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.

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.

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.

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 $x$.)

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.

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.

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.

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.

"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.

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$.\blacksquare$

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