Difference between revisions of "2014 USAJMO Problems/Problem 6"

Revision as of 17:13, 4 January 2016

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. [Diagram and additional $\LaTeX$ 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$", C + (-0.4,0), dir(90)); [/asy]$

(a)

[b]Solution 1:[/b] We will prove this 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 $x = \angle A/2 = \angle IAE$ and $y = \angle C/2 = \angle ICA$ . 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, $\angle MQC = \angle QCA = y$ . Furthermore, because $AI$ is an angle bisector of triangle $AFE$ , it is also an altitude of the triangle; combining this with $\angle QIA = x + y$ 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).

[b]Solution 2:[/b] First we show that the intersection $V'$ of $MP$ with the internal angle bisector of $C$ is the same as the intersection $V''$ of $EF$ with the internal angle bisector of $C.$ Let $D$ denote the intersection of $AB$ with the internal angle bisector of $C,$ and let $a,b,c$ denote the side lengths of $BC, AC, AB.$ By Menelaus on $V'', F, E,$ with respect to $\triangle ADC,$ $\frac{AF}{FD}\cdot \frac{DV''}{V''C}\cdot \frac{CE}{EA}=-1$ $\frac{DV''}{V''C}=-\frac{\frac{bc}{a+b}+\frac{a-b-c}{2}}{\frac{a+b-c}{2}}=\frac{b-a}{b+a}.$ Similarly, $\frac{BP}{PD}\cdot \frac{DV'}{V'C}\cdot \frac{CM}{MB}=-1$ $\frac{DV'}{V'C}=-\frac{\frac{ac}{a+b}-\frac{c}{2}}{\frac{c}{2}}=\frac{b-a}{b+a}.$ Since $V'$ and $V''$ divide $CD$ in the same ratio, they must be the same point. Now, since $\frac{b-a}{b+a}>-1,$ $I$ lies on ray $CV.$ $\blacksquare$

(b)

[b]Solution 1:[/b] Using a similar argument to part (a), point U lies on line $BI$ . Because $\angle MVC = \angle VCA = \angle MCV$ , 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.

[b]Solution 2:[/b] Let $I_A$ , $I_B$ , and $I_C$ be the excenters of $ABC$ . Note that the circumcircle of $ABC$ is the nine-point circle of $I_AI_BI_C$ . Since $AX$ is the external angle bisector of $\angle BAC$ , $X$ is the midpoint of $I_BI_C$ . Now $UV$ and $I_BI_C$ are parallel since both are perpendicular to the internal angle bisector of $\angle BAC$ . Since $IX$ bisects $I_BI_C$ , it bisects $UV$ as well.

@@ Line 71: / Line 71: @@
 </asy></center>
-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).
+(a)
-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.
+[b]Solution 1:[/b] 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).
+[b]Solution 2:[/b] 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>
+(b)
+[b]Solution 1:[/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 <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.
+[b]Solution 2:[/b] 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.

Page

Toolbox

Search

Difference between revisions of "2014 USAJMO Problems/Problem 6"

Revision as of 17:13, 4 January 2016

Problem

Solution