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Difference between revisions of "2014 USAJMO Problems/Problem 6"
(→‎Solution)
Line 9: Line 9:
 
'''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.'''
  
−
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 = <A/2 = <IAE and y = <C/2 = <ICA. We know that <math>MP // AC</math> because <math>MP</math> is a midsegment of triangle <math>ABC</math>; thus, by alternate interior angles (A.I.A) <MVE = <FEA = (180° - 2x) / 2 = 90° - x, because triangle <math>AFE</math> is isosceles. Also by A.I.A, <MQC = <QCA = y. 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 <QIA = x + y from the Exterior Angle Theorem gives <FRC = 90 - x - y. <VRQ = <FRC = 90 - x - y because they are vertical angles; however, .... This completes part (a).
+
<center>
 +
<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></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 = <A/2 = <IAE and y = <C/2 = <ICA. 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, <MQC = <QCA = y. 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 <QIA = x + y 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 = <VCA = <MCV, triangle <math>VMC</math> is isosceles. Similarly, triangle <math>BMU</math> is isosceles, from which we derive that VM = MC = MB = MU. Hence, triangle <math>VUM</math> is isosceles.
 
Now, we attempt part (b). Using a similar argument to part (a), point U lies on line <math>BI</math>. Because <MVC = <VCA = <MCV, triangle <math>VMC</math> is isosceles. Similarly, triangle <math>BMU</math> is isosceles, from which we derive that VM = MC = MB = MU. 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 X lies on ray <math>ID</math>.
+
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>.
  
−
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>MD // IA</math>. Furthermore, we have <VMD = <UMD = x 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 22:35, 30 November 2014

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.

[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]

We will first prove part (a) 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 = <A/2 = <IAE and y = <C/2 = <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, <MQC = <QCA = y. Furthermore, because $AI$ is an angle bisector of triangle $AFE$, it is also an altitude of the triangle; combining this with <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).

Now, we attempt part (b). Using a similar argument to part (a), point U lies on line $BI$. Because <MVC = <VCA = <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.

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