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Difference between revisions of "2016 USAMO Problems/Problem 5"

(Created page with "==Problem== An equilateral pentagon <math>AMNPQ</math> is inscribed in triangle <math>ABC</math> such that <math>M\in\overline{AB},</math> <math>Q\in\overline{AC},</math> and ...")
 
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Prove that <math>\overline{OI}</math> is parallel to <math>\ell,</math> where <math>O</math> is the circumcenter of triangle <math>ABC,</math> and <math>I</math> is the incenter of triangle <math>ABC.</math>
 
Prove that <math>\overline{OI}</math> is parallel to <math>\ell,</math> where <math>O</math> is the circumcenter of triangle <math>ABC,</math> and <math>I</math> is the incenter of triangle <math>ABC.</math>
 
==Solution==
 
==Solution==
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{{solution}}
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 +
Write <math>\angle{JKL} = K</math> for all <math>J,K,L</math> chosen as distinct vertices of triangle <math>ABC</math>. Define <math>a, b, c</math> as sides opposite to angles <math>A, B</math>, and <math>C</math>, respectively. Place the triangle in the Euclidean plane with <math>A</math> at the origin and <math>C</math> on the positive x-axis. Assume without loss of generality that C is acute.
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Consider the sides of the pentagon as vectors and note that <cmath>\overrightarrow{AM} + \overrightarrow{MN} + \overrightarrow{NP} = \overrightarrow{AQ} + \overrightarrow{QP} \qquad (1)</cmath>
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Define <math>\delta</math> and <math>\gamma</math> as the angles made between the positive x-axis and <math>\overrightarrow{MN}</math> and <math>\overrightarrow{QP}</math>, respectively. Considering the x and y coordinates of the vectors in <math>(1)</math>, it follows that
 +
<cmath>\cos \delta - \cos \gamma = 1 - \cos A - \cos C    \qquad (2)</cmath>
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<cmath>\sin \delta - \sin \gamma = \sin C - \sin A        \qquad (3)</cmath>
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 +
Suppose <math>\sin C - \sin A = 0</math>. Then <math>A = C</math>, and the triangle is isosceles. In this case, it is clear by symmetry that <math>\overline{OI}</math> is vertical. Further, since point <math>S</math> exists, <math>\delta \neq \gamma</math>, so <math>\delta + \gamma = 180</math> and <math>\overrightarrow{MN} + \overrightarrow{QP}</math> must be vertical as well.
 +
 
 +
For the remainder of the proof, assume <math>\sin C \neq \sin A</math>. Note that <cmath>\frac{\cos y – \cos x}{\sin x - \sin y} = \frac{ (\cos y - \cos x)(\sin x + \sin y)}{ (\sin x - \sin y)(\sin x + \sin y)} = \frac {\sin x + \sin y}{\cos x + \cos y} </cmath> whenever <math>x, y \in \mathbb{R}</math> and <math>\sin x \neq \sin y</math>. Note further that the slope of the line defined by the vector formed by summing vectors <math>(\cos x, \sin x)</math> and <math>(\cos y, \sin y)</math> is this expression. Since <math>\ell</math> is parallel to <math>\overrightarrow{MN} + \overrightarrow{QP}</math>, the slope of <math>\ell</math> can be formed by dividing expressions in <math>(2)</math> and <math>(3)</math> and inverting the sign: <cmath>\frac{\cos A + \cos C – 1}{\sin C - \sin A} \qquad (4)</cmath>
 +
 
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Determine the coordinates of <math>I</math> by drawing perpendiculars from <math>I</math> to the sides and vertices of the triangle. By exploiting congruence between pairs of right triangles that share a vertex, one can partition <math>a,b,c</math> into <math>p+q, p+r, q+r</math> where <math>p,q,r</math> are bases of these triangles that lie on the sides of triangle <math>ABC</math>. From here it is clear that <math>I = \left(\frac{b + c – a}{2} , \frac{b + c – a}{2}\tan(A / 2)\right)</math>.
 +
 
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To find the coordinates of <math>O</math>, note that <math>\angle{OJK} + \angle{OJL} = \angle{J}</math> and that <math>\angle {OJK} = \angle{OKJ}</math> in any acute triangle <math>JKL</math>. It easily follows that <math>\angle{OJK} = 90 – L</math>. Note also that the perpendicular from <math>O</math> to <math>\overline{JK}</math> bisects <math>\overline{JK}</math>. Hence, <cmath>O = \left(\frac{b}{2},\frac{b}{2} \cot B\right) \qquad (5)</cmath> if triangle <math>ABC</math> is acute.
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If triangle <math>JKL</math> is obtuse at <math>\angle J</math>, then it can be similarly shown that <math>\angle{OKL} = \angle{OLK} = J – 90</math> but that the remaining angles of this form are still <math>90-L</math> and <math>90-K</math>. It easily follows that <math>(5)</math> holds if <math>\angle A</math> is obtuse. If <math>\angle B</math> is obtuse then <math>\angle OAC = B – 90</math> and the <math>y</math> coordinate of <math>O</math> is <math>-\frac{b}{2} \tan{B-90}</math>. From this, <math>(5)</math> follows in this case as well.
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We can conclude the slope of <math>\overline {OI}</math> is <cmath>\left(\frac{b \cot {B} – (b + c – a) \tan(A/2)}{a – c}\right) = \left( \frac {\cos B – \tan{(A/2)}\sin B}{\sin A - \sin C}\right) + \tan(A/2)  \quad (6)</cmath> by the Law of Sines and rearrangement.
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Setting <math>(6) = (4)</math> is equivalent to <cmath>1 - \cos A - \cos C = \cos B + \tan(A/2)(\sin A - \sin B - \sin C)</cmath>
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 +
Since <math>\tan(A/2) = \frac{\sin A}{1 + \cos A}</math>, this equation is equivalent to <cmath>(1 + \cos A)(\cos A + \cos B + \cos C – 1) = (\sin A)(\sin B + \sin C - \sin A)</cmath>
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This equation is equivalent to <cmath>\cos(A + B) + \cos(A + C) + \cos B + \cos C = 0</cmath> which is evident.
  
 
{{MAA Notice}}
 
{{MAA Notice}}
 
==See also==
 
==See also==
 
{{USAMO newbox|year=2016|num-b=4|num-a=6}}
 
{{USAMO newbox|year=2016|num-b=4|num-a=6}}

Revision as of 23:14, 11 July 2016

Problem

An equilateral pentagon $AMNPQ$ is inscribed in triangle $ABC$ such that $M\in\overline{AB},$ $Q\in\overline{AC},$ and $N, P\in\overline{BC}.$ Let $S$ be the intersection of lines $MN$ and $PQ.$ Denote by $\ell$ the angle bisector of $\angle MSQ.$

Prove that $\overline{OI}$ is parallel to $\ell,$ where $O$ is the circumcenter of triangle $ABC,$ and $I$ is the incenter of triangle $ABC.$

Solution

Write $\angle{JKL} = K$ for all $J,K,L$ chosen as distinct vertices of triangle $ABC$. Define $a, b, c$ as sides opposite to angles $A, B$, and $C$, respectively. Place the triangle in the Euclidean plane with $A$ at the origin and $C$ on the positive x-axis. Assume without loss of generality that C is acute.

Consider the sides of the pentagon as vectors and note that \[\overrightarrow{AM} + \overrightarrow{MN} + \overrightarrow{NP} = \overrightarrow{AQ} + \overrightarrow{QP} \qquad (1)\]

Define $\delta$ and $\gamma$ as the angles made between the positive x-axis and $\overrightarrow{MN}$ and $\overrightarrow{QP}$, respectively. Considering the x and y coordinates of the vectors in $(1)$, it follows that \[\cos \delta - \cos \gamma = 1 - \cos A - \cos C \qquad (2)\] \[\sin \delta - \sin \gamma = \sin C - \sin A \qquad (3)\]

Suppose $\sin C - \sin A = 0$. Then $A = C$, and the triangle is isosceles. In this case, it is clear by symmetry that $\overline{OI}$ is vertical. Further, since point $S$ exists, $\delta \neq \gamma$, so $\delta + \gamma = 180$ and $\overrightarrow{MN} + \overrightarrow{QP}$ must be vertical as well.

For the remainder of the proof, assume $\sin C \neq \sin A$. Note that \[\frac{\cos y – \cos x}{\sin x - \sin y} = \frac{ (\cos y - \cos x)(\sin x + \sin y)}{ (\sin x - \sin y)(\sin x + \sin y)} = \frac {\sin x + \sin y}{\cos x + \cos y}\] whenever $x, y \in \mathbb{R}$ and $\sin x \neq \sin y$. Note further that the slope of the line defined by the vector formed by summing vectors $(\cos x, \sin x)$ and $(\cos y, \sin y)$ is this expression. Since $\ell$ is parallel to $\overrightarrow{MN} + \overrightarrow{QP}$, the slope of $\ell$ can be formed by dividing expressions in $(2)$ and $(3)$ and inverting the sign: \[\frac{\cos A + \cos C – 1}{\sin C - \sin A} \qquad (4)\]

Determine the coordinates of $I$ by drawing perpendiculars from $I$ to the sides and vertices of the triangle. By exploiting congruence between pairs of right triangles that share a vertex, one can partition $a,b,c$ into $p+q, p+r, q+r$ where $p,q,r$ are bases of these triangles that lie on the sides of triangle $ABC$. From here it is clear that $I = \left(\frac{b + c – a}{2} , \frac{b + c – a}{2}\tan(A / 2)\right)$.

To find the coordinates of $O$, note that $\angle{OJK} + \angle{OJL} = \angle{J}$ and that $\angle {OJK} = \angle{OKJ}$ in any acute triangle $JKL$. It easily follows that $\angle{OJK} = 90 – L$. Note also that the perpendicular from $O$ to $\overline{JK}$ bisects $\overline{JK}$. Hence, \[O = \left(\frac{b}{2},\frac{b}{2} \cot B\right) \qquad (5)\] if triangle $ABC$ is acute.

If triangle $JKL$ is obtuse at $\angle J$, then it can be similarly shown that $\angle{OKL} = \angle{OLK} = J – 90$ but that the remaining angles of this form are still $90-L$ and $90-K$. It easily follows that $(5)$ holds if $\angle A$ is obtuse. If $\angle B$ is obtuse then $\angle OAC = B – 90$ and the $y$ coordinate of $O$ is $-\frac{b}{2} \tan{B-90}$. From this, $(5)$ follows in this case as well.

We can conclude the slope of $\overline {OI}$ is \[\left(\frac{b \cot {B} – (b + c – a) \tan(A/2)}{a – c}\right) = \left( \frac {\cos B – \tan{(A/2)}\sin B}{\sin A - \sin C}\right) + \tan(A/2) \quad (6)\] by the Law of Sines and rearrangement.

Setting $(6) = (4)$ is equivalent to \[1 - \cos A - \cos C = \cos B + \tan(A/2)(\sin A - \sin B - \sin C)\]

Since $\tan(A/2) = \frac{\sin A}{1 + \cos A}$, this equation is equivalent to \[(1 + \cos A)(\cos A + \cos B + \cos C – 1) = (\sin A)(\sin B + \sin C - \sin A)\]

This equation is equivalent to \[\cos(A + B) + \cos(A + C) + \cos B + \cos C = 0\] which is evident.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

See also

2016 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
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