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2014 USAJMO Problems/Problem 6

Revision as of 22:35, 30 November 2014 by Pinetree1 (talk | contribs) (→‎Solution)

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.

[asy] unitsize(5cm); import olympiad; pair A, B, C, I, M, N, P, E, F, U, V, X, R; A = dir(190); B = dir(120); C = dir(350); I = incenter(A, B, C); label("$A$", A, W); label("$B$", B, dir(90)); label("$C$", C, dir(0)); dot(I); label("$I$", I, SSE); draw(A--B--C--cycle); real r, R; r = inradius(A, B, C); R = circumradius(A, B, C); path G, g; G = circumcircle(A, B, C); g = incircle(A, B, C); draw(G); draw(g); label("$\Gamma$", dir(35), dir(35)); label("$\gamma$", 2/3 * dir(125)); M = (B+C)/2; N = (A+C)/2; P = (A+B)/2; label("$M$", M, NE); label("$N$", N, SE); label("$P$", P, W); E = tangent(A, I, r, 1); F = tangent(A, I, r, 2); label("$E$", E, SW); label("$F$", F, WNW); U = extension(E, F, M, N); V = intersectionpoint(P--M, F--E); label("$U$", U, S); label("$V$", V, NE); draw(P--M--U--F); X = dir(235); label("$X$", X, dir(235)); draw(X--I, dashed); draw(C--V, dashed); [/asy]

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 $\overline{MP} \parallel \overline{AC}$ because $MP$ is a midsegment of triangle $ABC$; thus, by alternate interior angles (A.I.A) $\angle MVE = \angle FEA = (180^\circ - 2x) / 2 = 90^\circ - x$, 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 $\angle FRC = 90 - x - y$. Also, $\angle VRQ = \angle FRC = 90 - x - y$ because they are vertical angles. This completes part (a).

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 $BC$. Let D be the midpoint of $UV$; our goal is to prove that points X, D, and I are collinear, which equates to proving $X$ lies on ray $ID$.

Because $MD$ is also an altitude of triangle $MVU$, and $MD$ and $IA$ are both perpendicular to $EF$, $\overline{MD} \parallel \overline{IA}$. Furthermore, we have $\angle VMD = \angle UMD = x$ because $APMN$ is a parallelogram.

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