Difference between revisions of "2014 USAJMO Problems/Problem 6"
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==Solution== | ==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.''' | + | '''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.''' [Diagram and additional <math>\LaTeX</math> by pinetree1] |
<center> | <center> | ||
Line 65: | Line 65: | ||
draw(X--I, dashed); | draw(X--I, dashed); | ||
draw(C--V, dashed); | draw(C--V, dashed); | ||
− | + | draw(A--I); | |
+ | label("$x^\circ$", A + (0.2,0), dir(90)); | ||
+ | label("$y^\circ$", C + (-0.4,0), dir(90)); | ||
</asy></center> | </asy></center> | ||
− | We will first prove part (a) via contradiction: assume that line <math>IC</math> intersects line <math>MP</math> at Q and line <math>EF</math> and R, with R and Q not equal to V. Let x = | + | We will first prove part (a) via contradiction: assume that line <math>IC</math> intersects line <math>MP</math> at <math>Q</math> and line <math>EF</math> and <math>R</math>, with <math>R</math> and <math>Q</math> not equal to <math>V</math>. Let <math>x = \angle A/2 = \angle IAE</math> and <math>y = \angle C/2 = \angle ICA</math>. We know that <math>\overline{MP} \parallel \overline{AC}</math> because <math>MP</math> is a midsegment of triangle <math>ABC</math>; thus, by alternate interior angles (A.I.A) <math>\angle MVE = \angle FEA = (180^\circ - 2x) / 2 = 90^\circ - x</math>, because triangle <math>AFE</math> is isosceles. Also by A.I.A, <math>\angle MQC = \angle QCA = y</math>. Furthermore, because <math>AI</math> is an angle bisector of triangle <math>AFE</math>, it is also an altitude of the triangle; combining this with <math>\angle QIA = x + y</math> from the Exterior Angle Theorem gives <math>\angle FRC = 90 - x - y</math>. Also, <math>\angle VRQ = \angle FRC = 90 - x - y</math> 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 <math>BI</math>. Because <MVC = | + | Now, we attempt part (b). Using a similar argument to part (a), point U lies on line <math>BI</math>. Because <math>\angle MVC = \angle VCA = \angle MCV</math>, triangle <math>VMC</math> is isosceles. Similarly, triangle <math>BMU</math> is isosceles, from which we derive that <math>VM = MC = MB = MU</math>. Hence, triangle <math>VUM</math> is isosceles. |
− | Note that X lies on both the circumcircle and the perpendicular bisector of segment <math>BC</math>. Let D be the midpoint of <math>UV</math>; our goal is to prove that points X, D, and I are collinear, which equates to proving <math>X</math> lies on ray <math>ID</math>. | + | Note that <math>X</math> lies on both the circumcircle and the perpendicular bisector of segment <math>BC</math>. Let <math>D</math> be the midpoint of <math>UV</math>; our goal is to prove that points <math>X</math>, <math>D</math>, and <math>I</math> are collinear, which equates to proving <math>X</math> lies on ray <math>ID</math>. |
Because <math>MD</math> is also an altitude of triangle <math>MVU</math>, and <math>MD</math> and <math>IA</math> are both perpendicular to <math>EF</math>, <math>\overline{MD} \parallel \overline{IA}</math>. Furthermore, we have <math>\angle VMD = \angle UMD = x</math> because <math>APMN</math> is a parallelogram. | Because <math>MD</math> is also an altitude of triangle <math>MVU</math>, and <math>MD</math> and <math>IA</math> are both perpendicular to <math>EF</math>, <math>\overline{MD} \parallel \overline{IA}</math>. Furthermore, we have <math>\angle VMD = \angle UMD = x</math> because <math>APMN</math> is a parallelogram. |
Revision as of 22:09, 1 December 2014
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. [Diagram and additional by pinetree1]
![[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); draw(A--I); label("$x^\circ$", A + (0.2,0), dir(90)); label("$y^\circ$", C + (-0.4,0), dir(90)); [/asy]](http://latex.artofproblemsolving.com/3/b/5/3b518f7233f90f68bacb15dc1dcf2b5f1c3e8f21.png)
We will first prove part (a) via contradiction: assume that line intersects line
at
and line
and
, with
and
not equal to
. Let
and
. 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,
. Furthermore, because
is an angle bisector of triangle
, it is also an altitude of the triangle; combining this with
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
, triangle
is isosceles. Similarly, triangle
is isosceles, from which we derive that
. Hence, triangle
is isosceles.
Note that lies on both the circumcircle and the perpendicular bisector of segment
. Let
be the midpoint of
; our goal is to prove that points
,
, and
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.