2014 USAJMO Problems/Problem 6

Problem

Let $ABC$ be a triangle with incenter $I$ , incircle $\gamma$ and circumcircle $\Gamma$ . Let $M,N,P$ be the midpoints of sides $\overline{BC}$ , $\overline{CA}$ , $\overline{AB}$ and let $E,F$ be the tangency points of $\gamma$ with $\overline{CA}$ and $\overline{AB}$ , respectively. Let $U,V$ be the intersections of line $EF$ with line $MN$ and line $MP$ , respectively, and let $X$ be the midpoint of arc $BAC$ of $\Gamma$ .

(a) Prove that $I$ lies on ray $CV$ .

(b) Prove that line $XI$ bisects $\overline{UV}$ .

Solution

Extra karma will be awarded to the benefactor who so kindly provides a diagram and additional LATEX for this solution. Before then, please peruse your own diagram.

We will first prove part (a) via contradiction: assume that line $IC$ intersects line $MP$ at Q and line $EF$ and R, with R and Q not equal to V. Let x = <A/2 = <IAE and y = <C/2 = <ICA. We know that $MP // AC$ because $MP$ is a midsegment of triangle $ABC$ ; thus, by alternate interior angles (A.I.A) <MVE = <FEA = (180° - 2x) / 2 = 90° - x, because triangle $AFE$ is isosceles. Also by A.I.A, <MQC = <QCA = y. Furthermore, because $AI$ is an angle bisector of triangle $AFE$ , it is also an altitude of the triangle; combining this with <QIA = x + y from the Exterior Angle Theorem gives <FRC = 90° - x - y. <VRQ = <FRC = 90° - x - y because they are vertical angles; however, .... This completes part (b).

Now, we attempt part (b). Using a similar argument to part (a), point U lies on line $BI$ . Because <MVC = <VCA = <MCV, triangle $VMC$ is isosceles. Similarly, triangle $BMU$ is isosceles, from which we derive that VM = MC = MB = MU. Hence, triangle $VUM$ is isosceles.

Note that X lies on both the circumcircle and the perpendicular bisector of segment $BC$ . ....

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2014 USAJMO Problems/Problem 6

Problem

Solution