1987 OIM Problems/Problem 2

Problem

On a triangle $ABC$ , $M$ and $N$ are the respective midpoints of sides $AC$ and $AB$ , and $P$ is the centroid of $\triangle ABC$ . Prove that, if is possible to inscribe a circle in the quadrilateral $ANPM$ , then triangle $ABC$ is isosceles.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

By the Pitot Theorem, it is possible to inscribe such a circle if and only if $AN+MP=PN+MA$ . As a result, we can substitute. We know that $AN=\frac{1}{2}c$ and $MA=\frac{1}{2}b$ . Furthermore, the formula for the length of the median from $A$ of any triangle $\triangle ABC$ is $\frac{1}{2}\sqrt{2b^2+2c^2-a^2}$ Using this and the $2:1$ ratio property of the median, we must have $PN=\frac{1}{6}\sqrt{2a^2+2b^2-c^2}$ $MP=\frac{1}{6}\sqrt{2a^2+2c^2-b^2}$ Combining the above, we find that $\frac{1}{2}c+\frac{1}{6}\sqrt{2a^2+2c^2-b^2}=\frac{1}{6}\sqrt{2a^2+2b^2-c^2}+\frac{1}{2}b$ which is equivalent to $\sqrt{2a^2+2c^2-b^2}-\sqrt{2a^2+2b^2-c^2}=3(b-c)$ Let $x=\sqrt{2a^2+2c^2-b^2}$ and $y=\sqrt{2a^2+2b^2-c^2}$ for simplicity. Then $x-y=3(b-c)$ . If $b>c$ , then analyzing the square roots yields $x<y$ , but the right-hand side is positive, a contradiction. Similarly, if $c>b$ , then $x>y$ , but the right-hand side is negative, also a contradiction. Thus we must have $b=c$ , and testing this case shows that it does indeed work, finishing the proof.

~ eevee9406

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1987 OIM Problems/Problem 2

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