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2011 USAMO Problems/Problem 5: Difference between revisions

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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>.


==Solution==
==Solution 1==
{{solution}}
Lemma. If <math>AB</math> and <math>CD</math> are not parallel, then <math>AB, CD, Q_1 Q_2</math> are concurrent.


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>.
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>.
==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>.




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.
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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By the terms of the problem, <math>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}</math>.  This involves utilizing the fact that if two subangles of an angle of the quadrilateral are equal, then their complements at that quadrilateral angle are equal as well.   
By the terms of the problem, <math>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}</math>.  (If two subangles of an angle of the quadrilateral are equal, then their complements at that quadrilateral angle are equal as well.)    




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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>.
== Solution 3 ==
<i><b>Case 1</b></i> The lines <math>AB</math> and <math>CD</math> are not parallel. Denote <math>E = AB \cap CD.</math>
<cmath>\angle Q_1 BC = \angle ABP, \angle Q_1 CB = \angle DCP \implies</cmath>
Point <math>Q_1</math> isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle EBC \implies</math>
<math>EP</math> and <math>EQ_1</math> are isogonals with respect to <math>\angle BEC.</math>
Similarly point <math>Q_2</math> isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle EAD \implies</math>
<math>EP</math> and <math>EQ_2</math> are isogonals with respect to <math>\angle BEC.</math>
Therefore points <math>E, Q_1, Q_2</math> lies on the isogonal <math>EP</math> with respect to <math>\angle BEC \implies</math>
<math>Q_1Q_2</math> is not parallel to <math>AB</math> or <math>CD.</math>
<i><b>Case 2</b></i>  <math>AB||CD.</math> We use <i><b>The isogonal theorem in case parallel lines</b></i> and get
<cmath>AB||Q_1Q_2 || CD. \blacksquare</cmath>
'''vladimir.shelomovskii@gmail.com, vvsss'''


==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}}
[[Category:Olympiad Geometry Problems]]
{{MAA Notice}}

Latest revision as of 17:22, 15 February 2025

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 $S=\frac{|PC|}{|PB|}\frac{|PA|}{|PD|}\frac{|PD|}{|PC|}\frac{|PB|}{|PA|}=1$.

Solution 3

Case 1 The lines $AB$ and $CD$ are not parallel. Denote $E = AB \cap CD.$

\[\angle Q_1 BC = \angle ABP, \angle Q_1 CB = \angle DCP \implies\] Point $Q_1$ isogonal conjugate of a point $P$ with respect to a triangle $\triangle EBC \implies$

$EP$ and $EQ_1$ are isogonals with respect to $\angle BEC.$

Similarly point $Q_2$ isogonal conjugate of a point $P$ with respect to a triangle $\triangle EAD \implies$

$EP$ and $EQ_2$ are isogonals with respect to $\angle BEC.$

Therefore points $E, Q_1, Q_2$ lies on the isogonal $EP$ with respect to $\angle BEC \implies$

$Q_1Q_2$ is not parallel to $AB$ or $CD.$

Case 2 $AB||CD.$ We use The isogonal theorem in case parallel lines and get \[AB||Q_1Q_2 || CD. \blacksquare\]

vladimir.shelomovskii@gmail.com, vvsss

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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