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

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==Solution 3==
 
==Solution 3==
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Let <math>A', B',</math> and <math>C'</math> be the arc midpoints of <math>BC, CA, AB,</math> respectively. Let <math>E</math> be crosspoint of <math>AI</math> and <math>B'C'.</math> Therefore <math>O</math> is the circumcenter of triangle <math>A'B'C',</math> points <math>A, I, E,</math> and <math>A'</math> are collinear, <math>\angle A'EB' = \frac {\overset{\Large\frown} {AC'}+\overset{\Large\frown} {B'C} +\overset{\Large\frown} {CA'}}{2} = 90^\circ \implies AA' \perp B'C' \implies I</math> is orthocenter of <math>\triangle A'B'C' \implies IO</math> is the Euler line of <math>\triangle A'B'C'.</math>
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[[File:2016 USAMO 5a.png|500px|right]]
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Let <math>A', B',</math> and <math>C'</math> be the arc midpoints of <math>BC, CA, AB,</math> respectively. Let <math>E</math> be crosspoint of <math>AI</math> and <math>B'C'.</math> Therefore <math>O</math> is the circumcenter of triangle <math>A'B'C',</math> points <math>A, I, E,</math> and <math>A'</math> are collinear, <math>\angle A'EB' = \frac {\overset{\Large\frown} {AC'}+\overset{\Large\frown} {B'C} +\overset{\Large\frown} {CA'}}{2} = 90^\circ \implies AA' \perp B'C'</math>
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<math>\implies I</math> is orthocenter of <math>\triangle A'B'C' \implies</math>
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<math>IO</math> is the Euler line of <math>\triangle A'B'C'.</math>
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Let <math>G</math> be the centroid of <math>\triangle A'B'C' \implies G </math> lies on line <math>IO</math>
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<math>\implies \overline {OG} = \frac {\overline{OA'} + \overline{OB'} + \overline{OC'}}{3}</math> is paraller to <math>\overline{OI}.</math>
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<math>\overline{OB'} \perp \overline{AC} \implies \overline{OB'} \perp \overline{QA}.</math>
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Similarly <math>\overline{OC'} \perp \overline{AM}, \overline{OA'} \perp \overline{NP},</math> rotation from <math>\overline{OA'}</math> to <math>\overline{NP}</math>, from <math>\overline{OB'}</math> to <math>\overline{QA},</math> and from <math>\overline{OC'}</math> to <math>\overline{AM}</math> is in clockwise direction, <math>|\overline{QA}|=| \overline{AM}| =|\overline{NP}|, |\overline{OA'}| = |\overline{OB'}| = |\overline{OC'}| \implies</math>
 +
 
 +
<math>\overline{QA} + \overline{AM} + \overline{NP} = \overline{QM} + \overline{NP}</math> is perpendicular to <math>|\overline{OI}.</math>
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<math>|\overline{MN}| = |\overline{QP}|</math> therefore in accordance with <i><b>Claim</b></i> <math>\overline{MN} + \overline{QP}</math> is parallel to <math>\overline{OI}.</math>
 +
This sum is parallel to <math>\ell,</math>, so we are done.
  
 
==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 00:36, 11 October 2022

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 1

Let $D$ be the intersection of line $AI$ and the circumcircle of $\Delta ABC$ (other than $A$), then $OD\perp BC$. Let $R$ be the point such that $NPQR$ is a rhombus. It follows that $OD\perp QR$.

Since $AM=AQ$, $AI\perp MQ$, or $DI\perp MQ$. It follows that $\angle ODI=\angle RQM$.

Since $BO=OD$, $MA=AQ$, $\angle BOD=\angle MAQ$, it follows that $\Delta BOD\sim\Delta MAQ$, so $AQ/MQ=OD/BD$.

It is given that $AQ=NP=RQ$, and by basic properties of the incenter, $ID=BD$. Therefore, $RQ/MQ=OD/ID$, so $\Delta RQM\sim\Delta ODI$. Since the rotation between the two triangles in 90 degrees, $OI\perp MR$. However, $l$ is parallel to the bisector of $MNR$, which is perpendicular to $MR$, so we are done.

Solution 2

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

Solution 3

2016 USAMO 5a.png

Let $A', B',$ and $C'$ be the arc midpoints of $BC, CA, AB,$ respectively. Let $E$ be crosspoint of $AI$ and $B'C'.$ Therefore $O$ is the circumcenter of triangle $A'B'C',$ points $A, I, E,$ and $A'$ are collinear, $\angle A'EB' = \frac {\overset{\Large\frown} {AC'}+\overset{\Large\frown} {B'C} +\overset{\Large\frown} {CA'}}{2} = 90^\circ \implies AA' \perp B'C'$ $\implies I$ is orthocenter of $\triangle A'B'C' \implies$ $IO$ is the Euler line of $\triangle A'B'C'.$

Let $G$ be the centroid of $\triangle A'B'C' \implies G$ lies on line $IO$ $\implies \overline {OG} = \frac {\overline{OA'} + \overline{OB'} + \overline{OC'}}{3}$ is paraller to $\overline{OI}.$ $\overline{OB'} \perp \overline{AC} \implies \overline{OB'} \perp \overline{QA}.$ Similarly $\overline{OC'} \perp \overline{AM}, \overline{OA'} \perp \overline{NP},$ rotation from $\overline{OA'}$ to $\overline{NP}$, from $\overline{OB'}$ to $\overline{QA},$ and from $\overline{OC'}$ to $\overline{AM}$ is in clockwise direction, $|\overline{QA}|=| \overline{AM}| =|\overline{NP}|, |\overline{OA'}| = |\overline{OB'}| = |\overline{OC'}| \implies$

$\overline{QA} + \overline{AM} + \overline{NP} = \overline{QM} + \overline{NP}$ is perpendicular to $|\overline{OI}.$ $|\overline{MN}| = |\overline{QP}|$ therefore in accordance with Claim $\overline{MN} + \overline{QP}$ is parallel to $\overline{OI}.$ This sum is parallel to $\ell,$, so we are done.

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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