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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
polygon's area doesn't add much when combined with its centric symmetry
mathematics2003   7
N 3 minutes ago by sttsmet
Source: 2021ChinaTST test4 day2 P2
Find the smallest real $\alpha$, such that for any convex polygon $P$ with area $1$, there exist a point $M$ in the plane, such that the area of convex hull of $P\cup Q$ is at most $\alpha$, where $Q$ denotes the image of $P$ under central symmetry with respect to $M$.
7 replies
mathematics2003
Apr 14, 2021
sttsmet
3 minutes ago
a^2-bc square implies 2a+b+c composite
v_Enhance   41
N 6 minutes ago by cursed_tangent1434
Source: ELMO 2009, Problem 1
Let $a,b,c$ be positive integers such that $a^2 - bc$ is a square. Prove that $2a + b + c$ is not prime.

Evan o'Dorney
41 replies
v_Enhance
Dec 31, 2012
cursed_tangent1434
6 minutes ago
IMO Shortlist 2008, Geometry problem 2
April   42
N 13 minutes ago by s27_SaparbekovUmar
Source: IMO Shortlist 2008, Geometry problem 2, German TST 2, P1, 2009
Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE = \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$.

Proposed by Charles Leytem, Luxembourg
42 replies
April
Jul 9, 2009
s27_SaparbekovUmar
13 minutes ago
3^n + 61 is a square
VideoCake   25
N 19 minutes ago by endless_abyss
Source: 2025 German MO, Round 4, Grade 11/12, P6
Determine all positive integers \(n\) such that \(3^n + 61\) is the square of an integer.
25 replies
VideoCake
Monday at 5:14 PM
endless_abyss
19 minutes ago
Problem 5
blug   3
N 25 minutes ago by Jt.-.
Source: Czech-Polish-Slovak Junior Match 2025 Problem 5
For every integer $n\geq 1$ prove that
$$\frac{1}{n+1}-\frac{2}{n+2}+\frac{3}{n+3}-\frac{4}{n+4}+...+\frac{2n-1}{3n-1}>\frac{1}{3}.$$
3 replies
blug
May 19, 2025
Jt.-.
25 minutes ago
Inequality with xy+yz+zx=1
Kimchiks926   14
N an hour ago by math-olympiad-clown
Source: Baltic Way 2022, Problem 4
The positive real numbers $x,y,z$ satisfy $xy+yz+zx=1$. Prove that:
$$ 2(x^2+y^2+z^2)+\frac{4}{3}\bigg (\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\bigg) \ge 5 $$
14 replies
Kimchiks926
Nov 12, 2022
math-olympiad-clown
an hour ago
Problem 7
SlovEcience   7
N an hour ago by GreekIdiot
Consider the sequence \((u_n)\) defined by \(u_0 = 5\) and
\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]a) Prove that there exist infinitely many positive integers \(n\) such that \(u_n > 2020n\).

b) Compute
\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]
7 replies
SlovEcience
May 14, 2025
GreekIdiot
an hour ago
D1033 : A problem of probability for dominoes 3*1
Dattier   1
N an hour ago by Dattier
Source: les dattes à Dattier
Let $G$ a grid of 9*9, we choose a little square in $G$ of this grid three times, we can choose three times the same.

What the probability of cover with 3*1 dominoes this grid removed by theses little squares (one, two or three) ?
1 reply
Dattier
May 15, 2025
Dattier
an hour ago
Inequality
lgx57   1
N 2 hours ago by sqing
Source: Own
$a,b,c \in \mathbb{R}^{+}$,$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1$. Prove that
$$a^abc+b^bac+c^cab \ge 27(ab+bc+ca)$$
1 reply
lgx57
2 hours ago
sqing
2 hours ago
Easy but Nice 12
TelvCohl   2
N 2 hours ago by AuroralMoss
Source: Own
Given a $ \triangle ABC $ with orthocenter $ H $ and a point $ P $ lying on the Euler line of $ \triangle ABC. $ Prove that the midpoint of $ PH $ lies on the Thomson cubic of the pedal triangle of $ P $ WRT $ \triangle ABC. $
2 replies
TelvCohl
Mar 8, 2025
AuroralMoss
2 hours ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   8
N 2 hours ago by atdaotlohbh
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
8 replies
OgnjenTesic
May 22, 2025
atdaotlohbh
2 hours ago
Balkan Mathematical Olympiad
ABCD1728   1
N 2 hours ago by ABCD1728
Can anyone provide the PDF version of the book "Balkan Mathematical Olympiads" by Mircea Becheanu and Bogdan Enescu (published by XYZ press in 2014), thanks!
1 reply
ABCD1728
May 24, 2025
ABCD1728
2 hours ago
Unexpecredly Quick-Solve Inequality
Primeniyazidayi   2
N 2 hours ago by sqing
Source: German MO 2025,Round 4,Grade 11/12 Day 2 P1
If $a, b, c>0$, prove that $$\frac{a^5}{b^2}+\frac{b}{c}+\frac{c^3}{a^2}>2a$$
2 replies
Primeniyazidayi
4 hours ago
sqing
2 hours ago
cute geo
Royal_mhyasd   0
2 hours ago
Source: own(?)
Let $\triangle ABC$ be an acute triangle and $I$ it's incenter. Let $A'$, $B'$ and $C'$ be the projections of $I$ onto $BC$, $AC$ and $AB$ respectively. $BC \cap B'C' = \{K\}$ and $Y$ is the projection of $A'$ onto $KI$. Let $M$ be the middle of the arc $BC$ not containing $A$ and $T$ the second intersection of $A'M$ and the circumcircle of $ABC$. If $N$ is the midpoint of $AI$, $TY \cap IA' = \{P\}$, $BN \cap PC' = \{D\}$ and $CN \cap PB' =\{E\}$, prove that $NEPD$ is cyclic.
PS i'm not sure if this problem is actually original so if it isn't someone please tell me so i can change the source (if that's possible)
0 replies
Royal_mhyasd
2 hours ago
0 replies
Pal triangles
v_Enhance   11
N Nov 15, 2024 by Mathandski
Source: USA January TST for IMO 2013, Problem 1
Two incongruent triangles $ABC$ and $XYZ$ are called a pair of pals if they satisfy the following conditions:
(a) the two triangles have the same area;
(b) let $M$ and $W$ be the respective midpoints of sides $BC$ and $YZ$. The two sets of lengths $\{AB, AM, AC\}$ and $\{XY, XW, XZ\}$ are identical $3$-element sets of pairwise relatively prime integers.

Determine if there are infinitely many pairs of triangles that are pals of each other.
11 replies
v_Enhance
Jul 30, 2013
Mathandski
Nov 15, 2024
Pal triangles
G H J
Source: USA January TST for IMO 2013, Problem 1
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v_Enhance
6877 posts
#1 • 4 Y
Y by bobjoe123, HamstPan38825, Adventure10, Rounak_iitr
Two incongruent triangles $ABC$ and $XYZ$ are called a pair of pals if they satisfy the following conditions:
(a) the two triangles have the same area;
(b) let $M$ and $W$ be the respective midpoints of sides $BC$ and $YZ$. The two sets of lengths $\{AB, AM, AC\}$ and $\{XY, XW, XZ\}$ are identical $3$-element sets of pairwise relatively prime integers.

Determine if there are infinitely many pairs of triangles that are pals of each other.
Z K Y
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polya78
105 posts
#2 • 5 Y
Y by langkhach11112, Aryan-23, Adventure10, Mango247, and 1 other user
I believe there are an infinite number of pals. If we complete the parallelogram, we see that the triangle with side lengths $AB, 2*AM, AC$ has twice the area of each triangle. So we are done if we can find and infinite number of relatively prime triples $(a,b,c)$ such that $a<b<c$ and triangles with side lengths $(a,2b,c),(2a,b,c)$ have the same area.

Using the area formula $\sqrt{s(s-x)(s-y)(s-z)}$ and crunching away, we wind up with the formula $2c^2=5(a^2+b^2)$. Now if $(n,m)$ satisfy $(2n+1)^2+1=2m^2$ (and there are an infinite number of such pairs), then it follows that we can let $a=2n+3,b=4n+1,c=5m$. Since $c$ is approximately $7n$, the relevant triangle inequalities are satisfied. The only case where where we don't have relatively prime numbers is when the common divisor is 5, and we can just forget about those cases or reduce by a factor of 5.
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Idio-logy
206 posts
#3
Y by
There are an infinite number of pals. Let the common three-element set be $\{a,b,c\}$. By Stewart's theorem, if we choose two of them (say $b$ and $c$) as two sides and $a$ be the median, the remaining side of the triangle has length $l=\sqrt{2(b^2+c^2-2a^2)}.$ So by the area formula
\begin{align*}
    \Delta &= \frac14\sqrt{-l^4-b^4-c^4+2l^2b^2+2b^2c^2+2c^2l^2}\\
    &= \frac14\sqrt{-4(b^2+c^2-2a^2)^2-b^4-c^4+4(b^2+c^2)(b^2+c^2-2a^2) + 2b^2c^2}\\
    &= \frac14\sqrt{-16a^4-b^4-c^4+8a^2b^2+8a^2c^2+2b^2c^2}.
\end{align*}Without loss of generality, two triangles have $\{a,b\}$ and $\{b,c\}$ as sides respectively. Then $$-16a^4-b^4-c^4+8a^2b^2+8a^2c^2+2b^2c^2 = -16c^4-b^4-a^4+8c^2b^2+8a^2c^2+2b^2a^2$$which is equivalent to $2b^2=5a^2+5c^2$.
Consider any solution to $(10u+3)^2-10v^2=-1$ (there are infinitely many by recursion), and take $(a,b,c) = (6u+1, 5v, 8u+3)$. It's not hard to verify that the resulting triangles satisfy the triangle inequality, and $a,b,c$ are pairwise coprime.
This post has been edited 1 time. Last edited by Idio-logy, Dec 27, 2020, 4:11 AM
Reason: fix
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naman12
1358 posts
#4
Y by
I'm posting this because memorizing random formulas from the AMC actually helps.

The answer is affirmitive. Note that at least one side of $\triangle ABC$ and $\triangle XYZ$ are equal - let $AB=XY=x,AC=y,XZ=z$ without loss of generality. Then, by 2, we get that $AM=z$ and $XW=y$. Thus, using the median formula, we get
\[2z=2AM=\sqrt{2(AB^2+AC^2)-BC^2}=\sqrt{2(x^2+y^2)-z^2}\implies BC=\sqrt{2x^2+2y^2-4z^2}\]By symmetry, we get that
\[YZ=\sqrt{2x^2+2z^2-4y^2}\]We have the following lemma:
Lemma. For a general triangle with side lengths $a,b,c$, we have the area of that triangle is also given by
\[\dfrac 12\sqrt{a^2c^2-\left(\dfrac{a^2+c^2-b^2}{2}\right)^2}\]Proof. We just need to show that this is equivalent to Heron's formula. We get that by the difference of squares formula
\begin{align*}
    \dfrac 12\sqrt{a^2c^2-\left(\dfrac{a^2+c^2-b^2}{2}\right)^2}&=\dfrac 14\sqrt{4a^2c^2-(a^2+c^2-b^2)^2}\\
    &=\dfrac 14\sqrt{(a^2+c^2+2ac-b^2)(a^2+c^2-2ac-b^2)}\\
    &=\dfrac 14\sqrt{((a+c)^2-b^2)((a-c)^2-b^2)}\\
    &=\dfrac 14\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}
\end{align*}which is Heron's formula. $\blacksquare$

Now, we get that
\[[ABC]=\dfrac 12\sqrt{x^2y^2-\left(\dfrac{x^2+y^2-(2x^2+2y^2-4z^2)}{2}\right)^2}=\dfrac 14\sqrt{4x^2y^2-\left(4z^2-x^2-y^2\right)^2}\]By symmetry, we get
\[[XYZ]=\dfrac 14\sqrt{4x^2z^2-\left(4y^2-x^2-z^2\right)^2}\]Equating, we get
\[4x^2(y^2-z^2)=4(x^2y^2-x^2z^2)=\left(4z^2-x^2-y^2\right)^2-\left(4y^2-x^2-z^2\right)^2=-5(3z^2+3y^2-2x^2)(y-z)(y+z)\]Assuming $y\neq z$, we get
\[-4x^2=15z^2+15y^2-10x^2\]which means that
\[2x^2=5y^2+5z^2\]Let $z=1$, and then we get
\[y^2-10\left(\dfrac x5\right)^2=1\]which has infinitely many solutions as a Pell equation. Thus, we can take any solution of
\[k^2-10\ell^2=1\]and use $(x,y,z)=(5\ell,k,1)$. We still need that $x^2+y^2\geq z^2$ and $x^2+z^2\geq y^2$. As $x$ and $y$ are positive integers, the first is obviously satisfied. For the second, we note that it is equivalent to
\[25\ell^2+1\geq k^2=1+10\ell^2\]which is definitely true. Thus, taking the pair mentioned above indeed is a solution, and as there are an infinitely many $(k,\ell)$, the problem statement is affirmitive.

Remark. I don't believe it! This information (the Lemma) was only memorized for the AMC! I'm actually surprised that it's on the TST - I'm glad for the AMC now (for the first time). Also, I pity those who don't know this formula.
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cyshine
236 posts
#5 • 1 Y
Y by PRMOisTheHardestExam
I gave this problem to my students last week, and I realized that the solutions here choose one of the elements of the set $\{AB,AC,AM\}=\{XY,XZ,XW\}$ equal to $1$, or state that since $c\approx 7n$ the triangle inequality is satisfied (which is not true because $a+b=6n+4<7n$.)
So here's a solution that works, without ever using Pell's equation. First, let me explain why setting an element equal to $1$ does not work. First, no triangle with integer sides can have one of them equal to $1$ unless it is isosceles (otherwise if $m>n$ are the other sides then $m\ge n+1$, violating the triangle inequality). Now, in most cases the median cannot be $1$: if you complete the parallelogram $ABDC$ (that is, define $D=B+C-A$), you'll see that you can construct a triangle with two sides $AB$, $AC$ and $A$-median $AM$ iff you can construct a triangle $\Delta$ with three sides $AB$, $AC$ and $2AM$ (both constructions are half of the parallelogram, and by the way, both triangles have the same area, so you can skip all the computations with medians.) If the median is $1$, then one side of $\Delta$ is $2$, and you either have an isosceles triangle with sides $2,a,a$ and area $\sqrt{a^2-1}$ or a triangle with sides $2,a,a+1$ and area $\frac1{16}\sqrt{3(4a^2+4a-3)}$, both of which are increasing with $a$.
For the sake of completion, let me rewrite the computations: let $AB=XY=x$, $AC=XW=y$ and $AM=XZ=z$ (otherwise the triangles are congruent). Then using the expanded form of Heron's formula $16S=2a^2b^2 + 2a^2c^2 + 2b^2c^2 - a^4 - b^4 - c^4$ for $(a,b,c)=(x,y,2z)$ and $(x,2y,z)$ you can find
$2x^2y^2+8x^2z^2+8y^2z^2 - x^4 - y^4 - 16z^4 = 8x^2y^2 + 2x^2z^2 + 8y^2z^2 - x^4 - 16y^4 - z^4\iff 2x^2(z^2-y^2) = 5(z^2-y^2)(z^2+y^2)
\iff 2x^2 = 5y^2+5z^2.$
Since $5\mid x$, let $x=5t$, so $y^2+z^2=10t^2$. Then $y,z<\sqrt{10}t<5t=x$, and the largest side of each of the triangles with sides $x,y,2z$ or $x,2y,z$ is either $x$ or $2z$ in the first one, and either $x$ or $2y$ in the second one. We need then $x < y+2z$, $y+x > 2z$, $x<2y+z$, and $z+x > 2y$. That's why setting $y=1$ or $z=1$ is a bad idea: if $y=1$ then $z\approx \sqrt{10}t$ and $2z \approx 2\sqrt{10}t = \sqrt{40}t > 5t + 1 = x+y$, violating the triangle inequality.
What to do then? One neat trick you can do to try and keep $y$ and $z$ somewhat close to each other is taking advantage of the fact that $10=3^2+1^2$ and set $y=u+3v$ and $z=3u-v$ (we'll take care of signs later). Then
$y^2+z^2 = 10t^2\iff (u+3v)^2 + (3u-v)^2 = 10t^2 \iff u^2+v^2=t^2,$
and we can use Pythagorean triples! We can use the general form $(2mn; m^2-n^2; m^2+n^2)$, but to keep things simple we set $n=1$ and $m=2a$: $u=4a^2-1$, $v=4a$, and $t=4a^2+1$. Then $x=5t=5(4a^2+1)=20a^2+5$, $y=u+3v=4a^2+12a-1$, and $z=3u-v=12a^2-4a-3$. Then $x+y=24a^2+12a+4 = 2(12a^2+6a+2)>2z$, $x+z=32a^2-4a+2=2(16a^2-2a+1)>2z$, $2y+z=20a^2+20a-5>x$, and $y+2z=28a^2+4a-7>x$.
It remains to check whether $\gcd(x,y,z)=1$. Let $d=\gcd(4a^2+1,4a^2+12a-1)$. Then $d$ is odd and $d\mid (4a^2+12a-1-(4a^2-1))\iff d\mid 12a+2\iff d\mid 6a+1$. Then $d\mid 2a(6a+1)-3(4a^2+1)\iff d\mid 2a-3$, and $d\mid 6a+1-3(2a-3)\iff d\mid 10\iff d\mid 5$. We can just choose $a$ such that $4a^2+12a-1\not\equiv 0\pmod 5\iff -a^2+2a-1\not\equiv0\pmod 5\iff a\not\equiv 1\pmod 5$ (which also takes care of the factor $5$ in $x$.)
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anser
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cyshine wrote:
I gave this problem to my students last week, and I realized that the solutions here choose one of the elements of the set $\{AB,AC,AM\}=\{XY,XZ,XW\}$ equal to $1$, or state that since $c\approx 7n$ the triangle inequality is satisfied (which is not true because $a+b=6n+4<7n$.)

If WLOG $AB = XY = c$, $AM = XZ = a$, and $AC = XW = b$, then we get $2c^2 = 5a^2+5b^2$. By the triangle inequality on $\triangle AMC$ and $\triangle AMB$, $a+b > \frac{1}{2}BC > |b-a|$ and $c+a > \frac{1}{2}BC > c-a$. So $a+b > \frac{1}{2}BC > c-a$ and $2a+b > c$. One can also verify that as long as $2a+b > c$, $\frac{1}{2}BC = \sqrt{\frac{1}{2}b^2 + \frac{1}{2}c^2 - a^2}$ satisfies the triangle inequalities (so it's a necessary and sufficient condition). Analogously in $\triangle XYZ$ we need $2b + a > c$. So I believe that we can't set $a$ or $b$ equal to 1, but the first solution is correct.
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Idio-logy
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Quote:
First, let me explain why setting an element equal to $1$ does not work.
Indeed, my original solution was wrong. Triangle inequality implies that one would need $a+b > 2c > b-a$ and $b+c > 2a > b-c$ for the construction to work, but my original construction does not satisfy them. However, one could still use Pell equations (along the same lines as the first solution): consider any solution to $(10u+3)^2-10v^2=-1$ (there are infinitely many by recursion), and take $(a,b,c) = (6u+1, 5v, 8u+3)$. It's not hard to verify that they satisfy the size constraint and are pairwise coprime. Thanks for pointing out the mistake.
This post has been edited 1 time. Last edited by Idio-logy, Dec 27, 2020, 4:12 AM
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CyclicISLscelesTrapezoid
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Let $BC=a$, $CA=b$, $AB=c$, $YZ=x$, and $AM=m$. After analyzing config issues, we assume WLOG that $XY=m$, $XW=c$, and $XZ=b$. There's a way to derive $2b^2=5(c^2+m^2)$ without Heron's. Notice that since the areas of triangles $ABM$ and $XYW$ are the same, $\sin \angle BAM=\sin \angle YXW$. It's easy to check that $\angle BAM \neq \angle YXW$, so $\angle BAM+\angle YXW=180^\circ$. Then, we use LOC on $\triangle ABM$ and $\triangle YXW$ to get \[\frac{a^2}{4}=c^2+m^2-2cm \cos \angle BAC\]and \[\frac{x^2}{4}=c^2+m^2-2cm \cos \angle YXW=c^2+m^2+2cm \cos \angle BAC.\]Add these equations to get $a^2+x^2=8c^2+8m^2$. Then, plug in the median length formulas for $AM$ and $XW$ to prove that.
This post has been edited 1 time. Last edited by CyclicISLscelesTrapezoid, Dec 29, 2021, 5:19 AM
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IAmTheHazard
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The answer is yes, there do exist infinitely many pairs.

Suppose a triangle $PQR$ has sides $x,y$, and corresponding median with length $z$. Then the third side has length $\sqrt{2(x^2+y^2-2z^2)}$ by the midpoint formula. Then, expanding Heron's formula out and substituting, we find that
$$[PQR]=\frac{1}{4}\sqrt{-16z^4-x^4-y^4+8z^2x^2+8z^2y^2+2x^2y^2}.$$Now, WLOG let $AB=XY=a$, so $AC=XW=b$ and $AM=XZ=c$. The condition that $[ABC]=[XYZ]$ becomes
$$-16b^4-a^4-c^4+8b^2a^2+8b^2c^2+2a^2c^2=-16c^4-a^4-b^4+8c^2a^2+8b^2a^2+2a^2b^2,$$which simplifies to $2a^2=5b^2+5c^2$. Consider the equation $x^2-2y^2=-1$, over the nonnegative integers, which is a Pell equation that has infinitely many solutions, with minimal solution $(x,y)=(1,1)$. Evidently, any solution $(x,y)$ has $x$ odd, so we may take $(a,b,c)=(5y,2x-1,x+2)$ which we can verify satisfies the equation.
We now have to check that the triangle inequality holds. We can find that $BC=\sqrt{2((5y)^2+(2x-1)^2-2(x+2)^2)}$ and likewise $YZ=\sqrt{2((5y)^2+(x-1)^2-2(2x-1)^2)}$, and after some brute force computation we can show that we indeed get triangles with these values. Finally, we have to prove that infinitely many such $(a,b,c)$ are pairwise relatively prime. Throw out all $(x,y)$ such that $x \equiv 3 \pmod{5} \implies 5 \mid y$, which leaves an infinite number of $(x,y)$ still. Then $\gcd(2x-1,x+2)=\gcd(5,x+2)=1$. Further, we have $\gcd(x+2,5y)=\gcd(x+2,y)$, and if some prime $p$ divides both $x+2$ and $y$, then $x^2-2y^2 \equiv 4$, so $p \mid 5 \implies p=5$, contradiction. Likewise, $\gcd(2x-1,5y)=\gcd(2x-1,y)$, and if some prime $p$ divides them both then we have
$$4x^2-8y^2=-4 \implies 1 \equiv -4 \pmod{p} \implies p=5,$$which yields the same contradiction. Hence there are an infinite number of triples $(a,b,c)$ such that $\gcd(a,b)=\gcd(b,c)=\gcd(c,a)=1$, so we're done. $\blacksquare$

Remark: awful problem.
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HamstPan38825
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"You can't put computational problems on an olympiad..." —someone, probably

The answer is yes. Without loss of generality we may set $AB=XY=a$, $AM=XZ=b$, $AC=XN=c$. Now, the median formula gives the following two equalities:
\begin{align*}
2c^2+2a^2-BC^2 &= 4b^2 \\
2a^2+2b^2-YZ^2 &= 4c^2.
\end{align*}Now, equating both sides of the area of the two triangles yields $$a^2c^2 - \left(\frac{4b^2-a^2-c^2}2\right) = a^2b^2-\left(\frac{4c^2-a^2-b^2}2\right).$$This eventually simplifies to $$a^2 = \frac 52(b^2+c^2).$$Now, we may fix $c$ to get some solution to the Pell equation $$x^2-10y^2=1.$$It can be verified that these solutions also satisfy the triangle inequality.
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Martin2001
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Let $c<b<a,$ and WLOG let $AB=XY=a, AM=XZ=b,AC=XN=c.$ Note that $BC^2=2a^2+2c^2-4b^2,YZ^2=2a^2+2b^2-4c^2$ by Stewart's, so after a long computation with Heron we get that $2a^2=5(b^2+c^2).$ Now, let $(a,b,c)=(5m, 4n+1, 2n+3),$ where $(2n+1)^2+1=2m^2,$ which we have infinite solutions by Pell. Now, note that if we flip the triangle through $M,$ and let $A'=D,$ we have that triangle ABD has sidelengths either $a,2b,c$ or $a,b,2c,$ which both work by Triangle inequality as $a \approx 5\sqrt{2}n \approx=7n,$ and finally the only way for $a,b,c$ to not be coprime is when there is a common factor of $5,$ which we can either discard or divide out from all$.\blacksquare$
This post has been edited 1 time. Last edited by Martin2001, Jul 22, 2024, 5:48 PM
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Mathandski
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