The weight of a medium apple can be closely approximated by the normal distribution with a mean of 200 grams and standard deviation of 12 grams.

What is the weight of an apple in the 80th percentile?
In a group of 50,000 apples, how many will weigh less than 180 grams?
What weights bracket the middle 50% of the population?

When posting routine homework problems, please, show what you've done yourself and where you got stuck. Then you may expect a hint about what to do next :)


function
search
calculus
calculus computations

let $ f$ a numeral function , such that : $ f(x)\equal{} \sin(x)$

prove that $ f$ doesn't exept a limit when $ x$ tends to $ \plus{} \infty$

I never thought I'd be saying this but 'use the search function'

This has been asked a MILLION times.

No nonzero periodic function does (except constant functions).  Look at the definition of a limit to infinity.


[hide](i'm assuming you mean "simplify" and not "solve")
the expression can be written as $\frac{\frac{\cos 2006}{\sin 2006}-1}{\frac{1}{\sin 2006}-\frac{1}{\cos 2006}}$

so we multiply the top and bottom by $\sin 2006\cos2006$:

$\frac{\cos^{2}2006-\sin 2006\cos2006}{\cos 2006-\sin 2006}$
$\frac{\cos 2006(\cos 2006-\sin 2006)}{\cos 2006-\sin 2006}=\cos 2006$
[/hide]

[quote="perfectnumber628"]I'm assuming you mean "simplify" and not "solve"[/quote]

Yeah, I did.  *edits*

[quote="lingomaniac88"]Simplify: $\frac{\cot2006-1}{\csc2006-\sec2006}$[/quote]

This one is in my trig book...i know the answer is $\cos 2006$


calculus
integration
logarithms
inequalities
function
calculus computations

[quote="kunny"]Let $0<a<b$.Prove the following inequaliy.

\[\frac{1}{b-a}\int_a^b \left(\ln \frac{b}{x}\right)^2 dx<2\][/quote]

Directly evaluating the integral and with few little computations, I solved the problem
but I wish to prove this problem by treating the left hand side of the inequality as the average of the function $\left( {\ln \frac{b}{x}} \right)^2$ in [a, b].  Then I will post my solution.

hello kunny, we obtain
$\int_{a}^{b}\ln^2\left(\frac{b}{x}\right)\,dx=-\ln^2\left(\frac{b}{a}\right)a-2\ln\left(\frac{b}{a}\right)a-2a+2b$. Thus we get
$\frac{-\ln^2\left(\frac{b}{a}\right)a-2\ln\left(\frac{b}{a}\right)a-2a+2b}{b-a}<2 \Leftrightarrow \frac{\ln\left(\frac{b}{a}\right)a\left(\ln\left(\frac{b}{a}\right)+2\right)}{a-b}<0$ and this is true since $0<a<b$.
Sonnhard.


:D
 Find, with proof, the last four digits of the number $ 3^{2005}$. I know the answer is 0243. I think its enough to prove that $ 3^{1000}\equal{}1\mod 10000$.

I have posted a proof ,sorry if it is wrong ,did it in a hurry .
We have $ \gcd(3,10000) = 1$ ,by Eulers totient theorem we get $ 3^{\varphi(10000)} = 3^{4000}\equiv 1 \bmod{10000} \implies (3^{2000} + 1 )(3^{2000} - 1 ) \equiv 0 \bmod{10000}$ .

We have, $ 10000 \mid (3^{2000} + 1 )(3^{2000} - 1 )$ ,since $ 10000 = 5^{4} * 2^{4}$  we  have

each of $ 5^{4}$ and $ 2^{4}$ dividing $ (3^{2000} + 1 )(3^{2000} - 1 )$ . It is easy to observe that

 $ \gcd(5 , 3^{2000} + 1) = 1$ by fermats little theorem . So we have $ 5^{4} \mid 3^{2000} - 1$

and by the eulers theorem we can prove that $ 2^{4} \mid 3^{2000} - 1$ ,from these two results and

fact that $ \gcd( 2^{4} ,5^{4}) = 1$ we get $ 10000 \mid 3^{2000} - 1 \implies 3^{2000} \equiv 1
\bmod{10000} \implies 3^{2005} \equiv 243 \bmod{10000}$ and the conclusion easily follows that the last four digits are $ 0243$ .

[quote="hpoincare"]In fact, let $ p$ be a prime, $ p\notin\{2,5\}$. Then prove that $ p^{1000} \equal{} 1\mod 10000$.
Show that (<==) is false.[/quote]
We can proceed in the same way as I did in the first post by factorising $ p^{2000} \minus{} 1$ as $ (p^{1000} \plus{} 1 )(p^{1000} \minus{} 1 )$ as $ p \neq 2 ,5$

ABCD is a cyclic quadrilateral with A and D on X axis .AB is along the line y=√3(x+1) and CD is along the line x+y=6.If C lies along the line 2y-7=0, the co ordinates of B arejavascript:opener.emoticon(':spider:')javascript:opener.emoticon(':gathering:')


geometry
3D geometry
tetrahedron
vector
linear algebra
matrix

Given a tetrahedron $ OABC$ such that $ OA\equal{}3,\ OB\equal{}4,\ OC\equal{}5$ and $ \angle{AOB}\equal{}\angle{AOC}\equal{}45^\circ,\ \angle{BOC}\equal{}60^\circ$.
Find the volume of the tetrahedron.

See [url=http://en.wikipedia.org/wiki/Tetrahedron#Volume_of_any_tetrahedron]here[/url] for a simple vector derivation of a directly applicable formula (volume given the three side-lengths and angles from a vertex).

$ V \equal{} \frac{1}{6} \begin{vmatrix}5 & 0 & 0 \\ 2 & 2\sqrt{3} & 0 \\ \sqrt{3} & \sqrt{3} & \sqrt{3} \end{vmatrix} \equal{} \frac{1}{6} \cdot 5 \begin{vmatrix} 2\sqrt{3} & 0 \\ \sqrt{3} & \sqrt{3} \end{vmatrix} \equal{} 5$


For what value(s) of x, $0^\circ \le x < 360^\circ$ is $\cos 42^\circ cos 32^\circ+\sin 42^\circ sin 32^\circ = \sin x?$

[hide]
$\cos a \cos b+\sin a \sin b =\cos (a-b)$
$\cos 10$
80, 100.[/hide]

[hide]Applying the difference of angles formula inverse, we have

$\cos42^\circ\cos32^\circ+\sin42^\circ\sin32^\circ=\cos(42-32)^\circ$

Thus, $\sin(x)=\cos10^\circ$

$x=100\text{ or }80$.[/hide]


hi ramtin  i think that friends had cought it.

i have :

sum  a^2/b^2+c^2>=3


inequalities
function
inequalities proposed

If $ x,y,z,k \in \mathbb{R}^\plus{}$

Prove that

$ \sum_{cyc} \frac{x}{\sqrt{x^2\plus{}kyz}} \ge \frac{3}{\sqrt{k\plus{}1}}$

[quote="Ronald Widjojo"]If $ x,y,z,k \in \mathbb{R}^ \plus{}$

Prove that

$ \sum_{cyc} \frac {x}{\sqrt {x^2 \plus{} kyz}} \ge \frac {3}{\sqrt {k \plus{} 1}}$[/quote]

This is false for $ k\le8$:
Set $ x\equal{}1$ and $ y\equal{}z\equal{}1/n$, and let $ n$ tend to infinity.

[quote="Ronald Widjojo"]If $ x,y,z,k \in \mathbb{R}^ \plus{}$

Prove that

$ \sum_{cyc} \frac {x}{\sqrt {x^2 \plus{} kyz}} \ge \frac {3}{\sqrt {k \plus{} 1}}$[/quote]

By Holder inequality we have
\[ \left( {\sum {\frac {a}{{\sqrt {a^2 \plus{} kbc} }}} } \right)^2 \left( {\sum {a\left( {a^2 \plus{} kbc} \right)} } \right) \ge \left( {\sum a } \right)^3
\]
It suffices us show that
\[ \left( {k \plus{} 1} \right)\left( {\sum a } \right)^3 \ge 9\left( {\sum {a\left( {a^2 \plus{} kbc} \right)} } \right)
\]

\[ \Leftrightarrow \left( {k \plus{} 1} \right)\left( {\sum {a^3 } \plus{} 3\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)} \right) \ge 9\left( {\sum {a^3 } \plus{} 3kabc} \right)
\]

\[ \Leftrightarrow \left( {k \minus{} 8} \right)\sum {a^3 } \plus{} 3\left( {k \plus{} 1} \right)\left( {\sum a } \right)\left( {\sum {ab} } \right) \ge \left( {3\left( {k \plus{} 1} \right) \plus{} 27k} \right)abc
\]
which is true since
\[ \left( {k \minus{} 8} \right)\sum {a^3 } \ge 3\left( {k \minus{} 8} \right)abc
\]

\[ 3\left( {k \plus{} 1} \right)\left( {\sum a } \right)\left( {\sum {ab} } \right) \ge 27\left( {k \plus{} 1} \right)abc
\]

[b]Solution:[/b]
Consider a function:

$ f(x) \equal{} \frac {1}{\sqrt {x}}$.It is clearly convex,i.e. $ f''(x) > 0$.

We can WLOG assume that $ x \plus{} y \plus{} z \equal{} 1$ and apply Jensen inequality for considered function,as follows:

$ \sum_{x,y,z}\frac {x}{\sqrt {x^2 \plus{} kyz}}\geq\sqrt {\frac {1}{x^3 \plus{} y^3 \plus{} z^3 \plus{} 3\cdot k\cdot xyz}}$.

We need to prove that:

$ \sqrt {\frac {1}{x^3 \plus{} y^3 \plus{} z^3 \plus{} 3\cdot k\cdot xyz}}\geq\frac {3}{\sqrt {k \plus{} 1}}$.

Or

$ 9(x \plus{} y \plus{} z)^3 \plus{} (k \minus{} 8)\geq 9(x^3 \plus{} y^3 \plus{} z^3) \plus{} 27\cdot k\cdot xyz$.

As we know by am-gm:

$ 9(x \plus{} y \plus{} z)^3 \plus{} (k \minus{} 8)\geq 9(x^3 \plus{} y^3 \plus{} z^3) \plus{} (k \minus{} 8) \plus{} 27\cdot 8\cdot xyz$.

So it is enough to show that

$ (k \minus{} 8)\geq 27\cdot(k \minus{} 8) xyz$.

But the last inequality easily follows from am-gm inequality.

$ \blacksquare$.


[hide]Note that the product of the roots, $ab=b$ and the sum of the roots, $a+b=-a$.  Thus, we have

$a=1$ and $b=-2$.[/hide]

[quote="#H34N1"]Suppose that $a$ and $b$ are nonzero real numbers, and that the equation $x^{2}+ax+b$ has solutions $a$ and $b$. What is the pair $(a,b)$?[/quote]

[hide="solution"]By veta's theorem,

$-a=a+b$
$b=ab$

so by the second equation, $a=1$ and plugging this into the first equation, $b=-2$

Therefore, $(1,-2)$[/hide]


trigonometry
MATHCOUNTS
ARML
geometric series
number theory
modular arithmetic

What is the value of the remainder when $\tan^{2005}{\frac{\pi}{12}}+\cot^{2005}{\frac{\pi}{12}}$ is divided by $9$?

Mathcounts doesn't have trig.... This was an ARML problem I think.

Oh well, (they don't have trig!?!!?) can someone do it anyway?

Well tan(pi/12)=(2-sqrt[3])sqrt[3]...or if you like, 2sqrt[3]-3.  That to the 2005 is
(2sqrt[3])^2005-(3((2sqrt[3])^2004))(2005C2)+...-3^2005.  Every two terms, you're multiplying the term behind the coefficient of 2005Cx by 9/16...  So then you could probably make two geometric series on top of two binomial expansion series, then add it up.

[hide]I got tan(pi/12)=2-sqrt(3). I don't know what's going on there, but assuming what I got was right, the expression becomes:
$(2-\sqrt{3})^{2005}+(2+\sqrt{3})^{2005}$
In the binomial expansion of the two terms, we can see that any time $\sqrt{3}$ or $-\sqrt{3}$ is raised to an odd power, the terms containing them differ only by sign and cancel out. So we know it's an integer and we only have to consider when $\sqrt{3}$ is raised to an even power. But all terms with said even power higher than 2 are divisible by 9, so we only need to consider $2^{2006}+2^{2004}\cdot 3$. After a lot of tortured modular arithmetic manipulations, I get that $2^{2006}$ is 4 mod 9 and $2^{2004}$ is 1 mod 9, so my answer is 7. That was probably completely wrong, but...[/hide]

Looks good.  Finding $2^{2006}+2^{2004}\cdot 3 \mod 9$ can be calculated fairly simply though .  Factor to get $ 2^{2004}(4+3)$  We know that $2^6=64 \equiv 1 \mod 9$.  Thus $ 2^{2004}(7) \equiv 7$ since 2004 is divisible by 6.

Wow...much simpler than the twisted stuff I did. *Feels stupid*


Okay, the numbers may look a little ugly, so I won't blame you if you use a computer program/calculator etc.. but use it on your own risk!  (and only to get a hint.. see if you can solve it without a computer/calculator -or at least not use all brute force!).

Also, not to spoil others fun,  hide your answer.

given: $ n$ is non-negative integer, and

${ a_0 = \frac {11} 2}$   and $ a_1 = \frac {61}{11}$

$ a_{n + 1} = 2020 - \frac {22129 - \frac {60270}{a_{n - 1}}}{a_n}$

What is the value of $ a_{2009}$ rounded to the nearest integer?

[hide="This is pretty contrived"]Write $ a_n \equal{} \frac{b_{n\plus{}1}}{b_n}$ so that $ b_0 \equal{} 2, b_1 \equal{} 11, b_2 \equal{} 61$. After some algebra on the recurrence, this becomes $ b_{n\plus{}3} \equal{} 2020b_{n\plus{}2} \minus{} 22129b_{n\plus{}1} \plus{} 60270b_n$. The characteristic polynomial of this equation has roots $ 5,6,2009$, which you can guess by factorizing $ 60270$ and seeing how to pick factors with sum $ 2020$. Then we can verify $ b_n \equal{} 5^n \plus{} 6^n$. \[ a_{2009} \equal{} \frac{5^{2010} \plus{} 6^{2010}}{5^{2009}\plus{}6^{2009}} \equal{} \frac{5 \cdot \left( \frac{5}{6} \right)^{2009} \plus{} 6}{\left( \frac{5}{6} \right)^{2009} \plus{} 1}.\]
Approximating $ \left( \frac{5}{6} \right)^{2009}$ as $ 0$ gives $ a_{2009}$ close to $ 6$.[/hide]

MellowMelon answer is, of course, correct. Yes it was a little contrived, [hide]My point in all this, (what I hoped), if anyone tried to do this with a computer, brute force method, but without thinking,  in all likely hood  the answer would have comeout to be ~ 2009 (instead of ~6) . .[/hide]


do p(1) for the sum of the coefficients, and do a roots of unity filter to get the sum of the coefficients in the form x^{5k}


Nope... still wrong, try with $(a,b,c)~(0,0,-1)$.

hix ... sorry everybody, I have made mistake ...  :(
Let $x,y,z$ be positive real numbers. Prove that:
\[x^{6}+y^{6}+z^{6}+xy^{5}+yz^{5}+zx^{5}\ge 2(x^{5}y+y^{5}z+z^{5}x)\]

What do you think about it  :roll: For real numbers, Is it right?  :?:

[quote="zaizai-hoang"]hix ... sorry everybody, I have made mistake ...  :(
Let $x,y,z$ be positive real numbers. Prove that:
\[x^{6}+y^{6}+z^{6}+xy^{5}+yz^{5}+zx^{5}\ge 2(x^{5}y+y^{5}z+z^{5}x) \]
What do you think about it  :roll: For real numbers, Is it right?  :?:[/quote]
Try $x=1.3,$ $y=1$ and $z=0.$ :wink:

ok as the title of this post is Simple  lets solve the following 2 simple problemZ! :D

$\text{1.}$ For positive integer $n$ prove that:

$n\left((n+1)^{\frac{1}{n}}-1\right)<1+\frac{1}{2}+\cdots+\frac{1}{n}$

$\text{2.}$For positive reals $x,y,z$ such that $xyz=1$ prove that:

$\frac{xy}{x+1}+\frac{yz}{y+1}+\frac{zx}{z+1}\geq \frac{3}{2}$

 :)

[quote="arqady"][quote="zaizai-hoang"]Let $a,b,c$ be real numbers. Prove that
\[a(a-b)^{2}+b(b-c)^{2}+c(c-a)^{2}\ge 0 \]
[/quote]
Maybe $a(a+b)^{3}+b(b+c)^{3}+c(c+a)^{3}\geq0$ $?$  It's true and nice.  :wink:[/quote]

For your problem arqady, does it have anything to do with this form(with couple of substitutions):

$(\frac{a^{2}+b^{2}+c^{2}+d^{2}}{3})^{2}\geq 3abcd$ with $a+b+c=0$? :maybe:

EDIT: for the 2nd problem of sinajackson, put $x=\frac{a}{b}$ etc., and you get nesbit... :P

EDIT#2: and the 1st problem of sinajackson is really giving me a hard time... :oops:

[quote="delegat"]For second inequality of [b]sinajackson[/b], when $xyz=1$ I think that it is enough to take:
$x=\frac{a}{b}$  $y=\frac{b}{c}$ and $z=\frac{c}{a}$

then inequality reduces to well known Nesbits inequality.[/quote]

Look at my post above..  :wink:

[quote]Look at my post above.. [/quote]

Ups Sorry :oops: (Nisam vidio).

[quote="sinajackson"]
$\text{1.}$ For positive integer $n$ prove that:

$n\left((n+1)^{\frac{1}{n}}-1\right)<1+\frac{1}{2}+\cdots+\frac{1}{n}$

[/quote]

well this is my solution for this one, its really a beautiful problem:

by AM-GM:

$\sqrt[n]{2\times\frac{3}{2}\times\frac{4}{3}\cdots\times\frac{n+1}{n}}\leq\frac{\sum^{n}_{i=1}\frac{i+1}{i}}{n}\Longrightarrow\sqrt[n]{n+1}\leq\frac{n+\sum^{n}_{i=1}\frac{1}{i}}{n}$

$=1-\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}\Longrightarrow n\left(\sqrt[n]{n+1}-1\right)<1+\frac{1}{2}+\cdots+\frac{1}{n}$

 :)


limit
trigonometry
algebra proposed
algebra

Let $ x_1=y_1=\sqrt{3},x_{n+1}=x_n+\sqrt{1+x_n^2},y_{n+1}=\frac{y_n}{1+\sqrt{1+y_n^2}}$. Prove that: $ 2<x_ny_n\leq 3$ and $ \lim \limits_{n\rightarrow\infty} y_n=0$


Computing Combinations Part 2

We find a formula for combinations, writing C(n,r) in terms of factorials.  We do so by examining a few examples first, then using these as a guide to find our formula.  We relate our formula to the problem of counting arrangements of a particularly conveniently chosen word.

The formula that I typically use for
$\binom{N}{R}=\frac{N*N-1*N-2.....(\text{R times})}{R!}$

The other way you could express this as:
$\binom{N}{R}=\frac{N!}{N-R!*R!}$

Let's use some examples here, using both of the formulas!

$\binom{10}{3}$

Using the first formula, we get:
$\frac{10*(10-1)*(10-2)}{3!}\implies \frac{10*9*8}{3!}\implies \frac{10*9*8}{3*2*1}\implies \frac{\cancel{10*9}*8}{\cancel{3*2}}\implies 3*5*8= \boxed{120}$

Using the second formula we get:
$\binom{10}{3}=\frac{10!}{10-3!*3!}\implies \frac{10!}{7!3!}$

We know $\frac{10!}{7!}\implies 10*9*8$, because we have: $\frac{10*9*8*7*6*5*4*3*2}{7*6*5*4*3*2}\implies \frac{10*9*8*\cancel{7*6*5*4*3*2}}{\cancel{7*6*5*4*3*2}}\implies 10*9*8$

We have: $\frac{10*9*8}{3*2*1}$ which is what we got last time, and we know it equals $\boxed{120}$ from the calculations in the first proof.


A square piece of toast ABCD of side length 1 and centre O
is cut in half to form two equal pieces ABC and CDA. If the
triangle ABC has to be cut into two parts of equal area, one
would usually cut along the line of symmetry BO. However,
there are other ways of doing this. Find, with justi¯cation,
the length and location of the shortest straight cut which
divides the triangle ABC into two parts of equal area.

FROM the British Olympiad

[hide]There are four ways of intersection.
1)We can cut From a point of $CB$, perpendicular (the shortest way) to $AB$
The length of the cut $x$ will thus be: $0<x<AC$
2)We can cut from point $C$ to a point on $AB$: $AC<x<BC$
3)We can cut on the line $OA$: $x=\frac{1}{\sqrt{2}}$
4)We can cut on a line, parallel to $BC$ thus $0<x<\sqrt{2}$

I'm too lazy to calculate it all, and it's way too warm, so I stop here...
another something: look at 2) $\rightarrow$ the values are greater then 1), so 2) can be discarded.

Your shortest way is wrong .... i think because look at your 3) thing its smaller I calculated
 :D Jan please try and go on the topic (really toast) in this forum because i dont know what to do? :blush:

If you set the angle the line is at, there is obviousally only one cut at this angle that will divide it in two. So get this line, and get the lenght. Then we have the distance of the line in terms of an angle. So how do we do that!?!

i cant understand what you are talking about?
i have given all the question in the attachment? :D


Determine the value of a, b, and c so that the following equalities become true for any value of x.

           (ax + b)( x - 1) = 2x^2 - x - 1

 need help asap...

c can be anything, because it's not in the equation.

he probably ment (cx-1) instead of (x-1)
in order to solve you just multiply the (ax+b)(?c?x-1) and then compare it to the given equality

One of  easy, but less known method is given below:  ( look  at it  [b]only[/b] after you have solved it on your own)

 [hide]in $(ax+b)(x-1) = 2x^2 -x -1$
put x =0
we get $  (0+b)(-1) = 0+0 -1$  or   $b=1$
also put x=2 and we get $2a+1 = 5 $ which gives $a=2$

You choose value of x such that expression is simplest ...:)
 [/hide]

[quote="math_rules"]Determine the value of a, b, and c so that the following equalities become true for any value of x.

           $(ax + b)( x - 1) = 2x^2 - x - 1$

 need help asap...[/quote]

How about try to solve this one:

Determine the value of a, b, and c so that the following equalities become true for any value of x.

$(ax + b)(cx - 1) = 2x^2 - x - 1$

Hint:
[hide]b=1, but there are two pairs of (a, c)[/hide]

[quote="mathsidiot"](ax+b)(cx-1)=acx 2 +(bc-a)x-b=2x 2 -x-1
Compare Coeffients to get b=1, ac=2 and c-a=-1
c=a-1       a(a-1)=2 a=2 and c=1[/quote]

$a(a-1)=2$ have two roots.  You missed the second solution:

[hide]$a(a-1)=2$

$a^2-a-2=0$

$(a-2)(a+1)=0$

$a=2, a=-1$

$a=2, c=1$

$a=-1, c=-2$[/hide]


Using the digits: $ 1,2,3,4,5,$ players $ A$ and $ B$ compose a $ 2005$-digit number $ N$ by selecting one digit at a time: $ A$ selects the first digit, $ B$ the second, $ A$ the third and so on. Player $ A$ wins if and only if $ N$ is divisible by $ 9$. Who will win if both players play as well as possible?

a number is divisible by 9 if and only if the sum of its digits is divisible by 9, so consider the ongoing sum as A and B add on numbers/digits.

A adds the last number, he will be able to make a multiple of 9 iff the previous sum=4,5,6,7,8 (mod 9), thus B will win if on his last turn he gives a sum=0,1,2,3 (mod 9). B will always be able to do this unless the sum from the previous turn = 3 (mod 9)

by similar logic if A wants to obtain a sum = 3 (mod 9) he must have a sum =6 (mod 9) on his previous turn, and if he wants a sum = 6 (mod 9) he needs a sum=0 (mod 9) on the previous turn. and there the pattern repeats.

so if we reverse the sequence of turns, If A is to win the sum on each of A's turns must be 0,3,6,0,3,6.... (mod 9), the 1003th number in this sequence, which corresponds to A's first turn is 0, however there is no way that can have a sum =0 (mod 9) after choosing just one of the elements of {1,2,3,4,5} therefore A cannot win so B will be the winner.


Let $ ABCD$ be a convex quadrilateral such that $ AD\parallel{}BC$ and $ AC$ is perpendicular to $ BD$. Let the angle bisectors of $ \angle A,\angle B$ intersect at $ Y$, those at $ \angle C,\angle D$ intersect at $ X$. Prove that $ XY\parallel{}AD$.

Let $ M$ and $ N$ be the points where the inner angle bisectors of $ \angle BAD$ and $ \angle ADC$ meet the sideline $ BC.$ It's easy to see that $ \angle AYX$ and $\angle DXC$ are right due to

$ \angle DAB \plus{} \angle ABC \equal{} 180^{\circ}$ and $ \angle ADC \plus{} \angle DCB \equal{} 180^{\circ}.$

Hence, triangles $\triangle ABN$ and $\triangle CDN$ are isosceles with bases $\overline{ AN}$ and $\overline{ DM}$ $ \Longrightarrow$ $ X,Y$ are midpoints of $ DM,AN$ $ \Longrightarrow$ $ X,Y$ lie on the midparallel of the trapezoid.


If ABCD is a four-digit positive integer in base 10 that is divisible by 33, what is the remained when 13(BCDA+3) is divided by 33?

[hide]WLOG, let $ABCD=3333$.  Thus we must find the remainder when $12(3336)$ is divided by $33$.  So $\frac{12(3336)}{33}=\frac{12(1112)}{11}$.  Multiplying that out, then dividing it is not hard, and we get a remainder of 2.  But we divided out that 3, so we must multiply by it at the end (convince yourself of this using smaller examples), giving a remainder of $3\cdot2=\boxed{6}$.

If you don't see why we multiply by that 3 at the end, look at the division algorithm:

$a=bq+r$

When we divide the $a$ by 3, we also divide the $r$ by 3.  That's why we multiply by it later.[/hide]

[hide="Without WLOG"] $33$ is, of course, really $3$ and $11$.

$3$:  $A+B+C+D \equiv 0 \bmod 3$
$11$:  $A+C \equiv B+D \bmod 11$

Hence $13(BCDA+3)$ leaves a remainder of $0$ upon division by $3$ and $6$ upon division by $11$, so by CRT our remainder is $\boxed{6}$. [/hide]

I understand the division rule for determining if a number is divisible by three. How do you check if it is divisible by 11?

[hide="One line"] $d_{0}+10 d_{1}+100d_{2}+...+10^{k}d_{k}\equiv d_{0}-d_{1}+d_{2}-d_{3}+... \pm d_{k}\bmod 11$ [/hide] [hide="More explanation"] The logic is really quite similar to the logic for division by $9$.  Let the digits of a number $n$ be

$n = \sum_{i=0}^{k}10^{i}d_{i}$

Then, since $10 \equiv 1 \bmod 9$ it follows that $10^{i}\equiv 1 \bmod 9$, which means that

$n \equiv \sum_{i=0}^{k}d_{i}\bmod 9$

Now, since $10 \equiv-1 \bmod 11$ it follows that $10^{i}\equiv (-1)^{i}\bmod 11$ and so

$n \equiv \sum_{i=0}^{k}(-1)^{i}d_{i}\bmod 11$ [/hide]


function
inequalities
real analysis
real analysis unsolved

1.f is continuous on the non-negative reals. For each positive k the sequence f(k), f(2k), f(3k), ... tends to zero. Is it necessarily true that f(x) → 0 as x → ∞.

2.Find a real valued function f on the real line such that f(x)/x2 → 1 as x → ∞, but f '(x)/(2x) does not tend to 1 as x → ∞. Prove that such an f cannot be convex.
(then is is true that the second lim doesn't exist?)

3.Is there a non-trivial continuous real-valued function defined on the real numbers such that f(x) + f(2x) + f(3x) = 0 for all x?

4.Define x0, x1, x2, ... by x0 = 1, xn = xn-1 + 1/xn-1 (for n > 0). Prove xn → ∞. How fast? [In other words, find upper and lower bounds for xn.]
(how to find these upper and lower limits?)

5.All members of the sequence a1, a2, a3, ... are positive and an < (an-2 + an-1)/2 for n > 2. Prove the sequence converges.
(where to start from)

10x very much.all the problems are from
http://www.kalva.demon.co.uk/seminar/sem11-20.html

Hint for #1: [hide="Choose a positive epsilon."] The sets $A_{n}: =\{x\in\mathbb{R}\colon \forall m\ge n \ |f(mx)|\le\epsilon\}$ are closed and their union is $\mathbb{R}$. Apply the Baire category theorem.[/hide]

This is my solution for #5:

Since $a_{n}< (a_{n-2}+a_{n-1})/2$ for $n > 2$,
we can say there are 3 possibilities:
either $a_{n}< a_{n-2}$, $a_{n}< a_{n-1}$,
or $a_{n}< a_{n-2}$ but $a_{n}> a_{n-1}$
or $a_{n}< a_{n-1}$ but $a_{n}> a_{n-2}$
Since the members are all positive, for $a_{n}< a_{n-2}, a_{n}< a_{n-1}$, the sequence will definitely converge since it's decreasing and the lower bound is $0$.

If $a_{n}< a_{n-2}$ but $a_{n}> a_{n-1}$, then $a_{n+1}< a_{n-1}, a_{n+1}> a_{n}$. But since $a_{n}> a_{n-1}$, there's a contradiction. so not possible. Same goes for the 3rd possibility  :D

[hide="an idea for 5"] $a_{n}< \frac{a_{n-1}+a_{n-2}}{2}< \frac{3a_{n-2}+a_{n-3}}{4}< \frac{5a_{n-3}+3a_{n-4}}{8}< ...$
In this way $a_{n}$ is less than some convex combination of $a_{0}$ and $a_{1}$, hence bounded.
Bolzano-Weierstrass: a bounded sequence in $R^{n}$ has a convergent subsequence.
My idea (not rigorous yet) is to find such a sequence. Say it's limit is $L$. Then we can find $a_{n}, a_{m}$ so that $n, m, n-m$ are all large and $a_{n}$ and $a_{m}$ are close to $L$. Then, using the recursive inequality again, hopefully we can then deduce that $a_{n+1}$ is also near $L$, or something? Maybe $a_{n-1}$? Maybe a little more complicated, such as finding three terms close to $L$ and bounding an arbitrary other one above and below using those? [/hide]

[quote="mlok"][b]violetcraze[/b]: different possibilities may be realized by different values of $n$.

[hide="Hint for #5"]$a_{m}\le \max(a_{n},a_{n-1})$ for all $m\ge n$.[/hide][/quote]

Hmm...i don't get it. The condition is $a_{n}< (a_{n-1}+a_{n-2})/2$ for all $n>2$, so doesn't that mean that it holds for all values throughout the sequence?  :(

[quote="violetcraze"]The condition is $a_{n}< (a_{n-1}+a_{n-2})/2$ for all $n>2$, so doesn't that mean that it holds for all values throughout the sequence?  :([/quote]
Yes, it does. But when you consider three different cases, you implicitly assume that the same case occurs for all values of $n$.

Such sequences certainly exist: $a_{n}=1/n$ or any other decreasing sequence.

Would someone please solve #3 for me?  :maybe:

[quote="violetcraze"]For #5, does such a function exist at all? Seems impossible to me ...  :huh:[/quote]

sorry, my question was, for #2. was in a hurry  :blush:

[hide="An example for #2"] $f(x)=x^{2}+\sin(x^{2})$[/hide]

[hide="Answer to the parenthetical question in #2"] No, the lack of convexity does not imply that the second limit does not exist - convexity is a condition on $f''$, so it can be destroyed by a perturbation of $f$ that leaves $f$ and $f'$ essentially unchanged.[/hide]

Re #4: Since $x_{n}^{2}=x_{n-1}^{2}+2+1/x_{n-1}^{2}$, it follows that $x_{n}\ge\sqrt{2n+1}$. To get an upper bound, use the inequality $x_{n}\le x_{n-1}+1/\sqrt{2n+1}$.


If 4 different balls are placed randomly in 3 different boxes, what is the probability that non of the boxes are empty. Assume there is enough room in each box for 4 balls???

Anybody got a clue?

[hide]Let the ordered triple $(a,b,c)$ represent the number of balls in the boxes.
There are three ways to make it so that every box has at least one ball: $(2,1,1)$, $(1,2,1)$, and $(1,1,2)$.

Now, label the balls W, X, Y, Z.  How many ways can we put balls in the boxes that have one each?  The first 1-box has 4 possibilities, and the other 1-box has 3 possibilities.  That's 12 ways.

So the probability is $\frac{36}{81}$, which simplifies to $\frac{4}{9}$.[/hide]

I did not quite understand how you got that answer...I tried and I did it this way...is this wrong then?

The 3 digits numbers below represent the way the balls could end up in the boxes.  For example, 0 in the first box, 0 in the second box, 4 in the 3rd box.

004, 013, 022, 031, 040, 103, 112, 121, 130, 202, 211, 220, 301, 310, 400

Since each ball is placed randomly, each of the above possibilities are equally likely.  Only 3 of the 15 possible outcomes have at least 1 ball in all 3 boxes.  Thats 20%.

kirha, your answer is correct. There is a way to arrive at 15 combinations of balls in boxes without brute force, namely
[hide="this"]
Consider arrangements of 4 $*$'s and 2 $|$'s where the $*$'s represent balls and the $|$'s separate the boxes. (i.e $**|*|*$ means 2 balls in box 1, 1 ball in each of boxes 2 and 3, while $***\parallel *$ means 3 balls in box 1, none in 2, and 1 in 3). Each arrangement corresponds to a way of placing balls into boxes. So we just need to count the arrangements of 2 $|$'s within 6 characters, which is $\binom{6}{2}=\frac{6\cdot 5}{2\cdot 1}=15$.
[/hide]
(Of course, this problem is relatively easy to brute force, but if you were given, say, 5 boxes and 20 balls...)


Let $\left\{x_{n}\right\}_{n\in\mathbb{N}}$ be a Cauchy sequence.  Prove that there exists $B\in\mathbb{Q}^{+}$ such that $\left|x_{n}\right|<B$ for all $n\in\mathbb{N}$.

A cauchy sequence converges so it is bounded .

And of course we can choose a rational bound .  :maybe:

I basically need to prove this using the definition of a Cauchy sequence (rather than just knowing it converges).

$\exists\,N$ such that if $n,m>N,$ then $|x_{n}-x_{m}|<1.$ (All $1$ is doing here is being some convenient specific number.)

Find a rational number $A$ such that

$A\ge \max\{x_{1},x_{2},\dots,x_{N},x_{N+1}\}$

The maximum of a finite set exists, and we can always find an rational number (an interger, if we want it to be) that is greater than a given real number. (That's the Archimedean property.)

Let $B=A+1.$ Then for $1\le n\le N,$ we have $x_{n}\le A<B.$

For $n>N,$ we have

$x_{n}\le x_{N+1}+|x_{n}-x_{N+1}| <A+1=B.$

One can vary the main idea of this proof to get that in any metric space, a Cauchy seqence is always bounded. Note that I did not use anything about any limit of the sequence - the argument did not depend on completeness.


How would you go about solving this equation? It can't be too difficult, as the test it was on was sort of simple, but I can't figure it out. $ x, y, z\in\mathbb{Z}$

$ xyz\equal{}4(x\plus{}y\plus{}z)$
$ x<y<z$

I was thinking of dividing it into cases, using the fact that if the product of the three numbers is divisible by four:

Two are even and not divisible by four
One is divisible by four
Two are divisible by four
All three are divisible by four

But this seems...wrong. Help?

I made an arithmetic error on this problem when I did that test and got it wrong, I was quite mad.

[hide]

Anyway, its not too hard to see that if x >= 3 then there are no solutions.

So that leaves x=1 and x=2.

Case 1, x=1:

yz = 4 + 4y +4z

z = (4+4y)/(y-4)

So (y-4)|(4+4y), and (y-4)|(4y-16)  so (y-4)|20

So y can be 5,6,8,9,14,24

But since y<z we only get the triples (x,y,z)=(1,5,24),(1,6,14),(1,8,9)
__________
Case 2:  x=2

2yz= 8 + 4y + 4z

z = (8+4y)/(2y-4)

So (2y-4)|(8+4y) and (2y-4)|(4y-8)  So combining these and dividing by two gives (y-2)|8.  So y can only be 3, 4,6,10

But again y<z so our triples here are (2,3,10) and (2,4,6).[/hide]

[quote="nateharman1234"]I made an arithmetic error on this problem when I did that test and got it wrong, I was quite mad.

Anyway, its not too hard to see that if x >= 3 then there are no solutions.

So that leaves x=1 and x=2.

Case 1, x=1:

yz = 4 + 4y +4z

z = (4+4y)/(y-4)

So (y-4)|(4+4y), and (y-4)|(4y-16)  so (y-4)|20
[/quote]So, in terms of variables, this line could be rewritten as $ a|b$ and $ a|4a$ so $ a|4a\minus{}b$ if $ a\equal{}y\minus{}4$ and $ b\equal{}4\plus{}4y$? Are you allowed to treat these situations as if | were = and add up or subtract corresponding sides?

eh i just used $ x<3$ and SFFT. worked fine, took like 2 minutes. imo, it was the most straight forward method.

SImon's Favorite Factroing Trick. Look up in the Wiki.


What fraction of a foot is equal to nine inches? Express your answer as a common fraction.

A foot is $ 12$ inches, so $ 9$ inches is $ \frac{9}{12} \equal{} \frac{3}{4}$ of a foot.


let a be the largest solution of the below equation:
$ x^3 \minus{}3x^2 \plus{}1\equal{}0$
prove that :
$ {both } [a^{1804}] { and } [a^{2004}]$ are divisible by 17

[hide="hint"]
$ f(x)\equal{}x^3\minus{}3x^2\plus{}1$
$ f'(x)\equal{}3x^2\minus{}6x$
Easy to check it has 3 real solution such that
$ \minus{}1<x_1<0<x_2<1<x_3$
Cosider the sequence
$ a_n\equal{}x_1^n\plus{}x_2^n\plus{}x_3^n$
[/hide]

Let $ \minus{}1<x_1<0<x_2<1$ another roots, then $ a\plus{}x_1\plus{}x_2\equal{}3,2<a<3$. Therefore $ x_1\plus{}x_2>0$ and $ [a^n]\equal{}a_n\minus{}1$, were
$ a_n\equal{}x_1^n\plus{}x_2^n\plus{}a^n\in N$. We have $ a_0\equal{}3,a_1\equal{}3,a_2\equal{}3^2\equal{}9$ and $ a_{n\plus{}3}\equal{}3a_{n\plus{}2}\minus{}a_n$.
We can calculate $ b_n\equal{}a_n\mod 17$. It give $ b_0\equal{}3,b_1\equal{}3,b_2\equal{}9,b_3\equal{}7,b_4\equal{}1,b_5\equal{}11,b_6\equal{}9,b_7\equal{}9,b_8\equal{}16,b_9\equal{}5,...$
Obviosly $ b_n$ is periodic for $ n\ge T$ with period $ T<17^3$.
$ x^3\minus{}3x^2\plus{}1\equal{}(x\minus{}4)^2(x\plus{}4)\mod 17$, therefore $ b_n\equal{}(bn\plus{}c)4^n\plus{}d(\minus{}4)^n\mod n$ and period $ T\le 16*17$.


inequalities
function
inequalities unsolved

[quote="highly_effective_teens"]prove that if F is a function satisfying  F" >= 0 ( '  F is a convex function '  ) on R , all real a, b, c have
F(a) + F(b) + F(c) + F((a+b+c)/3) >= 4/3 (f((a+b)/2) + f((b+c)/2) + f((a+c)/2)[/quote]
Use $ f(\frac{a\plus{}b}{2}) \le \frac{f(a)\plus{}f(b)}{2}$ e.t.c
Then you must show $ f(\frac{a\plus{}b\plus{}c}{3}) \le \frac{f(a)\plus{}f(b)\plus{}f(c)}{3}$
This is Jensen inequality for  $ a,b,c$

[quote="kunny"]The inequality is famous, but I can't remember the name. :([/quote]
Its name is popovicu's inequality? Right? :maybe:

You Kunny, in my mind the name of the inequality that you don't remember is Chebyshev inequality, highly known in Baltic countries but poor known in the others because of the polytical reasons. Thanks.

Note, one can say it's popovicu inequality, but i think it's wrong.

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21786 .  I believe it is Popoviciu's Inequality after all.

madshiet:  what inequality do you mean by "Chebyshev's inequality"?  The one I know is [url=http://mathworld.wolfram.com/ChebyshevSumInequality.html]this one[/url], which is very well-known outside of the Baltic countries.

me?

So i'm wrong, you know, nowadays i'm very confused; living at the huge city like that, you think it's easy to escape from the stress?

This is not Popoviciu, nor is this Chebyshev, nor is the Chebyshev inequality unknown in the West for any political reasons. See http://www.mathlinks.ro/Forum/viewtopic.php?t=69833 . It follows from Popoviciu but it is weaker than Popoviciu.

For your second question: Go into the forest, behave like a troll.

  darij


calculus
derivative
trigonometry
calculus computations

A curve is defined by the parametric equations x = cos2t, y = sin2t where 0 < t < pi/2

(i) Find the equation of the tangent when t = pi/8

(ii) Show that sin2t(d2y/dx2)  + (dy/dx)^2 = k where k is an integer stating the value of k.

A curve is defined by the parametric equations $ x = \cos{2t}, \,y = \sin{2t}$ where $ 0 < t < \frac{\pi}{2}$

$ (i)$ Find the equation of the tangent when $ t = \frac{\pi}{8}$

$ (ii)$ Show that $ \sin{2t}\frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{2}= k$ where k is an integer stating he value of k.


Here's how this works. One person posts a problem, a second posts a solution [b](not just an answer, a solution)[/b] and posts a second problem, a third posts a solution and posts a third problem, etc. These problems should be all GEOMETRY. I shall start.

[b][color=olive]Problem 1[/color][/b]
$ B$ and $ C$ are trisectors of segment $ AD$, with $ AB > AC$. The line $ EF$ is parallel to line $ AC$. Simplify the following ratio:
\[ \frac {2*[ABE]*[ACF]*[BFE]}{[CFE]*[BDF]^2}
\]
Note: $ [ABE]$ means the area of triangle $ ABE$.

[hide]Let $ h$ be the distance between lines $ AD$ and $ EF$, and let $ d \equal{} AC \equal{} BC \equal{} BD$. Then the ratio becomes
$ \frac {2(\frac {(AB)(h)}{2})(\frac {(AC)(h)}{2})(\frac {(EF)(h)}{2})}{(\frac {(EF)(h)}{2})(\frac {(BD)(h)}{2})^2} \equal{} \frac {2(AB)(AC)(EF)}{(EF)(BD)^2} \equal{} 2(2d)(d)/(d)^2 \equal{} 4$
[/hide]

[color=olive][b]Problem 2[/b][/color]
In triangle $ ABC$, $ D$ is the midpoint of $ BC$. If $ m\angle C \equal{} m\angle DAB$ and $ m\angle CDA \equal{} 45^{\circ}$, find the measure of angle $ C$.

since mBAC=135-mC+mBAD=135
we have t.ADB similar to t.ABC
BD/AB=AB/BC
2BD^2=AB
BD/AB=1/sq.root(2)
 also, by sine... BD/AB= sinmC/sin135
thus, sinmC= 1/2
since mC is less than 45 thus mC is 30.

problem..easy but same flavour.
In triangle ABC, D is the midpoint of AB and E is the point of
trisection of BC closer to C. Given that ang.ADC = ang.BAE, find ang.BAC
hope the marathon continues...[/url]


geometry
rhombus
trapezoid
floor function
AMC
AIME
trigonometry

In triangle $ABC,$ it is given that angles $B$ and $C$ are congruent. Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB},$ respectively, so that $AP=PQ=QB=BC.$ Angle $ACB$ is $r$ times as large as angle $APQ,$ where $r$ is a positive real number. Find the greatest integer that does not exceed $1000r.$

[hide="ugly brute force angle chasing"][asy]defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60), S; S=intersectionpoint(Q--C,P--B); draw(A--B--C--A);draw(B--P--Q--C--R--Q);draw(A--R--B);draw(P--R--S); label("A",A,(0,1));label("B",B,(-1,-1));label("C",C,(1,-1));label("P",P,(1,1)); label("Q",Q,(-1,1));label("R",R,(1,0));label("S",S,(-1,0));[/asy]Let point $ R$ be in $ \triangle ABC$ such that $ QB \equal{} BR \equal{} RP$, thus $ PQBR$ is a rhombus and $ APRB$ is an isosceles trapezoid.
Let $ \angle BAC \equal{} \angle QBR \equal{} \angle QPR \equal{} 2x$ and $ \angle ABC \equal{} \angle ACB \equal{} y$, which means $ x \plus{} y \equal{} 90$.
$ \triangle QBC$ is isosceles with $ QB \equal{} BC$, so $ \angle BCQ \equal{} 90 \minus{} \frac {y}{2}$.
Let $ S$ be the intersection of $ QC$and $ BR$. Since $ \angle BCQ \equal{} \angle BQC \equal{} \angle BRS$, $ BCRS$ is cyclic, which means $ \angle RBS \equal{} \angle RCS \equal{} x$.
Since $ APRB$ is an isosceles trapezoid, $ BP \equal{} AR$, but since $ AR$ bisects $ \angle BAC$, $ \angle ABR \equal{} \angle ACR \equal{} 2x$.
Therefore we have that $ \angle ACB \equal{} \angle ACR \plus{} \angle RCS \plus{} \angle QCB \equal{} 2x \plus{} x \plus{} 90 \minus{} \frac {y}{2} \equal{} y$.
We solve the simultaneous equations $ x \plus{} y \equal{} 90$ and $ 2x \plus{} x \plus{} 90 \minus{} \frac {y}{2} \equal{} y$ to get $ x \equal{} 10$ and $ y \equal{} 80$.
$ \angle APQ \equal{} 180 \minus{} 4x \equal{} 140$, $ \angle ACB \equal{} 80$, so $ r \equal{} \frac {80}{140} \equal{} \frac {4}{7}$.
$ \left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor \equal{} \boxed{571}$
[/hide]When did Kalva start updating again? Last time I checked the 2000 AIME I wasn't up, but now it's updated to 2004..

Sorry for revive. New solution... trig bash time! (Actually not too bad, only took me about 5 minutes.)
[hide="Solution"]Since angle $ B$ is congruent to angle $ C$, and since we have $ AP=QB$, we also have $ AQ=CP$. Now let $ AP=QP=QB=BC=a, AQ=CP=b$. Let $ \angle A=\alpha$, and draw line $ BP$. Do a easy angle chase to find $ \angle QPA=180-2\alpha, \angle PQB=180-\alpha, \angle C= 90-\frac{\alpha}{2}, \\
\angle QBP=\frac{\alpha}{2},\angle PBC=90-\alpha, \angle BPC=\frac{3\alpha}{2}.$
Then by law of sines on triangle $ AQP$, we have $ \frac{\sin 2\alpha}{b}=\frac{\sin \alpha}{a} \iff \frac{a}{b}=\frac{\sin \alpha}{\sin 2\alpha}=\frac{1}{2 \cos \alpha}$. Also, by law of sines on triangle $ BPC$, we have $ \frac{\cos \alpha}{b}=\frac{\sin \frac{3\alpha}{2}}{a} \iff \frac{a}{b}=\frac{\sin \frac{3\alpha}{2}}{\cos \alpha}$.
Equating the two values, we see that $ \sin \frac{3\alpha}{2}=\frac{1}{2}$, or $ \frac{3\alpha}{2}=30,150,390,\ldots \Rightarrow \alpha=20,120,260,\ldots$. But $ \angle APQ=180-2\alpha>0$, so $ \alpha<90$. Then we have $ \alpha=20, \angle APQ=140, \angle C=80$, and our desired answer is ${ \left\lfloor 1000(\frac{80}{140})\right\rfloor}=\boxed{571}$.[/hide]

Where is this problem from? A past AIME? (It looks kind of familiar)

[quote="azjps"][hide="Painless   (from wiki); edit: this is linked from resources; its 2000 aime I #14"][asy]defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60); draw(A--B--C--A);draw(P--Q);draw(A--R--B);draw(P--R);D(R--C,dashed); label("\(A\)",A,(0,1));label("\(B\)",B,(-1,-1));label("\(C\)",C,(1,-1));label("\(P\)",P,(1,1)); label("\(Q\)",Q,(-1,1));label("\(R\)",R,(1,0));[/asy]
Let point $ R$ be in $ \triangle ABC$ such that $ QB \equal{} BR \equal{} RP$. Then $ PQBR$ is a rhombus, so $ AB \parallel PR$ and $ APRB$ is an isosceles trapezoid. Since $ \overline{PB}$ bisects $ \angle QBR$, it follows by symmetry in isosceles trapezoid $ APRB$ that $ \overline{RA}$ bisects $ \angle BAC$. Thus $ R$ lies on the perpendicular bisector of $ \overline{BC}$, and $ BC \equal{} BR \equal{} RC$. Hence $ \triangle BCR$ is an equilateral triangle.

Now $ \angle ABR \equal{} \angle BAC \equal{} \angle ACR$, and the sum of the angles in $ \triangle ABC$ is
\[ \angle ABC \plus{} \angle ACB \plus{} \angle BAC \equal{} 3\angle BAC \plus{} 120^{\circ} \equal{} 180^{\circ} \, \Longrightarrow\,\angle BAC \equal{} 20^{\circ}.
\]
Then $ \angle APQ \equal{} 140^{\circ}$ and $ \angle ACB \equal{} 80^{\circ}$, so the answer is $ \left\lfloor 1000 \cdot \frac {80}{140} \right\rfloor \equal{} \left\lfloor \frac {4000}{7} \right\rfloor \equal{} \boxed{571}$.
[/hide][/quote]

What is the intuition for choosing auxilary points like $R$ over another point?

I would also like an answer to sunny2000's question. How do you get that idea?

[quote=Not_a_Username]Bump; I would also like an answer to sunny2000's question[/quote]

Well known technique. Triangle with two congruent sides, we have to create quadrilateral. :)

[hide="An Interesting Solution"]
Note that $AQ=PC=a$; let $AP=1$. Then $P$ has barycentric coordinates $(a:0:1)$ and $Q$ has barycentric coordinates $(1:a:0)$. Thus, $\vec{PQ}= (1-a:a:-1)$. Applying the distance formula, $$a+(a+1)^2(1-a)+(a+1)^2(a-1)a=(a+1)^2$$
$$a=(a+1)^2(1-a(a-1)-(1-a))\Longrightarrow 1=(a+1)^2(2-a)\Longrightarrow a^3-3a=1$$
Now, we are motivated by the identity $4\cos^3(x)-3\cos(x)=\cos(3x)$. Plugging in $a\mapsto 2a$,
$$8a^3-6a=1 \Longrightarrow 4a^3-6a=\frac{1}{2}\Longrightarrow \cos(3x)=\frac{1}{2}$$ which means $a=2\cos\frac{\pi}{9}$.

Let $\angle AQP=\theta$; angle chasing gives $\angle APQ=\pi-2\theta$ and $\angle ACB=\frac{\pi-\theta}{2}$. By the Law of Cosines on triangle $APQ$, $$\bigg(2\cos\frac{\pi}{9}\bigg)^2=2+2\cos2\theta\Longrightarrow \cos^2\frac{\pi}{9}=\cos^2(\theta)$$ by the double angle identity $\cos(2\theta)=2\cos^2(\theta)-1$, so then $\theta=\frac{\pi}{9}$. This gives $\angle APQ=\frac{7\pi}{9}$, $\angle ACB=\frac{4\pi}{9}$ so our answer is $$\bigg[1000\cdot \frac{4}{7}\bigg]=\boxed{571.}$$
[/hide]

[quote=EulerMacaroni]
[hide=An Interesting Solution]
Note that $AQ=PC=a$; let $AP=1$. Then $P$ has barycentric coordinates $(a:0:1)$ and $Q$ has barycentric coordinates $(1:a:0)$. Thus, $\vec{PQ}= (1-a:a:-1)$. Applying the distance formula, $$a+(a+1)^2(1-a)+(a+1)^2(a-1)a=(a+1)^2$$$$a=(a+1)^2(1-a(a-1)-(1-a))\Longrightarrow 1=(a+1)^2(2-a)\Longrightarrow a^3-3a=1$$Now, we are motivated by the identity $4\cos^3(x)-3\cos(x)=\cos(3x)$. Plugging in $a\mapsto 2a$,
$$8a^3-6a=1 \Longrightarrow 4a^3-6a=\frac{1}{2}\Longrightarrow \cos(3x)=\frac{1}{2}$$which means $a=2\cos\frac{\pi}{9}$.

Let $\angle AQP=\theta$; angle chasing gives $\angle APQ=\pi-2\theta$ and $\angle ACB=\frac{\pi-\theta}{2}$. By the Law of Cosines on triangle $APQ$, $$\bigg(2\cos\frac{\pi}{9}\bigg)^2=2+2\cos2\theta\Longrightarrow \cos^2\frac{\pi}{9}=\cos^2(\theta)$$by the double angle identity $\cos(2\theta)=2\cos^2(\theta)-1$, so then $\theta=\frac{\pi}{9}$. This gives $\angle APQ=\frac{7\pi}{9}$, $\angle ACB=\frac{4\pi}{9}$ so our answer is $$\bigg[1000\cdot \frac{4}{7}\bigg]=\boxed{571.}$$ [/hide][/quote] I also tried doing this the first time I saw it, but couldn't figure out what to do with the cubic :(

[hide=painless trig]
Set $\angle{APQ} = \theta$. Angle chase with isosceles quadrilaterals to get $\angle{PBC} = \theta/2$. Then $\angle{BPC} = 180 - (45 + 3\theta/4)$. Now using law of sines on APQ and BPC (after labelling $PC = AQ = y$ and $AP = QB = x$) finishes the problem. We find $\theta = 140$ and just compute the rest.
[/hide]

[quote=azjps][hide="Painless  :) (from wiki); edit: this is linked from resources; its 2000 aime I #14"][asy]defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60); draw(A--B--C--A);draw(P--Q);draw(A--R--B);draw(P--R);D(R--C,dashed); label("\(A\)",A,(0,1));label("\(B\)",B,(-1,-1));label("\(C\)",C,(1,-1));label("\(P\)",P,(1,1)); label("\(Q\)",Q,(-1,1));label("\(R\)",R,(1,0));[/asy]
Let point $ R$ be in $ \triangle ABC$ such that $ QB \equal{} BR \equal{} RP$. Then $ PQBR$ is a rhombus, so $ AB \parallel PR$ and $ APRB$ is an isosceles trapezoid. Since $ \overline{PB}$ bisects $ \angle QBR$, it follows by symmetry in isosceles trapezoid $ APRB$ that $ \overline{RA}$ bisects $ \angle BAC$. Thus $ R$ lies on the perpendicular bisector of $ \overline{BC}$, and $ BC \equal{} BR \equal{} RC$. Hence $ \triangle BCR$ is an equilateral triangle.

Now $ \angle ABR \equal{} \angle BAC \equal{} \angle ACR$, and the sum of the angles in $ \triangle ABC$ is
\[ \angle ABC \plus{} \angle ACB \plus{} \angle BAC \equal{} 3\angle BAC \plus{} 120^{\circ} \equal{} 180^{\circ} \, \Longrightarrow\,\angle BAC \equal{} 20^{\circ}.
\]
Then $ \angle APQ \equal{} 140^{\circ}$ and $ \angle ACB \equal{} 80^{\circ}$, so the answer is $ \left\lfloor 1000 \cdot \frac {80}{140} \right\rfloor \equal{} \left\lfloor \frac {4000}{7} \right\rfloor \equal{} \boxed{571}$.
[/hide][/quote]

Sorry for the revive but, what is the motivation of point R's construction? Or does it come from practice?

Ahh how big a difference two years makes.

[hide=Problem would be better as a proof question]
Let $\angle BCA=\theta$, so $\angle BAC=180^\circ-2\theta$, $\angle APQ=4\theta-180^\circ$, and $\angle QBP=90^\circ-\theta$. Consider a rotation at $Q$ sending $P$ to $B$, and denote by $R$ the image of $A$ with this rotation applied.
[asy]
size(9cm);
defaultpen(fontsize(9pt));
pair A=(-15tan(80pi/180),0), B=(0,-15), C=(0,15), P=(-55.5,5.2), Q=(-29.5,-9.8), R=(26,0);
draw(A--B--C--cycle, linewidth(0.5));
draw(P--Q, linewidth(0.5));
draw(Q--R--B, linewidth(0.4)+grey);
draw(A--R--C, linewidth(0.4)+grey);
draw(P--B, linewidth(0.4)+grey);
dot("$A$", A, (-1,0));
dot("$B$", B, (0.5,-1));
dot("$C$", C, (0.5,1));
dot("$P$", P,  NW);
dot("$Q$", Q, SW);
dot("$R$", R, (1,0));
[/asy]
Observe that with this rotation comes a spiral similarity sending $\triangle APR$ to $\triangle PQB$; hence,
$$\angle QRA=\angle QBP=90^\circ-\theta\implies \angle ARB+\angle RBC=90^\circ.$$
This, in turn, implies that $R$ lies on the perpendicular bisector of $\overline{BC}$, so $RC=RB=BC$ and $\triangle RBC$ is equilateral. It follows that
$$3\theta-180^\circ=\angle CBR=60^\circ\implies \theta=80^\circ,$$
and therefore
$$\lfloor 1000r\rfloor=\left\lfloor\frac{4000}{7}\right\rfloor=\boxed{571}.$$
[/hide]
[hide=Motivation for above]
It's kinda dumb, but obviously the first thing you would want to deal with is the equal lengths condition: one of my favorite ways of doing so is by utilizing "cut and paste", where you move equal lengths together in hopes of attaining something good. Furthermore, the only elementary tools that come to the top of my head (for relating lengths and angles) are similarity, congruence, and circles. I basically tried combining these ideas and angle chasing to see if something worked, and the spiral similarity solution was successful.
[/hide]

This could be one of my favorite AIME problems to date.

Set $\angle ABC = \angle ACB = x$, and WLOG $QB=QP=BC=AP=1$. With some angle chasing one obtains $\angle BQP = 2x$ and $\angle BCP = 360^{\circ}-4x$. Construct $R$ on $\overline{PC}$ such that $\overline{QR} \parallel \overline{BC}$. Observe that by the Law of Cosines on $\triangle BQC$, $$QC = \sqrt{2-2\cos x},$$ and Ptolemy's on isosceles trapezoid $QRCB$ gives us $$QR + 1 = 2-2 \cos x \iff QR = 1-2\cos x.$$ Furthermore, Law of Sines on $\triangle QPR$ yields $$\frac{QR}{QP} = \frac{\sin(360^\circ-4x)}{\sin x} = -\frac{\sin 4x}{\sin x} = -4\cos 2x \cos x.$$ Therefore, expanding with cosine double angle, $$-4\cos x(2\cos^2 x-1) = 1-2\cos x \iff 8\cos^3x - 6\cos x + 1 = 0.$$ By the third Chebyshev polynomial for cosine, $$8\cos^3 x - 6 \cos x = 2\cos 3x,$$ so $\cos 3x = -\frac 12$. Note that $x=40^\circ$ and $x=120^\circ$ both fail, so $x = 80^\circ$.

Now for the answer extraction: $\angle APQ = 140^\circ$ and $\angle ACB = 80^\circ$, so the answer is $\left \lfloor 1000 \cdot \frac 57 \right \rfloor = \boxed{571}$.

[quote=Python54][hide=Hint]Let $\angle BAC=20^{\circ}$. [/hide]

[hide=Very Fast and Low-Tech Solution]
Let $\angle BAC=20^{\circ}$. Thus, $\angle ACB=\angle ABC=80^{\circ}$. Furthermore, because $\triangle APQ$ is isosceles, $\angle APQ=140^{\circ}$. So, $$r=\frac{\angle ACB}{\angle APQ}=\frac{80}{140}=.571... \implies \lfloor 1000r \rfloor=\boxed{571}$$[/hide]

[hide=Motivation] In the problem, we are never given specific angle measures for any of the angles (there aren't even restrictions for the angle measures!). So, we use this to our benefit and plug in any convenient angle measure for one angle and then use it find the measure of the other angles, converting this AIME #14 problem into a basic geometry exercise! I used $20^{\circ}$ as the measure of $\angle BAC$, but you could have also used $\angle ACB=30^{\circ}$, for example, or even (if you wanted to make your life difficult) $\angle QPC=131.498$. The point is, as long as the angle you substitute is between $0$ and $180$ (otherwise you wouldn't even have a triangle), any angle measure can substituted for almost any angle! [/hide][/quote]

This doesn't work, you simply got the correct value of $\angle BAC$ by luck.

[hide=Cool pure synthetic solution sketch, don't think it has been posted]
Note that $AQ = CP$. This motivates rotating $ABC$ 180 degrees about the midpoint of $AC$, then reflecting about its vertical line of symmetry. Then, $\triangle PP'Q$ is equilateral, and it is a trivial angle chase after this.
[/hide]


[quote=Edward_Tur]In convex 7-gon $ ABCDEFG$
$ AC \parallel EF, BD \parallel FG, CE \parallel GA, DF \parallel AB,EG \parallel BC, FA \parallel CD$.
Proof, that $ GB \parallel DE$.[/quote]

Using complex coordinates we have
\begin{align*}
\frac{a-c}{f-e}\in\mathbb{R}\\
\frac{a-c}{f-e}=\frac{\overline{a}-\overline{c}}{\overline{f}-\overline{e}}\\
(a\overline{f}-\overline{a}f)+(e\overline{a}-\overline{e}a)+(f\overline{c}-\overline{f}c)+(c\overline{e}-\overline{c}e)=0.
\end{align*}
Analogously
\begin{align*}
(b\overline{g}-\overline{b}g)+(f\overline{b}-\overline{f}b)+(g\overline{d}-\overline{g}d)+(d\overline{f}-\overline{d}f)=0\\
(c\overline{a}-\overline{c}a)+(g\overline{c}-\overline{g}c)+(a\overline{e}-\overline{a}e)+(e\overline{g}-\overline{e}g)=0\\
(d\overline{b}-\overline{d}b)+(a\overline{d}-\overline{a}d)+(b\overline{f}-\overline{b}f)+(f\overline{a}-\overline{f}a)=0\\
(e\overline{c}-\overline{e}c)+(b\overline{e}-\overline{b}e)+(c\overline{g}-\overline{c}g)+(g\overline{b}-\overline{g}b)=0\\
(f\overline{d}-\overline{f}d)+(c\overline{f}-\overline{c}f)+(d\overline{a}-\overline{d}a)+(a\overline{c}-\overline{a}c)=0.
\end{align*}
Summing up these six equations gives
\begin{align*}
-(g\overline{e}-\overline{g}e)-(d\overline{g}-\overline{d}g)-(e\overline{b}-\overline{e}b)-(b\overline{d}-\overline{b}d)=0\\
\frac{g-b}{e-d}=\frac{\overline{g}-\overline{b}}{\overline{e}-\overline{d}}\\
\frac{g-b}{e-d}\in\mathbb{R}.
\end{align*}
So $GB\parallel DE$. In particular the convexity of $ABCDEFG$ is not needed.

$S(GBD)=S(BDF)=S(DFA)=S(FAC)=S(ACE)=S(CEG)=S(EGB)\Rightarrow S(GBD)=S(EGB)\Rightarrow GB\parallel DE$.


there are 10 photos, and there are three men in each photo:
in the middle is a man, standing at the left side of him is his son, and for the right side is his brother. Now knowing that the men standing in the middle of each photo are all different, find the minimum of people that appear in the photos.

because there obviously are finite people in all 10 photoes so there exist the youngest and the oldest who can't stand in the middle of any photoes . Furthermore , there are atleast 10 differential ppl so the minimum of people are 12 .
It is done if we can point out a case in which there are 12 ppl . Now let K is the one in middle of the Kth photoes , K-1 and K+1 consecutively are his son and his brother . Consider the case :
(2;1;0);(3;2;1);(4;3;2);(5;4;3);(6;5;4);(7;6;5);(8;7;6);(9;8;7);(10;9;8);(11;10;9) which has 12 ppl , so we are done  :)

i think there is a mistake in your solution,considering
(2;1;0);(3;2;1),

2 is the son of 1, how can 2 be the brother of 1 in the second group?

hmm actually K+1 is K's brother and K-1 is K's son , but hmm ...my solution still didn't work because then 2 is 1's father and also 2 is a brother of 1 .


analytic geometry
geometry
graphing lines
slope
perpendicular bisector

1. [b]Prove that the equation to the circle of which the points(x1,y1)and (x2,y2)are the ends of a chord of a segment containing an angle "theta" is,

(x-x1)(x-x2)+(y-y1)(y-y2)+or-cot(theta)[(x-x1)(y-y2)-(x-x2)(y-y1)]=0 :|

2.Find the equations to the circles in which the line joining the points(a,b) and(b,-a)is a chord subtending an angle of 45 degree at any point on its circumference.[/b]

[quote="joshisunny16"]1. [b]Prove that the equation to the circle of which the points(x1,y1)and (x2,y2)are the ends of a chord of a segment containing an angle "theta" is,

(x-x1)(x-x2)+(y-y1)(y-y2)+or-cot(theta)[(x-x1)(y-y2)-(x-x2)(y-y1)]=0 :| [/quote]

We have two circles that satisfy the wanted condition, not just one.

[hide="Hit"]
Use the angle $ 2\theta$ and the equation of the perpendicular bisector of $ (x_1,y_1),(x_2,y_2)$ of to get the coordinates of its center $ O(h,k)$

Compute the distance from $ O(h,k)$ to $ (x_1,y_1)$ or $ (x_2,y_2)$ to get the radius

Use $ (x \minus{} h)^2 \plus{} (y \minus{} k)^2 \equal{} r^2$

This procedure yields to possible equations of the circle.[/hide]

Hi joshisunny16.

My solution is exactly the same as KennyO's, I'm afraid.

As the chord subtends an angle $ \theta$ at the circumference, it subtends an angle $ 2\theta$ at the centre. We can find the mid-point of the chord. Joining the end-points of the chord to the centre and the mid-point of the chord to the centre, we have two congruent right-angled triangles with an angle $ \frac{2\theta}{2} \equal{} \theta$ at the centre. We know the length of half the chord: $ \sqrt{(\frac{x_{1}\minus{}x_{2}}{2})^{2} \plus{} (\frac{y_{1}\minus{}y_{2}}{2})^{2}}$. This enables us to find $ h$, the length of the side joining the mid-point to the centre, and $ r$ the radius of the circle, in terms of half the length of the chord and $ \cot \theta$. We know the slope of the chord so we have the slope of its mediator (perpendicular bisector) and we also know the coordinates of a point on the mediator (the mid-point of the chord $ (\frac{x_{1}\plus{}x_{2}}{2}, \frac{y_{1}\plus{}y_{2}}{2}$). We can then solve between the equations of these two lines to find the centre of the circle whose equation we then have.

I hope this is helpful.


analytic geometry
geometry unsolved
geometry

Let $w$ the brocard angle in a triangle ABC(exist P such that $\angle PAB=\angle PCA=\angle PBC=w$), and $K$ the Lemoinne point(isogonal conjugate of the centroid). Prove that the distance of $K$ to $AB$ is $\frac{AB}{2}.tanw$.

NOTE:I have an operative solution using:
[hide]Tripolar coordinates[/hide]
But i want a short way.        Sorry for my english :blush:

Use these lemma: ( $G$ is the centroid of $\triangle ABC$ and $K_{c}$ is the projection of $K$ to $AB$ )- I think all of them are well know
$cot w = cotA+cotB+cotC = \frac{a^{2}+b^{2}+c^{2}}{4S_{\triangle ABC}}\\cot \angle BAK = cot \angle CAG = \frac{CA^{2}+AG^{2}-CG^{2}}{\frac{4}{3}S_{\triangle ABC}}= \frac{3b^{2}+c^{2}-a^{2}}{4S_{\triangle ABC}}\\ KA = \frac{bc\sqrt{2b^{2}+2c^{2}-a^{2}}}{b^{2}+c^{2}}, K_{c}A= \frac{b^{2}c}{a^{2}+b^{2}}$
The distance between $K$ and $AB$ is $K_{c}A.tan \angle KAB$
We need to prove that:
$\frac{AB}{2}tan w = K_{c}A .tan \angle KAB \\ \Leftrightarrow \frac{AB}{K_{c}A}= 2\frac{ cot w}{cot \angle KAB}\\ \Leftrightarrow \frac{a^{2}+b^{2}}{a^{2}}= \frac{2(a^{2}+b^{2}+c^{2})}{3b^{2}+c^{2}-a^{2}}$
May be I was wrong... but now I think it's not true  :(


For what value of $ x$ is the expression $ \frac{2x^3\plus{}3}{x^2\minus{}20x\plus{}100}$ not defined?

[hide]For it to be undefined, we need $ x^2\minus{}20x\plus{}100$ to be $ 0$. We quickly notice that it is a perfect square of $ (x\minus{}10)^2$. Thus, $ x\equal{}10$.[/hide]


$a_{n}$ :$a_{0},a_{1}\in N$ $a_{n+2}=|\frac{a_{n}a_{n+1}}{a_{n}-a_{n+1}}|$
when $a_{n}\not=a_{n+1}$ if for an integer $k$ $a_{k+1}=a_{k}$
then show that both $a_{k},a_{k+1}$are integer.

I think this problem isn't very hard,but interesting. :)

I think we also can prove that for any $a_{0},a_{1}$ there is an integer $k$ such that $a_{k}=a_{K+1}$


Mock AMC F is ready!!!! I haven't decided when is the official time yet but it's going to sometime in the Thanksgiving break.... You can always take it early!

Okay I think I will make the official time on Sunday, Nov. 28th, at 4:00pm to 5:15pm EST.

This test is AMC12 level. There are about 17 to 19 original problems...


inequalities
logarithms
calculus
derivative
inequalities proposed

$ a,b,c>0,a^2\plus{}b^2\plus{}c^2\equal{}1$.Pove that
$ \frac{a}{ab\plus{}3}\plus{}\frac{b}{bc\plus{}3}\plus{}\frac{c}{ca\plus{}3}<\equal{}3/4$

[quote="tronghieu"]$ a,b,c > 0,a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.Pove that
$ \frac {a}{ab \plus{} 3} \plus{} \frac {b}{bc \plus{} 3} \plus{} \frac {c}{ca \plus{} 3} < \equal{} 3/4$[/quote]
It is the problem in my ebook, it must be $ a^2\plus{}b^2\plus{}c^2\equal{}3$ . :)

[quote="can_hang2007"][quote="tronghieu"]$ a,b,c > 0,a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.Pove that
$ \frac {a}{ab \plus{} 3} \plus{} \frac {b}{bc \plus{} 3} \plus{} \frac {c}{ca \plus{} 3} < \equal{} 3/4$[/quote]
It is the problem in my ebook, it must be $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ . :)[/quote]

A similar one,not easy:

$ \frac a{a^2\plus{}3}\plus{}\frac b{b^2\plus{}3}\plus{}\frac c{c^2\plus{}3} \le \frac3{4}$  for $ a,b,c>0$ such that $ abc\equal{}1$.

[quote="zaizai-hoang"]I don't think it is hard  :) Using the following inequality:
\[ f(x) \equal{} \frac {1}{4} \plus{} \frac {\ln{x}}{8}\ge \frac {x}{x^2 \plus{} 3}
\]
[/quote]
It is not always true, Hoang, for example $ x\equal{}0.001$

[quote="dduclam"]$ \frac a{a^2 \plus{} 3} \plus{} \frac b{b^2 \plus{} 3} \plus{} \frac c{c^2 \plus{} 3} \le \frac3{4}$  for $ a,b,c > 0$ such that $ abc \equal{} 1$.[/quote]
We have the nice and stronger is
\[ \left( \frac{a\minus{}3b}{a\plus{}3b}\right)^2\plus{} \left( \frac{b\minus{}3c}{b\plus{}3c}\right)^2\plus{}\left( \frac{c\minus{}3a}{c\plus{}3a}\right)^2 \ge \frac{3}{4}\]
for all $ a,b,c \ge 0$ :)

[quote="zaizai-hoang"]I don't think it is hard  :) Using the following inequality:
\[ f(x) \equal{} \frac {1}{4} \plus{} \frac {\ln{x}}{8}\ge \frac {x}{x^2 \plus{} 3}
\]

\[ \Leftrightarrow f'(x) \equal{} \frac {(x^2 \plus{} 6x \minus{} 3)(x \plus{} 3)(x \minus{} 1)}{8x(x^2 \plus{} 3)^2}
\]
Easy to check it out:
\[ \Leftrightarrow min f(x)_{x\in (0, \plus{} \infty)} \equal{} f(1) \equal{} 0
\]
We are done![/quote]
Sorry for my hastiness  :blush: We have:
\[ f'(x) \equal{} \frac {(x^2 \plus{} 6x \minus{} 3)(x \plus{} 3)(x \minus{} 1)}{8x(x^2 \plus{} 3)^2}\ge 0\]
$ \iff 0\le x\le 2\sqrt{3}\minus{}3$ or $ 1\le x$. Who can complete my solution?

[quote="dduclam"][quote="can_hang2007"][quote="tronghieu"]$ a,b,c > 0,a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.Pove that
$ \frac {a}{ab \plus{} 3} \plus{} \frac {b}{bc \plus{} 3} \plus{} \frac {c}{ca \plus{} 3} < \equal{} 3/4$[/quote]
It is the problem in my ebook, it must be $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ . :)[/quote]

A similar one,not easy:

$ \frac a{a^2 \plus{} 3} \plus{} \frac b{b^2 \plus{} 3} \plus{} \frac c{c^2 \plus{} 3} \le \frac3{4}$  for $ a,b,c > 0$ such that $ abc \equal{} 1$.[/quote]

Anyway, it is always easy to prove by p,q,r technique .  :wink: Expanding is the best way  :D
\[ \frac a{a^2 \plus{} 3} \plus{} \frac b{b^2 \plus{} 3} \plus{} \frac c{c^2 \plus{} 3} \le \frac3{4}
\]

\[ \iff 27\sum a^2 \plus{} 84 \plus{} 9\sum a^2b^2\ge 36\sum a \plus{} 12\sum ab(a \plus{} b) \plus{} 4\sum ab
\]
Transforming into $ p \equal{} a \plus{} b \plus{} c,q \equal{} ab \plus{} bc \plus{} ca,r \equal{} abc$ we have:

\[ \iff 27\sum a^2 \plus{} 84 \plus{} 9\sum a^2b^2\ge 36\sum a \plus{} 12\sum ab(a \plus{} b) \plus{} 4\sum ab
\]

\[ \iff f(q) \equal{} 27p^2 \minus{} (54 \plus{} 12q)p \plus{} 9q^2 \minus{} 58q \plus{} 120\ge 0
\]
We have:
\[ \frac {9}{p}\le q\le \frac {9 \plus{} p^3}{4p}
\]
There are two case:
$ \bullet$ The first case: $ 18q\ge 58 \plus{} 12p$ we have $ f'(q)\ge 0$ so
\[ f(p,q)\ge f\left(p,\frac {9}{p}\right) \equal{} \frac {3(p \minus{} 3)(9p^3 \plus{} 9p^2 \plus{} 31p \minus{} 81)}{p^2}\ge 0,\forall p\ge 3
\]
$ \bullet$ The first case: $ 18q\le 58 \plus{} 12p$ we have $ f'(q)\le 0$ so

\[ f(p,q)\ge f\left(p,\frac {9 \plus{} p^3}{4p}\right)\]
\[ \equal{} \frac {(p \minus{} 3)(9p^5 \minus{} 21p^4 \plus{} 137p^3 \minus{} 291p^2 \plus{} 615p \minus{} 243)}{16p^2}\ge 0,\forall p\ge 3\]
In fact:
\[ p^5 \minus{} 21p^4 \plus{} 137p^3 \minus{} 291p^2 \plus{} 615p \minus{} 243
\]

\[ \equal{} (p \minus{} 1)^4(9p \plus{} 15) \plus{} (p \minus{} 1)^2(143p \minus{} 59) \plus{} 405p \minus{} 199\ge 0,\forall p\ge 3
\]
We are done  :D

[quote="zaizai-hoang"]I don't think it is hard  :) Using the following inequality:
\[ \frac {1}{4} \plus{} \frac {\ln{x}}{8}\ge \frac {x}{x^2 \plus{} 3}
\]
[/quote]
We can complete as this one :)
The above inequality holds for all $ x \ge \frac {1}{3}$. Hence
Assume $ a \ge b \ge c$, we consider two cases
Case 1. If $ c \ge \frac {1}{3}$, using the above inequality we can get the result.
Case 2. If $ c \le \frac {1}{3}$, with note that
\[ \frac {a}{a^2 \plus{} 3} \le \frac {1}{2\sqrt {3}}
\]

\[ \frac {b}{b^2 \plus{} 3} \le \frac {1}{2\sqrt {3}}
\]
and since $ c \le \frac {1}{3}$, we get
\[ \frac {c}{c^2 \plus{} 3} \le \frac {1/3}{(1/3)^2 \plus{} 3} \equal{} \frac {3}{28}
\]
Hence
\[ LHS \le \frac {3}{28} \plus{} \frac {1}{\sqrt {3}} < \frac {3}{4}
\]
:)

By the way, we can prove the original inequality just by the AM-GM Inequality. :)

Why you know my result is correct for all $ x\ge \frac{1}{3}$ ?

[quote="can_hang2007"][quote="zaizai-hoang"]Why you know my result is correct for all $ x\ge \frac {1}{3}$ ?[/quote]
Well, the secret come from the second case indeed. :)[/quote]
Can you explain more clearly?  :P

[quote="dduclam"][quote="can_hang2007"][quote="tronghieu"]$ a,b,c > 0,a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.Pove that
$ \frac {a}{ab \plus{} 3} \plus{} \frac {b}{bc \plus{} 3} \plus{} \frac {c}{ca \plus{} 3} < \equal{} 3/4$[/quote]
It is the problem in my ebook, it must be $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ . :)[/quote]

A similar one,not easy:

$ \frac a{a^2 \plus{} 3} \plus{} \frac b{b^2 \plus{} 3} \plus{} \frac c{c^2 \plus{} 3} \le \frac3{4}$  for $ a,b,c > 0$ such that $ abc \equal{} 1$.[/quote]
That is my solution.
Using[b] Cauchy-Schwarz Inequality[/b].we have:
$ \frac{1}{3}\sum a  \minus{} \sum {\frac{a}{{a^2  \plus{} 3}}}  \equal{} \sum {\frac{{a\,^3 }}{{3a^2  \plus{} 9}}}  \ge \frac{{(a^2  \plus{} b^2  \plus{} c^2 )^2 }}{{3\sum\limits_{sym} {a^3 }  \plus{} 9\sum\limits_{sym} a }}$
We need prove$ \frac{{(a^2  \plus{} b^2  \plus{} c^2 )^2 }}{{3\sum\limits_{sym} {a^3 }  \plus{} 9\sum\limits_{sym} a }} \ge \frac{1}{3}\sum a  \minus{} \frac{3}{4}$
setting p=a+b+c;q=ab+cbc+ca,we 'll prove
$ f(q) \equal{} \frac{9}{4}p^3  \minus{} p^2 q \plus{} 4q^2  \plus{} \frac{{15}}{4}p \plus{} \frac{{27}}{4} \minus{} \frac{{27}}{4}pq \minus{} 3p^2  \ge 0$
$ f'(q) \equal{}  \minus{} p^2  \minus{} \frac{{27}}{4}p \le 0$ so $ f(q)$ is a convex fuction.So
$ f(q) \ge f(\frac{{p^2 }}{3}) \equal{} (p \minus{} 3)(\frac{{p^3 }}{3} \plus{} \frac{{p^2 }}{3} \minus{} 2p \minus{} \frac{9}{4}) \ge 0$
the inequality is trivial. :)

[quote="zaizai-hoang"]I don't think your solution was correct  :wink:
In this problems, $ f'(q) \equal{} \minus{} p^2 \plus{} 8q \minus{} \frac {27p}{4}$  :wink:[/quote]
I dont think so. :(   :wink:
I think that's $ f'(q)$

What do you mean? I am sure that your solution is wrong. Because maybe you have made mistake in calculating derivative of $ f(q)$. We don't have $ f(p^2/3)\ge 0,\forall p\ge 3$. Please check your solution again!


vector
linear algebra
linear algebra unsolved

I have an unsolved problem.

Let be V vector space with scalar product and (v1,...vn) linearly independent family , calculate number of orthonormals (e1,...,en) wich we can build using Gram–Schmidt if

span{v1,...,vi}=span{e1,...,ei}, for all i=1,n.

my teaher told me that the answer is 2^n but i don't really know how to build those vectors.

The most powerful problem solving technique that I've ever learned is this: look at small cases!

In this case, you will get the greatest insight from considering the problem when $ n\equal{}1$.  There are two choices: what are they?  When you are sure you understand that, move on to $ n\equal{}2$.  Draw a picture.  When you see the $ 2^2\equal{}4$ possibilities, you should understand the original problem completely. :)

This should specify that it's the real version. In a complex space, there are infinitely many options.


find all functions $ f : \mathbb R \rightarrow \mathbb R$  $ (x\plus{}y)(f(x)\minus{}f(y))\equal{}(x\minus{}y)f(x\plus{}y)$

$ \frac{f(x)\minus{}f(y)}{x\minus{}y}\equal{}\frac{f(x\plus{}y)}{x\plus{}y}$
if $ f$ is continous then $ y\to x$ gives $ 2xf'(x)\equal{}f(2x)$


calculus
derivative
function
limit
calculus computations

Construct a multivariable function, all of whose directional derivatives are 0 at the origin, but which, nevertheless, is not differentiable at the origin.

Here's an idea: set $f(x,y)=0$ for all points except those with $x>0$ and $0<y<x^{2}$. This already guarantees that all directional derivatives are $0$ at the origin. It remains to mess the function up in the remaining part of the plane.

Consider $f(x,y)=\frac{xy^{3}}{x^{2}+y^{6}}$ at every point except the origin, with $f(0,0)=0.$ This has partial derivatives equal to zero at the origin. Now take a direction $(a,b)$ that's not along either axis. The directional derivative at the origin is

$\lim_{t\to0}\frac{f(ta,tb)}{t}= \lim_{t\to0}\frac{ab^{3}t^{4}}{a^{2}t^{3}+b^{6}t^{7}}=0.$

But $f$ is not even continuous at the origin, much less differentiable.


Please help me to solve these two problems.
1. Prove that there are no positive integers $ a,b,n > 1$ such that
$ (a^n \minus{} b^n)|(a^n\plus{}b^n)$
2. If $ a$ and $ b > 2$ are any positive integers, prove that $ 2^n \plus{} 1$ is not
divisible by $ 2^b \minus{} 1$.


geometry
circumcircle
symmetry
angle bisector
geometry unsolved

[size=167]ABC is a acute triangle,CD is perpendicular to AB(the perpendicular foot is D).M is the midpoint of AB.line $ l$ passed through the point M. s.t., $ l$ and CA intersect at K;  $ l$ and CB intersect at L. CK=CL. Point S is the circumcenter of triangle CKL. Proof that SD=SM.[/size]

 :)
Thanks!

Pick $ A'$ on ($ CB$ such that $ CA' \equal{} CA$ , denote $ V$  the midpoint of the arc $ AB$ that not contains $ C$ on the circle (ABC); obviously, $ CV$ is the angle bisector of $ \angle ACB$ and, due to symmetry, $ A'V \equal{} AV$ , but $ AV \equal{} BV$, so $ A'V \equal{} BV$ (*). It's clear also that $ AA'$ is perpendicular to $ CV$, hence $ AA' \parallel ML$ and, $ M$ is the midpoint of $ AB$, $ L$ is the midpoint of $ A'B$; the triangle $ A'VB$ is isosceles - see (*) - it follows that $ VL$ is its altitude. The same way prove that $ VK$ and $ AC$ perpendicular, hence $ CKVL$ is a cyclic quadrangle , and the circle of diameter $ CV$ is its circumcircle. $ DM$ is the orthogonal projection of the segment $ CV$ on $ AB$, $ D$ and $ M$ are symmetric about the midpoint of $ CV$ which is also midpoint of $ AB$ , then $ SD \equal{} SM$ . Our proof is completed

Let $ P, Q, N$ be the midpoints of $ MD, CL, KC.$
Applying [b]Carnot[/b]'s theorem, we will show that $ AP^2 \minus{} BP^2 \plus{} QB^2 \minus{} QC^2 \plus{} NC^2 \minus{} NA^2 \equal{} 0$ and complete the proof.
$ \Leftrightarrow AP^2 \minus{} BP^2 \plus{} (QL \minus{} LB)^2 \minus{} QC^2 \plus{} NC^2 \minus{} (AK \plus{} KN)^2 \equal{} 0$
$ \Leftrightarrow AP^2 \minus{} BP^2 \plus{} LB^2 \minus{} 2QL.LB \minus{} AK^2 \minus{} 2AK.KN \equal{} 0 (*)$
Applying [b]Menelaus[/b]'s theorem, we can easy prove that $ AK \equal{} BL$
Therefore $ (*) \Leftrightarrow AP^2 \minus{} BP^2 \equal{} CL.LB \plus{} AK.KC \equal{} 2AK.KC$
$ \Leftrightarrow (AD \minus{} DP)^2 \minus{} (BD \plus{} DP)^2 \equal{} 2AK.KC \equal{} \frac {1}{2}(AD^2 \minus{} DB^2)$
$ \equal{} \frac {1}{2}(AC^2 \minus{} BC^2) (1)$
But $ AK \equal{} KC \plus{} BC \equal{} \frac {1}{2}(AC \plus{} BC)$
So $ 2AK.KC \equal{} (AC \plus{} BC)(AC \minus{} \frac {AC \plus{} BC}{2})$
$ \equal{} \frac {1}{2}(AC \plus{} BC)(AC \minus{} BC) \equal{} \frac{1}{2}(AC^2 \minus{} BC^2) (2)$
From $ (1)$ and $ (2)$ we are done.


How many pairs $ (x,y)$ of positive integers with $ x\leq y$ satisfy $ gcd(x,y) \equal{} 5!$ and $ lcm(x; y) \equal{} 50!$?

[hide="Answer"]$ 2^{14}$[/hide]

[hide="Solution"]There are 15 primes less than 50: $ 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47$. Each of these primes has a different exponent in the prime factorization of $ 5!$ and $ 50!$. We require that one of $ x$ and $ y$ have the exponent for $ 5!$ and the other have it for $ 50!$. Then $ (x,y) \equal{} 5!$ and $ [x,y] \equal{} 50!$ with respect to that prime. So we have 2 choices for assigning exponents for each of the 15 primes, so $ 2^{15}$. But now we have to consider $ x\leq y$. Since clearly $ x\neq y$ and $ (a,b)$ is a solution if and only if $ (b,a)$ is a solution, taking all pairs $ (x,y)$ that work for which $ x\leq y$ is simply cutting the set in half. So the answer is $ 2^{15}/2 \equal{} 2^{14}\equal{} 16384$.[/hide]


Hi, I am Alan Zhao from Evergreen Valley High School in San Jose, California.  I am asking around to see if there is any high school or program in the San Francisco Bay Area with an open slot for the AMC 12B test.  My school's administration is not at all receptive to the idea of math contests of math club in general (I am president) and flatly refused to register us for the AMC B test because the test takes place over a school break.  As a junior, I really want to  have at least a shot at qualifying for USAMO (I got a 4 on AIME last year).  I am willing to pay the entire registration fee for the test.  Feel free to email me at alanzzhao@yahoo.com or AIM me at xprometheasx.

Thanks!

Wow, I'm amazed at the activity of this forum.  Thirty minutes after my original post, I already have 3 offers to take it at other schools. Art of Problem Solving is awesome!

Haha.  Of course.  :roll:   Then I chose the right time to post.  Oh and don't hesitate to send me a pm just because I have 3 offers already.  I'd like to get as many options as possible  :) .

[quote="now a ranger"]This may be due to the AMCs having been just a day before this day.  :D[/quote]

And because the OP lives in California?


linear algebra
matrix
vector
linear algebra unsolved

Supposed $ A$ is a matrix representation of the linear transformation $ T$ so that $ T(B)\equal{}B^T$ for all $ B \in M_n(C)$ determine all the possible eigenvalues of $ A$. Thank you for the solution before.

Since $ T^2$ is the identity operator, the only possible eigenvalues are the roots of the equation $ x^2\minus{}1\equal{}0.$

The eigenspace of the eigenvalue $ 1$ consists of the symmetric matrices, and has dimension $ \frac{n(n\plus{}1)}2.$

The eigenspace of the eigenvalue $ \minus{}1$ consists of the skew-symmetric matrices, and has dimension $ \frac{n(n\minus{}1)}2.$

As those dimensions add up to $ n^2,$ the dimension of the whole vector space, the operator is diagonalizable.

One note: when I say symmetric, I mean precisely that, not Hermitian. And the operator that sends $ B$ to $ B^*$ (the Hermitian adjoint, or conjugate transpose) is not linear over $ \mathbb{C}.$

If we were to change the underlying field from $ \mathbb{C}$ to a field of characteristic $ 2,$ then there would be only one eigenvalue ($ 1$, which is the same thing as $ \minus{}1$) and the operator $ T$ would fail to be diagonalizable.


combinatorics solved
combinatorics
Sets
Enumeration
duplicate

$a$ and $b$ are positive coprime integers. A subset $S$ of the non-negative integers is called [i]admissible[/i] if $0$ belongs to $S$ and whenever $k$ belongs to $S$, so do $k + a$ and $k + b$. Find $f(a, b)$, the number of admissible sets.

[color=#cc0000][mod says: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=219750& ][/color]

for a solution go to:

http://www.kalva.demon.co.uk/short/soln/sh00c6.html

cheers!


I'm gonna take AMC 8. Does anybody have any advice?
What can I do for Prep.?
Thanks so much! :lol:

You could study the past AMC8s to get an idea.But overall i gess that the AoPs books r useful.

AMC8 is pretty easy. It only has a tiny bit of logic. You can get better in AMC8 just by studying your text book. It's the only contest I have taken that has textbook problems.

It does not include as much problem solving as MATHCOUNTS, as varunronks said.

For practice, you should first learn the basic curriculum up to Grade 8 that is taught in school, then take some practice tests, look for weaknesses, and study them. Then, you should start improving your time. Also, look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=281776]here[/url].

Also in the Mathcounts resources here forum there are people willing to trade past amc 8 tests

Make sure you have good test taking skills so you don't mess up on easy problems

DOUBLE CHECK! That's the biggest tip I have. Do not mess up on the problems you solve, as the AMC 8 is very easy, and so missing a single question will drop you in the rankings.

There are some previous competition problems here:
https://www.artofproblemsolving.com/Wiki/index.php/AMC_8_Problems_and_Solutions

The AMC 8 is a piece of cake. Trust me!
Just practice past contests!

umm... Yeah... Like Pink Panther said, the AMC8 is like a warm-up... AMC10-AIME is what I'm studying for...

This is what killed me though: silly mistakes

WATCH OUT!!! If you accidentally forget a negative or something, it could totally mess you up...

That's quite insulting to people who don't score quite as high on the AMC 8 as you do.

To the OP: As the test is multiple choice, sometimes you can just look at the choices to find the correct answer.  Also, the multiple choice factor sometimes helps you if you get careless.  For example, if your answer was 360, and two answer choices are 360 and 720, check to make sure you don't need to multiply by 2 at the end.


For which non-negative integers $ a<2007$ the congruence	$ x^2\plus{}a \equiv 0 \mod 2007$ has got exactly two different non-negative integer solutions?
That means, that there exist exactly two different non-negative integers $ u$ and $ v$ less than $ 2007$, such that $ u^2\plus{}a$ and $ v^2\plus{}a$ are both divisible by $ 2007$.

It seems that I have made mistakes in my solution as I got a strange result.
[hide]The problem is equivalent to
  Find $ 0 < a < 2007$ such that
[b](i)[/b] $ 9|a$  and  $ 9k^2 \equal{} \minus{} \frac {a}{9} (mod\; 223)$ or
[b](ii)[/b] $ 223|a$  and  $ 223k^2 \equal{} \minus{} \frac {a}{223}(mod\; 9)$ for some integers $ k$
where (i) seems impossible to be simplified further.

[b][i]Proof:[/i][/b]
For $ u,v \neq 0$ , $ u^2 \equal{} (2007 \minus{} u)^2(mod 2007)$ so there must be at least an integer $ v$ such that $ u^2 \equal{} v^2$ for any fixed $ u$.
$ u^2 \equal{} v^2 (mod\;2007)\Leftrightarrow u^2 \equal{} v^2(mod \;9) and u^2 \equal{} v^2(mod \;223)$
$ \Leftrightarrow v \equal{} u \; or\; \minus{} u (mod\;9)$   and   $ v \equal{} u\; or \; \minus{} u(mod\;223)$
$ \Leftrightarrow v_1 , v_2 ,v_3$ are the only three distinct possible values of $ v$ where $ v_1 \equal{} \minus{} u(mod \;2007) ;v_2 \equal{} u(mod\;9), \minus{} u(mod\;223);v_3 \equal{} \minus{} u(mod \;9),u(mod \;223)$ unless $ u \equal{} v \equal{} 0(mod \;9)$  or  $ u \equal{} v \equal{} 0(mod \;223)$, in these two case $ v$ has one value only (as $ v_1 \equal{} v_2$ and $ v_3\equal{}u$ in 1st case,$ v_1\equal{}v_3$ and $ v_2\equal{}u$ in 2nd case)
The case $ u \equal{} v \equal{} 0(mod\;9)$ is [b](i)[/b] and the case $ u \equal{} v \equal{} 0(mod\;223)$ is [b](ii)[/b].[/hide]


inequalities
geometry
college
geometric inequality
inequalities proposed

[quote="hungkhtn"]Hi, Gemath. Following inequality is dedicated to you. First I intend to send this problem to Mathematics and Youth Magazines, but when I see you in this forum (and I did know you when I am at 10th Class  :lol: ), I have to present it (it is the only one geometric inequality in my book  :lol: ). I must say that I am not so good at Geometric inequality, so maybe the following is not hard (but I think it is very nice).

$\bigstar$ Problem.
In every non-obtuse triangle $ABC$ which has the area $1$, find the triangle make the following expression minial
\[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}. \]
@ Gemath. I want to confirm some information. Are you Luong The Vinh? If that, do you know Nguyen Duc Thinh? He took part in IMO 45 with me in Greece and he said that you are so good at geometric. I believe that he is the member who learn Geometry the best in Vietnamese team at that time. He also say he does not fear any one but Luong The Vinh  :blush:  So if you are Luong The Vinh (and you must study in England, mustn't you?), nice to meet you. If not, nice to meed you, too  :D[/quote]

Thanks for nice present  :D hungkhtn I will try it if I can , but I am not any men that you remind above !

Ok here is my solution we can prove the lemma given $x,y,z>0$ and
$xy+yz+zx=1$ we have $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{5}{2}$
Equality holds iff $x=y=1,z=0$ and others permution.
I know a solution maybe of Mathcircle Berkeley but I won't post it I want to find nice solution from others mathlinkers!
Now let $x+y=a,y+z=b,z+x=c$ note that $xz+yz+zx=r(4R+r)=1$ we have problem
[b]Problem[/b]:In any triangle such that $r(4R+r)=1$ minimum of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is $\frac{5}{2}$ iff $a=b=1,c=2$( the triangle can degrade).
Now with non-obtuse triangle $ABC$ we always have $a'=a^{2},b'=b^{2},c'=c^{2}$ are three side of a triangle with area of it $S'=r(4R+r)=1$ are three side of a tringle apply the above problem we have  minimum of $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}$ is $\frac{5}{2}$ iff $a'=a^{2}=b'=b^{2}=1,c'=c^{2}=2\Leftrightarrow a=b=1,c=\sqrt{2}\Leftrightarrow \triangle ABC$ is right isoceles with side $1,1,\sqrt{2}$.
Actually we can get the inequalities sharper than my lemma we will get the problem find minimum of $\frac{1}{a^{k}}+\frac{1}{b^{k}}+\frac{1}{c^{k}},k\ge 0$ when $S=1$

Yes, it is exactly the right answer. The minimum is $\frac{5}{4}$, holds for the right isoceles $ABC$. I think this result is quite interesting for   a problem that many of you will believe that the answer only belongs to equilateral triangle.


Aside from natural variances in the test from year-to-year, there were several intentional changes in the difficulty level of the competitions over the years.

In 1983, the AHSME got significantly "easier".  In 2000, the AHSME split into the AMC-10 and AMC-12.  The AMC-10 tended to pick up the easier questions, while the AMC-12 tended to pick up the "harder" questions.

Read much more is in this interesting history:  [url]http://www.math.uncc.edu/~hbreiter/AHSME/Steve&Leo.html[/url]

[quote]In 1983, the AHSME got significantly "easier".[/quote]

This was because the AIME was introduced in 1983, I believe.  As a result, the AHSME no longer needed to feature problems hard enough to select USAMO-caliber people.

[quote="beta"]Which is harder, AHSME or AMC12?

I find AHSME to be a lot easier, I did 1995 AHSME as a practice and I solved 27 problems in one and half hour(but of course, I got 2 wrong), but I only solved 15 problems on the 2004 AMC12( and still got 3 wrong).

Is AHSME really that much easier or is it just something wrong with me?[/quote]

The 1994 and 1995 AHSMEs were uncharacteristically easy compared the exams from the several years before.  In 1994 there were over 100 perfect scores and well over a dozen in 1995.  This compared with about 1-3 per year for the several years prior.

I think that the AMC committee has allowed the exam to evolve every five or six years and gradually change into a problem solving exam that covers more and more current mathematical ideas, though not stretching into the obscure (except maybe 1 problem per exam).  I think these changes come with it a volatility in testing results.  I think the overall difficulty gradually rises, but the exams from the early 90s seem to me to be more difficult than last year's exams, but approximately on par with this year's AMCs12s.


I have two nephews Sanjay (Nickname can be S) and Pratik (nick name can be P).  You can  take it as given that both are good in math.  I once  selected 2 positive integers, both  greater than 1 (equal or greater than 2), and less than 50.  Let the sum of these numbers be S and Product of these numbers be P. I gave sum of these numbers (S) to S(anjay), and product of these numbers (P) to P(ratik).

Now S calls P on phone and  tells; "You can never guess the  sum S"
Little later P Calls S , " Yes, now I do know the  value of S"
Little later S calls P . " Big deal, now I also know the value of P"

What were the numbers I selected?

Cute -- I'll need a little bit of time, though.  This is a lot like the "how old are my children?" question, only tougher.

Okay, so, from the first statement I get [hide]that we know that S is not the sum of two prime numbers.  So in particular we know that it is odd.[/hide] I'll keep working from there.

you actually know a tiny tiny bit stronger than that  eg [hide] 2 and 4 aren't both prime, but those two adding together to 6 won't work either (since 1 and 8 isn't a possibility)[/hide]

[hide]

like JBL said, the sum cannot be the sum of two primes, so it must be from the list 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, etc.  Now, we must look at each possible sum, in a guess and check method to see what works.  We know that P is able to tell what the two numbers are, which means of all pairs (a,b) a and b positive integers > 1 and a :ge: b, with a*b = product, all but one pair has the property that a+b can be written as the sum of two primes (thus he can eliminate all but one).  When S now knows the two numbers, this tells us that of all pairs (a,b) a and b positive integers > 1 and a :ge: b, with a+b = sum, if we consider a*b, only one of the pairs would have led P to know what the two numbers are.

I'll leave this up to you, but it is easy to show that 11 does not work, and the next possible sum is 17.

17 = (2+15),(3+14),(4+13),(5+12),(6+11),(7+10),(8+9), so we must go through all of the products, which are 30, 42, 52, 60, 66, 70, 72 respectively.  Now, 30 = (2*15), (3*10), (5*6), but note that 2+15 = 17, and 5+6 = 11 both cannot be written as the sum of two primes, so 30 is not possible.  42 = (2*21),(3*14),(6*7), but 2+21=23 and 3+14=17 both cannot be written as the sum of two primes.  If i continue in this fashion, 30,42,60,66,70,72 will not work, but 52 will.  52 = (2*26),(4*13), and (4,13) is the only pair with sum that cannot be written as the sum of two primes, so that means that P would get it, since only one pair of (a*b) couldn't be written as the sum of two primes, and S would get it because only 52 worked.

Therefore, the two numbers are 4 and 13.

[/hide]


quadratics
inequalities
number theory proposed
number theory

Find all the $ (a,b,c)\in \mathbb{N}\times \mathbb{N}\times \mathbb{N}$ such that $ abc\plus{}ab\plus{}c\equal{}a^3$ .

Obviously $ a|c$ so let $ c \equal{} ak$ and then divide the equation by $ a$. We have $ abk \plus{} b \plus{} k \equal{} a^{2}$. Examine as quadratic towards $ a$ we have $ D \equal{} b^{2}k^{2} \plus{} 4b \plus{} 4k$ and $ D$ must be perfect square. If $ b,k\ge 2$ then $ \left(bk \plus{} 1\right)^{2} > b^{2}k^{2} \plus{} 4b \plus{} 4k > \left(bk\right)^{2}$. So the solutions are $ \left(a,b,c\right) \equal{} \left(2,1,2\right),\left(3,1,6\right),\left(3,2,3\right),\left(2,2,1\right)$.

Actually I noticed a mistake in my solution. The inequality $ \left(kb \plus{} 1\right)^{2}>k^{2}b^{2} \plus{} 4k \plus{} 4b$ is not true only for one of $ b$ and $ k$ smaller than 2. It coulde be transformed in $ 2\left(k \minus{} 2\right)\left(b \minus{} 2\right) > 7$. Check all the possibilities till $ b$ or $ k$ is smaller or equal to five and we get solutions $ \left(a,b,c\right) \equal{} \left(4,3,4\right),\left(4,1,12\right),\left(5,4,5\right),\left(5,1,20\right),\left(6,5,6\right),\left(6,1,30\right)$. If $ b > 5$ (respectively if $ k > 5$) we have $ k \equal{} \{1,2\}$. So $ D \equal{} 4\left(b^{2} \plus{} b \plus{} 2\right)$ and $ D \equal{} \left(b \plus{} 2\right)^{2}$. In the first case since $ b > 5$ we have $ \left(k \plus{} 1\right)^{2} > k^{2} \plus{} k \plus{} 2 > k^{2}$. In the second case we get infinitely many solutions : $ \left(a,b,c\right) \equal{} \left(m \plus{} 1,m,m \plus{} 1\right),\left(m \plus{} 1,1,m^{2} \plus{} m\right)$ for every $ m\in \mathbb{N}$.


logarithms
inequalities
function
inequalities proposed

Given $ a, b, c > 0$ and $ abc \equal{} 1$. Prove that: $ 81(1 \plus{} a^2)(1 \plus{} b^2)(1 \plus{} c^2) \leq\ 8(a \plus{} b \plus{} c)^4$

@arqady: It's an old result of hungkhtn and you can see a long proof by mixing variable in some books.
But now, I found an easy proof by Am-Gm for it. Hope that you'll take care of it, my dear friend :)

Besides, the general problem is also an open question in some forums  :)

Can you post a proof using mixing variables dear arqady???

Thank you.

Now, here is my proof  :)
Because $ abc \equal{} 1$, so easily to see that $ a^2 \plus{} 1 \equal{} \sqrt [3]{a^6} \plus{} \sqrt [3]{a^2b^2c^2} \equal{} \sqrt [3]{a^2}\left(\sqrt [3]{a^4} \plus{} \sqrt [3]{b^2c^2}\right)$

$ b^2 \plus{} 1 \equal{} \sqrt [3]{b^2}\left(\sqrt [3]{b^4} \plus{} \sqrt [3]{c^2a^2}\right)$ ; $ c^2 \plus{} 1 \equal{} \sqrt [3]{c^2}\left(\sqrt [3]{c^4} \plus{} \sqrt [3]{a^2b^2}\right)$

So, $ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \equal{} \sqrt [3]{a^2b^2c^2}(\sqrt [3]{a^4} \plus{} \sqrt [3]{b^2c^2})(\sqrt [3]{b^4} \plus{} \sqrt [3]{c^2a^2})(\sqrt [3]{c^4} \plus{} \sqrt [3]{a^2b^2})$
$ \equal{} (\sqrt [3]{a^4} \plus{} \sqrt [3]{b^2c^2})(\sqrt [3]{b^4} \plus{} \sqrt [3]{c^2a^2})(\sqrt [3]{c^4} \plus{} \sqrt [3]{a^2b^2})$ (because of $ abc \equal{} 1$)

Now, our ineq can be write into $ 81(\sqrt [3]{a^4} \plus{} \sqrt [3]{b^2c^2})(\sqrt [3]{b^4} \plus{} \sqrt [3]{c^2a^2})(\sqrt [3]{c^4} \plus{} \sqrt [3]{a^2b^2}) \le 8(a \plus{} b \plus{} c)^4$
This is an homogeous ineq, so don't need the condition $ abc \equal{} 1$.
Now, assume that $ a \plus{} b \plus{} c \equal{} 3$. We need to prove $ (\sqrt [3]{a^4} \plus{} \sqrt [3]{b^2c^2})(\sqrt [3]{b^4} \plus{} \sqrt [3]{c^2a^2})(\sqrt [3]{c^4} \plus{} \sqrt [3]{a^2b^2}) \le 8$

Using Am-Gm's ineq with the form $ \sqrt [3]{xyz} \le \dfrac{x \plus{} y \plus{} z}{3}$, we obtain $ \sqrt [3]{a^4} \plus{} \sqrt [3]{b^2c^2} \equal{} \sqrt [3]{a^2.a.a} \plus{} \sqrt [3]{bc.bc.1} \le \frac {a^2 \plus{} a \plus{} a}{3} \plus{} \frac {bc \plus{} bc \plus{} 1}{3}$.

Or $ \sqrt [3]{a^4} \plus{} \sqrt [3]{b^2c^2} \le \dfrac{a^2 \plus{} 2bc \plus{} 2a \plus{} 1}{3}$

Similarly, we also have $ \sqrt [3]{b^4} \plus{} \sqrt [3]{c^2a^2} \le \dfrac{b^2 \plus{} 2ca \plus{} 2b \plus{} 1}{3}$ ; $ \sqrt [3]{c^4} \plus{} \sqrt [3]{a^2b^2} \le \dfrac{c^2 \plus{} 2ab \plus{} 2c \plus{} 1}{3}$

So,
$ (\sqrt [3]{a^4} \plus{} \sqrt [3]{b^2c^2})(\sqrt [3]{b^4} \plus{} \sqrt [3]{c^2a^2})(\sqrt [3]{c^4} \plus{} \sqrt [3]{a^2b^2}) \le$
$ \le \dfrac{(a^2 \plus{} 2bc \plus{} 2a \plus{} 1)(b^2 \plus{} 2ca \plus{} 2b \plus{} 1)(c^2 \plus{} 2ab \plus{} 2c \plus{} 1)}{27} \equal{} P$

Using Am-Gm for three numbers again with the form $ xyz \le \dfrac{(x \plus{} y \plus{} z)^3}{27}$, we have

$ P \le \dfrac{(a^2 \plus{} 2bc \plus{} 2a \plus{} 1 \plus{} b^2 \plus{} 2ca \plus{} 2b \plus{} 1 \plus{} c^2 \plus{} 2ab \plus{} 2c \plus{} 1)^3}{27^2} \equal{}$
$ \equal{} \dfrac{[a \plus{} b \plus{} c)^2 \plus{} 2(a \plus{} b \plus{} c) \plus{} 3]^3}{27^2} \equal{} 8.$

The proof completed!

PS: It's also a nice problem in our new book "Inequalities With Beautiful Solutions" (Vasile Cirtoaje - V.Q.B.Can - T.Q.Anh, GIL Publishing House 2009)  :)

[quote="nguoivn"]Given $ a, b, c > 0$ and $ abc \equal{} 1$. Prove that: $ 81(1 \plus{} a^2)(1 \plus{} b^2)(1 \plus{} c^2) \leq\ 8(a \plus{} b \plus{} c)^4$
[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3427747#p3427747

We note $ p=a+b+c,q=ab+bc+ca,r=abc=1 $, result $ p\ge 3,q\ge 3 $
Inequality returns to:
1)     $ 81(p-1)^2+81(q-1)^2\le 8p^4 $
I will use the well-known inequality, namely:
$ a,b,c\ge 0\Rightarrow a^2+b^2+c^2+2abc+1\ge 2(bc+ca+ab)\Leftrightarrow (a+b+c)^2+2abc+1\ge 4(bc+ca+ab) $
$ \Leftrightarrow p^2+3\ge 4q\Leftrightarrow p^2-1\ge 4(q-1) $
2)     $ q-1\le\frac{p^2-1}{4} $
It is sufficient that:
3)      $ 81(p-1)^2+81\frac{(p^2-1)^2}{16}\le 8p^4\Leftrightarrow 81s^2[4+(1+s)^2]\le 8(1+2s)^4 $
where $ s=\frac{p-1}{2},s\ge 1 $
3')      $ 81s^2(s^2+2s+5)\le 6(1+8s+24s^2+32s^3+16s^4) $
$ \Leftrightarrow 47s^4+94s^3-213s^2+64s+8\ge 0\Leftrightarrow $
$  (s-1)(47s^3+141s^2-72s-8)\ge 0 $
The last inequality is true because $ s\ge 1 $ and function $ h(s)=47s^3+141s^2-72s-8 $
is increasing, therefore $ h(s)\ge h(1)>0 $
____
Marin Sandu

let $a+b+c=p,ab+bc+ca=q$, the inequality becomes
$$p^4-6p^2+24p-12q^2+12q-27 \ge0$$
which is a quadratic function of q.
since$q\ge 3>1/2$, we only need to prove the inequality when q is maximal.
Schur gives $q\le \frac{p^3+9}{4p}$ so it remains
$$p^4-6p^2+24p-27-\frac{3(p^6+18p^3+81)}{4p^2}+\frac{3(p^3+9)}{p} \ge0$$
$$(p-3)(p^5+3p^4-3p^3+33p^2-9p+81)\ge0$$
which is obvious since $p\ge 3$.

[quote=Vasc]A similar inequality is the following:
If $a,b,c$ are positive real numbers such that $abc=1$, then
$$3(2a^2+1)(2b^2+1)(2c^2+1)\le (a+b+c)^4.$$[/quote]
Generality ([url = "http://artofproblemsolving.com/community/c6h460911"]see[/url])
[quote=Nguyenhuyen_AG]Let $a,b,c>0$ such that $abc=1$ and $k\ge 0.$ Prove that \[(a+b+c)^4(k+1)^3\ge 81(a^2+k)(b^2+k)(c^2+k).\][/quote]