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

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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.'''
 
'''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).
+
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).
  
 
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.
  
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Note that X lies on both the circumcircle and the perpendicular bisector of segment <math>BC</math>. ....
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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>.
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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.

Revision as of 12:16, 18 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 (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 $EF$, $MD // IA$. Furthermore, we have <VMD = <UMD = x because $APMN$ is a parallelogram.

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