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.
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) <MVE = <FEA = (180° - 2x) / 2 = 90° - x, 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 <FRC = 90° - x - y. <VRQ =
= 90° - x - y because they are vertical angles; however, .... 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 X lies on ray
.
Because is also an altitude of triangle
, and
and
are both perpendicular to
,
. Furthermore, we have <VMD = <UMD = x because
is a parallelogram.