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

(Created page with "==Problem== Let <math>ABC</math> be a triangle with incenter <math>I</math>, incircle <math>\gamma</math> and circumcircle <math>\Gamma</math>. Let <math>M,N,P</math> be the mid...")
 
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==Solution==
 
==Solution==
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'''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.'''
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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 (b).
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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.
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Note that X lies on both the circumcircle and the perpendicular bisector of segment <math>BC</math>. ....

Revision as of 10:47, 17 May 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.

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 $MP // AC$ because $MP$ is a midsegment of triangle $ABC$; thus, by alternate interior angles (A.I.A) <MVE = <FEA = (180° - 2x) / 2 = 90° - 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 <FRC = 90° - x - y. <VRQ = <FRC = 90° - x - y because they are vertical angles; however, .... This completes part (b).

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 $BC$. ....

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