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Iberoamerican Mathematical Olympiad

Geometry Difficulty 5.5 AIME, harder Prove it Ibero-American Mathematical Olympiad

Problem:

In the triangle $ABC$ , the midpoints of $AC$ and $AB$ are $M$ and $N$ respectively. $BM$ and $CN$ meet at $P$ . Show that if it is possible to inscribe a circle in the quadrilateral $AMPN$ (touching every side), then $ABC$ is isosceles.

Solution

Solution:

If the quadrilateral has an inscribed circle then $AM + PN = AN + PM$ (consider the tangents to the circle from $A, M, P, N$ ). But if $AB > AC$ , then $BM > CN$ (see below). We have $AN = AB / 2$ , $PM = BM / 3$ , $AM = AC / 2$ , $PN = CN / 3$ , so it follows that $AM + PN < AN + PM$ . Similarly, $AB < AC$ implies $AM + PN > AN + PM$ , so the triangle must be isosceles.

To prove the result about the medians, note that $BM^2 = BC^2 + CM^2 - 2 BC \cdot CM \cos C = (BC - CM \cos C)^2 + (CM \sin C)^2$ . Similarly, $CN^2 = (BC - BN \cos B)^2 + (BN \sin B)^2$ . But $MN$ is parallel to $BC$ , so $CM \sin C = BN \sin B$ . But $AB > AC$ , so $BN > CM$ and $B < C$ , so $\cos B > \cos C$ , hence $BN \cos B > CM \cos C$ and $BC - CM \cos C > BC - BN \cos B$ . So $BM > CN$ .

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