trigonometry
geometry
angle bisector
perpendicular bisector
geometry unsolved

Let $ ABC$ be a triangle in which $ AB \equal{} AC$ and let $ I$ be its in-centre. Suppose $ BC \equal{} AB \plus{} AI$. Find $ \angle{BAC}$

[hide]Let $ AB\equal{}AC\equal{}x$, $ \angle ABC\equal{}\angle ACB\equal{}\beta$. We have $ \angle ABI\equal{}\beta/2$, $ \angle BAI\equal{}90^{\circ}\minus{}\beta$, $ \angle AIB\equal{}90^{\circ}\plus{}\beta/2$. So $ \frac{AI}{\sin\frac{\beta}2}\equal{}\frac{AB}{\sin(90^{\circ}\plus{}\frac{\beta}2)}$ hence $ AI\equal{}x\tan\frac{\beta}2$. We also have $ BC\equal{}2x\cos\beta$. Thus $ BC\equal{}AB\plus{}AI$, which gives $ 2x\cos\beta\equal{}x\plus{}x\tan\frac{\beta}2$ and $ 2\cos\beta\equal{}1\plus{}\tan\frac{\beta}2$, so $ 2(2\cos^2\frac{\beta}2\minus{}1)\equal{}1\plus{}\tan\frac{\beta}2$. Let $ \tan\frac{\beta}2\equal{}k$. We get $ 2\left(\frac{2}{1\plus{}k^2}\minus{}1\right)\equal{}1\plus{}k$

\[ \frac{1\minus{}k^2}{1\plus{}k^2}\equal{}\frac{1\plus{}k}2 \\ 2\minus{}2k^2\equal{}1\plus{}k\plus{}k^2\plus{}k^3 \\ k^3\plus{}3k^2\plus{}k\minus{}1\equal{}0 \\ (k\plus{}1)(k^2\plus{}2k\minus{}1)\equal{}0 \\ k\equal{}\sqrt{2}\minus{}1\equal{}\tan(22.5)\] So $ \beta\equal{}45^{\circ}$ and hence $ \angle BAC\equal{}90^{\circ}$.[/hide]

Let $ AI$ hit $ BC$ at $ X$, so $ r\equal{}\frac{AX\cdot BX}{AB\plus{}BX}$. From angle bisector $ BI$ we have $ \frac{AB}{2BX\minus{}AB}\equal{}\frac{BX}{r}\equal{}\frac{AB\plus{}BX}{AX}$ or
\[ AB\cdot AX\equal{}2BX^2\plus{}AB\cdot BX\minus{}(AX^2\plus{}BX^2)\equal{}BX^2\plus{}AB\cdot BX\minus{}AX^2\\\\
\implies(BX\minus{}AX)(BX\plus{}AX\plus{}AB)\equal{}0\implies BX\equal{}AX\implies\angle BAC\equal{}90^\circ.\]

Produce $ BA$ to $ D$ such that $ AD \equal{} AI$. From the given condition, we have that $ BD \equal{} BC$. An easy angle chasing gives that $ AICD$ is cyclic. Since $ AI \equal{} AD$ , $ \angle ICA \equal{} \angle ACD$
Hence, $ \angle BCD \equal{} \frac {3C}{2}$. Sum of angles in triangle $ BCD \equal{} C \plus{} \frac {3C}{2} \plus{} \frac {3C}{2} \equal{} 4C \equal{} 180^\circ$ and Hence, $ \angle A \equal{} 90^\circ$

[img]http://img163.imageshack.us/img163/2796/39933455.png[/img]

take a point D on the interior of BC such that AB = BD and join A to D
so, AC = BI. A simple angle chasing gives angle BIA = angle ADC which is obtuse
so, in traingles AIB and ADC, using SSA CONGRUENCY FOR OBTUSE ANGLED TRIANGLES, we get
Angle A=2angle C or A=90 degree

I found at least three other solutions.

[hide]First Solution

Extend BA such that AK = AI.
First notice that CBK is isosceles and I is its circumcenter.
Therefore KIB is isosceles and  $ \angle IKA \equal{} \alpha \equal{} IBK$.
But IAK is isosceles by construccion, then by External angle, $ \angle BAI \equal{} 2\ IKA \equal{} 2\ \alpha$.
Now in triangle BAC we have: $ \angle ABC \equal{} 2\alpha \equal{} \angle ACB \equal{} \frac {1}{2} \angle BAC$. Then $ \angle BAC \equal{} 90^{\circ}$.

Sorry now I see that my approach is very close to lajanugen's but it was the first natural thing a I thought. :D
[/hide]

[hide]Second Solution.
Let WLOG, $ BC \equal{} 2$, $ AB \equal{} x$.
Let F be the orthogonal projection of I over AB. Then triangles IFA and BAM are similar (M is the midpoint of BC). $ AI \plus{} AB \equal{} BC$ so $ AI \equal{} 2 \minus{} x$. Then

$ \frac {2 \minus{} x}{x} \equal{} \frac {AI}{AB} \equal{} \frac {AF}{AM} \equal{} \frac {x \minus{} 1}{\sqrt {x^2 \minus{} 1}} \equal{} \sqrt {\frac {x \minus{} 1}{x \plus{} 1}}$.

Then solving for $ x$ we find $ x \equal{} \sqrt {2}$, and follows (the only possible root).
[/hide]

[hide]Third solution.
Choose an Cartesian reference such that $ B \equal{} ( \minus{} 1,0)$, $ C \equal{} (1,0)$ and $ A \equal{} (0,m)$, $ m > 0$.
Then AB: $ y \equal{} m(x \plus{} 1)$ and CA: $ y \equal{} m(x \minus{} 1)$.
It's easy to find that $ I \equal{} (0,i)$, $ i \equal{} \frac {\sqrt {1 \plus{} m^2} \minus{} 1}{m}$.
Using condition $ AB \plus{} AI \equal{} BC$ gives  $ m \equal{} 1$ and we are done.
[/hide]

You can also use some of the ideas in the second solution to find a trigonometric one.

lajanugen
thanx a lot for ur solution,that was pleasing.

[hide="My solution to question 1....another trigonometry approach"]

Let $ \angle BCA \equal{} \angle ABC \equal{} x$
Now $ AI \equal{} 4R sin \frac {B}{2} sin \frac {C}{2} \equal{} 4R sin^2 \frac {x}{2}$ (This is quite known otherwise it can be derived by using $ r \equal{} 4R sin \frac {A}{2} sin \frac {B}{2} sin \frac {C}{2}$)

Also by extended rule of sines,
$ AB \equal{} 2R sin x , BC \equal{} 2R sinA \equal{} 2R sin2x$

Plugging these into $ BC \equal{} AB \plus{} AI$
$ sin 2x \equal{} sinx \plus{} sin^2 \frac {x}{2}$
Bringing sin$ x$ to the other side and using the product identity and then cancelling, we have
$ cos \frac {3x}{2} \equal{} sin \frac {x}{2}$
$ cos \frac {3x}{2} \equal{} cos (90 \minus{} \frac {x}{2})$
Now since both the sides are positive and none of them is in the 4th quadrant
$ \frac {3x}{2} \equal{} 90 \minus{} \frac {x}{2}$
and so BLAH[/hide]

i sort of suck at geometry,  so i converted it into a cubic eqn in sinA/2 and then solved it by guessing 90

Let $\{D \}=AI\cap BC$, $AB=AC=b$, $BD=CD=a$, $AD=h$ and, obviously, $DI=r$. By angle bisector theorem, $\frac{AI}{DI}=\frac{AB}{BD}$, but $DI=r=\frac{S}{s}=\frac{ah}{a+b}$, hence $AI=\frac{bh}{a+b}$, therefore from the given relation we get $\frac{bh}{a+b}+b=2a$, or $b(b+h)=a(2a+b)$ $(*)$.
Next, take $M \in (BA$ produced so that $AM=AD$ and $N \in (BC$ produced so that $CN=CA$. From the isosceles triangles $AMD$ and $CAN$ we get $m(\widehat {AMD})=\frac{m(\widehat{BAD})}{2}$ and $m(\widehat {ACN})=\frac{m(\widehat{ACB})}{2}$, but from $(*)$ the quad $MADN$ is cyclic, hence $\widehat{BAD}=\widehat{ACB}$, i.e. $\widehat{ACB}=45^\circ$.

Best regards,
sunken rock

Hi , is this method ok to you ?

Let P be the point of tagency of side AB so we got a right-angled triangle AIP .

Let Q be a point of BC such that BQ = AI , since BP = BC/2 , we have :

BC = AB + AI = (AP + BC/2 ) + AI  implies  BC/2 = AP + AI  , also let R be the foot of perpendicular bisector of BC . We have BC/2 = BR = AP + AI = AP + BQ  then  QR = AP and triangle AIP conguent to triangle QIR so angle BAR = angle RQI , A,I,Q,B, are concylic but AI = BQ so it is also an isosceles trapzeium then  angle BAR = angle ABR = 90/2 = 45 . finally angle BAC = 90 (degrees) .

Here's a coordinate bash(not fully detailed though):-
Extend $AI$ to meet $BC$ at $P$(say).As $AB=AC$ $AP$ is perpendicular to $BC$..
This clarifies our choice of the origin..(Set $P$ as the origin and $AP$ and $BC$ as the coordinate axes.)
Set $C:(c,0)$ and $A:(0,a)$.Clearly $B:(-c,0)$..
Using the formula for the INCENTRE we get $I:(0$, $\frac {ac}{c+ \sqrt {a^2+c^2}})$.
Using $AB+AI=BC$ we get $c^2 - a^2=0$(after solving)..
$\angle BAC = \frac{2ac}{c^2 - a^2}$.
So $tanA$=$n.d.$
thus $\angle A=90$
Hope you like it.

Extend $BA$ to $D$ such that $AD=AI.$
$\angle AIC=90^\circ +\frac {B}{2}$
$BD=BC \implies \angle ADC=90^\circ -{B}{2}$
$\implies$ $ABCD$ is a $cyclic$ $quadrilateral$
$\angle ICA=x=\angle IDC$
$\angle AID=y=\angle AID$
$AI=ID \implies x=y$
Adding angles of $\Delta ADC$ we get $A=90^\circ$

[quote=lajanugen]Produce $ BA$ to $ D$ such that $ AD \equal{} AI$. From the given condition, we have that $ BD \equal{} BC$. An easy angle chasing gives that $ AICD$ is cyclic. Since $ AI \equal{} AD$ , $ \angle ICA \equal{} \angle ACD$
Hence, $ \angle BCD \equal{} \frac {3C}{2}$. Sum of angles in triangle $ BCD \equal{} C \plus{} \frac {3C}{2} \plus{} \frac {3C}{2} \equal{} 4C \equal{} 180^\circ$ and Hence, $ \angle A \equal{} 90^\circ$

[/quote]

What angles do you start from when angle chasing?

Dear Mathlinkers,

this result

https://artofproblemsolving.com/community/c6t48f6h1923157_a_parallel_to_bc

help to have a proof with any calculation...

Sincerely
Jean-Louis

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Perles%202.pdf  p. 20...

Sincerely
Jean-Louis

Take $D$ on $BC$ such that $CA=CD$. Hence $DB=AI$. I claim that $\angle BAC=90$, in order to prove this, we'll prove that $DI || AB \Leftrightarrow IDBA$ cyclic $(i)$, because this would improve (by simple angle chase) that if $\angle IBA=a$, then $a=90-3a$, because $IDBA$ would be an isosceles trapezoid. To prove $(i)$ just notice that $AI \perp BC$, and $CI\perp AD$ (because of bisector being perpendicular bisector in isosceles triangles), hence $I$ is the orthocenter of $ADC$, thus $\angle DIC=90+a$, $\angle IDA=\angle ADC-\angle IDC=90-a-(90-2a)=a$ and the conclusion follows


logarithms geometry Euler calculus integration limit

Here is a problem a friend gave me, but I was at a loss of words...

$\text{Find the smallest integer } n \text{ such that } 1 + \displaystyle\frac{1}{2}  + \displaystyle\frac{1}{3} + \displaystyle\frac{1}{4} + ... \displaystyle\frac{1}{n} > 4$

The only solution I proposed to him was to use your calculator, but I'm sure there is a quicker method... and fyi, the answer is:

[hide]31[/hide]

Just for future reference, is there an easy way to find sums of either infinite or partial haromic sequences??

You know the proof that the harmonic series diverges, right?  Well, that gives you an upper bound and you should be able to simply test from there.  Also, the harmonic series grows very much like the natural logarithm, so that can give you another approximation.  In the end, though, I don't think you can get around simply adding up a lot of fractions.

Of course you can also get a   fairly accurate value  as

$(1+1/2+1/3+.... 1/n) $ =  $ln(n) + 0.577215665 + 1/(2n) - 1/(12n^2) + 1/(120n^4) .... $
(edited later : typo was edited)

If n is small  you can just directly add the terms, if n is large,  above  formula  can give a fairy  accurate value up to several decimal points. .
For example even for n=10  you can have  6-7 figure accuracy. Its more accurate  for larger values of n)

The above may be a little advance  for high school math but the formula is  nice  (may be  a little less known)
 so I am putting here. ..

Of course, simplified form of above (as  JBL  mentions)  is    $ln(n) + 0.577 $  (approx when n is large)

BTW  if one took $ e^{(4-.577 )}$    you get something like 30.8 which rounds up to 31.

Hi - This is a fairy well known, you may like to google/look up in standard text books   under 'Bernoulli Numbers' or  'Euler-Maclaurin ' formula ...  Basically it comes from:

-  Approximate the area under the curve  (y=1/x)  which gives ln(x). One can easily  get upper and lower bound for the summation (within 1/2n)

-   Assume  the expansion is given by S(n) =  ln(n) +  A + B/n +C/n^2 +  ... and then  use the fact S(n+1) - S(n) = 1/(n+1)  and equate the terms by expanding both sides... this will give the values for B,  C etc..

(This can also be  seen in detail   under "Euler - Maclaurin  method for  estimationg the difference between an integral and   summation of series)

Sorry if I am not going into too much details but do hope  that you will be able to look it up where it is explained much better than I can do it here....

The formula looks like:
$(1+1/2+1/3+..1/n) = ln(n) +\gamma -   \sum {\frac{B_k}{k  n^k}}$

$B_k$ are Bernoullli coeffcients and  $\gamma$ is euler constant.

[quote="AntonioMainenti"][quote="pkothari13"][quote="Gyan"]Of course you can also get a   fairly accurate value  as

$(1+1/2+1/3+.... 1/n) $ =  $ln(n) + 0.577215665 + 1/(2n) - 1/(12n^2) + 1/(720n^4) .... $

[/quote]

how were you able to derive that equation?? what is the actual general form of it?[/quote]I'm not sure where he got part of that (edit: see Gyan's post above for that part), but I know Euler discovered that \[\lim_{n\to\infty}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}-\ln{n}\right)=.57721566490153286\ldots,\]
also called Euler's Constant, or $\gamma$. (edit 2: as a sidenote, it is surprisingly still not known whether or not $\gamma\in\mathbf{Q}$.)[/quote]

Did not realize that  I had a[b] typo[/b] (corrected in the original)  $B_4 = -1/30$ so the last term should read $1/(120n^4)$ and not $1/(720n^4)$
BTW - About  "still unknown if $\gamma$ is rational or not,  it was rumored that Hardy once said that he will give his chair to anyone who proved that  it was rational . (Read somewhere that If it is rational, and can be written  as  a/b  then b must have more than a few hundred thousands digits :)


function
ceiling function
rotation
exterior angle

A billiard ball strikes ray $BC$ at point $C$, with angle of incidence $\alpha$ as shown. The billiard ball continues its path, bouncing off line segments $AB$ and $BC$ according to the rule: angle of incidence equals angle of reflection. If $AB=BC$, determine the number of times the ball will bounce off the two line segments (including the first bounce, at $C$). Your answer will be a function of $\alpha$ and $\beta$.

I have a solution using angle calculations, but there is supposedly a reflectional/symmetry solution that operates by reflecting the path into a straight line (according to Zeitz).

[hide]
Every time the billiard ball bounces, the angle of incidence increases by $\beta$ (by the exterior angle theorem).

When the ball begins to bounce backwards, we can interpret this as an obtuse angle of incidence. When the angle exceeds $180-\beta$, the ball will not hit the walls again. Suppose the ball bounces $k$ times, including the first bounce at $C$. Then we want the minimum $k$ such that
$\alpha+(k-1) \beta \geq 180-\beta \\ k \beta \geq 180-\alpha \\ k \&\geq \frac{180-\alpha}{\beta}\\ k = \boxed{\left\lceil \frac{180-\alpha}{\beta}\right\rceil}$[/hide]

Rotate the upper side CW for $k\beta$ and find the intersections of the incidence line with the new rays. Or just see the diagram, which is much clearer than words.

Now Phelpedo's formula becomes obvious.


Let $ F$ be an increasing real function on $ 0\leq x \leq 1$ such that

(i) $ F(x/3)\equal{}F(x)/2$
(ii) $ F(1\minus{}x)\equal{}1\minus{}F(x)$

for all $ x$. Find $ F(\frac{173}{1993})$ and $ F(\frac{1}{13})$.


quadratics
algebra
polynomial
calculus
real analysis
real analysis unsolved

Greetings everyone, i'm new to this, i really liked it here.

Well, i was in solving this problem (concerning numerical methods) but i couldn't do it, so please if anyone can help me.... thanks in advance :)

""I need to solve the van der Pol equation using second order Taylor method""

here's the equation of van der pal :

y'' + :mu: (y :^2: - 1 )y' + y = 0

I could very well be wrong, but wouldn't this method of proof simply involve treating y as a quadratic polynomial in x, or something of the sort?

In any event, you probably want to put it in the calculus forum.

well, yes, the beginning must be in quadratic form (2nd order of taylors' series) , anyway, thanx pal, i'll put it in the calculus forum too.


a, b, c are positive real numbers.
Prove that
  a/(b+c)  +  b/(c+a)   +  c/(a+b) >= 3/2

It is Nesbitt ineq.
Hint
use rearengement or Chebychev.

[b]Chebyshev's Inequality[/b]
Let $a_1\geq a_2\geq ...\geq a_n$ and $b_1\geq b_2\geq ...\geq b_n$, then
$a_1b_1+a_2b_2+...+a_nb_n\geq \frac{(a_1+a_2+...+a_n)(b_1+b_2+...b_n)}{n}\geq a_1b_n+a_2b_{n-1}+...+a_nb_1$
Equality holds when $a_1=a_2=...=a_n$ or $b_1=b_2=...=b_n$
By the way, if you want more solutions, you can refer to
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=51580[/url].

AM-HM proof:
$\sum \frac{a}{b+c}=\sum\frac{a+b+c}{b+c}-3$

Since: $(\sum \frac{a+b+c}{b+c})\cdot (\sum \frac{b+c}{a+b+c}) = (\sum \frac{a+b+c}{b+c}) \cdot 2 \geq 9$.

So we get: $\sum \frac{a}{b+c}=\sum \frac{a+b+c}{b+c}-3 \geq \frac{9}{2}-3=\frac{3}{2}$.

By the way: this should be moved to the Inequality theorem section.

You can use Cauchy Schwalz's inequality extended version.

We can also use Jensen's Inequality:

First homogenize by setting a+b+c=1.

Now set $f(x)=\frac{x}{1-x}$.  It is easy to show this is convex.  Now apply Jensen's.

Try this too:

For a,b,c,d prove:

$\sum_{cyc} \frac{a}{b+c+d} \ge \frac{4}{3}$

AM-HM still works: $(\sum \frac{a+b+c+d}{b+c+d})\cdot(\sum \frac{b+c+d}{a+b+c+d}) \geq 16$.

I think the generialization should be:
$\frac{x_1}{x_2+x_3+x_4+\cdots+x_n} + \frac{x_2}{x_1+x_3+x_4\cdots+x_n}+\cdots+\frac{x_n}{x_1+x_2+x_3+\cdots+x_{n-1}} \geq \frac{n}{n-1}$
How do I re-write it using sigma notation?

[quote="shobber"]I think the generialization should be:
\[\frac{x_1}{x_2+x_3+x_4+\cdots+x_n} + \frac{x_2}{x_1+x_3+x_4\cdots+x_n}+\cdots+\frac{x_n}{x_1+x_2+x_3+\cdots+x_{n-1}} \geq \frac{n}{n-1}\]
How do I re-write it using sigma notation?[/quote]
Your inequality can still be proven by your previous method.
For the sigma notation, I think this one may be okay.
$\sum_{cyc} \frac{x_1}{x_2+x_3+x_4+\cdots+x_n}\geq\frac{n}{n-1}$

You could also write it as $\sum_{i = 1}^n \frac{x_i}{S - x_i} \geq \frac{n}{n - 1}$ where $S = \sum_{i = 1}^nx_i$ or $\sum_{j = 1}^n\left(\frac{x_j}{\displaystyle \sum_{i \neq j}x_i}\right) \geq \frac{n}{n - 1}.$


This problem must be confusing people because the bound is so weak.

Note first that f(x)=x^2n is convex for all x.  Therefore, since a+b+c=2, (a,b,c) majorizes (2/3,2/3,2/3) implying a^2n+b^2n+c^2n>=3(2/3)^2n>1/(2^2n - 2).

Oh yes. Or just aply power means.

  Darij

I don't believe the power mean inequality can be used here.  It only applies when all the variables are positive, which is not necessarily the case here.  Of course, since the bounds are so weak, I'm sure it would be easy to force all the variables to be positive.

[quote="Michael Lipnowski"]I don't believe the power mean inequality can be used here.  It only applies when all the variables are positive, which is not necessarily the case here.[/quote]

Yes, sorry, I have overlooked it. But in the case when some variables are negative, you can actually consider |a|, |b|, |c| instead, so that you have $\left| a\right| +\left| b\right| +\left| c\right| \geq 2$. Then, applying the Power Mean, you still get $\left| a\right| ^{2n}+\left| b\right| ^{2n}+\left| c^{2n}\right| >\frac{1}{2^{2n}-2}$, what is, of course, equivalent to $a^{2n}+b^{2n}+c^{2n}>\frac{1}{2^{2n}-2}$ (obviously, it's for that reason the powers must be even).

  Darij


[hide]$V=12=lwh=(2w)wh=2w^2h \Rightarrow w^2h=6.$

Minimum surface area $\Rightarrow w=h=\sqrt[3]{6}.$[/hide]

that problem is in my algebra 2 book...or a really similiar one

also, chess, you did not minimize the surface area because the top is cut off...
$(2w)(w)(h)(12) = 12$
$w^{2} = \frac{6}{h}$
$2(w^{2}) +  6(w)(h) = SA$
$\frac{12}{h} + 6\sqrt{6}\sqrt{h}$  minimizing this requires calculus
anyway the answer is $\frac{2 \sqrt[3]{9}}{3}$

volume=w*2w*h=2(w^2) * h=12
h=6/w^2
Surface area A=w*2w + 2*(w*h + 2w*h)=2w^2 +6*6/w=2w^2 + 36/w
For max. or min surface area
dA/dw=0
4w-36/w^2=0
w^3=9
w=9^(1/3)

At this w, we have min area(after calcu which I have skipped).
h=6/9^(2/3)

Well is there some errror :?:

[quote]A rectangular box having no top is to be twice as long as it is wide and is to hold 12 cubic feet. Find the height of the box that has the minimum surface area[/quote]

[hide]

 You start out with the basic formula for finding the volume of a rectangular box.

[tex]V = L [/tex] x [tex] W[/tex] x [tex]H[/tex]

let the length = [tex]x[/tex]
     the width = [tex]\frac {x}{2}[/tex]
     the height = [tex]h[/tex]

The volume of the box would then be [tex]\frac {x^2h}{2}[/tex]

Since the given volume is [tex]12 \ ft^3[/tex], we have [tex]\frac {x^2h}{2} = 12[/tex]

We can now get [tex]h[/tex] in terms of [tex]x[/tex] so [tex]h = \frac {24}{x^2}[/tex]

We now need to find the total surface area of the box.

[tex]A = 3xh + \frac {x^2}{2} = \frac {72}{x} + \frac {x^2}{2}[/tex]

You may have to use Calculus to find the answer.

[tex]\frac {dA}{dx} = x - \frac {72}{x^2}[/tex]

by finding the minimum area, we must set [tex]\frac {dA}{dx}[/tex] equal to [tex]0[/tex]

so [tex]x - \frac {72}{x^2} = 0; \ x^3 - 72 = 0[/tex] so [tex] x = \sqrt [3] {72}[/tex] or [tex]2 \sqrt[3] {9}[/tex]

so [tex]h = \frac {24}{(2 \sqrt[3] {9})^2} = \frac {6}{\sqrt[3] {81}}[/tex] or a height of approximately [tex]1.378 \ ft[/tex]

[/hide]


I, an adult, was walking on a sidewalk and came upon a Jewish religious leader.  He wanted to demonstrate his magic act to
me, but I thought silently to myself that I'm too old to be seeing it.  Besides, he was not acting at all serious with me; in fact he was just the opposite of being serious.

Do you know what I told him before I walked on?

Here is my answer:

[hide]Silly Rabbi, tricks are for kids![/hide]


inequalities
calculus
derivative
inequalities open

Hello,

Let $a,b,c$ be three  non-negative real numbers such that: $a+b+c=1$. Prove the following:

\[
2\leq \sqrt{{{1-a}\over{1+a}}}+\sqrt{{{1-b}\over{1+b}}}+\sqrt{{{1-c}\over{1+c}}}\leq
1+{{2}\over{\sqrt{3}}}.
\]

For the left one, we do the substitution, $a=x+y$ and so on, then it's equivalent to prove $\sum \sqrt{\frac{a}{b+c}} \ge 2$ with $a+b+c=2$. But then it's simple to prove that $\sqrt{\frac{a}{b+c}} \ge a$, with equality holds when $a=1$ or $a=0$. So the left inequality is true with the equality case $(a,b,c)=(1,1,0)$.

[quote="Michael"]I think that the problem can be generalized to $n$ real numbers $a_1,a_2,\cdots,a_n$ such that $a_1+a_2+\cdots+a_n=1$, and then the LHS (resp. RHS) will be: $n-1$ (resp. $n-2+{{2}\over{\sqrt{3}}}$).

However, I still cannot find a nice approach to solve this problem. Does anyone have an idea. Thanks.[/quote]

here's a starting idea (lhs):

compute the second derivative of $f(x)=\sqrt{\frac{1-x}{1+x}}$ and see its root is $\frac12$.

so on $\left[0,\frac12 \right]$, $f$ is convex and on $\left[ \frac12,1 \right]$ is concave

we have two cases:

$a_1,a_2,\ldots,a_n \leq \frac12$ - in this case we can make equal all $a_i$s to $\frac1n$

$a_1,\ldots,a_{n-1} \leq \frac12 < a_n$ - in this case we can do: $a_1=a_2=\ldots=a_{n-1} < \frac12$.

you have now $(n-1)a+b=1$ and have to minimize $(n-1) \sqrt{\frac{1-a}{1+a}} + \sqrt{\frac{1-b}{1+b}}$

this can be reduced to a one variable ineq (with $0 < a < \frac12$), and looks doable, but it is very ugly

for the rhs, i think that a more complicated procedure could be used: we use karamata to move variables apart, towards $0$ or $\frac12$. ( i learned this from a solution of mildorf ) anyway, i'm not sure if this actually works , but i'll work on it and tell you tomorrow the result :D

The problem:

Let and $a_1,\ldots,a_n$ be positive reals such that $a_1+\ldots+a_n=1$, where $n \in \mathbb{N}$, $n \geq 2$. Then we have \[ \sqrt{\frac{1-a_1}{1+a_1}} + \ldots + \sqrt{\frac{1-a_n}{1+a_n}} \leq n-2+\frac{2 \sqrt{3}}{3} . \]

The (i hope) correct solution:

First, the case $n=2$: (we replace $a_1$ by $a$ and $a_2$ by $b$) \[ \sqrt{\frac{1-a}{1+a}} + \sqrt{\frac{1-b}{1+b}} \leq \frac{2 \sqrt{3}}{3} . \]

Using the fact that $a+b=1$, we rewrite the inequality:

\[ \Leftrightarrow \sqrt{\frac{1-a}{1+a}} + \sqrt{\frac{a}{2-a}} \leq \frac{2 \sqrt{3}}{3}  \]

\[ \Leftrightarrow \frac{\sqrt{(1-a)(2-a)}+\sqrt{a(1+a)}}{\sqrt{(1+a)(2-a)}} \leq \frac{2 \sqrt{3}}{3}  \]

\[ \Leftrightarrow \sqrt{(1-a)(2-a)}+\sqrt{a(1+a)} \leq \sqrt{(1+a)(2-a)} \cdot \frac{2 \sqrt{3}}{3}  \]

(all is positive, so square it)

\[ \Leftrightarrow (1-a)(2-a)+a(1+a)+2\sqrt{(1-a)(2-a)a(1+a)} \leq (1+a)(2-a) \cdot \frac{4}{3}  \]

\[ \Leftrightarrow 6 \sqrt{(1-a)(2-a)a(1+a)} \leq 10a(1-a)+2  \]

\[ \Leftrightarrow 3 \sqrt{(1-a)(2-a)a(1+a)} \leq 5a(1-a)+1  \]

(again all is positive, so square again)

\[ \Leftrightarrow 0 \leq 1-8a+24a^2-32a^3+16a^4=(2a-1)^4 , \] which is trivially true :)

For $n=2$, equality holds iff $a_1=a_2=\dfrac12$.

Yeah, it's too complicated. But you see, i'm not good at solving ineqs. quite on the contrary :(

emm.. for the left part you simply need $\sqrt{\frac{1-a}{1+a}} \ge 1-a$
For the right part, I think your solution is overtroublesome, because you actually need to prove the case for n=2 only. Observe that $\frac{1-a}{1+a} \le 1$, then....

[quote="silouan"]Then what?Please go to finish[/quote]
isn't it obvious? Take the first (n-2) variables we have $\sum \sqrt{\frac{1-x_i}{1+x_i}} \le n-2$. Then for the remaining two variables we use the result proved for n=2. Then done.

[quote="silouan"]Why none replies?[/quote]

to be honest, i didn't understand what you wanted to say in your previous post; anyway, to my mind, siuhochung's soln is OK


I am just curious to know how many users are there in Mathlinks from India? And how many are moderators? Can anyone help me in that?

I'm not sure if there is a "proper" way to do this.

Oh, I don't participate in the Indian forum much, but I am Indian.


Eric drived 3 hours and 30 minutes, or 210 minutes.

and Eric was driving at least 50 mph during 6:30~8:30, or 120 minutes.

$ \frac{120}{210}\equal{}\frac{4}{7}$

which is about 57.1428%.

answer : 57%


Call a [b]word [/b] forming by alphabetical small letters $a,b,c,\ \cdots x,y,z$ and [b]a periodic word[/b] arranging by a certain word more than two times repeatedly.For example kyonkyon is eight-letter periodic word. Given two words $W_1,\ W_2$ which have the same number of letters and have a different first letter,
if you will remove the letter, $W_1$ and $W_2$ will be same word.Prove that either $W_1$ or $W_2$ is not periodic.

It's a simple conclusion of the fact that every word has a Minimal Period Lenght like $T$ such that every other period length $T'$ is a multiple of $T$.
You can prove this like this :
If $T$ and $T'$ are two period length's then let $a_i$ be the $i$th letter of the word.
We want to prove that $(T,T')$ is also a period length; Because of bezout theorem there are $a,b$ such that $aT-bT'=(T,T')$
Now we have $a_i=a_{i+aT}=a_{i+aT-bT'}=a_{i+(T,T')}$ so $(T,T')$ is also a period length.
(If the indices are out of range then just put the word next to itself many times to form an infinite sequence of periodic letters)

It is also even more so a simple consequence of the celebrated Wilf-Fine Theorem.


We can suppose $a^2+b^2+c^2=3$, then the ineq becomes $\sum \frac{4a^3}{a^2+3}\ge3.$ We can apply the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64933]Half concave and half convex theorem (theorem 1')[/url] to the function $f(x)=\frac{4x\sqrt{x}}{x+3}$ (where $x=a^2$) which yields $f(x)+f(y)+f(z)\ge f(1,1,1)=3$ for $x+y+z=3$. Done :roll:

[quote="yookon88"]How can you suppose a^2+b^2+c^2=3? I'm quite new to solving inequalities with these methods and I would like to learn.
Thank you[/quote]

That is what we call hommogenizing/unhommogenizing technique.

but by doing this isn't he restricting a,b,c


Let A, B, and C represent different digits greater than zero.  Determine the minimum value of the expression:
$\frac{ABC}{A+B+C}$
Note:A is the humdreds digit, B is the tens digit, and C is the ones digit

is there a nice way to do this problem?
[hide]
heres how i proceeded:

(100A + 10B + C)/(A+B+C) = 1 + (99A + 9B)/(A+B+C).  C drops out of the numerator and is just in the denominator so we can max it out to C = 9.  Now clearly on the top A should be as small, to A = 1 (but my question -- how do we know that the drop int he denominator won't be as significant ?).  Then playing around with B through the numbers 2 through 8 gives a minimum of B = 8 so ABC = 189.  Still i dont feel that comfortable setting A to 1.  Anyone have any ideas?
[/hide]


Divisibility
modular arithmetic
remainder
IMO Shortlist
number theory

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in $\LaTeX$. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[b]Case 1: n odd[/b]

Let $n=p_1^{a_1}\ldots p_k^{a_k}$, where $p_i$ are distinct odd primes. Then $P_n=\prod_{S\subseteq [k]}\alpha_S$, where for $S\subseteq[k],\ \alpha_S$ is congurent to $1\pmod{p_i^{a_i}},\ \forall i\in S$, and congruent to $-1\pmod{p_j^{a_j}},\ \forall j\in[k]\setminus S$ ($[k]$ is just a compact notation for $\{1,2,\ldots,k\}$). Since for each $i\in[k]$, there are $2^{k-1}$ subsets $S\subseteq[k]$ for which $\alpha_S\equiv-1\pmod{p_i^{a_i}}$, it means that $P_n\equiv (-1)^{2^{k-1}}\pmod {p_i^{a_i}},\ \forall i$, and thus $P_n\equiv(-1)^{2^{k-1}}\pmod n$. In other words, if $n$ is odd, the answer is $1$ if $n$ has at least $2$ prime divisors, and $-1$ if it's a prime power.

[b]Case 2: n even, but not a power of $2$[/b]

Let $n=2^ap_1^{a_1}\ldots p_k^{a_k}$, and let $\tilde n=\frac n{2^n}$. The possibilities for $x\pmod{\tilde n}$ are those from the preceding paragraph, while those for $2^a$ are $x\equiv 1\pmod 2$ if $a=1$, $x\equiv \pm1\pmod 4$ if $a=2$, and $x\equiv \pm 1,2^{a-1}\pm 1\pmod{2^a}$. Since $\pm 1,2^{a-1}\pm 1$ give $1\pmod{2^a}$ when squared, and each situation appears $2^k$ times, so an even number of times (since $k\ge 1$), it means that $P_n\equiv 1\pmod{2^a}$, no matter what $a$ is.

When $a\ge 2$, each residue $\pmod{\tilde n}$ appears an even number of times, so $P_n\equiv 1\pmod n$ whenever $n=2^ap_1^{a_1}\ldots p_k^{a_k},\ k\ge 1,a\ge 2$.

When $a=1$, $P_n$ gives $-1\pmod{\tilde n}$ if $k=1$, and $1\pmod{\tilde n}$ if $k\ge 2$, so $P_n\equiv -1\pmod n$ if $k=1$, and $1\pmod n$ if $k\ge 2$.

[b]Case 3: n is a power of $2$[/b]

If $n=2$, then the answer is $1$, if $n=4$ it's $-1$, and if $n=2^a,a\ge 3$, then we have $P_n\equiv 1\cdot(-1)\cdot(2^{a-1}-1)\cdot(2^{a-1}+1)\pmod{2^a}$, so $P_n\equiv 1\pmod n$.
I hope it's Ok.

The answer can be phrased as $P_n = n-1$ when $n=2, 4, p^k, 2p^k$ for some odd prime $p$ and positive integer $k$; and $P_n=1$ otherwise.

Recall that only the numbers of the form $2, 4, p^k, 2p^k$ have primitive roots. Is there a relationship between the two results?

yes there is, the groups formed by the residues of those numbers over multiplication only have 2 elements of order $2$, while the others have more than $2$, and it makes the problem a simple corollary from some abstract algebra result!

This problem was used as problem 6 on the final exam of the 3rd TST of Taiwan 2005.

I think I have a nicer solution for this. (Previously posted in my blog, [url=http://www.artofproblemsolving.com/blog/72010]here[/url].)

Notice that, if $x^2 \equiv 1 \pmod n$ then $(n-x)^2 \equiv 1 \pmod n$.
Now, $\frac n2^2 \not \equiv 1 \pmod n$ for any $n>2$, so we can assume $x\not \equiv (n-x) \pmod n$
From now on , all congruences are taken $\pmod n$ unless otherwise stated.
Now, \[P_n \equiv \prod_{i=1}^{i^2 \equiv 1, i<n} i \equiv \prod_{i=1}^{i^2 \equiv 1, i<\frac n2} i(n-i) \equiv \prod_{i=1}^{i^2 \equiv 1, i<\frac n2}(-1) \pmod n\]
Let $f(n)$ be the number of $0<i<n$'s such that $i^2 \equiv 1 \pmod n$.
Then $P_n \equiv (-1)^{\frac{f(n)}2} \pmod n$.
Now, we will prove some lemmas to find $f(n)$.

[u]Lemma 1:[/u] $f(n)$ is multiplicative.
Proof:
Let $p,q$ be positive integers such that $p^2 \equiv 1 \pmod x$ and $q^2 \equiv 1 \pmod y$ with $p<x,q<y$. Then , if  $(x,y)=1$, we get $(pq)^2 \equiv 1 \pmod {xy}$ and $pq < xy$.
So, $f(xy)=f(x)f(y)$ for $(x,y)=1$.

[u]Lemma 2:[/u] $f(p^i)=2 \forall \text{ prime }p >2$

Proof:
Let $x^2 \equiv 1 \pmod {p^i}$.
So $p^i \mid (x+1)(x-1)$
Let $p^k \mid x+1 , p^{i-k} \mid x-1$.
But $(x+1)-(x-1)=2$, so one of $k,i-k$ must be $0$.
So $x+1 \equiv \pmod {p^i}$ or $x-1 \equiv \pmod {p^i}$.
The first part implies $x=p^i-1$, the second part implies $x = 1$.
So $f(p^i)=2$ for prime $p >2$.

It is easy to see that $f(2)=1,f(4)=2$ and $f(2^j)=4$ for $j>2$.

So we get from lemma 1,2, and the fact stated above that,

If $n  = \prod {p_i}^{e_i}$, then $f(n) = \prod f({p_i}^{e_i})$,with
$f(p^x) = 2$ for prime $p>2$,
$f(2^x)=4$ for $x>2$, and
$f(2)=1,f(4)=2$.
So, we get from \[P_n \equiv (-1)^{\frac{f(n)}2} \pmod n\]  that

\[P_n=\{\begin{matrix}
n-1 & \text{for } n=p^k \text{ or } n=2p^k\\
1 & \text{for all other x}
\end{matrix}\]

And, we are done.

[quote="Mahi"]

[u]Lemma 1:[/u] $f(n)$ is multiplicative.
Proof:
Let $p,q$ be positive integers such that $p^2 \equiv 1 \pmod x$ and $q^2 \equiv 1 \pmod y$ with $p<x,q<y$. Then , if  $(x,y)=1$, we get $(pq)^2 \equiv 1 \pmod {xy}$ and $pq < xy$.
So, $f(xy)=f(x)f(y)$ for $(x,y)=1$.

[/quote]
How did you get $(pq)^2 \equiv 1 \pmod {xy}$ ?

I have split the problem into cases:

[b]Case 1[/b]:n is odd

Let $n=\prod_{i=1}^{r}{{p_i}^{\alpha_i}}$ where the $p_i$ are odd primes.Then it is easy to see that the only incongruent numbers satisfying $x^2 \equiv 1\pmod{{p_i}^{\alpha_i}}$ are $x \equiv -1,1\pmod{{p_i}^{\alpha_i}}$.This occurs for for all the $r$ powers of form ${p_i}^{\alpha_i}$.Now we will group the residues as follows:replace $1$ by $-1$ and vice versa.For example for $r=3$ the residue group $(1,-1,-1)$ is changed to $(-1,1,1)$.Note that if $x_0$ is a solution to the following congruences
$x \equiv 1 \pmod{{p_1}^{\alpha_1}}$
$x \equiv -1\pmod{{p_2}^{\alpha_2}}$
$x \equiv -1\pmod{{p_3}^{\alpha_3}}$

and $x_1$ is a solution to the congruences
$x \equiv -1\pmod{{p_1}^{\alpha_1}}$
$x \equiv 1 \pmod{{p_2}^{\alpha_2}}$
$x \equiv 1 \pmod{{p_3}^{\alpha_3}}$

then $x_0x_1 \equiv -1\pmod{n}$.This occurs in general for two solutions of $x_0$ and $x_1$ of the complementary residue sets.There are $2^r$ total residue sets,so if we group them in this way there would be $2^{r-1}$ pairs.Hence the product of all the elements will be $(-1)^{2^{r-1}}$ which is $1$ for $r \ge 2$ and $-1$ for $r=1$.

[b]Case 2[/b]:$n$ is even but not a power of $2$

Here $n$ can be written as $2^k \prod_{i=1}^{r}{{p_i}^{\alpha_i}}$.Let $k \ge 3$.Then $x^2 \equiv 1\pmod{2^k}$ has four incongruent solutions:namely $1,-1,2^{k-1}-1,2^{k-1}+1$(Its easy to check that they are distinct).We pair the residue sets as follows:replace $2^{k-1}-1$ with $2^{k-1}+1$,$1$ with $-1$ and vice versa modulo $2^k$ and for the other primes replace $1$ with $-1$ and vice versa.Then it is clear that if $x_0$ is a solution of a residue set and $x_1$ is a solution of its complementary residue set we have $x_0x_1 \equiv -1\pmod{n}$.There can be $2^{r+2}$ residue sets and hence $2^{r+1}$ pairs.Consequently $P_n=(-1)^{2^{r+1}}=1$.
If $k=1$ then there are $2^{r-1}$ residue set pairs and so $P_n=1$ if $r \ge 2$ and $P_n=-1$ if $r=1$
If $k=2$ then there are $2^r$ residue set pairs and $P_n=1$

[b]Case 3[/b]:$n$ is a power of $2$

This is a trivial case.If $n=2^k$ then $P_n=1$ if $k \ge 3$ and $P_n=-1$ if $k$ is $1$ or $2$

Combining these three cases we get $P_n=n-1$ if $n=2,4,p^k,2p^k$ where $p$ is an odd prime and $P_n=1$ otherwise.

[quote=orl]Given an integer ${n>1}$, denote by $P_{n}$ the product of all positive integers $x$ less than $n$ and such that $n$ divides ${x^2-1}$. For each ${n>1}$, find the remainder of $P_{n}$ on division by $n$.

[i]Proposed by John Murray, Ireland[/i][/quote]

The answer is $\boxed{P_n \equiv -1 \pmod n \text{ if n has a primitive root and 1 otherwise.}}$
Denote $S_n=[x \in N|x \le n,n|x^2-1]$
Observe that if $x \in S$ so is $n-x$ so $P_n \equiv x(n-x) \equiv (-1)^m$ where $S_n$ has $2m $ elements.
Now it is just casework:-
[b][color=#f00]Case #1:[/color][/b] $n$ is odd.
Let $n=\prod_{i=1}^{r}{{p_i}^{\alpha_i}}$ where the $p_i$ are odd primes.
By CRT,choose $x_0$ to be a solution to the following congruences
$x \equiv 1 \pmod{{p_1}^{\alpha_1}}$
$x \equiv -1\pmod{{p_2}^{\alpha_2}}$
$x \equiv -1\pmod{{p_3}^{\alpha_3}}$
and $x_1$ be a solution to the congruences
$x \equiv -1\pmod{{p_1}^{\alpha_1}}$
$x \equiv 1 \pmod{{p_2}^{\alpha_2}}$
$x \equiv 1 \pmod{{p_3}^{\alpha_3}}$
then $x_0x_1 \equiv -1\pmod{n}$.Since this occurs for any two numbers which are in the residue class modulo $n$ and there are $2^r$ total residue sets,it follows that there would be $2^{r-1}$ pairs.Hence the product of all the elements will be $(-1)^{2^{r-1}}$ which is $1$ for $r \ge 2$ and $-1$ for $r=1$.
[b][color=#f00]Case #2:-[/color][/b] $n$ has a prime factor other than $2$
Write $n=2^k \prod_{i=1}^{r}{{p_i}^{\alpha_i}}$
Similar to the previous case,pairing residue classes gives $2^{r+1}$ pairs.
[b][color=#f00]Case #3:-[/color][/b] $n=2^k$
Here,$P_n=1$ if $k \ge 3$ and $P_n=-1$ if $k$ is $1$ or $2$
Combining the above possibilities,we get the result.$\blacksquare$

Let $S$ denote the set of all $x<n$ such that $n$ divides $x^2-1.$ Fuethermore, let $S^*$ be the set of $x$ in $S,$ strictly smaller than $n/2.$ Assume that $n>2.$ Then, $n/2\notin S,$ so $S^*\cap(S\setminus S^*)=\emptyset.$ The trick is to see that $s\in S^*\iff n-s\in S\setminus S^*.$ It follows that \[\prod_{s\in S}s=\prod_{s\in S^*}s(n-2)\equiv \prod_{s\in S^*}-s^2\equiv (-1)^{|S^*|/2}\bmod{n}.\]We used that if $s\in S^*\subset S$ then $n\mid s^2-1$ so $s\equiv 1\bmod{n}.$ Fuethermore, observe that $|S^*|/2=|S|/4.$

Now, let $n=2^t\cdot p_1^{k_1}\cdot...\cdot p_m^{k_m}.$ Observe that each $s\in S$ is the solution to one of the following congruences, for a suitable choice of $\pm1$  \[s\equiv \pm 1\bmod{2^t}, \quad 2\equiv\pm 1\bmod{p_1^{k_1}} \quad ... \quad s\equiv \pm 1\bmod{p_m^{k_m}}.\]Now, by the CRT, each choice of plusses and minuses yields a unique $s\in S.$ Hence, $|S|$ is equal to the number of choices of $\pm 1$ in the above system of congruences. That number is equal to $f(t)\cdot 2^m.$ where $f(i)=1$ if $i=0,1$ and $f(i)=2$ otherwise.

That little change comes because $-1$ and $+1$ are, in fact, the same residue modulo $2$ so changing the sign won't yield a different result. Now, to finish, note that $|S|=2^m\cdot f(t)=2$ only for $n=p^k$ or $2p^k.$ Otherwise, $4\mid 2^m\cdot f(t).$

Therefore, the answer is: $-1$ if $p^k$ or $2\cdot p^k$ and $+1$ for all other $n$. (In lesser words, $-1$ if $n$ admits a primitive root and $+1$ otherwise)

Solved with [b]megarnie[/b].

Firstly, note that if $x$ works, then $n-x$ works. Then $x(n-x)\equiv -x^2\equiv -1\pmod n$, so if we just pair the possible $x$'s up, the problem reduces to finding the remainder when the number of possible $x$ is divided by $4$. Particularly, if $x=\frac n2$, then $$n|(x^2-1)\rightarrow (2x)|(x^2-1)\rightarrow x|(x^2-1)\rightarrow x=1\rightarrow n=2,$$ where $P_2$ is just $1$. So, define $f(n)$ to be the number of $x$ that work for a given $n$.

We first examine when $4|n$. If $n=4$, then $f(4)=2$. Otherwise, there exists quadruples $\left(x,\frac n2-x,\frac n2+x,n-x\right)$ for $x<\frac n4$ such that if one of them work, all four of them work. Note that if $x=\frac n4$, then $$n|(x^2-1)\rightarrow (4x)|(x^2-1)\rightarrow x|(x^2-1)\rightarrow x|1\rightarrow x=1\rightarrow n=4,$$ contradiction. So $4|f(4a)$ for positive integers $a>1$.

[b]Lemma 1.[/b] For odd primes $p$ and positive integers $r$, $f\left(p^r\right)=2$.
[b]Proof.[/b] Note that $p^r|(x+1)$ or $p^r|(x-1)$, since $p\nmid\gcd(x+1,x-1)$ from the Euclidean algorithm. $1$ and $p^r-1$ clearly work, so $f\left(p^r\right)=2$.

[b]Lemma 2.[/b] For odd $n\ge 3$, $f(n)=f(2n)$.
[b]Proof.[/b] Consider the set $S$ of all $x$ that work for $n$. Then the set that work for $2n$ has to be a subset of $T=S\cup (S+n)$. However, for any $x\in S$, note that $\{x,x+n\}$ forms a complete residue set mod $2$. So one of them has an odd square, and the other one has an even square. Since both clearly have a square that is $1$ mod $n$, exactly one of them work. So the number of $x$ that work are actually equivalent for $n$ and $2n$, which is what we wanted.

[b]Lemma 3.[/b] For odd prime $p$, positive integer $r$, and positive odd $n>2$ such that $p\nmid n$, $4|f(p^r n)$.
[b]Proof.[/b] Consider the set $S$ of all $x$ that work for $n$. Then the set that work for $p^r n$ has to be a subset of $T=S\cup (S+n)\cup\dots\cup(S+(p^r-1)n)$. However, for any $x\in S$, note that $\{x,x+n,\dots,x+(p^r-1)n\}$ forms a complete residue set mod $p^r$, which must have $f(p^r)$ of them square to $1$ mod $p^r$. Since everything in the set clearly have a square that is $1$ mod $n$, $f(p^r)$ of them work by CRT. But since there are $|S|$ different such $x$'s, $f(p^r n)=|S|f(p^r)=f(n)f(p^r)$, and since both factors on the RHS are even, we have what we wanted.

Thus, by repeatedly applying these three lemmas with $r=1$, we can see that for squarefree numbers $n$, $$f(n)\equiv\begin{cases}
1&\qquad\text{if }n=2 \\
2&\qquad\text{if }n\text{ has }1\text{ odd prime factor} \\
0&\qquad\text{if }n\text{ has }2\text{ or more odd prime factors} \\
\end{cases}\pmod{4}.$$

Our final case is then if $n$ is not squarefree and not a multiple of $4$. From lemma 2, we can just consider such odd $n$ and extend it to $2n$ later.

Using $r>1$ cases from lemmas 1 and 3, when $n$ is not squarefree and not divisible by $4$, $$f(n)\equiv\begin{cases}
f\left(\frac n2\right)&\qquad\text{if }2|n \\
2&\qquad\text{if }n\text{ is a power of an odd prime} \\
0&\qquad\text{if }n\text{ is odd but not a power of an odd prime} \\
\end{cases}\pmod{4}.$$

So, finally, combining all of these cases, here are our answers.
$$\boxed{P_n=\begin{cases}
n-1&\qquad\text{if }n\in\{4,p^k,2p^k\}\text{ for odd primes }p\text{ and positive integers }k \\
1&\qquad\text{otherwise}
\end{cases}}$$

Idk why but the solutions to all these look complicated. I will just give the idea of my solution.  If $n=2$, then the answer is $1$.

Call $x$ good iff $n \mid x^2-1$.

Now $x$ is good iff $-\frac1x$ is good. And also see that if $n>2$, then $x \not \equiv -\frac1x \pmod n$, and so we just have to count the number of self inverses modulo $n$ and the pair them up.

See that we can easily do it by CRT, since if for some $q \mid n$ where $q$ is an odd prime, then we have \[q^{v_q(n)} \mid x^2-1 \implies q^{v_q(n)} \mid x+1 \text{ or } q^{v_q(n)} \mid x-1\]b Let $k$ is the number of odd prime factors of $n$.
\begin{enumerate}
\ii If $n$ is odd, then the number of self inverses is $2^k$.
\ii If $v_2(n)=1$ and $n>2$, then again it is $2^k$.
\ii If $v_2(n)=2$, then it is $2^{k+1}$.
\ii And lastly if $v_2(n) \geq 3$, then it is $2^{k+2}$. \end{enumerate}
Now pairing them up we get the answer as \[\boxed{P(n) = \begin{cases} -1 \text{ when } n=4, \text{ } p^a \text{ or } 2p^b \text{ where $p$ is an odd prime and } a,b \in \mathbb{N} \\ +1 \text{ otherwise} \end{cases}}\]


The odd positive integers are grouped as follows: \[ \{1\}, \{3,5\}, \{7,9,11\},\{13,15,17,19\},...\] Prove that the sum of the numbers in the $ n$th group is $ n^3$.

[hide]The sum of the first $ m$ odd positive integers is $ m^2$. Furthermore, the $ l$th triangular number $ T_l$ can be generated by $ T_l = \frac {l(l + 1)}{2}$. As we can see, the last number in the $ n$th group is  the $ T_n$th odd integer, due to the fact that there is 1 odd integer in the first group, 2 in the second, 3 in the third and so on.

This means that our formula to generate the sum of the $ n$th group is the difference between the sums of the odd integers from $ 1$ to $ T_n$ and the odd integers from $ 1$ to $ T_{n-1}$, or:
\begin{align*} & \left(\frac {n(n + 1)}{2}\right)^2 - \left(\frac {(n - 1)n}{2}\right)^2 \\
= & \left(\frac {n^2}{4}\right)((n^2 + 2n + 1) - (n^2 - 2n + 1)) \\
= & \left(\frac {n^2}{4}\right)(4n) \\
= & n^3 \end{align*}[/hide]

[hide="Slightly different"]
Let $ T_n$ denote the $ n^{\text{th}}$ triangular number, and $ n^{\text{th}}$ be the nth set of the odd numbers.

Note that the greatest term in $ S_n$ is $ T_n^{\text{th}}$ odd number, or $ 2\frac{(n)(n\plus{}1)}{2}\minus{}1\equal{}n^2\plus{}n\minus{}1$. Seeing that the terms in $ S_n$ form an arithmetic sequence of $ n$ terms with a common difference of 2, we deduce that the smallest term in $ S_n$ is $ n^2\plus{}n\minus{}1\minus{}2(n\minus{}1)\equal{}n^2\minus{}n\plus{}1$. Adding the two terms and dividing by two, we find the average of the elements in $ S_n$ to be $ \frac{2n^2\plus{}n\minus{}n\minus{}1\plus{}2}{2}\equal{}n^2$. Therefore, the average of the terms in $ S_n$ is $ n^2$.

The sum of the terms in an arithmetic sequence can be found by multiplying the average by the number of terms. Since there are $ n$ terms in $ S_n$, we multiply that by our previously found average to get $ n^3$. Therefore, the sum of the numbers in the $ n^{\text{th}}$ group is $ n^3$. Q.E.D.
[/hide]

[hide]Look at the differences between the last terms between consecutive groups.

It looks like this: 4, 6, 8, 10, etc.

Thus, we can see that there exists an equation in the form of $ ax^2 \plus{} bx \plus{} c$ which gives us $ 1, 5, 11, 19$, etc. for $ x\equal{} 1, 2, 3, ...$

Since $ a \plus{} b \plus{} c \equal{} 1$ and $ 4a \plus{} 2b \plus{} c \equal{} 5,$ $ 3a \plus{} b \equal{} 4$.

Also, $ 8a \plus{} 2b \equal{} 10$ by using x=3 and x=1. So $ 4a \plus{} b \equal{} 5$.
This gives $ a \equal{} 1, b \equal{} 1, c \equal{} \minus{} 1$.

$ x^2 \plus{} x \minus{} 1$ gives the last term of each set.
Similarly, $ x^2 \minus{} x \plus{} 1$ gives the first term. Since there are $ x$ terms for each set, we get the sum is $ x^2*x \equal{} x^3$[/hide]
Pretty much zephyredx's way. You can't just say it's the $ T_{n}^{th}$ odd number though, or can you...?

"Hey look there is a pattern" isn't a legitimate method of proof. You need to explain why the differences are the way that they are. But I'm not sure why one would look at the differences when you can just say that the last term of the nth set is $ 1\plus{}2\plus{}...\plus{}n\equal{}T_n$ or whatever. That is clear enough. I feel like to justify that the differences between the last terms forms 2,4,..., you might as well just use the other argument.


Putnam
trigonometry
quadratics
calculus
calculus computations

[b]A clock's minute hand has length 4 and its hour hand length 3. What is the distance between the tips at the moment when the distance separating them is increasing most rapidly?[/b]

[hide="solution"]Let $ \theta$ be the angle between the hour hand and some arbitrary point on the clock, then we're looking to maximize $ d'(\theta)$ where
\[ d(\theta)^2 \equal{} (4\cos \theta \minus{} 3\cos 12 \theta)^2 \plus{} (4\sin\theta \minus{} 3\sin 12\theta)^2 \equal{} 25 \minus{} 24\cos 11\theta
\]
Differentiate and simplify to find
\[ d''(\theta) \equal{} \minus{} 1452d(\theta)^{ \minus{} 3}\left(12 \plus{} 12\cos^2 11\theta \minus{} 25\cos 11\theta\right)
\]
$ d'(\theta_m)$ is a maximum where $ d''(\theta_m) \equal{} 0$ so we solve the quadratic for $ \cos 11 \theta_m$ to find $ \cos 11\theta_m \equal{} \frac {3}{4}$, and hence $ \boxed{d(\theta_m) \equal{} \sqrt {7}}$[/hide]

What year Putnam was this?

This question has been asked before so you can find some more solutions

[quote="adrenalinefix36"]Where did the 12(theta) come from in the first equation?[/quote]
The minute hand is moving at 12 times the rate of the hour hand. The implied unit of time being used in JoeBlow's solution is "per 12 hours."


We put an easy non-elastic rope on a cyllinder as in the diagram. The rope gripps on to the cyllinder if the following inequality is true for the forces at the two ends:
$\frac12 F_{A}\leq F_{B}\leq 2F_{A}$
Using the inequality find the coefficient of friction between the rope and the cyllinder.


Complementary Counting Part 1

We "count what we don't want" to count how many 3-digit license plates have at least one 8.  We discuss clues for when to try this strategy, and we discover an interesting algebraic relationship.

I tested this with numbers from 1 and 2 to 100 and 101 and they work!!!

Here's a proof of why:

$ (x\plus{}1)^3\minus{}x^3\equal{}3x^2\plus{}3x\plus{}1$
$ x^3\plus{}3x^2\plus{}3x\plus{}1\minus{}x^3\equal{}3x^2\plus{}3x\plus{}1$
$ 3x^2\plus{}3x\plus{}1\equal{}3x^2\plus{}3x\plus{}1$
$ Q.E.D.$

For all the tests, I used this Python code:

#!/usr/bin/env python
import time
x = 1
while x <101:
    y=x+1
    print x, y
    if y**3-x**3==3*x**2+3*x+1:
        print "Works!"
    x+=1
    time.sleep(1)


Here's yet another weird word game. The goal is to transform one word into another through a series of intermediate steps with each step being a word as well. Standard Scrabble-type no proper nouns, etc. rules apply (but one-letter words are ok). Each step, you may add a letter, remove a letter, or replace one letter with another. At the end of your post, post a transformation for the next person to do. Here's an example:

[quote]night->day[/quote]
[b]night[/b]
sight
sigh
sig
sag
say
[b]day[/b]

Start with cable -> print.

[b]CABLE[/b]
TABLE
TALE
TILE
TILT
TINT
PINT
[b]PRINT[/b]

[b]SMALL[/b]
MALL
MAIL
MAIN
GAIN
GAIT
WAIT
WIT
BIT
BIG
BUG
HUG
[b]HUGE[/b]

$\boxed{\text{while}\; \to \; \text{man}}$

[quote="The Zuton Force"]
$\boxed{\text{while}\; \to \; \text{man}}$[/quote]

[b]WHILE[/b]
WHINE
WHIN
WIN
WAN
[b]MAN[/b]

$\boxed{HUT \to TANK}$

[b]WHILE[/b]
WILE
MILE
MALE
MANE
[b]MAN[/b]

Also,
[b]SMALL[/b]
MALL
MILL
RILL
RILE
RULE
RUE
RUG
HUG
[b]HUGE[/b]
is faster.

[b]HUT[/b]
HUNT
RUNT
RANT
RANK
[b]TANK[/b]

[b]FOUR[/b] --> [b]FIVE[/b]

Yes miyomiyo, but I felt the urge to go through "big" before getting to "huge"  :wink: I was going to put medium in there too but I didn't have the patience.

[b]HUT[/b]
HAT
RAT
RAN
TAN
[b]TANK[/b]

[b]Edit:[/b] Beaten to it by lingo, so here's the solution to his.

[b]FOUR[/b]
FUR
FUN
FIN
FINE
[b]FIVE[/b]

I'll allow the next poster to be creative, since I'm running low on creativity at the moment  :)

Same amount of steps
[b]HUT[/b]
HAT
PAT
PAN
TAN
[b]TANK[/b]

$\boxed{\text{words}\rightarrow \text{numbers}}$

This is a lot harder than the previous ones...

[hide="Progress?"]WORDS
WORD
WARD
WAR
JAR
JAB
NAB
NUB
NUMB <<< that's too bad
DUMB
DUMA
DUMAS

...

CHAMPER
CHAMBER
CAMBER
CUMBER
UMBER
UMBERS
NUMBERS[/hide]

[hide="solution"]
WORDS
WORD
WARD
WAR
JAR
JAB
NAB
NUB
NUMB
DUMB
DUB
DUD
BUD
BUN
RUN
RUNS
SUNS
SONS
TONS
TOES
TOE
ROE
ROW
RAW
RAM
HAM
SHAM
SHAME
SAME
CAME
CAMP
CAP
CAPE
CAPER
CAMPER
CHAMPER
CHAMBER
CAMBER
CUMBER
UMBER
UMBERS
NUMBERS[/hide]

Zuton Force did a few of them.

a simple one:
HOSE to GROWS

[quote="1=2"][hide="solution"]
WORDS
WORD
WARD
WAR
JAR
JAB
NAB
NUB
NUMB
DUMB
DUB
DUD
BUD
BUN
RUN
RUNS
SUNS
SONS
TONS
TOES
TOE
ROE
ROW
RAW
RAM
HAM
SHAM
SHAME
SAME
CAME
CAMP
CAP
CAPE
CAPER
CAMPER
CHAMPER
CHAMBER
CAMBER
CUMBER
UMBER
UMBERS
NUMBERS[/hide][/quote][hide="Simpler"][b]NUMBERS[/b]
NUMBER
UMBER
AMBER
AMBLER
AMBLE
ABLE
ALE
HALE
WHALE
WHOLE
WHORE
WORE
[b]WORD[/b][/hide]

Ah right, missed ambler  :stink:

[b]hose[/b]
hoe
how
row
grow
[b]grows[/b]

[b]water[/b]
later
late
hate
hat
bat
ban
band
bond
blond
[b]blood[/b]

$\boxed{\text{fire}\rightarrow\text{water}}$

Fire
wire
ware
wade
wader
[b]water[/b]

Salt --> Pepper

[quote="Nerd_of_the_Ages"]since noone is posting any, heres one: rhythm -> chaos.
is this one allowed? Anakin -> Vadar[/quote]

anakin
vanakin <-- dont ask me
vadakin <-- son of vadars =)
vadarin <-- vadar in somewhere
vadari <-- plural of vadar
vadar

cud it be like this?

anakin
ankin (company name)
andin (another company name)
anden
ander
ader
vader

mouse-cat

[b]mouse[/b]
douse
dose
done
cone
con
can
[b]cat[/b]

heres one im doing for fun (first time trying this)

[b]calm[/b]
lack
lock
cold
sold
sod
sad
said
aid
maid
[b]mad[/b]

wait, your aloud to switch around the order of the words, right?

here's one to work on

sister ---> annoying

How did you go from [b]calm[/b] to [b]lack[/b]?

[b]calm[/b]
cam
ram
rad
[b]mad[/b]

um, you take of the m and add a k, then change the order of the letters. Are you allowed to do that? That's also how I got from lock to cold.

Meanwhile, another one for fun

[b]soft[/b]
loft
loot
cloot (a cloven hoof)
clout
cloud
[b]loud[/b]

these are harder than they seem o.O

You can't change the order of the letters, that's in another game.

oh, sorry. I'll make sure i dont do that next time.

I'm not sure that's possible, but that's probably cuz i suck at this game. How would you get the igh?

Told you i suck at this game  :blush:
im a bit better at stretch, i think

[quote="i_like_pie"][b]LEFT[/b]
LIFT
SIFT
SIT
SIG
SIGH
SIGHT
[b]RIGHT[/b][/quote]

I was actually thinking more or less the same thing, but I don't think "sig" is actually a word...

Check the first post in this thread.  SIG was used there.


Let A, B, C denote three distinct points with integer coordinates in $ \mathbb{R}^{2}$. Prove that if $ (|AB|\plus{}|BC|)^{2}< 8[ABC]\plus{}1$ (where |AB| denotes the length of AB, and [ABC] denote the area of triangle ABC), then A, B, and C are three vertices of a square.

EDIT: I put this in the wrong forum. This should probably be in either the Olympiad-Geometry or Olympiad-Inequalities Forum. Mods, feel free to move it.

[hide]
Let $ B$ be the origin, $ C$ have coordinates $ (c,0)$, and $ A$ have coordinates $ (a,b)$ (where all are at least 0).

Then, the LHS = $ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}2c\sqrt{a^{2}\plus{}b^{2}}$, and the RHS = $ 4bc\plus{}1$. (Using Pythagorean for LHS and 1/2bh for RHS)

If $ a > 0$, we have $ b^{2}\plus{}c^{2}\ge 2bc$, $ 2c\sqrt{a^{2}\plus{}b^{2}}> 2bc$, and $ a^{2}\ge 1$. Therefore, we have $ \text{LHS}> 4bc\plus{}1$, a contradiction.

If $ a \equal{} 0$ and $ b\neq c$, we have $ b^{2}\plus{}c^{2}> 2bc$, so $ \text{LHS}\equal{} b^{2}\plus{}c^{2}\plus{}2bc > 4bc$, and since they're all integers, $ \text{LHS}\ge 4bc\plus{}1$, a contradiction.

Therefore, $ a\equal{}0$, $ b\equal{}c$, so $ \triangle ABC$ is an isosceles right triangle and therefore $ A,B,C$ are three vertices of a square.

[/hide]

Two questions about that proof.

1. When supposing $ a > 0$, how did you conclude $ a^{2}\geq 1$?

2. How do you know $ a < 0$ is impossible?

EDIT: Thanks to below poster.

[quote="MellowMelon"]Two questions about that proof.

1. When supposing $ a > 0$, how did you conclude $ a^{2}\geq 1$?

2. How do you know $ a < 0$ is impossible?[/quote]

1. $ a$ is an integer as the problem stated

2. Negating the value of $ a$ will not change the equation.

[quote="not_trig"][hide]
Let $ B$ be the origin, $ C$ have coordinates $ (c,0)$, and $ A$ have coordinates $ (a,b)$ (where all are at least 0).

Then, the LHS = $ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}2c\sqrt{a^{2}\plus{}b^{2}}$, and the RHS = $ 4bc\plus{}1$. (Using Pythagorean for LHS and 1/2bh for RHS)

If $ a > 0$, we have $ b^{2}\plus{}c^{2}\ge 2bc$, $ 2c\sqrt{a^{2}\plus{}b^{2}}> 2bc$, and $ a^{2}\ge 1$. Therefore, we have $ \text{LHS}> 4bc\plus{}1$, a contradiction.

If $ a \equal{} 0$ and $ b\neq c$, we have $ b^{2}\plus{}c^{2}> 2bc$, so $ \text{LHS}\equal{} b^{2}\plus{}c^{2}\plus{}2bc > 4bc$, and since they're all integers, $ \text{LHS}\ge 4bc\plus{}1$, a contradiction.

Therefore, $ a \equal{} 0$, $ b \equal{} c$, so $ \triangle ABC$ is an isosceles right triangle and therefore $ A,B,C$ are three vertices of a square.

[/hide][/quote]

I think its wrong to set the problem up as you have. The problem said that the coordinates of A, B, and C were integers in $ \mathbb{R}^{2}$. This doesn't mean that any of the sides of triangle ABC necessarily have integer length. But in your choice of coordinates you've set it up so that the length of side BC is an integer c.

However, the a=0 portion still holds; if it's a right triangle with all points integer coordinates, we may re-orient the triangle such that b, c are integers :wink:

I think it can be proven that for every triangle of the form $ (0,0), (x,y), (u,v)$, where the variables are integer, a coordinate system can be chosen so that the triangle transforms into $ (0,0), (a,0), (b,c)$ - the unit of measurement will be different, of course.


give a, b, c is three positive real  such that: abc=1
prove that
1/a +1/b +1/c +6/(a+b+c) >= 5

I think you must post it in

Forum index » Olympiad Section » Inequalities » Inequalities Unsolved Problems

Your problem posted before here
   http://www.mathlinks.ro/Forum/viewtopic.php?t=141305  :wink:


my school celebrated mole day at 6:02 in my chem teacher's room today. i didn't go, but i have soph chem in that class first thing in the morning. they had oMOLEts, and there were some leftover eggs, so there were a few oMOLEts left. we brought all this food to celebrate in class. our teacher dressed in a mole suit...

Happy mole day. I celebrated by making Moleso.

I didn't realise there was a mole day. I'll have to celebrate with a mole of potatoes tonight, I suppose. A mole of Avoragdo numbers?

Avagadro's Number yes. I forgot about this  :D .


limit
logarithms
calculus
calculus computations

[hide="Click to open the hidden chest"]For $k \approx 0$,

$\frac{\ln{(\ln{(1+k^{4})})}}{\ln{(\ln{(1+k^{2})})}}= \frac{2\ln k+\ln\left( 1-\frac{k^{2}}{2}+\cdots \right)}{4\ln k+\ln\left( 1-\frac{k^{4}}{2}+\cdots \right)}= \frac{2+\frac{1}{\ln k}\ln\left( 1-\frac{k^{2}}{2}+\cdots \right)}{4+\frac{1}{\ln k}\ln\left( 1-\frac{k^{4}}{2}+\cdots \right)}$.

So this converges to $\frac{1}{2}$.[/hide]

[quote="misan"]did you mean to say $2$?  :huh:[/quote]

Oh, my god...  :blush:  You're right. I fixed it:

[hide="Click to unlock the rusty chest"]
$\frac{\ln{(\ln{(1+k^{4})})}}{\ln{(\ln{(1+k^{2})})}}= \frac{4\ln k+\ln\left( 1-\frac{k^{4}}{2}+\cdots \right)}{2\ln k+\ln\left( 1-\frac{k^{2}}{2}+\cdots \right)}= \frac{4+\frac{1}{\ln k}\ln\left( 1-\frac{k^{4}}{2}+\cdots \right)}{2+\frac{1}{\ln k}\ln\left( 1-\frac{k^{2}}{2}+\cdots \right)}$.
[/hide]


can any genius crack this one!! p.s. dont forget to show the work!!!!!

A criminal is driving a 2005 Ford Mustang. A police cruiser is able to reach a top speed of 125 mph and the Mustang can go 145 mph. How much further ahead would the Mustang be after 10 minutes of driving at top speed if they both started from the same spot?

2.5 miles

3 miles

3 1/3 miles

4.5 miles

This problem is pockmarked with holes. If they started from the same spot, they must accelerate to their top speed, which is another thing in itself. Assuming they have the same acceleration, the only difference is after 125 mph, but still a relatively small amount.

[hide="At a constant speed"]Ten minutes is one sixth of an hour. SO, $ \frac{125}{6}\equal{}D_{po\minus{}po}$ and $ \frac{145}{6}\equal{}D_{POS waste of life}$

or just $ D_{1}$ and $ D_{2}$ respectively if you're sensitive.
$ \left|D_{1}\minus{}D_{2}\right| \equal{}\frac{10}{3}$ miles.

The answer is [b]C[/b], or $ 3$and $ \frac{1}{3}$ miles.[/hide]

First, before you post here: redefine genius. Second, post this nonsense in the introductory physics forum, which can be found under the arts and sciences panel.

Also, you did not give the rate of acceleration.


geometry
area
geometric inequality
IMO Shortlist
Dusan Djukic
IMO 2006
IMO

Assign to each side $b$ of a convex polygon $P$ the maximum area of a triangle that has $b$ as a side and is contained in $P$. Show that the sum of the areas assigned to the sides of $P$ is at least twice the area of $P$.

My proof goes by another way, the main idea is that you can "cancel" parallel sides.
Pardon me that I am too tired to type it now.
And thank Valentin for posting the problems  :)

[quote="nmtruong1986"]I am strictly sure that we must prove there exists one point $O$ inside $P$ such that for each side $AB$ (corresponds an assigned area $S$), the area of OAB is not greater than a half of $S$. Whenever one can prove with only $N=3,4,5,6$ ($P$ is an $N$_polygon), one uses the Helly's lemma. That is clear? I will check and post if it is really true. :D[/quote]

This is FALSE! You can find a counterexample for $n = 6$.

[quote="Tiks"]Guys what do you think,is this a problem of geometry or combinatorics?I mean in the short list.[/quote]
 I think that is geometry because of one simple reason:On 99,9 percents of newer IMO (last 15 years) there were two problems from Geometry part of shortlist.

There are two solutions on the official IMO website.

On this problem, only 10 IMO contestants scored $\geq 5$! They are

CHN 4
CHN 6
DEU 6
FRA 5
KOR 6 (6 marks)
LVA 3 (5 marks)
MDA 1
POL 3
RUS 1
RUS 4

So this problem is probably more difficult than this one: http://www.mathlinks.ro/Forum/viewtopic.php?p=352683#p352683

Actually, there exists a fairly short solution. Draw the lines through all vertices of the polygon that divide the area in half. Look at the corresponding "butterflies". They cover the entire polygon (that is true for any set of lines). Now, for each butterfly ABOCD
[code]
 A __ B
  \  /
   \/ O
   /\
 C/__\ D
[/code]
(Here AB is not necessarily parallel to CD, but I was too lazy to create a PNG picture), we clearly, have $|OA|\cdot |OB|=|OC|\cdot |OD|$ (here we use the fact that both AD and BC  divide the area of the polygon into 2 equal parts). It follows from here that either $|OB|\ge |OC|$ or $|OA|\ge |OD|$. Thus either the area of BDC or that of ACD is at least twice that of OCD and similarly for the other side. Thus the sum of the areas of the triangles corresponding to $AB$ and $CD$ is at least twice the area of the butterfly. Now it is clear that it doesn't really matter whether to consider the whole sides of the original polygon or to subdivide each side into several smaller sides: the sum in question remains the same. The end.

[quote="fedja"]we clearly, have $|OA|\cdot |OB|=|OC|\cdot |OD|$ (here we use the fact that both AD and BC  divide the area of the polygon into 2 equal parts).[/quote]
I fail to see why this is true.

[quote="fedja"]The areas of the triangles $OAB$ and $OCD$ must be equal.[/quote]
why? the polygon isn't regular.

Because $AB$ and $CD$ divide the area in two equal parts. Now look at what is to the "right" of $AB$ and what is to the "right" of $CD$ on the picture. You have a big common part to the right of the broken line $BOD$ plus one triangle in one case and the other one in the other case. But the total area in both cases must be half of that of the polygon. So, disregarding the area of the common part, we see that the triangles must have the same area.

why there must be lines that divides the polygons area by 2. this isnt regular i think

[quote="A. S. K."]fedja, have you been on IMO 2006? You should get a special prize for your solution...[/quote]

If I'm not mistaken, he's a professor.

If I remember correctly, the first official solution is very similar to fedja's solution. In fact, the subdivision of the sides at the end of the solution more or less generates the fake vertices you need for fedjas solution.

The second solution mentioned here with minkowski sums is more or less the second official solution - in fact, the "associate" as they call it there is the minkowski sum $P+P^{t}$.

At IMO, I didn't see the thing with the minkowski sum but my solution is quite similar to the second official solution - I do some things to get vertices at extreme positions and cancel parallel sides.

Since solutions of the german team will be available at some time in the internet and I had to write down the solution to number 6, I have a .pdf-file of my solution. I'm attaching it in German and English.

Peter

Here are 2 problems looking very useful to P6. Both are known in Russia.

1. (SPbMO-88) Let $a_{1}, a_{2},\ldots , a_{n}$ be sides and also side-lengths of a convex polygon $P$. Let $d_{1}, d_{2},\ldots , d_{n}$ be lengths of projections of the polygon to $a_{1}, a_{2},\ldots , a_{n}$ respectively. Prove that \[\sum_{i=1}^{n}\frac{a_{i}}{d_{i}}\le 4.\] 2. (SPb math circles) Let $ABCD$ be a convex quadrilateral with a point $O$ in the interior and points $P,Q,R,S$ on the sides $AB,BC,CD,DA$ respectively such that $APOS\text{ and }CROQ$ are parallelogramms. Prove that $\sqrt{S_{BQOP}}+\sqrt{S_{DSOR}}\le \sqrt{S_{ABCD}}.$

1 contains an idea to reduce an arbitrary polygon to central-symmetric. 2 is used to explain Brunno-Minkowski inequality. So we (team Russia) could solve it much better.

I started to read Scholze's solution to the problem 6, and there is something I don't understand. In the case 1 there was said that "for every side $b$ of $P$ the line $\bar b$ which is parallel to $b$, meets $P$ in some point and has greatest distance to $P$". But in metric topology, I have been taught that if a line $l$ meets a set $A$, then $d(l,A)=0$. Is therefore $\bar b$ an arbitrary line which meets $P$?

Sorry, just a stupid typo: replace the last $P$ by $b$.

On the excellent Russian site http://www.problems.ru this problem
is recognized as the middle-level problem (with the mark 5+, while
the maximal complexity is 10). However, the solution posted there
is pretty unclear (at least, for me). Look
[url]http://www.problems.ru/view_problem_details_new.php?id=110751[/url]

What do you think?

It's also unclear for me. I'll post here the Russian solution, and my translation (it isn't good, and words aren't translated correctly!)

Воспользуемся соображением:

1) если двигать одну из вершин (трех)(много)угольника с постоянной скоростью, то его площадь меняется тоже с постоянной скоростью.
2) линейная комбинация линейных функций тоже линейна.
3) линейная функция достигает максимума (минимума) на границе отрезка.

Пусть P - вершина M, и число сторон M не меньше 4, R, T - соседние с P вершины. Будем двигать P параллельно [RT]. Тогда при движении в любом из двух направлений вершина M выйдет на продолжение стороны P.

Применив эти соображения, сведем задачу к случаю, когда P лежит на продолжении одной из сторон M, т.е. один из углов M равен π.

Но в этом случае дело сводится к многоугольнику с меньшим числом вершин и завершается индукционным спуском.

We use the following:

1) When moving from one vertex to another with constant velocity, the area(???) is also changing at constant velocity

2) A linear combination of linear functions is also linear

3) A linear function reaches its maximum (minimum) on the boundaries of its line

Let P be a vertex of M, with the number of sides of M not less than 4, and R and T are neighboring vertexes with P. Move P parallel to RT. Then, when moving in any direction, vertex of M will be the extension of sides of P (???)
Applying 1),2),3)  we reduce the problem to the one when P lies on the extension of one of the sides of M, that is when one of the angles is $ \pi$. Then it reduces to a problem with less vertexes, which ends in an inductive descent.

OK, there are things I don't understand.
What are M and P (later in the solution, it seems like they aren't vertex and polygon).
1)?
And the second ???

This is a very nice solution (partially understood), but can anyone explain the remaining for me?

I came up with a beautiful proof, but then noticed it had a gaping gap!  Fortunately Kai Neergard engineered a bridge.
See http://camoo.freeshell.org/IMO_2006_6.pdf
I also included a short proof based on the Minkowski sum.
Laura

The very nice and simple solution in larkl's preceding post is based on a beautiful idea which to my knowledge was never advanced before in relation to this problem. This idea is to consider a linearly parametrised family of polygons XX(t) which contains several polygons relevant to the problem: XX(0) ~= XX(1) ~= PP, where PP is the polygon of the IMO problem, XX(1/2) ~= 1/2 QQ, where QQ is the 'associate' in the language of the shortlist's Solution 2 (which according to Peter Scholze is also Charlydif's F), and in the limits t --> +- oo asymptotically XX(t) ~= |t| RR traversed twice, where RR's side vectors are those of PP in the order of increasing line direction modulo pi when we assume they are traversed in PP in the order of increasing vector direction modulo 2 pi so that PP has positive oriented area [PP]. The factors in the congruences are scale factors. Since [XX(t)] is a quadratic in t, the areas [PP], [QQ] and [RR] are related. In particular [RR] < 0 implies [QQ] >= 4 [PP], which is eqivalent to the IMO problem's assertion.

I supplied the proof that indeed [RR] < 0. larkl cites an analytic version of it. I prefer, myself, the following geometric one. Note that the areas [TT_i] introduced below are the terms in the final sum in the analytic calculation.

Let RR = A_0...A_{n-1} and A_n = A_0. For i = 0, ..., n - 1, let m_i be the line A_iA_{i+1}, and let l be a line through A_0 within the least positive angle from m_{n-1} to m_0. Since this angle and the least positive angles from m_i to m_{i+1} for i = 1, ..., n - 1 sum to pi, the direction of l is outside any of the angles mentioned most recently. For i = 0, ..., n - 1, let B_i be the point of intersection of l and m_i. We notice B_0 = B_{n-1} = A_0. The area [RR] is the sum of [TT_i] over i = 0, ..., n - 2, where TT_i = B_iA_{i+1}B_{i+1}. For i = 0, ..., n - 2, since the direction of l is outside the least positive angle from m_i to m_{i+1}, the oriented angle \angle B_iA_{i+1}B_{i+1} is congruent to this angle modulo 2 pi, so [TT_i] < 0. Therefore [RR] < 0.

The relation between [PP], [QQ] and [RR] is 2[PP] = [QQ]/2 + [RR]. Instead of considering the quadratic [XX(t)], one may derive this relation using vector algebra. To understand the following, notice the relation between PP and RR, which is elucidated very nicely in larkl's text: The polygon RR is composed of the sequences of sides of PP with a common most distant vertex V in PP. These sequences appear in RR in the order so that, in PP, each sequence begins at the V of the preceding sequence and ends at that of the next one. Therefore, RR is got from PP by a parallel displacement of each of its sides with the vector OV, where O is an arbitrary constant point and V its most distant vertex in PP. Also note that [QQ]/2 is the sum of the maximal triangular areas of the IMO problem. Therefore, if P is a point on the side with the vector s, we have

[RR] = 1/2 sum ( OP + OV ) x s = 1/2 sum ( 2 OP - VP ) x s = 2 [PP] - [QQ]/2 .

Alternatively, let s_i = A_iA_{i+1} in the notation above. Then

2 [RR] = sum_{i<j} s_i x s_j .

Let S and T be the sets of such indices i that if u is a vector on l, the angle from u to s_i is in the interval (-pi,0) and (0,pi), respectively. If, in the previous notation, P and V are associated with the side with index j, then if j is in S, the vector VP is the sum of the vectors s_i with i in T and i > j or i in S and i < j. Simlarly if j is in T with S and T interchanged. Thus

2 ([QQ]/2) = sum' s_i x s_j ,

where the prime indicates that the sum is restricted to the pairs (i,j) just described. Collecting the terms, one finds that 2 ( [RR] + [QQ]/2 ) is the sum of s_i x s_j over the following (i,j).

(i) i,j are both in S or both in T, and i < j. These (i,j) occur twice.

(ii) i is in S , j is in T or i is in T , j is in S.

The terms with (i,j) of the type (ii) sum to U x (-U) + (-U) x U = 0, where U = sum_{i in S} s_i = - sum_{i in T} s_i.

The area 2 [PP] is the sum of s_i x s_j over such (i,j) that i,j are both in S or both in T, and i < j, or i is in S and j in T. The terms with i in S and j in T sum to U x (-U) = 0. Thus

[RR]+  [QQ]/2 = 2 [PP] .

[quote=fmasroor]Can somebody rigorously explain fedja's solution please?[/quote] I finally take the time to write it up properly myself, after having been told the solution during Taiwan IMO camp in 2014.

We say a polygon in [i]almost convex[/i] if all its angles are at most $180^{\circ}$.

Note that given any convex or almost convex polygon, we can take any side $b$ and add another vertex on it, and the sum of the labels doesn't change (since the label of a side is the length of the side times the distance of the farthest point).

[b][color=red]Lemma:[/color][/b]  	Let $N$ be an even integer. 	Then any almost convex $N$-gon with area $S$ 	should have an inscribed triangle with area at least $2S/N$.

The main work is the proof of the lemma.

Label the polygon $P_0 P_1 \dots P_{N-1}$. Consider the $N/2$ major diagonals of the almost convex $N$-gon, $P_0 P_{N/2}$, $P_1 P_{N/2+1}$, et cetera. A [i]butterfly[/i] refers to a self-intersecting quadrilateral $P_i P_{i+1} P_{i+1+N/2} P_{i+N/2}$. An example of a butterfly is shown below for $N=8$.

[asy] pair P_0 = (-1,-3); pair P_1 = ( 0,-3); pair P_2 = (2.5,-3); pair P_3 = (3,0); pair P_4 = (1.8,1.2); pair P_5 = (1.1,1.9); pair P_6 = (0.3,2.7); pair P_7 = (-2,0.4); filldraw(P_0--P_2--P_3--P_6--P_7--cycle, invisible, blue);

filldraw(P_0--P_4--P_5--P_1--cycle, palered, red); draw(P_1--P_5, red); draw(P_2--P_6, red); draw(P_3--P_7, red);

dot("$P_0$", P_0, dir(P_0)); dot("$P_1$", P_1, dir(P_1)); dot("$P_2$", P_2, dir(P_2)); dot("$P_3$", P_3, dir(P_3)); dot("$P_4$", P_4, dir(P_4)); dot("$P_5$", P_5, dir(P_5)); dot("$P_6$", P_6, dir(P_6)); dot("$P_7$", P_7, dir(P_7));

/* TSQ Source:

P_0 = (-1,-3) P_1 = ( 0,-3) P_2 = (2.5,-3) P_3 = (3,0) P_4 = (1.8,1.2) P_5 = (1.1,1.9) P_6 = (0.3,2.7) P_7 = (-2,0.4) P_0--P_2--P_3--P_6--P_7--cycle 0.1 lightcyan / blue

P_0--P_4--P_5--P_1--cycle 0.2 lightred / red P_1--P_5 red P_2--P_6 red P_3--P_7 red

*/ [/asy]

[b][color=red]Claim:[/color][/b]  Every point $X$ in the polygon is contained in the wingspan of some butterfly.

[i]Proof.[/i]  Consider a windmill-like process which [list] 	[*]starts from some oriented red line $P_0 P_{N/2}$, 	oriented to face $P_0 P_{N/2}$ 	[*]rotates through $P_0 P_{N/2} \cap P_1 P_{N/2+1}$ 	to get line $P_1 P_{N/2+1}$, 	[*]rotates through $P_1 P_{N/2+1} \cap P_2 P_{N/2+2}$ 	to get line $P_2 P_{N/2+2}$, 	[*]\dots et cetera, until returning to line $P_{N/2} P_0$, 	but in the reverse orientation. [/list] At the end of the process, every point in the plane has switched sides with our moving line. The moment that $X$ crosses the moving red line, we get it contained in a butterfly, as needed. $\blacksquare$

[b][color=red]Claim:[/color][/b]  	If $ABDC = P_i P_{i+1} P_{i+1+N/2} P_{i+N/2}$ is a butterfly, 	one of the triangles $ABC$, $BCD$, $CDA$, $DAB$ 	has area at least that of the butterfly.

[i]Proof.[/i]  	Let the diagonals of the butterfly meet at $O$, 	and let $a = AO$, $b = BO$, $c = CO$, $d = DO$. 	If we assume WLOG $d = \min(a,b,c,d)$ 	then it follows $[ABC] = [AOB] + [BOC] \ge [AOB] + [COD]$, as needed. $\blacksquare$

Now, since the $N/2$ butterflies cover an area of $S$, it follows that one of the butterflies has area at least $S / (N/2) = 2S/N$, and so that butterfly gives a triangle with area at least $2S/N$, completing the proof of the lemma.

\bigskip

[b]Main proof[/b]: Let $a_1$, \dots, $a_n$ be the numbers assigned to the sides. Assume for contradiction $a_1 + \dots + a_n < 2S$. We pick even integers $m_1$, $m_2$, \dots, $m_n$ such that \begin{align*} 	\frac{a_1}{S} &< \frac{2m_1}{m_1 + \dots + m_n} \\ 	\frac{a_2}{S} &< \frac{2m_2}{m_1 + \dots + m_n} \\ 	&\vdotswithin\le \\ 	\frac{a_n}{S} &< \frac{2m_n}{m_1 + \dots + m_n}. \end{align*} which is possible by rational approximation, since the right-hand sides sum to $2$ and the left-hand sides sum to strictly less than $2$.

Now we break every side of $P$ into $m_i$ equal parts to get an almost convex $N$-gon, where $N = m_1 + \dots + m_n$.

The main lemma then gives us a triangle $\Delta$ of the almost convex $N$-gon which has area at least $\frac{2S}{N}$. If $\Delta$ used the $i$th side then it then follows the label $a_i$ on that side should be at least $m_i \cdot \frac{2S}{N}$, contradiction.

To each side $a$ of $P$, let $f(a)$ be the unique line parallel to $a$, on the same side of $a$ as $P$, and such that the intersection of $f(a)$ with $P$ has zero area (is either a point or a segment). Let $s(P)$ denote the set of edges of $P$, and let $\ell(a)$ denote the length of side $a\in s(P)$. Then, we see that the sum of the areas assigned to each side is \[\beta(P):=\frac{1}{2}\sum_{a\in s(P)}\ell(a)\cdot d(a,f(a)),\] where $d(x,y)$ is the distance between two parallel lines $x$ and $y$.

The first step is to reduce the problem to the case where $P$ has opposite sides that are parallel. Indeed, call a side $a\in s(P)$ [i]good[/i] if there is some other $b\in s(P)$ such that $a\parallel b$. We proceed by induction on the number of bad sides. We'll solve the case where all sides are good later, so we may assume for this induction that the base case of $0$ bad sides is given.

Suppose we have $P$ with at least one bad side. Let $Q$ be the polygon formed by all the lines $f(a)$ for $a\in s(P)$.

[b]Lemma:[/b]  There is some vertex of $Q$ that is outside of $P$.

[i]Proof:[/i]
[hide][img]https://lh3.googleusercontent.com/-9khT8u276zM/XuCpIMej0tI/AAAAAAAAFyI/b3UmelygOOYJQVEMNUeRzPA_4y2_soZtQCK8BGAsYHg/s0/2020-06-10.jpg[/img][/hide]
Suppose to the contrary that all vertices of $Q$ were inside $P$ (here $P$ includes its boundary). If any vertex of $Q$ is not a vertex of $P$, then some side of $Q$ intersects $P$ in the interior, which is bad since the sides of $Q$ are segments on the lines $f(a)$. Thus, all vertices of $Q$ are vertices of $P$. However, if any vertex of $P$ is not a vertex of $Q$, then it is easy to see that again, one of the sides of $Q$ intersects the interior of $P$, which is bad.

Thus, the vertices of $P$ and $Q$ are identical, so $P=Q$. This implies that every side of $P$ is good, so we have our desired contradiction. $\blacksquare$

Say $X$ is a vertex of $Q$ that is outside $P$. Let $a$ and $b$ be sides of $P$ such that the edges of $Q$ incident to $X$ are from $f(a)$ and $f(b)$.

Let $A$ be the vertex of $P$ that is on $f(a)$ and let $B$ be the vertex of $P$ that is on $f(b)$ (note that these may not be unique if one of $a$ or $b$ is good, and we will deal with this issue later). Let $I_1,\ldots,I_m$ be the vertices of $P$ that are on the same side of $AB$ as $X$ is such that $I_0=A,I_1,\ldots,I_m,B=I_{m+1}$ are consecutive vertices in $P$. If there were two possible choices for $A$, then choose $A$ such that $A,I_1,\ldots,I_m,B$ are indeed consecutive.

If there was any edge $c$ between $a$ and $b$ (on the other side as $A,I_1,\ldots,I_m,B$), then $f(c)$ would intersect the interior of the convex region $\triangle ABX$, which would imply that $X$ was not a vertex of $Q$, which is a contradiction. Thus, $a$ and $b$ are consecutive in $P$, and suppose they are both incident to $U$. We have the following key claim.

[b]Claim:[/b]  We have $U\in f(I_jI_{j+1})$ for all $0\le j\le m$, and $f(c)\not\in\{a,b,I_0I_1,\ldots,I_mI_{m+1}\}$ for $c\in s(P)\setminus\{a,b,I_0I_1,\ldots,I_mI_{m+1}\}$.

[i]Proof:[/i]
[hide][img]https://lh3.googleusercontent.com/-DP3_FgYlqp4/XuCpHWgjEuI/AAAAAAAAFyE/R3Cs2v7ZZX4c2CiqTdpkmLbppY6c5gEBgCK8BGAsYHg/s0/2020-06-10.jpg[/img][/hide]
We first prove that $U\in f(I_jI_{j+1})$ for all $0\le j\le m$. For some $j$, let $\ell$ be the line through $U$ parallel to $I_jI_{j+1}$. It suffices to show that $\ell$ doesn't intersect the interior of $P$. This is obvious by some directed angles.

Now, suppose we had $f(c)\in\{a,b,I_0I_1,\ldots,I_mI_{m+1}\}$ for some $c\in s(P)\setminus\{a,b,I_0I_1,\ldots,I_mI_{m+1}\}$. Then, $f(c)$ would intersect the interior of the convex region $\triangle ABX$, which is absurd as we saw before. This completes the proof of the claim. $\blacksquare$

The key to the induction is to take $P'$ to be the polygon that is the convex hull of $P$ and $X$. Note that the vertices of $P'$ are the vertices of $P$ with the $I_j$ removed and $X$ added. We have the following claim which shows that reducing $P$ to $P'$ is valid under the inductive monovariant.

[b]Claim:[/b]  The number of bad edges of $P'$ is less than the number of bad edges of $P$.

[i]Proof:[/i]  The number of vertices of $P'$ is the number of vertices of $P$ plus $1-m$. If both $a$ and $b$ are bad, then clearly $m\ge 0$, if just one of them is bad, then $m\ge 1$, and if both are good, then $m\ge 2$. We see that both $a$ and $b$ become good when we replace $P$ with $P'$, and the goodness of all edges in $s(P)\setminus\{a,b,I_0I_1,\ldots,I_mI_{m+1}\}$ is unaffected. We lose the bad edges $\{I_0I_1,\ldots,I_mI_{m+1}\}$ entirely, which are replaced with the two good edges $AX$ and $BX$. Furthermore, the edges $a$ and $b$ become good if they weren't before. All this implies that the number of bad edges goes down (this is just casework on the goodness of $a$ and $b$), so we're done. $\blacksquare$

Thus, we may assume that the problem is true for $P'$. By construction, the areas written down for the edges in $s(P)\setminus\{I_0I_1,\ldots,I_mI_{m+1}\}$ remain unchanged. The area written on the edge $I_jI_{j+1}$ is $[I_jI_{j+1}U]$ (for $P$), and the areas written on the edges $AX$ and $BX$ (for $P'$) are $[AXU]$ and $[BXU]$. Thus, \[\beta(P')-\beta(P)=[AXU]+[BXU]-[AI_1\cdots I_mBU]=[AI_1\cdots I_mBX]=\alpha(P')-\alpha(P),\] where $\alpha(R)$ is the area of $R$. Thus, \[[\beta(P)-2\alpha(P)] - [\beta(P')-2\alpha(P')]= \alpha(P')-\alpha(P)\ge 0,\] so since $\beta(P')-2\alpha(P')$, we must have $\beta(P)-2\alpha(P)\ge 0$, as desired. This completes the induction part of our proof.

Now we have to show the base case, i.e. the case where all the edges of $P$ are good. Let the vertices of $P$ be $A_1A_2\cdots A_{2n}$ with indices taken mod $2n$. Let $a_i$ be $A_iA_{i+1}$, and let $d_i=d(a_i,a_{i+n})$. Note that \[\beta(P) = \sum_{i=1}^n [\ell(a_i)+\ell(a_{i+n})]\cdot\frac{d_i}{2}.\] Let $m_i$ be the line halfway between $a_i$ and $a_{i+n}$. The key idea is to move the $m_i$ (while preserving the relative distances between $a_i,m_i,a_{i+n}$). As long as the polygon remains convex, the value of $\beta(P)$ does not change under this movement. However, the value of $\alpha(P)$ may change. We have the following key lemma.

[b]Lemma:[/b]  Suppose we have oriented non-parallel lines $m_1,\ldots,m_n$ that can move while maintaining their direction. We have fixed $r_1,\ldots,r_n$, and define $a_i$ to be the line exactly $r_i$ to the left of $m_i$ (this makes sense as the lines are oriented) and $a_{i+n}$ to be the line exactly $r_i$ to the right of $m_i$. Let $A_i=a_i\cap a_{i-1}$ for $1\le i\le 2n$. For an appropriate global area orientation, the signed area $[A_1A_2\cdots A_{2n}]$ attains a maximum value when all the $m_i$ are concurrent.

[i]Proof:[/i]  The proof proceeds by induction on $n$. The case $n=2$ is trivial, as the area $[A_1A_2A_3A_4]$ is actually always fixed. So suppose the result is true for some $n-1\ge 2$. We'll show it's true for $n$. Here is a diagram for $n=4$.
[hide][img]https://lh3.googleusercontent.com/-CbsBF5Mp5SE/XuCpGP1ebhI/AAAAAAAAFyA/ZaTtfRmzuD0MMnq5DddSP3G37GD9lUWiQCK8BGAsYHg/s0/2020-06-10.jpg[/img][/hide]
Let $X=a_{n-1}\cap a_{n+1}$ and $Y=a_{2n-1}\cap a_1$. Fix $m_1,\ldots,m_{n-1}$, while varying $m_n$. Then, we have the signed area equality \[[A_1A_2\cdots A_{2n}] = [YA_2\cdots A_{n-1}XA_{n+2}\cdots A_{2n-1}] - [YA_1A_{2n}] - [XA_{n+1}A_n].\] In the following diagram, the highlighted parallelogram is fixed as $m_n$ varies, and furthermore, its dimensions and orientation is only dependent on the fixed directions of $m_1,\ldots,m_{2n}$ and the fixed distances $r_1,\ldots,r_n$.
[hide][img]https://lh3.googleusercontent.com/-8cqH9WYVFfI/XuCpFbFZ2WI/AAAAAAAAFx8/twutVAxgs8MWmYVtSglaMMPU53TsZnJEQCK8BGAsYHg/s0/2020-06-10.jpg[/img][/hide]
As $m_n$ varies, we see that \[[YA_1A_{2n}] + [XA_{n+1}A_n] = k(p^2+q^2)\] where $p$ and $q$ represent the signed distances $A_1A_{2n}$ and $A_nA_{n+1}$, so $p+q:=s$ is fixed, and $k$ is some fixed proportionality constant only dependent on the fixed directions of $m_1,\ldots,m_{2n}$ and the fixed distances $r_1,\ldots,r_n$. Thus, $[YA_1A_{2n}] + [XA_{n+1}A_n]$ attains the same minimum value $T=ks^2/2$ as $m_n$ varies, no matter where the $m_1,\ldots,m_{n-1}$ are. The area \[[YA_2\cdots A_{n-1}XA_{n+2}\cdots A_{2n-1}]\] is exactly the area from the $n-1$ case of the lemma with lines $m_1,\ldots,m_{n-1}$, so the area \[[A_1A_2\cdots A_{2n}] = [YA_2\cdots A_{n-1}XA_{n+2}\cdots A_{2n-1}]-T\] is maximized when $m_1,\ldots,m_{n-1}$ are in optimal position, which is concurrent by the base case.

In the case that $m_1,\ldots,m_{n-1}$ are concurrent (so the polygon $YA_2\cdots A_{n-1}XA_{n+2}\cdots A_{2n-1}$ is centrally symmetric) , $m_n$ is placed optimally when $A_1A_{2n}=A_nA_{n+1}$. It's easy to see that this implies that $m_n$ also passes through the same center, so $m_n$ is concurrent with $m_1,\ldots,m_{n-1}$, as desired. $\blacksquare$

Thus, the area $\alpha(P)$ is at most \[\sum_{i=1}^{2n}\frac{\ell(a_i)+\ell(a_{i+n})}{2}\cdot d_i = \sum_{i=1}^{n}[\ell(a_i)+\ell(a_{i+n})]\cdot d_i=2\beta(P),\] as desired. This is because if we translate the $m_i$s to make $P$ centrally symmetric, the (signed) area becomes \[\sum_{i=1}^{2n}\frac{\ell(a_i)+\ell(a_{i+n})}{2}\cdot d_i,\] and this is an upper bound for $\alpha(P)$ by the lemma.

[hide=Solution]We will prove the problem by induction on $n$ (the number of vertices).The base cases $ n=3,4 $ are trivial.
Let $A_{i-1},A_i,A_{i+1},A_{i+2}$ be consecutive vertices of $P$ and consider moving $ A_i $ along the line $ A_{i-1}A_i $. Due to convexity, it will move between $ A_{i-1} $ and $ X=A_{i-1}A_i\cap A_{i+1}A_{i+2} $. The segment $ XA_{i-1} $ will be split into smaller segments, where each segment corresponds to a set of maximal triangles associated to sides which involve $A_i$. The endpoints of these smaller segments are either $ A_{i-1} $ or $X$ or points where some maximal triangles change for the first time-that is there exists a side which has two options for the vertex with maximal area and $ A_i $ is either one of the options or belongs to the side. It's easy to see that if a side has two such options, then these two are adjacent and form a side parallel to the initial side. This means that the moments of change are points on $ XA_i $ which lie on some parallel line through $ A_{i+1} $ to a side of the polygon.

It's easy to see that by moving $A_i$ each individual area involving $ A_i $ and some other two vertices is a linear function of $ A_i $, hence the entire contribution of $A_i$ in the quantity (sum of areas minus twice the area of $P$) is a linear function (depending on the smaller segment $A_i$ varies on) and as we want the quantity to be minimal, we can assume $ A_i$ is the endpoint of the segment.
If $ A_i=X $ or $ A_i=A_{i-1} $ we are easily done by the inductive hypothesis on the new polygon (which is formed by deleting $A_i$ and, in the first case, replacing $A_{i+1}$ with $X$) so let us assume that $ A_i $ lies on a parallel in $ A_{i+1} $ to some side of the polygon.
This reasoning can be applied to all sides (note that by the way this transformation is defined it won't affect any two previous parallel sides) so for each side there is another parallel side (in particular $n$ is even).

The parallel pairs of sides have to be of the form $ A_iA_{i+1} $ and $ A_{i+\frac{n}{2}}A_{i+\frac{n}{2}+1} $ since, if we consider one such pair, the other pairs will have to be in opposite arcs, implying that the pair of sides splits the vertices equally.
With this condition, our problem is to prove that the sum of the $ \frac{n}{2} $ trapezoids $ A_iA_{i+1}A_{i+\frac{n}{2}}A_{i+\frac{n}{2}+1} $ is at least twice the area of $P$.

The final strategy will be to translate $ A_{1+\frac{n}{2}}\dots A_{1} $ on the direction of $ \vec{A_1A_{2}} $ and consider how the difference between the sum of the areas and the double of area of $P$ changes. It's not hard to compute that the difference doesn't change, which means we can perform the maximal translation which brings one (or both) of $ A_1,A_{\frac{n}{2}+1}$ to $ A_2,A_{\frac{n}{2}+2}$. Reducing the overlapping vertices will create a smaller polygon on which we can apply the inductive hypothesis and we are done. [/hide]

Approximate the polygon with a smooth convex closed curve that does not have zero curvature at any point[hide=(*)]One way to do this is to construct small circular arcs near each vertex of the polygon and then join the arcs together with very flat parabolas, using bump functions to stitch the two together. The notion of approximation needed here is that except near the vertices, the curve is quite flat, and that the curve never protrudes very far from the polygon. An alternative approach is to approximate the polygon with an equiangular polygon and use the discrete Fourier transform later, though this actually changes the later computations nontrivially.[/hide]. The boundary of this shape can be parametrized with a function $f \colon \mathbb{R}/2\pi\mathbb{Z} \to \mathbb{C}$, such that $f(\theta)$ is the point on the boundary where $f'(\theta)$ is a positive multiple of $e^{i\theta}$.

Now, by Shoelace the area is given by
\[A = \int_0^{2\pi} \frac{f'(\theta) \overline{f(\theta)} - \overline{f'(\theta)} f(\theta)}{4i}\,d\theta,\]
while the sum of areas in the problem is well-approximated by
\[B = \int_0^{2\pi} \frac{f'(\theta) \overline{f(\theta)-f(\theta+\pi)} - \overline{f'(\theta)} (f(\theta)-f(\theta+\pi))}{4i}\,d\theta,\]
so it suffices to evaluate these quantities and compare them. To do this, we write $f$ as a Fourier series. First, note that $e^{-i\theta} f'(\theta)$ is real, so it can be written as $\textstyle \sum_{k \in \mathbb{Z}} c_k e^{ik\theta}$ with $c_{-k} = \overline{c_k}$. Also note that $c_1 = 0$, since $f'(\theta)$ has a vanishing zeroth Fourier coefficient. Furthermore, we have
\[f(\theta) = \sum_{k\in\mathbb{Z}\setminus\{0\}} \frac{c_{k-1}}{ik} e^{ik\theta},\]
modulo some constant that we can WLOG assume is $0$ (we have no convergence issues anywhere since smoothness implies the coefficients decay superpolynomially). Using the fact that
\[\int_{0}^{2\pi} \left(\sum_{k\in\mathbb{Z}} a_k e^{ik\theta} \right) \overline{\left(\sum_{k\in\mathbb{Z}} b_k e^{ik\theta} \right)}\,d\theta = 2\pi \sum_{k \in \mathbb{Z}} a_k \overline{b_k},\]
the integrals above can be evaluated to be
\[A = \pi \sum_{k \in \mathbb{Z}\setminus\{-1\}} \frac{\lvert c_k \rvert^2}{k+1} \text{ and } B = 2\pi \sum_{k \in 2\mathbb{Z}} \frac{\lvert c_k \rvert^2}{k+1},\]
respectively. Hence
\[B - 2A = 2\pi \sum_{k \in (2\mathbb{Z} + 1)\setminus\{-1\}} -\frac{\lvert c_k \rvert^2}{k+1} = 2\pi \sum_{\text{odd }k \geq 3}\lvert c_k \rvert^2 \left(\frac{1}{k-1} - \frac{1}{k+1}\right) \geq 0,\]
which finishes upon taking better and better approximations.


Well Im working on my USAMTS solution and it really long so it just goes on and on and on and when I build the file, then I view it, it outputs the pdf ok but the thing is it only has only page and since my thing is so long, my document cuts off at the bottom of the page. Whats wrong?

It could be due to all sorts of reasons, wrong type of paper, margins ... To start with please post here your tex document preamble ie all the lines before \begin{document}.

I just used what the template had which is...

[hide="This is long and I included all the %%%% stuff"]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%                                          %%
%% USE THIS FILE TO SUBMIT YOUR SOLUTIONS   %%
%%                                          %%
%% You must have the usamts.tex file in     %%
%% the same directory as this file.         %%
%% You do NOT need to submit the usamts.tex %%
%% with your solutions.  You only need to   %%
%% submit this file and the PDF file with   %%
%% your solutions (and please attach any    %%
%% images you include)                      %%
%%                                          %%
%% DO NOT ALTER THE FILE usamts.tex         %%
%%                                          %%
%% If you have any questions or problems    %%
%% using this file, or with LaTeX in        %%
%% general, please go to the LaTeX          %%
%% forum on the Art Of Problem Solving      %%
%% web site, and post your problem.         %%
%%                                          %%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% DO NOT ALTER THE FOLLOWING LINES

\documentclass[11pt]{article}
\usepackage{amsmath,amssymb,amsthm}
\usepackage[pdftex]{graphicx}
\usepackage{fancyhdr}
\pagestyle{fancy}
\include{usamts}
%% DO NOT ALTER THE ABOVE LINES
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%% If you would like to use Asymptote within this document (which is optional),
%% you can find out how at the following UR%%
%%   http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration
%%
%% As explained there, you will want to uncomment the line below.  But be
%% sure to check the website because there are several other steps that must
%% be followed.
%% \usepackage{asymptote}

%% Enter your real name here
%% Example: \realname{David Patrick}
\realname{Kelvin Do}

%% Enter your USAMTS username here
%% Example: \username{DPatrick}
%% IMPORTANT
%% If your username contains one of the following characters:
%%      # \$ ~ _ ^ % { } &
%% then this character must be preceded by a backslash \
%% for example: if your user name is math_genius, then the line below should be
%% \username{math\_genius}

\username{Ketvoman}

%% Enter your USAMTS ID# here
%% Example: \usamtsid{9999}
\usamtsid{10679}

%% Make sure that the year and round number
%% are correct.
%% Year 20 is the academic year 2008-09
%% Round 1 is the October 14, 2008 due date
%% Round 2 is the November 24, 2008 due date
%% Round 3 is the January 12, 2009 due date
%% Round 4 is the March 9, 2009 due date
\usamtsyear{20}
\usamtsround{1}
[/hide]

There must be a problem then with the content. Can you post the tex document, or if you don't want to make it public (which is understandable), then can you replace your text and mathematics with nonsense, at least as much as should go onto a new page?

I replaced it with stuff.  Stuff just runs to the bottom of the page and gets cut off.  At the bottom it should say
1
2
3
4
5

But it just runs past the page.

Thanks for that. You are missing an important line in USAMTS Fake.tex which will cause the problem (and also lots of other errors).
After the line
\pagestyle{fancy}
you need the line
\include{usamts}

Also ensure that usamts.tex is in the [i]same directory[/i] as your tex document. Then USAMTS Fake.tex and your original document should compile properly.

Ok I left that out for a reason.  When I did that every time I compiled my document, no matter what the text was above, the output was my real soultion file for some reason.  If you could download the usamts file and write \include{usamts}, the output will still cut off.  Well it does for my real solution file.  And on my real solution file there is the \include {usamts} and the usamts file is in the same directory as the solution file.

When you compile your file (either the fake one or the real one) with the include line, does LaTeX report any errors?

Oh yeah every time theres always errors even from the very beginning. I can download the original Usamts include file and just compile that and theres 16 errors from the start.  Real or fake theres always errors.  A couple warnings, and no bad boxes though.

If you get errors, then you should examine and remove them before doing anything else.

You don't want to compile the include file, since it's not meant to be a stand-alone file.

When you compile your file, what are the errors reported?

Ok well if you've seen the usamts solution template, the part where it says \realname, and \username and stuff like that gets errors.  I compile it then hit f9 to find the errors.  Also, I think this may be crucial, one of the errors says \include cannot be nested.  What does that mean?

What is the first error that you get?  Can you copy-and-paste the error log here?

By the way, what did you name your solutions file and your include file?

Oh yes you are the best.  That was the problem all along was that I named my file USAMTS and my include file usamts include.  Phew, thanks so much.


geometry
power of a point
radical axis
geometry unsolved

suppose that E is a point in the median CD of triangle ABC.The circle C1 which passes through E and is tangent to AB at A meets AC in M, and circle C2 which passes through E and is tangent to AB at B meets CB at N.Prove that the tangents of the two circles in points M and N meet in the median

Since the tangent lengths DA = DB to the circles $\mathcal C_{1},\ \mathcal C_{2}$ from the midpoint D of AB are equal, D is on their radical axis. Their intersection E is also on their radical axis, hence the C-median $CD \equiv ED$ of the $\triangle ABC$ is their radical axis. Since C is on the radical axis, powers of C to the circles $\mathcal C_{1},\ \mathcal C_{2}$ are equal, $CA \cdot CM = CB \cdot CN.$ This means that quadrilateral ABMN is cyclic, let $\mathcal O$ be its circumcircle. Let S be the external similarity center of the circles $\mathcal C_{1},\ \mathcal C_{2}$, which lies on their common external tangent AB. Let SM meet the circle $\mathcal C_{1}$ again at P and the circle $\mathcal C_{2}$ at N', Q, the points P, M, N', Q following on the line SM either in this order or in the order M, P, Q, N'. Powers of the point S to the circles $\mathcal C_{1},\ \mathcal C_{2}$ are

$p(S,\ \mathcal C_{1}) = SA^{2}= SM \cdot SP,\ \ \ p(S,\ \mathcal C_{2}) = SB^{2}= SN' \cdot SQ,$

respectively. Since S is the external similarity center of the circles $\mathcal C_{1},\ \mathcal C_{2}$ with radii $r_{1},\ r_{2},$

$\frac{SA}{SB}= \frac{SM}{SQ}= \frac{SP}{SN'}= \frac{r_{1}}{r_{2}}.$

Combining,

$SA \cdot SB = \frac{SM}{SQ}\cdot SB^{2}= \frac{SM}{SQ}\cdot SN' \cdot SQ = SM \cdot SN',$

which means that the quadrilateral ABMN' is also cyclic, $N' \in \mathcal O.$ Since both N, N' lie on the circles $\mathcal C_{2},\ \mathcal O,$ which intersect at B and at one other point, the points $N' \equiv N$ are identical. As a result, MN passes through the external similarity center S of $\mathcal C_{1},\ \mathcal C_{2}.$ The triangle pairs $\triangle SAM \sim \triangle SBQ,$ $\triangle SAP \sim \triangle SBN$ are then centrally similar with similarity center S, hence $AM \parallel BQ,\ AP \parallel BN.$ Let the tangents $t_{M},\ t_{N}$ of $\mathcal C_{1},\ \mathcal C_{2}$ at M, N meet at F. Let these tangents meet AB at X, Y (this is only for notation purpose). Then $\angle AMX = \angle APM = \angle BNQ \equiv \angle BNM$ and $\angle BNY = \angle BQN = \angle AMP \equiv \angle AMN.$ Consequently,

$\angle FMN = 180^\circ-(\angle NMA+\angle AMX) =$

$= 180^\circ-(\angle YNB+\angle BNM) = \angle FNM,$

which means that the triangle $\triangle FMN$ is isosceles with FM = FN. Since the tangent lengths from F to the circles $\mathcal C_{1},\ \mathcal C_{2}$ are equal, the point F lies on their radical axis $ED \equiv CD,$ identical with the C-median of the $\triangle ABC.$

Let the tangents $t_{M},\ t_{N}$ of $\mathcal C_{1},\ \mathcal C_{2}$ at $M, N$ meet at $F$. Let these tangents meet $AB$ at $X, Y$
 $\angle CNM=\angle CAB$ (A,B,N,M are cyclic)
$\angle YNB=\angle YBN$
So$\angle MNY=180-\angle CAB-\angle YBN =\angle ACB$
Similarly $\angle XMN=\angle ACB$
Thus $\angle XMN=\angle YNM$
So triangle $FMN$ is isosceles with $FM = FN$. So $F$ is on the radical axis of $C_{1}$ and $C_{2}$.Thus $F$ is on $CD$. $q.e.d$


$18)$ What is the sum of the reciprocals of the roots of the equation

$\frac{2003}{2004}x+1+\frac{1}{x}=0$?

Also posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131309]here[/url] :wink: .

[hide]
Just multiply through the denominators, then notice that $\frac{1}{r_{1}}+\frac{1}{r_{2}}=\frac{\sum r_{i}}{\prod r_{i}}$ and so we can apply Vieta's.

Or you can reverse the coefficients after multiplying through, and the sum will just be the opposite of the coefficient on $x$. (This comes from the fact that if you reverse the coefficients of a polynomial, its roots are the reciprocals of the original's roots. [/hide]