Difference between revisions of "2014 USAJMO Problems/Problem 6"
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(a) | (a) | ||
− | + | '''Solution 1:''' We will prove this 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). | |
− | + | '''Solution 2:''' First we show that the intersection <math>V'</math>of <math>MP</math> with the internal angle bisector of <math>C</math> is the same as the intersection <math>V''</math> of <math>EF</math> with the internal angle bisector of <math>C.</math> Let <math>D</math> denote the intersection of <math>AB</math> with the internal angle bisector of <math>C,</math> and let <math>a,b,c</math> denote the side lengths of <math>BC, AC, AB.</math> By Menelaus on <math>V'', F, E,</math> with respect to <math>\triangle ADC,</math> <cmath>\frac{AF}{FD}\cdot \frac{DV''}{V''C}\cdot \frac{CE}{EA}=-1</cmath> | |
<cmath>\frac{DV''}{V''C}=-\frac{\frac{bc}{a+b}+\frac{a-b-c}{2}}{\frac{a+b-c}{2}}=\frac{b-a}{b+a}.</cmath> Similarly, <cmath>\frac{BP}{PD}\cdot \frac{DV'}{V'C}\cdot \frac{CM}{MB}=-1</cmath> <cmath>\frac{DV'}{V'C}=-\frac{\frac{ac}{a+b}-\frac{c}{2}}{\frac{c}{2}}=\frac{b-a}{b+a}.</cmath> Since <math>V'</math> and <math>V''</math> divide <math>CD</math> in the same ratio, they must be the same point. Now, since <math>\frac{b-a}{b+a}>-1,</math> <math>I</math> lies on ray <math>CV.</math> <math>\blacksquare</math> | <cmath>\frac{DV''}{V''C}=-\frac{\frac{bc}{a+b}+\frac{a-b-c}{2}}{\frac{a+b-c}{2}}=\frac{b-a}{b+a}.</cmath> Similarly, <cmath>\frac{BP}{PD}\cdot \frac{DV'}{V'C}\cdot \frac{CM}{MB}=-1</cmath> <cmath>\frac{DV'}{V'C}=-\frac{\frac{ac}{a+b}-\frac{c}{2}}{\frac{c}{2}}=\frac{b-a}{b+a}.</cmath> Since <math>V'</math> and <math>V''</math> divide <math>CD</math> in the same ratio, they must be the same point. Now, since <math>\frac{b-a}{b+a}>-1,</math> <math>I</math> lies on ray <math>CV.</math> <math>\blacksquare</math> | ||
(b) | (b) | ||
− | + | '''Solution 1:''' 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 <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>. | 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. (incomplete) |
− | + | '''Solution 2:''' Let <math>I_A</math>, <math>I_B</math>, and <math>I_C</math> be the excenters of <math>ABC</math>. Note that the circumcircle of <math>ABC</math> is the nine-point circle of <math>I_AI_BI_C</math>. Since <math>AX</math> is the external angle bisector of <math>\angle BAC</math>, <math>X</math> is the midpoint of <math>I_BI_C</math>. Now <math>UV</math> and <math>I_BI_C</math> are parallel since both are perpendicular to the internal angle bisector of <math>\angle BAC</math>. Since <math>IX</math> bisects <math>I_BI_C</math>, it bisects <math>UV</math> as well. |
Revision as of 18:14, 4 January 2016
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)
(a)
Solution 1: We will prove this 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).
Solution 2: First we show that the intersection of
with the internal angle bisector of
is the same as the intersection
of
with the internal angle bisector of
Let
denote the intersection of
with the internal angle bisector of
and let
denote the side lengths of
By Menelaus on
with respect to
Similarly,
Since
and
divide
in the same ratio, they must be the same point. Now, since
lies on ray
(b)
Solution 1: 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. (incomplete)
Solution 2: Let ,
, and
be the excenters of
. Note that the circumcircle of
is the nine-point circle of
. Since
is the external angle bisector of
,
is the midpoint of
. Now
and
are parallel since both are perpendicular to the internal angle bisector of
. Since
bisects
, it bisects
as well.