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>. | ||
− | ==Solution== | + | ==Solution 1== |
− | + | 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>. | ||
+ | *[[Isogonal conjugate]] | ||
+ | |||
+ | ==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}\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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− | So, <math>S=\frac{\sin \angle ABQ_1}{\sin \angle BAQ_2}\frac{\sin \angle CDQ_2}{\sin \angle DCQ_1}\frac{\sin \angle BCQ_1}{\sin \angle CBQ_1}\frac{\sin \angle DAQ_2}{\sin \angle ADQ_2}</math> | + | So, <math>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}</math> |
+ | |||
+ | |||
+ | 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.) | ||
+ | |||
+ | |||
+ | Rearranging yields <math>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}</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>. | ||
==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 01:04, 15 February 2024
Contents
Problem
Let be a given point inside quadrilateral
. Points
and
are located within
such that
,
,
,
. Prove that
if and only if
.
Solution 1
Lemma. If and
are not parallel, then
are concurrent.
Proof. Let and
meet at
. Notice that with respect to triangle
,
and
are isogonal conjugates (this can be proven by dropping altitudes from
to
,
, and
or
depending on where
is). With respect to triangle
,
and
are isogonal conjugates. Therefore,
and
lie on the reflection of
in the angle bisector of
, so
are collinear. Hence,
are concurrent at
.
Now suppose but
is not parallel to
. Then
and
are not parallel and thus intersect at a point
. But then
also passes through
, contradicting
. A similar contradiction occurs if
but
is not parallel to
, so we can conclude that
if and only if
.
Solution 2
First note that if and only if the altitudes from
and
to
are the same, or
. Similarly
iff
.
If we define , then we are done if we can show that S=1.
By the law of sines, and
.
So,
By the terms of the problem, . (If two subangles of an angle of the quadrilateral are equal, then their complements at that quadrilateral angle are equal as well.)
Rearranging yields .
Applying the law of sines to the triangles with vertices at P yields .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.