2014 USAJMO Problems/Problem 6
Problem
Let be a triangle with incenter
, incircle
and circumcircle
. Let
be the midpoints of sides
,
,
and let
be the tangency points of
with
and
, respectively. Let
be the intersections of line
with line
and line
, respectively, and let
be the midpoint of arc
of
.
(a) Prove that lies on ray
.
(b) Prove that line bisects
.
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]](http://latex.artofproblemsolving.com/9/a/f/9afeeedd983c1aaacf03d1d33b69c006bf49395f.png)
We will first prove part (a) via contradiction: assume that line intersects line
at Q and line
and R, with R and Q not equal to V. Let x = <A/2 = <IAE and y = <C/2 = <ICA. We know that
because
is a midsegment of triangle
; thus, by alternate interior angles (A.I.A)
, because triangle
is isosceles. Also by A.I.A, <MQC = <QCA = y. Furthermore, because
is an angle bisector of triangle
, it is also an altitude of the triangle; combining this with <QIA = x + y from the Exterior Angle Theorem gives
. Also,
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 . Because <MVC = <VCA = <MCV, triangle
is isosceles. Similarly, triangle
is isosceles, from which we derive that VM = MC = MB = MU. Hence, triangle
is isosceles.
Note that X lies on both the circumcircle and the perpendicular bisector of segment . Let D be the midpoint of
; our goal is to prove that points X, D, and I are collinear, which equates to proving
lies on ray
.
Because is also an altitude of triangle
, and
and
are both perpendicular to
,
. Furthermore, we have
because
is a parallelogram.