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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.

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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
Problem 3
EthanWYX2009   5
N 16 minutes ago by parkjungmin
Source: 2023 China Second Round P3
Find the smallest positive integer ${k}$ with the following properties $:{}{}{}{}{}$If each positive integer is arbitrarily colored red or blue${}{}{},$
there may be ${}{}{}{}9$ distinct red positive integers $x_1,x_2,\cdots ,x_9,$ satisfying
$$x_1+x_2+\cdots +x_8<x_9,$$or there are $10{}{}{}{}{}{}$ distinct blue positive integers $y_1,y_2,\cdots ,y_{10}$ satisfiying
$${y_1+y_2+\cdots +y_9<y_{10}}.$$
5 replies
EthanWYX2009
Sep 10, 2023
parkjungmin
16 minutes ago
Inspired by old results
sqing   1
N 32 minutes ago by sqing
Source: Own
Let $a,b,c $ be reals such that $a^2+b^2+c^2=3$ .Prove that
$$(1-a)(k-b)(1-c)+abc\ge -k$$Where $ k\geq 1.$
$$(1-a)(1-b)(1-c)+abc\ge -1$$$$(1-a)(1-b)(1-c)-abc\ge -\frac{1}{2}-\sqrt 2$$
1 reply
+1 w
sqing
an hour ago
sqing
32 minutes ago
Made from a well-known result
m4thbl3nd3r   0
an hour ago
1. Let $a,b,c>0$ such that $$\sqrt{(a+b)(a+c)}+\sqrt{(b+a)(b+c)}+\sqrt{(c+a)(c+b)}=3+a+b+c.$$Prove that $$\sqrt{\frac{a+b}{2}}+\sqrt{\frac{b+c}{2}}+\sqrt{\frac{c+a}{2}}\ge ab+bc+ca.$$2. Let $x,y,z$ be sidelengths of a triangle such that $$x^2+y^2+z^2+6=2(xy+yz+zx).$$Prove that $$2\sqrt{2x}+2\sqrt{2y}+2\sqrt{2z}+(x-y)^2+(y-z)^2+(z-x)^2\ge x^2+y^2+z^2.$$
0 replies
+1 w
m4thbl3nd3r
an hour ago
0 replies
Interesting inequalities
sqing   5
N an hour ago by sqing
Source: Own
Let $ a,b,c,d\geq  0 , a+b+c+d \leq 4.$ Prove that
$$a(kbc+bd+cd)  \leq \frac{64k}{27}$$$$a (b+c) (kb c+  b d+  c d) \leq \frac{27k}{4}$$Where $ k\geq 2. $
5 replies
sqing
Yesterday at 12:44 PM
sqing
an hour ago
Long and wacky inequality
Royal_mhyasd   5
N 2 hours ago by Royal_mhyasd
Source: Me
Let $x, y, z$ be positive real numbers such that $x^2 + y^2 + z^2 = 12$. Find the minimum value of the following sum :
$$\sum_{cyc}\frac{(x^3+2y)^3}{3x^2yz - 16z - 8yz + 6x^2z}$$knowing that the denominators are positive real numbers.
5 replies
1 viewing
Royal_mhyasd
May 12, 2025
Royal_mhyasd
2 hours ago
Polynomials algebra
Foxellar   2
N 2 hours ago by elizhang101412
\textbf{9.} The real root of the polynomial \( p(x) = 8x^3 - 3x^2 - 3x - 1 \) can be written in the form
\[
\frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c},
\]where \( a, b, \) and \( c \) are positive integers. Find the value of \( a + b + c \).
2 replies
Foxellar
4 hours ago
elizhang101412
2 hours ago
Collinear points
tenplusten   2
N 3 hours ago by Blackbeam999
Let $A,B,C$ be three collinear points and $D,E,F$ three other collinear points. Let $G,H,I$ be the intersection of the lines $BE,CF$ $AD,CF$ and $AD,CE$,respectively. If $AI=HD$ and $CH=GF$.Prove that $BI=GE$



I hope you will use Pappus theorem in your solutions.
2 replies
tenplusten
Jun 20, 2016
Blackbeam999
3 hours ago
Some number theory
EeEeRUT   4
N 3 hours ago by juckter
Source: Thailand MO 2025 P9
Let $p$ be an odd prime and $S = \{1,2,3,\dots, p\}$
Assume that $U: S \rightarrow S$ is a bijection and $B$ is an integer such that $$B\cdot U(U(a)) - a \: \text{ is a multiple of} \: p \: \text{for all} \: a \in S$$Show that $B^{\frac{p-1}{2}} -1$ is a multiple of $p$.
4 replies
EeEeRUT
May 14, 2025
juckter
3 hours ago
Simple Geometry
AbdulWaheed   0
3 hours ago
Source: EGMO
Try to avoid Directed angles
Let ABC be an acute triangle inscribed in circle $\Omega$. Let $X$ be the midpoint of the arc $\overarc{BC}$ not containing $A$ and define $Y, Z$ similarly. Show that the orthocenter of $XYZ$ is the incenter $I$ of $ABC$.
0 replies
AbdulWaheed
3 hours ago
0 replies
Graph Theory
ABCD1728   0
3 hours ago
Can anyone provide the PDF version of "Graphs: an introduction" by Radu Bumbacea (XYZ press), thanks!
0 replies
ABCD1728
3 hours ago
0 replies
Number theory for people who love theory
Assassino9931   3
N 4 hours ago by NamelyOrange
Source: Bulgaria RMM TST 2019
Prove that there is no positive integer $n$ such that $2^n + 1$ divides $5^n-1$.
3 replies
Assassino9931
Jul 31, 2024
NamelyOrange
4 hours ago
Interesting inequalities
sqing   0
4 hours ago
Source: Own
Let $ a,b> 0 ,   a+b+a^2+b^2=2.$ Prove that
$$ab+ \frac{k}{a+b+ab} \geq \frac{3-k+(k-1)\sqrt{5}}{2}$$Where $ k\geq 2. $
$$ab+ \frac{2}{a+b+ab} \geq \frac{1+\sqrt{5}}{2}$$$$ab+ \frac{3}{a+b+ab} \geq  \sqrt{5} $$
0 replies
sqing
4 hours ago
0 replies
Stability of Additive Cauchy Equation
doanquangdang   1
N 5 hours ago by jasperE3
Show that if $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies
$$
|f(x+y)-f(x)-f(y)-x y| \leq \varepsilon\left(|x|^p+|y|^p\right)
$$for some $\varepsilon>0,$ $p \in[0,1)$ and for all $x, y \in \mathbb{R}$, then there exists a unique solution $a: \mathbb{R} \rightarrow \mathbb{R}$ of the functional equation $a(x+y)=$ $a(x)+a(y)$ for all $x, y \in \mathbb{R}$ such that
$$
\left|f(x)-a(x)-\frac{1}{2} x^2\right| \leq \frac{2}{2-2^p} \varepsilon|x|^p
$$for all $x \in \mathbb{R}$.
1 reply
doanquangdang
Aug 16, 2024
jasperE3
5 hours ago
Polynomials with common roots and coefficients
VicKmath7   10
N 6 hours ago by math-olympiad-clown
Source: Balkan MO SL 2020 A3
Let $P(x), Q(x)$ be distinct polynomials of degree $2020$ with non-zero coefficients. Suppose that they have $r$ common real roots counting multiplicity and $s$ common coefficients. Determine the maximum possible value of $r + s$.

Demetres Christofides, Cyprus
10 replies
VicKmath7
Sep 9, 2021
math-olympiad-clown
6 hours ago
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.
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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.
Z K Y
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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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