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Difference between revisions of "2011 USAMO Problems/Problem 5"

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Let <math>P</math> be a given point inside quadrilateral <math>ABCD</math>.  Points <math>Q_1</math> and <math>Q_2</math> are located within <math>ABCD</math> such that <math>\angle Q_1 BC = \angle ABP</math>, <math>\angle Q_1 CB = \angle DCP</math>, <math>\angle Q_2 AD = \angle BAP</math>, <math>\angle Q_2 DA = \angle CDP</math>.  Prove that <math>\overline{Q_1 Q_2} \parallel \overline{AB}</math> if and only if <math>\overline{Q_1 Q_2} \parallel \overline{CD}</math>.
 
Let <math>P</math> be a given point inside quadrilateral <math>ABCD</math>.  Points <math>Q_1</math> and <math>Q_2</math> are located within <math>ABCD</math> such that <math>\angle Q_1 BC = \angle ABP</math>, <math>\angle Q_1 CB = \angle DCP</math>, <math>\angle Q_2 AD = \angle BAP</math>, <math>\angle Q_2 DA = \angle CDP</math>.  Prove that <math>\overline{Q_1 Q_2} \parallel \overline{AB}</math> if and only if <math>\overline{Q_1 Q_2} \parallel \overline{CD}</math>.
  
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==Solution==
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==Solution 1==
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Lemma. If <math>AB</math> and <math>CD</math> are not parallel, then <math>AB, CD, Q_1 Q_2</math> are concurrent.
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Proof. Let <math>AB</math> and <math>CD</math> meet at <math>R</math>. Notice that with respect to triangle <math>ADR</math>, <math>P</math> and <math>Q_2</math> are isogonal conjugates (this can be proven by dropping altitudes from <math>Q_2</math> to <math>AB</math>, <math>CD</math>, and <math>AD</math> or <math>BC</math> depending on where <math>R</math> is). With respect to triangle <math>BCR</math>, <math>P</math> and <math>Q_1</math> are isogonal conjugates. Therefore, <math>Q_1</math> and <math>Q_2</math> lie on the reflection of <math>RP</math> in the angle bisector of <math>\angle{DRA}</math>, so <math>R, Q_1, Q_2</math> are collinear. Hence, <math>AB, CD, Q_1 Q_2</math> are concurrent at <math>R</math>.
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Now suppose <math>Q_1 Q_2 \parallel AB</math> but <math>Q_1 Q_2</math> is not parallel to <math>CD</math>. Then <math>AB</math> and <math>CD</math> are not parallel and thus intersect at a point <math>R</math>. But then <math>Q_1 Q_2</math> also passes through <math>R</math>, contradicting <math>Q_1 Q_2 \parallel AB</math>. A similar contradiction occurs if <math>Q_1 Q_2 \parallel CD</math> but <math>Q_1 Q_2</math> is not parallel to <math>AB</math>, so we can conclude that <math>Q_1 Q_2 \parallel AB</math> if and only if <math>Q_1 Q_2 \parallel CD</math>.
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*[[Isogonal conjugate]]
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==Solution 2==
 
First note that <math>\overline{Q_1 Q_2} \parallel \overline{AB}</math> if and only if the altitudes from <math>Q_1</math> and <math>Q_2</math> to <math>\overline{AB}</math> are the same, or <math>|Q_1B|\sin \angle ABQ_1 =|Q_2A|\sin \angle BAQ_2</math>.  Similarly <math>\overline{Q_1 Q_2} \parallel \overline{CD}</math> iff <math>|Q_1C|\sin \angle DCQ_1 =|Q_2D|\sin \angle CDQ_2</math>.
 
First note that <math>\overline{Q_1 Q_2} \parallel \overline{AB}</math> if and only if the altitudes from <math>Q_1</math> and <math>Q_2</math> to <math>\overline{AB}</math> are the same, or <math>|Q_1B|\sin \angle ABQ_1 =|Q_2A|\sin \angle BAQ_2</math>.  Similarly <math>\overline{Q_1 Q_2} \parallel \overline{CD}</math> iff <math>|Q_1C|\sin \angle DCQ_1 =|Q_2D|\sin \angle CDQ_2</math>.
  
  
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If we define <math>S =\frac{|Q_1B|\sin \angle ABQ_1}{|Q_2A|\sin \angle BAQ_2}\cdot\frac{|Q_2D|\sin \angle CDQ_2}{|Q_1C|\sin \angle DCQ_1}</math>, then we are done if we can show that S=1.
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If we define <math>S =\frac{|Q_1B|\sin \angle ABQ_1}{|Q_2A|\sin \angle BAQ_2} \times \frac{|Q_2D|\sin \angle CDQ_2}{|Q_1C|\sin \angle DCQ_1}</math>, then we are done if we can show that S=1.
  
  
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Applying the law of sines to the triangles with vertices at P yields <math>S=\frac{|PC|}{|PB|}\frac{|PA|}{|PD|}\frac{|PD|}{|PC|}\frac{|PB|}{|PA|}=1</math>.
 
Applying the law of sines to the triangles with vertices at P yields <math>S=\frac{|PC|}{|PB|}\frac{|PA|}{|PD|}\frac{|PD|}{|PC|}\frac{|PB|}{|PA|}=1</math>.
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==Solution 2==
 
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Lemma. If <math>AB</math> and <math>CD</math> are not parallel, then <math>AB, CD, Q_1 Q_2</math> are concurrent.
 
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Proof. Let <math>AB</math> and <math>CD</math> meet at <math>R</math>. Notice that with respect to triangle <math>ADR</math>, <math>P</math> and <math>Q_2</math> are isogonal conjugates. With respect to triangle <math>BCR</math>, <math>P</math> and <math>Q_1</math> are isogonal conjugates. Therefore, <math>Q_1</math> and <math>Q_2</math> lie on the reflection of <math>RP</math> in the angle bisector of <math>\angle{DRA}</math>, so <math>R, Q_1, Q_2</math> are collinear. Hence, <math>AB, CD, Q_1 Q_2</math> are concurrent at <math>R</math>.
 
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Now suppose <math>Q_1 Q_2 // AB</math> but <math>Q_1 Q_2</math> is not parallel to <math>CD</math>. Then <math>AB</math> and <math>CD</math> are not parallel and thus intersect at a point <math>R</math>. But then <math>Q_1 Q_2</math> also passes through <math>R</math>, contradicting <math>Q_1 Q_2 // AB</math>. A similar contradiction occurs if <math>Q_1 Q_2 // CD</math> but <math>Q_1 Q_2</math> is not parallel to <math>CD</math>, so we can conclude that <math>Q_1 Q_2 // AB</math> if and only if <math>Q_1 Q_2 // CD</math>.
 
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{{MAA Notice}}
 
  
 
==See also==
 
==See also==
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{{USAMO newbox|year=2011|num-b=4|num-a=6}}
 
{{USAMO newbox|year=2011|num-b=4|num-a=6}}
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[[Category:Olympiad Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 00:04, 15 February 2024

Problem

Let $P$ be a given point inside quadrilateral $ABCD$. Points $Q_1$ and $Q_2$ are located within $ABCD$ such that $\angle Q_1 BC = \angle ABP$, $\angle Q_1 CB = \angle DCP$, $\angle Q_2 AD = \angle BAP$, $\angle Q_2 DA = \angle CDP$. Prove that $\overline{Q_1 Q_2} \parallel \overline{AB}$ if and only if $\overline{Q_1 Q_2} \parallel \overline{CD}$.

Solution 1

Lemma. If $AB$ and $CD$ are not parallel, then $AB, CD, Q_1 Q_2$ are concurrent.

Proof. Let $AB$ and $CD$ meet at $R$. Notice that with respect to triangle $ADR$, $P$ and $Q_2$ are isogonal conjugates (this can be proven by dropping altitudes from $Q_2$ to $AB$, $CD$, and $AD$ or $BC$ depending on where $R$ is). With respect to triangle $BCR$, $P$ and $Q_1$ are isogonal conjugates. Therefore, $Q_1$ and $Q_2$ lie on the reflection of $RP$ in the angle bisector of $\angle{DRA}$, so $R, Q_1, Q_2$ are collinear. Hence, $AB, CD, Q_1 Q_2$ are concurrent at $R$.

Now suppose $Q_1 Q_2 \parallel AB$ but $Q_1 Q_2$ is not parallel to $CD$. Then $AB$ and $CD$ are not parallel and thus intersect at a point $R$. But then $Q_1 Q_2$ also passes through $R$, contradicting $Q_1 Q_2 \parallel AB$. A similar contradiction occurs if $Q_1 Q_2 \parallel CD$ but $Q_1 Q_2$ is not parallel to $AB$, so we can conclude that $Q_1 Q_2 \parallel AB$ if and only if $Q_1 Q_2 \parallel CD$.

Solution 2

First note that $\overline{Q_1 Q_2} \parallel \overline{AB}$ if and only if the altitudes from $Q_1$ and $Q_2$ to $\overline{AB}$ are the same, or $|Q_1B|\sin \angle ABQ_1 =|Q_2A|\sin \angle BAQ_2$. Similarly $\overline{Q_1 Q_2} \parallel \overline{CD}$ iff $|Q_1C|\sin \angle DCQ_1 =|Q_2D|\sin \angle CDQ_2$.


If we define $S =\frac{|Q_1B|\sin \angle ABQ_1}{|Q_2A|\sin \angle BAQ_2} \times \frac{|Q_2D|\sin \angle CDQ_2}{|Q_1C|\sin \angle DCQ_1}$, then we are done if we can show that S=1.


By the law of sines, $\frac{|Q_1B|}{|Q_1C|}=\frac{\sin\angle Q_1CB}{\sin\angle Q_1BC}$ and $\frac{|Q_2D|}{|Q_2A|}=\frac{\sin\angle Q_2AD}{\sin\angle Q_2DA}$.


So, $S=\frac{\sin \angle ABQ_1}{\sin \angle BAQ_2}\cdot\frac{\sin \angle CDQ_2}{\sin \angle DCQ_1}\cdot\frac{\sin \angle BCQ_1}{\sin \angle CBQ_1}\cdot\frac{\sin \angle DAQ_2}{\sin \angle ADQ_2}$


By the terms of the problem, $S=\frac{\sin \angle PBC}{\sin \angle PAD}\cdot\frac{\sin \angle PDA}{\sin \angle PCB}\cdot\frac{\sin \angle PCD}{\sin \angle PBA}\cdot\frac{\sin \angle PAB}{\sin \angle PDC}$. (If two subangles of an angle of the quadrilateral are equal, then their complements at that quadrilateral angle are equal as well.)


Rearranging yields $S= \frac{\sin \angle PBC}{\sin \angle PCB}\cdot\frac{\sin \angle PDA}{\sin \angle PAD}\cdot\frac{\sin \angle PCD}{\sin \angle PDC}\cdot\frac{\sin \angle PAB}{\sin \angle PBA}$.


Applying the law of sines to the triangles with vertices at P yields $S=\frac{|PC|}{|PB|}\frac{|PA|}{|PD|}\frac{|PD|}{|PC|}\frac{|PB|}{|PA|}=1$.

See also

2011 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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