 1$. the area of triangle $ OPQ$ is $ (\frac {1}{2})(\sin {\alpha})$. The area of triangle $ OP'Q$ is $ (\frac {1}{2})(\frac {\sin {\alpha}}{\cos {\alpha}})$ the area of the sector is $ (\frac {1}{2})(\alpha)$. So from the figure you have $ \sin {\alpha} < \alpha < \frac {\sin {\alpha}}{\cos {\alpha}}$ which simplifies to $ 1 > \frac {\sin{\alpha}}{\alpha} < \cos {\alpha}$  but when i try it for $ \alpha \equal{} 0$ it doesn't make sense cause undefined ain't equal to one. why the heck did that happen?

EDIT : Never mind it approaches one doesn't actually reach one, whoops  :D  :D  :D like like emoticons

Make sure that you understand that one of the most important points about limits is that a limit is independent of how a function is defined at that actual value.

That is,
$ \lim_{x\rightarrow a}f(x)\neq f(a)$
in all cases.  If
$ \lim_{x\rightarrow a}f(x)\equal{}f(a)$
then we have continuity.


How weird are you and why? =P

Any examples of weird querky things you like to do that are highly amusing? =D

I have a whole bookshelf dedicated to mathematics... does that count if I am not eccentric in any other way???


function
geometry
geometric transformation
reflection
symmetry
calculus
integration

Let $\zeta(s)$ be the Riemann zeta function show that the non-trivial zeros has real part equal to $\frac{1}{2^{n}}$ for some integer $n$.

Isn't this equivalent to the Riemann hypothesis? By the functional equation, we can reflect zeros across the critical line, and only one in each such pair can satisfy that condition.

You mean, "prove the Riemann Hypothesis :what?:"? If what you say is true, then the only possible value for $n$ is $1$ because of the symmetry of non-trivial zeroes around the line $\text{Re\,}z=\frac{1}{2}$. If you can show us how it can be done, we'll all applaud but if you cannot do it, this will be considered one of the most useless posts on the site: everybody knows the problem, no one can solve it, so why bother posting? :?

$ \zeta(s) = \sum^{\infty}_{k=1}{k^{-s}}$
Let $ s=x+i\cdot y$, where $ x, y \in R$
So: $ \sum^{\infty}_{k=1}{k^{-x-i\cdot y}}=0$
$ \sum^{\infty}_{k=1}{k^{-x-i\cdot y}}=\sum^{\infty}_{k=1}{e^{-x\cdot ln{k}-i\cdot y \cdot ln{k}}}= \sum^{\infty}_{k=1}{e^{-x \cdot ln{k}}\cdot (cos{y\cdot ln{k}}-i \cdot sin{y \cdot ln{k}})}=0$
So: $ \sum^{\infty}_{k=1}{e^{-x \cdot ln{k}}\cdot cos{y \cdot ln{k}}= 0}$ and $ \sum^{\infty}_{k=1}{e^{-x \cdot ln{k}}\cdot sin{y\cdot ln{k}}= 0}$
We should find such $ x$ and $ y$. I don`t know how to prove that $ x=\frac{1}{2}$ :(

I am sorry, I wasnt clear on my question. My question was if $ \frac{1}{2^{n}}$ is a zero then show that $ n=1$. But it got answered already.

[quote="Yuriy Solovyov"]$ \zeta(s) = \sum^{\infty}_{k=1}{k^{-s}}$
[/quote]

Hello Yuriy,

The expression you gave is valid only when $ Re[s]>1$.  Try and view Zeta as the complex-contour integral function Riemann originally designed:

$ \zeta[s]=\Gamma[1-s]\frac{1}{2\pi i}\int_{\multimap}\frac{z^{s-1}}{e^{-z}-1}dz\overset{Re[s]>1}{\longrightarrow}\sum_{k=1}^{\infty}\frac{1}{k^{s}}$

with that contour being the Hankel contour.

So what's causing the zeros?  Surely it must be the integrand.  To me, it resembles a neuron with a wave-train along an axon (where?), the integrand being the cell body, and it's output, the value of the zeta function.  When the wave-train reaches a particular . . . frequency,  that of $ s=1/2+iy$, the real and imaginary components of the output synchronize giving rise to the (non-trivial) zeros.  So . . . one possible approach to looking at the zeta function in a novel way would be to look . . . where?

Here's a snap-shot of the wave-train on what I think is the Zeta axon.  More than likely it's nothing but still fun.  I included the Matheamtica code to anamate the wave in case anyone is interested.  Just execute the code. Mathematica generates 100 frames, then double-click on one of them to start the animation.  You can tell where it is by the commands I supplied.

[code]a = 1/2;
Table[Plot3D[Im[Exp[(s - 1)*Log[z]]/(Exp[-z] - 1) /.
     {z -> x + I*y, s -> a + b*I}], {x, -5, 5}, {y, -100, 0},
   PlotPoints -> 50, ViewPoint -> {3.0106779769522243,
     -0.8077010498277987, 1.3166005974485675},
   ViewVertical -> {-0.29555163663839273, 0.34835778598040135,
     2.223870616497281}, ClipFill -> None], {b, 1, 10, 0.1}]
[/code]

[img]http://img409.imageshack.us/img409/8121/zetawavetrainuw9.jpg[/img]


inequalities
function
three variable inequality

[b]PRO[/b]: let be given real numbers $a,b,c$ such that: $a+b+c\ne 0$.prove the inequality:
$1)$: $\sum{\frac{(b+c-a)^2}{a^2+(b+c)^2}}\ge{\frac{3}{5}}$
$2)$: $\sum{\frac{(2a+b+c)^2}{2a^2+(b+c)^2}\le8}$

I think you mean
 $\sum{\frac{(b+c-a)^2}{a^2+(b+c)^2}}\ge{\frac{3}{5}}$
Am I right?

i've just corrected it
now, solve pro.okie?

dear nguyenquockhach:

can you tell me the source of this ieq?? :)
it looks beautiful :D
and here my solution: :D

$\sum\frac{1}{\frac{1}{(1-a/(b+c))^2}+\frac{1}{((b+c)/a-1)^2}}\geq3/5$

so let $x=\frac{b+c}{a}$

so      $1/x=\frac{a}{b+c}$

and then jensen forever :D  :D  :D  :D  :D  :D  :D  :D

Pro 1) : Japanese 97 . It's a famous ineq and was posted on ML before .
Pro 2) : USAMO 03
 I know 5 solutions for this problem .
 We can suppose $ a+b+c = 3 $
 It's easy to show :
 $ f(x) = \frac{(x+3)^2}{2x^2+(3-x)^2} \leq \frac{4x+4}{3} $

first one
Let $s=a+b+c$
then
$f(x)={(s-2x)^2 \over x^2+(s-x)^2}$
we have to proof that $f''(x)\geq 0$
then from JENSEN
$L\geq 3*f({s \over 3})=3*{1 \over 5}$
second one is the same

:D  :D  :D  :D thanks for all solutionssss :D  :D  :D  :D

but i must be remember everyone:the condition is $a+b+c\ne{0}$, [b]but not a,b,c possitive [/b]

okie---------
for [b]Pro 1[/b]: iff $a,b,c$ possitive, we can assume: $a+b+c=3$
and use: $\frac{1}{2x^2-6x+9}\le{\frac{2x+3}{25}}$.

I don't understand,
what about if $a+b+c<0$ then we must assume that
$a+b+c=-3$

[quote="cuong"]my method always work althought a+b+c<>0 :D

to nguyenquockhanh:your solution looks like siuhochung way :D[/quote]

yes, that true iff we have a,b,c possitive

@tequila: are you sure: $f''\ge0$

[quote="nguyenquockhanh"]Let $ a,b,c$ be [b]real[/b] numbers  such that $ a + b + c\ne 0$. Prove that

$ \frac {(2a + b + c)^2}{2a^2 + (b + c)^2} + \frac {(2b + c + a)^2}{2b^2 + (c + a)^2} + \frac {(2c + a + b)^2}{2c^2 + (a + b)^2}\le8$.[/quote]
$ \left(2a^2 + 3b^2 + 3c^2 - 4bc + 2ca + 2ab\right)\left[2a^2 + (b + c)^2\right] - (2a + b + c)^2\left(a^2 + b^2 + c^2\right)$

$ = (b - c)^2\left[3a^2 - 2a(b + c) + 2(b + c)^2\right]\geq0\Longrightarrow$

$ \sum{\frac {(2a + b + c)^2}{2a^2 + (b + c)^2}\leq\sum\frac {2a^2 + 3b^2 + 3c^2 - 4bc + 2ca + 2ab}{a^2 + b^2 + c^2} = 8}$.

USAMO 2003/5 see: http://www.mathlinks.ro/Forum/viewtopic.php?t=48989

By the way, the Vasc inequality

$ \frac {(qa + b + c)^2}{2a^2 + (b + c)^2} + \frac {(qb + c + a)^2}{2b^2 + (c + a)^2} + \frac {(qc + a + b)^2}{2c^2 + (a + b)^2} \le \frac {(q + 2)^2}{2}$,

holds if and only if $ 0 \leq q \leq 4$; with equality if $ b = c = - 2a,q = 4$.

The inequality inversely holds if and only if $ q\leq - \frac {1}{2}$; with equality if $ b = c = - 2a,q = - \frac {1}{2}$.

For [b]nonnegative[/b] numbers, the inequality if and only if $ 0 \leq q \leq 2 + \sqrt 6$; with equality if $ a = 0,b = c,q = 2 + \sqrt 6$.

The inequality inversely holds if and only if $ q\leq 2 - \sqrt 6$; with equality if $ a = 0,b = c,q = 2 - \sqrt 6$.

see: http://www.mathlinks.ro/Forum/viewtopic.php?t=65711

[quote="nguyenquockhanh"]Let be given [b]real [/b]numbers $ a,b,c$ such that $ a \plus{} b \plus{} c\ne 0$. Prove the inequality

$ \frac {(b \plus{} c \minus{} a)^2}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {(c \plus{} a \minus{} b)^2}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {(a \plus{} b \minus{} c)^2}{c^2 \plus{} (a \plus{} b)^2}\geq \frac {3}{5}$.[/quote]$ 25\left(a^2 \plus{} b^2 \plus{} c^2\right)(b \plus{} c \minus{} a)^2 \minus{} \left[a^2 \plus{} (b \plus{} c)^2\right]\left(7a^2 \plus{} 4b^2 \plus{} 4c^2 \plus{} 40bc \minus{} 20ca \minus{} 20ab\right)$

$ \equal{} \frac {(b \plus{} c \minus{} 2a)^2(b \plus{} c \minus{} 3a)^2}{2} \plus{} \frac {(b \minus{} c)^2}{2}\left[41a^2 \minus{} 50a(b \plus{} c) \plus{} 41(b \plus{} c)^2\right]$

$ \geq 0\Longrightarrow$

$ \sum\frac {(b \plus{} c \minus{} a)^2}{a^2 \plus{} (b \plus{} c)^2}\geq\sum\frac {7a^2 \plus{} 4b^2 \plus{} 4c^2 \plus{} 40bc \minus{} 20ca \minus{} 20ab}{25\left(a^2 \plus{} b^2 \plus{} c^2\right)} \equal{} \frac {3}{5}$.

Japan MO 1997, problem #2 see: http://www.mathlinks.ro/Forum/viewtopic.php?t=146

OLYMPIC 30 /4 of Vietnam see: http://www.mathlinks.ro/Forum/viewtopic.php?t=85178

[quote="nguyenquockhanh"]Let be given [b]real [/b]numbers $ a,b,c$ such that $ a \plus{} b \plus{} c\ne 0$. Prove the inequality

$ \frac {(b \plus{} c \minus{} a)^2}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {(c \plus{} a \minus{} b)^2}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {(a \plus{} b \minus{} c)^2}{c^2 \plus{} (a \plus{} b)^2}\geq \frac {3}{5}$.[/quote]Using the identity $ a^2 \plus{} (b \plus{} c)^2 \equal{} \frac {(a \plus{} b \plus{} c)^2 \plus{} (b \plus{} c \minus{} a)^2}{2}$ etc, the inequality becomes

$ \sum{\frac {(b \plus{} c \minus{} a)^2}{(a \plus{} b \plus{} c)^2 \plus{} (b \plus{} c \minus{} a)^2}}\geq \frac {3}{10}$,

i.e.

$ \sum{\frac {(a \plus{} b \plus{} c )^2}{(a \plus{} b \plus{} c)^2 \plus{} (b \plus{} c \minus{} a)^2}}\leq \frac {27}{10}$.

Let $ b \plus{} c \minus{} a \equal{} x,c \plus{} a \minus{} b \equal{} y,a \plus{} b \minus{} c \equal{} z$, the inequality equivalent to the following nice form :

If $ x,y,z$ be [b]real [/b]numbers such that $ x \plus{} y \plus{} z \equal{} 1$, then

$ \frac {1}{1 \plus{} x^2} \plus{} \frac {1}{1 \plus{} y^2} \plus{} \frac {1}{1 \plus{} z^2}\leq \frac {27}{10}$,

with equality if and only if $ x \equal{} y \equal{} z \equal{} \frac {1}{3}$.

For [b]nonnegative[/b] numbers $ x,y,z$, the reverse inequality (due to Ting-Yan Tian)

$ \frac {1}{1 \plus{} x^2} \plus{} \frac {1}{1 \plus{} y^2} \plus{} \frac {1}{1 \plus{} z^2}\geq \frac {5}{2}$

holds for $ x \plus{} y \plus{} z \equal{} 1$.

[hide="Proof."]$ \frac {(x \plus{} y \plus{} z)^2}{(x \plus{} y \plus{} z)^2 \plus{} x^2} \minus{} \frac {x \plus{} 2y \plus{} 2z}{2(x \plus{} y \plus{} z)} \equal{} \frac {x(y \plus{} z)}{2(x \plus{} y \plus{} z)\left[(x \plus{} y \plus{} z)^2 \plus{} x^2\right]}\geq 0\Longrightarrow$

$ \sum{\frac {1}{1 \plus{} x^2}} \equal{} \sum{\frac {(x \plus{} y \plus{} z)^2}{(x \plus{} y \plus{} z)^2 \plus{} x^2}}\geq\sum{\frac {x \plus{} 2y \plus{} 2z}{2(x \plus{} y \plus{} z)}} \equal{} \frac {5}{2}.$[/hide]

By the way, if $ x,y,z$ be [b]real [/b]numbers such that $ x \plus{} y \plus{} z \equal{} 1$, then

$ \frac {1}{k \plus{} x^2} \plus{} \frac {1}{k \plus{} y^2} \plus{} \frac {1}{k \plus{} z^2}\leq \frac {27}{9k \plus{} 1}$

holds if and only if $ k\geq \frac {7}{18}$, with equality if $ k \equal{} \frac {7}{18},x \equal{} \frac {2}{3}, y \equal{} z \equal{} \frac {1}{6}$;

if $ x,y,z$ be [b]nonnegative[/b] numbers such that $ x \plus{} y \plus{} z \equal{} 1$, then

$ \frac {1}{k \plus{} x^2} \plus{} \frac {1}{k \plus{} y^2} \plus{} \frac {1}{k \plus{} z^2}\geq \frac {1}{k \plus{} 1} \plus{} \frac {2}{k}$

holds if and only if $ k\geq \frac {1}{2}$, with equality if $ k \equal{} \frac {1}{2},x \equal{} 0, y \equal{} z \equal{} \frac {1}{2}$.

陈计老师：您的配方是怎么想出来的？怎么这么巧？

[quote="Ji Chen"]

By the way, if $ x,y,z$ be [b]real [/b]numbers such that $ x \plus{} y \plus{} z \equal{} 1$, then

$ \frac {1}{k \plus{} x^2} \plus{} \frac {1}{k \plus{} y^2} \plus{} \frac {1}{k \plus{} z^2}\leq \frac {27}{9k \plus{} 1}$

holds if and only if $ k\geq \frac {7}{18}$, with equality if $ k \equal{} \frac {7}{18},x \equal{} \frac {2}{3}, y \equal{} z \equal{} \frac {1}{6}$;
[/quote]
The Vasc's Left Concave Function Theorem kills it.
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=238857


Putnam MIT college Stanford CMU Harvard Princeton

On my part, I never heard of the IMC until after I graduated (from Berkeley).

yeah, it´s what i thought about. At least over here even the national contest is not so well-known between the undergraduated students. I am not so sure, but i think that the national olympiad for undergrad. in brazil  is like a selection to the imc, but i think that this selection is much more about spnsering then the IMC for itself since i heard that there is almost no  restriction in participating in IMC.  Generally the top guys in the NO are in the IMC and the brazilian team is very big, i think that this year the team was formed by 10 or 12 students from many different universities but all of them were students from the most traditional universities over here, what is not any injustice, since these universities usually  have the best results in the NO.

NO=National olympiad!

[quote="joml88"]Very few people from the US go.  It's in Europe so it costs a lot to travel there.  I don't know of any schools here that form teams and help with the cost of going.  Additionally, it doesn't seem to have enough prestige, in the U.S. at least, to make many people want to undergo that cost.[/quote]

Joml's right. The schools here don't/.

[quote="joml88"]I don't know of any schools here that form teams and help with the cost of going.  Additionally, it doesn't seem to have enough prestige, in the U.S. at least, to make many people want to undergo that cost.[/quote]

I think that way of thinking is partially faulted as quite some colleges put time and effort in European recruiting efforts. Most of the top people who have a potential interest to go to US mostly only consider the usual suspects MIT, Stanford etc. Thus it might the chance for other top schools to stand out, e.g. UC Berkeley, CalTech, CMU, Michigan, Pennsylvania, Brown, Cornell.
But also for MIT, Harvard etc. who have more than sufficient financial resources it is a chance for them get recognition by Europeans who cannot decide between the leading ones, e.g. look at this thread which discusses the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=126254]strongest math programs in US.[/url] You know the professors are really good, thus you attend that college that offers the best and strongest peer experience. And this can be seen in the IMC ranking though it does not show the median level but rather the strongest students at that school which is important in the end. So far I only know of efforts by Princeton's students to go there but which were actually Romanians who used it as chance to visit relatives and friends in their home country.

The 14th International Mathematics Competition for University Students [url=http://www.imc-math.org/](IMC)[/url] takes place in Blagoevgrad, Bulgaria from 3rd until 9th August 2007.

If another countries closer to the Atlantic host the competition it would be much more well know in the USA and west Europe, what seems to be quite impossible since that if you see the competition results without much attemption, you would say that it is a East-European math competition!

PS: It´s nice to see the weird type of letters these east-european countries have!

I meant that individuals won't pay their own way to go there because it's not that important for them.

This trip would probably cost at least $\$$1000.  That seems like an unnecessary luxury for schools to pay for, especially when they could be funding an internship, foreign-exchange, etc with it.

It would be nice for schools to be able to send students to the IMC and it would reflect well upon that school.  But it's just not their foremost priority.

[quote="Brambila"]Princeton has a moral obligation of forming a team to IMC since they are one of the sponsors of IMC...[/quote]They won't be sponsoring it if they didn't see it as a useful competition. ;)

As for prestige, it neither has prestige here and still I went there on own costs. It's the best holidays I've had so far, and I have nice medal on my curriculum. Why would I not go?

But indeed, US has many competitions that we don't have. Perhaps that is why it fills here a gap in the math system which is nonexistent in the US.

If some one get a Medal in IMO he dont need to go to army (for 2 years!!!)
It was a rule but it is not now.
IMC has no advantages for students here, Government pay 3 university each year to send a team. Else they would not send any.
It is not worth...I think American students have rational behaviour. It is not IMC it is EMC (european mathematical competition!!!)
Princeton sends team becasue of a PhD student who is from Romania! (he persuades the university to send a team)

I think I could not say what I mean!
about advantages:
Here every body who get GOLD medal in National Olympiad can enrol to university with out taking intrace exam (entrace exam is Hard so it is worth to take national olympiad)
About IMC I have 2 first prizes in IMC but the university dont accept me for PhD or master program. I should take a stupid exam (entrace exam)
About American Students , I made a mistake...Fee is not a big money for them as it is here :wink:
About our government they send teams only to prove for people that Iranian students can get prizes in International contests! (although Iran is not a rich country) It makes people proud! Here Government Is trying to say that IRANIAN ARE THE MOST INTELIGENT PEOPLE :rotfl: ...I think It is a stupid reason
Passing the university exam (in which they choose team) is harder than  recieving a first prize!!!! becasue it is free and students can visit a new country, so all the students who took part in IMO take part in IMC... I am going to take part agian :wink:

[quote="Brambila"]i think that you could find some argues to show that IMC it´s not worthfull for the governement( but none of them convincent enough). But saying that the STUDENTS have no advantage in going to a new country ( almost nowone knows countries like romania, or bulgary), know new people from different countries and different way of thinking and also getting in touch with a nice mathematics and mathematicians or future mathematicians...
Remember also that you are not just 18 when you go to a competition like IMC.
I think you had said that in a hurry, about the students, but man, if you really thinks that, you should get out a little bit of your bedroom.[/quote]

yeah, this stuff about proving that one is smarter then the other is really silly, but is nice to know that you have more challenging stuuf  in your own country  Over here, master exam is very idiot and i think that the doctoral exam is also easy, to you have an idea i am an average stuent in math, if i take IMC today, i think i would get about 40 or 60, but the more reasonable result is near 40. I started to get seriously in math competitions about 2 or 3 months ago, since that my first work on it was interrupted by final exams. I started with the high school stuff, so, i started to dedicate to stuff about college just some little time ago, and for the entrance master degree over here, you have to do a 2 months course in Analysis and Linear Algebra, and they select the students with best grades between them, but they look at your curriculum before it.Man,  i took some classes of the Analysis course and me, that just ended the first year of college in the end of the last year, saw that this course was really a low level course, that wasn´t a serious analysis course man, i think that the course would be nice to guys from high school, but man, for people graduated in math that was ridiculous and believe, between the graduated guys, the most of them were suffering a lot with this idiot course. There are some universities more serious, but not THAT serious.In resume, the most challenging things that we have over here in my university at least ( which is one of the most "serious" of the country") is national olympiads for undergraduated students, IMC and an exam that you can take in the 3 year to study for 2 or 3 years in the french "Ecole's" ( what is fuc.. nice!) but i dont know that this exam is just for ecole polytechnique or all ecoles in general, but to you have an idea, the guys who goes there are considered very very good here, but after 3 years of university in brazil, they are send as freshmans in these Ecoles....
The governement wanna put everyone in university no matter what, even if  to do it, you have to fuc.. up all the undergraduation, master´s degree and doctoral degree system. Brazil is too stupid man, i wanna get out of here as soon as possible eheheheh.
PS: sorry if you didn´t like my complanings, but i just saw a oportunity to say all this stuff eheheh!

[quote="Brambila"]Princeton has a moral obligation of forming a team to IMC since they are one of the sponsors of IMC...[/quote]

Actually, that's not true. The princeton press isn't affiliated with Princeton university.

I've been running the Princeton Math team this year, and I can say that at first Princeton's participation in the IMC was started because of the Romanians and other Europeans here who are home during the summers and take it for fun. However, we just recently started competing (2004), and we did send a girl from Korea that year.

Additionally, there wasn't much incentive to attend before because the Putnam is the standard all the American universities use to compare undergraduate mathematics. Instead, we've used the IMC as more of a fun experience to compete, especially since we don't have enough resources to send our 'best' team possible. Also, most students are either interning or doing REU's at places away from home, and these locations aren't finalized until May or June right before IMC. Furthermore, transportation to Eastern European countries drives some students away due to safety concerns.

Overall, though, I've been trying to increase Princeton participation in the IMC. This past year, we had students from Singapore, Romania, the US, and Bulgaria, although some were in London, Moscow, and Austria at the time. The math department has the funds to let us compete, but there are results expected too, so we want to choose our competitions and participants wisely so that funds continue to trickle in.

[quote="Brambila"]I have a big doubt in my mind. Considering tha usually USA is always between the top 5 countries in the IMO, that they import great talents from from europe in general to study in their universities, and having an annual hard national competition for undergraduated studies ( Putnam), why do they not acchieve great results in IMC? For example, Brazil usually is top 25 top 30 in IMO, but i think that if you do an average of the brazilian results in IMC, they would be better ranked than USA!
Is the IMC's style? If is, is the IMC's style so different from the Putnam? I really can´t see big differences between them.[/quote]

Italian people never participate fo IMC (and the travel cost isn't a problem here)... I suppose I'll be the first one (in 2008)...


AMC
USA(J)MO
USAMO
AIME
AMC 10
AMC 12
AMC 8

i know this may sound dumb, but what are USAMO or AIME OR some other stuff that usually starts with an A, which one do u take first? and how do u prepare for it? And what stuff do u need to no for it?

and what do they test u in each test?

don't laugh :rotfl:

hmm you are laughing

but...

first you take AMC (10 or 12)

if you get a 120 or top 1% on AMC10, or a 100 or top 5% on AMC12, you qualify for AIME

then, to qualify for USAMO, you calculate 10*AIME + AMC to get your index, and if it's high enough, you go too MOP (which is, by the way, pretty tricky)

or, you could qualify for USAMO and fail, like david bay rush

back to the point: you prepare with old problems (which you can order from AMC (for everything) or find on http://kalva.demon.co.uk (for AIME, USAMO)) and books (AoPS, Art and Craft of Problem Solving by Paul Zeitz, Problem Solving Strategies by Engel, and other books)

orrrrrrr

you could just have a 1337 math teacher like zuming - and then you'll be pretty good

but DAVID BAY RUSH is spamming - i don't think zuming taught him that

All the contests you named are part of a series of tests, under the AMC umbrella.

The AMC 8 is for people in grades 8 or lower.
The AMC 10 is for people in gredes 10 or lower.
The AMC 12 is for people in grades 12 or lower.

If you do well on those tests, you can move on to the AIME.

Next is the USAMO.

Then the MOSP (a camp).

And then the olympiads.

[quote="not_trig"]
if you get a 120 or top 1% on AMC10, or a 100 or top 5% on AMC12, you qualify for AIME
[/quote]

So it's different every year.

Perfect scores, of course, automatically make it.

[quote="Karth"][quote="dingzhou"]how many 8th grader or below qualified for aime?[/quote]

There are many. In fact there are even a few middle schoolers across the nation (probably lesser than 50) who have qualified for the USAMO. There was even a 4th grader in Ontario who qualified for the USAMO (though this wouldn't really be in the USA).[/quote]

There were like 40 this year because the AMC 10 Index cutoff was ridiculously low and the AIME was ridiculously middle school favored.
eh I probably didn't deserve to make it with the scores I got but oh well I did almost ok on the USAMO so sure

Man, this is what FAQ's are for: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11693

wow.. i'm going into eighth grade, and the thing that sux is that our school doesnt make us take AMC10, so i can't do anything about it  :stretcher: , but i DO take this one course that helps prepare for AIME, and my AMC10 scores are ridiculous... (91.5 for first time!!!) :surrender:

and there is NO way that i'm going to qualify for USAMO..
i mean, i'm not even shooting for nationals in MATHCOUNTS.....
soooooooo yeah...
i'm not THAT good at math.. (at least compared 2 u guys..)

Don't degrade yourself :)

If your school doesn't have the AMC 10, check with your local high school or perhaps university. I was the only kid who took the AMCs at my school (I was an eighth grader in 06-07), so my high school sent me the contests.

There is NO way that you're going to qualify? Oh come on. You have a long time to work, and a 91.5 AMC 10 score isn't ridiculous at all.

Good luck :)

thanks..
you make me feel better..
i'm probably going 2 contact the teachers @ the high school in our district..
i hope to qualify for AIME.....


I've been working on some distribution problems lately. At times, you would place out 9 items, and choose 2 of the 8 of the gaps for divider spaces. Other times, you would put down 2 more items, and choose two of them. What difference does it make?

The former regards the restriction that there are no empty "urns"; what is in between each divider must have atleast one item.  In the latter, there is no restriction.


Let a triangle $ABC$  with  $\angle A=120$ .Let its circumcenter  $O$  and its bisector $AD$ with $D$ in $BC$ .Let also  $M$ the midpoint of $AC$ and let the parallel from $O$ to $AD$ which intersects $BM$ at $G$ .Prove that $G$ is the centroid of the triangle $ABC$

That's easy  :P

Let $E$ be the intersection of $AD$ with the circumcircle of $\Delta ABC$, and let $I$ be the midpoint of $BC$:
It's clear that triangle $BCE$ is equilateral and that $I,O$ and $E$ are collinear.
Now we see that $\Delta GIO$ and $\Delta AIE$ are similar.
So we know that $\frac{GI}{AI}=\frac{IO}{IE}=\frac{1}{3}$
So it is proven that $G$ is the centroid of $\Delta ABC$.


Consider a matrix $ \displaystyle A \equal{} \left( \begin{array}{cc} 1 \plus{} a & b \\
c & 1 \plus{} d \end{array} \right)$ such that  $ ad \equal{} bc\ (a,\ b,\ c,\ d\in{\mathbb{R}})$

Take an arbiterary point $ P$ on $ xy$ plane. Suppose that $ P$ is mapped to $ Q$, $ Q$ mapped to $ R$ by linear transformation expressed by $ A$. Prove that $ P,\ Q,\ R$ are collinear.

Write $ B \equal{} A \minus{} I$, where $ I$ is the $ 2 \times 2$ identity matrix. Then $ \det B \equal{} ad \minus{} bc \equal{} 0$, so $ B$ is not invertible. So $ B$ has rank at most 1.

Then $ Q \equal{} A(P) \equal{} P \plus{} B(P)$, while $ R \equal{} A(Q) \equal{} Q \plus{} B(Q)$. Rearranging, $ Q \minus{} P \equal{} B(P)$ and $ R \minus{} Q \equal{} B(Q)$. Since $ B$ has rank at most 1, the vectors $ Q\minus{}P$ and $ R\minus{}Q$ must be parallel; therefore $ R,P,Q$ are collinear.

The rank of a linear transformation is the dimension of the space spanned by vectors in its image. For example, the matrix

$ \left( \begin{array}{rr} 1 & 0 \\ 1 & 1 \end{array} \right)$ has rank 2, while

$ \left( \begin{array}{rr} 1 & 2 \\ 1 & 2 \end{array} \right)$ has rank 1. (All vectors in the column space are multiples of $ (1,1)^T$)

Thank you. My solution is almost same as yours, finally we will get $ \overrightarrow{PR} \equal{} (a \plus{} d\plus{}2)\overrightarrow{PQ}$ by a simple calculation.

I was working on something that broke it into two cases: $ B$ is nilpotent, meaning that $ B^2\equal{}0,$ or $ B$ is not nilpotent, in which case it has a nonzero eigenvalue and is diagonalizable. That's more complicated than necessary - Nukular's solution is better.

One note from this: $ B$ is nilpotent if and only if $ a\plus{}d\equal{}0.$ Notice the appearance of $ a\plus{}b$ in what kunny just said.


Triangle $ ABC$ is inscribed into a circle. Does there exist a point $ D$ on this circle such that $ ABCD$ is a circumscribed quadrilateral?  (A. Kukush)

[hide]
Suppose $ D$ is $ A$. Then $ AD \plus{} BC < AB \plus{} CD$. If $ D$ is $ C$, then $ AD \plus{} BC > AB \plus{} CD$. As we vary $ D$ from $ A$ to $ C$ along the arc that subtends $ \angle B$, $ AD$ and $ CD$ vary continuously, so somewhere we must have $ AD \plus{} BC \equal{} AB \plus{} CD$. Since $ ABCD$ is convex and the sums of opposite sides are equal, $ ABCD$ is inscriptable.
[/hide]

[quote="durt"]
... so somewhere we must have $ AD \plus{} BC \equal{} AB \plus{} CD$. Since $ ABCD$ is convex and the sums of opposite sides are equal, $ ABCD$ is inscriptable.
[/quote]

I think this is not correct for all quadrilaterals but you can easily prove this for cyclic quadrilaterals. Since $ ABCD$ is cyclic we are done!


trigonometry
geometry
ratio
trapezoid
parallelogram
AMC 10
AMC 10 A

Convex quadrilateral $ ABCD$ has $ AB\equal{}9$ and $ CD\equal{}12$.  Diagonals $ AC$ and $ BD$ intersect at $ E$, $ AC\equal{}14$, and $ \triangle AED$ and $ \triangle BEC$ have equal areas.  What is $ AE$?

$ \textbf{(A)}\ \frac{9}{2}\qquad \textbf{(B)}\ \frac{50}{11}\qquad \textbf{(C)}\ \frac{21}{4}\qquad \textbf{(D)}\ \frac{17}{3}\qquad \textbf{(E)}\ 6$

[hide]
Using the sine area formula on triangles $ AED$ and $ BEC$, we see that

$ (AE)(ED)\equal{}(BE)(EC)\Rightarrow \frac{AE}{EC}\equal{}\frac{BE}{ED}$.

Since $ \angle AEB\equal{}\angle DEC$, triangles $ AEB$ and $ DEC$ are similar.  Their ratio is $ \frac{AB}{CD}\equal{}\frac{3}{4}$.  Since $ AE\plus{}EC\equal{}14$, we must have $ AE\equal{}6$, $ EC\equal{}8$.  Answer is $ E$.
[/hide]

[quote="1=2"]This is one of those neat problems where you can assume that the quadrilateral is a trapezoid. :wink:[/quote]

Not just any trapezoid, but an isosceles trapezoid :D

^actually it doesnt have to be isosceles.

It just needs to parallel opposite sides (so the heights are the same)

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1044557#p1044557

This is a pretty similar probelm from 2008 AMC 10 A

[quote="Klebian"]am i wrong in stating that you can find it's a parallelogram? (using the equal heights on both sides)[/quote]Trapezoid. Not parallelogram. Opposite sides aren't equal.

[quote="CatalystOfNostalgia"][quote="E^(pi*i)=-1"][hide]
Using the sine area formula on triangles $ AED$ and $ BEC$, we see that

$ (AE)(ED) \equal{} (BE)(EC)\Rightarrow \frac {AE}{EC} \equal{} \frac {BE}{ED}$.

Since $ \angle AEB \equal{} \angle DEC$, triangles $ AEB$ and $ DEC$ are similar.  Their ratio is $ \frac {AB}{CD} \equal{} \frac {3}{4}$.  Since $ AE \plus{} EC \equal{} 14$, we must have $ AE \equal{} 6$, $ EC \equal{} 8$.  Answer is $ E$.
[/hide][/quote]

Sine area formula? Why don't you just use the fact that the triangles have the same height?  :wink:[/quote]

Because they don't . . .  :maybe:

I'm not actually using the sine area formula, just the fact that in two triangles with equal area and a pair of equal angles, the product of the sides containing that angle is the same.

yes it is interesting I did not know that if two triangles have the same area and share an angle they are similar

the triangles are similar because of SAS, so the alternate interior angles are congruent, thus AB||CD

Arghh, i missed this

[quote="brightzhu"]How do you prove that it is a trapezoid ($ \overline{AB}\parallel{}\overline{CD}$)? I realized it was similar to a past AMC problem, so I just assumed they were parallel, but how do you prove it?[/quote]See my solution above. Mathemonster is starting from the sine area formula, but you don't need anything high-tech at all.

There is a theorem,called the buterfly theorem, about the areas in a trapezoid, so i used that.

[quote=worthawholebean][hide="Solution"]Let $ [ABC]$ denote the area of triangle $ ABC$. $ [AED]\equal{}[BEC]$, so $ [ABD]\equal{}[AED]\plus{}[AEB]\equal{}[BEC]\plus{}[AEB]\equal{}[ABC]$. Since triangles $ ABD$ and $ ABC$ share a base, they also have the same height [b]and thus $ \overline{AB}\parallel{}\overline{CD}$ [/b]and $ \triangle{AEB}\sim\triangle{CED}$ with a ratio of $ 3: 4$. $ AE\equal{}\frac{3}{7}AD\equal{}6\rightarrow\boxed{\textbf{(E)}}$.[/hide][/quote]

I am confused about how this follows; I understand that triangles $ABD$ and $ABC$ share a common height to $AB$ but why is this height necessarily perpendicular to $CD$?

Because we know that $AED=BEC$, we could assume that $ABC=ABD$, since this is basically when $AEB$ is added to $AED$ and $BEC$. This suggests that $\overline{AB}||\overline{CD}$, because $ABD$ and $ABC$ share the base $AB$, so they must have the same height in order to have the same area.

[hide=Solution]Note that $[AED]=[BED]\implies [ACD]=[BDC].$ Since they share the same base $\overline{DC}$, the lengths of their altitudes from $A$ and $B$ must be equal. Hence, $\overline{AB} \parallel \overline{DC}$ and quadrilateral $ABCD$ is a trapezoid. We find that $$\frac{AE}{EC}=\frac{6}{8}$$ $$ \implies AE=\left(\frac{6}{14}\right)\cdot 14 = \boxed{6}$$[/hide]

[hide=#19][url=aops.com/community/user/350767][b]OlympusHero[/b][/url] · Yesterday at 9:25 PM [url=aops.com/community/p22578975](view)[/url][color=transparent]helo[/color]
[quote=worthawholebean][hide="Solution"]Let $ [ABC]$ denote the area of triangle $ ABC$. $ [AED]\equal{}[BEC]$, so $ [ABD]\equal{}[AED]\plus{}[AEB]\equal{}[BEC]\plus{}[AEB]\equal{}[ABC]$. Since triangles $ ABD$ and $ ABC$ share a base, they also have the same height and thus $ \overline{AB}\parallel{}\overline{CD}$ and $ \triangle{AEB}\sim\triangle{CED}$ with a ratio of $ 3: 4$. $ AE\equal{}\frac{3}{7}AD\equal{}6\rightarrow\boxed{\textbf{(E)}}$.[/hide][/quote]

Can someone tell me how $AB \parallel CD$ was concluded from this? I don't understand. This is probably me being silly as it's late, but I figured I might as well ask.

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[color=#5b7083][aops]x[/aops] 19[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]

I think they meant $AB\perp CD$.

does anyone mind drawing a pic to help me out here?

draw a trapezoid $ABCD$ with $AB<CD$ and $\overline{AB} \parallel \overline{CD}$. Obviously you have to prove what I just typed, but this is what the quadrilateral looks like.


Dear Sir,

I tried to attend mathjam class on 29 sept, 8.30 pm ET. But the classroom couldnot be downloaded.

I am a new member. My login name is bhaskar.

That Math Jam has been rescheduled to next week.

When you say it cannot be downloaded, what do you mean?  What specifically happened when you went to the Classroom page?