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

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==Solution==
 
==Solution==
 
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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 =\dfrac{|Q_1B|\sin \angle ABQ_1}{|Q_2A|\sin \angle BAQ_2}\dfrac{|Q_2D|\sin \angle CDQ_2}{|Q_1C|\sin \angle DCQ_1}</math>, then we are done if we can show that S=1.
  
 
==See also==
 
==See also==

Revision as of 15:13, 8 June 2011

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

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

See also

2011 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6
All USAMO Problems and Solutions