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Difference between revisions of "2019 USAMO Problems/Problem 2"

(Created page with "==Problem== Let <math>ABCD</math> be a cyclic quadrilateral satisfying <math>AD^2 + BC^2 = AB^2</math>. The diagonals of <math>ABCD</math> intersect at <math>E</math>. Let <ma...")
 
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==Solution==
 
==Solution==
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Realize that there is only one point <math>P</math> on <math>\overline{AB}</math> satisfying the conditions, because <math>\angle APD</math> decreases and <math>\angle BPC</math> increases as <math>P</math> moves from <math>A</math> to <math>B</math>. Therefore, if we prove that there is a single point <math>P</math> that lies on <math>\overline{AB}</math> such that <math>\angle APD \cong \angle BPC</math>, and that <math>PE</math> bisects <math>CD</math>, it must coincide with the point from the problem, so we will be done.
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Since <math>AD^2 + BC^2 = AB^2</math>, there is some <math>P</math> on <math>AB</math> such that <cmath>AD^2 = AP \times AB \text{ and } BC^2 = BP \times BA.</cmath> Thus, <math>\frac{AP}{AD} = \frac{AD}{AB}</math> and <math>\frac{BP}{BC} = \frac{BC}{BA}</math>. Thus we have that <math>\triangle APD \sim
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\triangle ADB</math> and <math>\triangle BPC \sim \triangle BCA</math>, meaning that <math>\angle APD = \angle ADB = \angle ACB = \angle BPC</math>.
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We next must show that <math>PE</math> bisects <math>CD</math>. Define <math>K</math> as the intersection of <math>AC</math> and <math>PD</math> and <math>L</math> as the intersection of <math>BD</math> and <math>PC</math>. We know that <math>APLD</math> and <math>BPKC</math> are cyclic, because
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<cmath>\angle'ADL = \angle'ACB = \angle'BPC = \angle'AP L,</cmath> where <math>\angle'</math> represents an angle which is measured <math>\text{mod } \pi</math>. Furthermore, quadrilateral <math>AKLB</math> is also cyclic, because
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<cmath>\angle'AKB = \angle'CKB = \angle'CP B</cmath> and <math>\angle'ALB = \angle'APD</math>, and these are equal.
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As the quadrilaterals are cyclic, we have that <math>\angle'KCD = \angle'ABD = \angle'ABL = \angle'AKL = \angle'CKL</math>, meaning that <math>CD \parallel KL</math>. Thus <math>CDKL</math> is a trapezoid whose legs intersect at <math>P</math> and whose diagonals intersect at <math>E</math>. Therefore, line <math>PE</math> bisects the bases <math>CD</math> and <math>KL</math>, as wished. ~ciceronii
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:03, 27 February 2020

Problem

Let $ABCD$ be a cyclic quadrilateral satisfying $AD^2 + BC^2 = AB^2$. The diagonals of $ABCD$ intersect at $E$. Let $P$ be a point on side $\overline{AB}$ satisfying $\angle APD = \angle BPC$. Show that line $PE$ bisects $\overline{CD}$.

Solution

Realize that there is only one point $P$ on $\overline{AB}$ satisfying the conditions, because $\angle APD$ decreases and $\angle BPC$ increases as $P$ moves from $A$ to $B$. Therefore, if we prove that there is a single point $P$ that lies on $\overline{AB}$ such that $\angle APD \cong \angle BPC$, and that $PE$ bisects $CD$, it must coincide with the point from the problem, so we will be done.

Since $AD^2 + BC^2 = AB^2$, there is some $P$ on $AB$ such that \[AD^2 = AP \times AB \text{ and } BC^2 = BP \times BA.\] Thus, $\frac{AP}{AD} = \frac{AD}{AB}$ and $\frac{BP}{BC} = \frac{BC}{BA}$. Thus we have that $\triangle APD \sim \triangle ADB$ and $\triangle BPC \sim \triangle BCA$, meaning that $\angle APD = \angle ADB = \angle ACB = \angle BPC$.

We next must show that $PE$ bisects $CD$. Define $K$ as the intersection of $AC$ and $PD$ and $L$ as the intersection of $BD$ and $PC$. We know that $APLD$ and $BPKC$ are cyclic, because \[\angle'ADL = \angle'ACB = \angle'BPC = \angle'AP L,\] where $\angle'$ represents an angle which is measured $\text{mod } \pi$. Furthermore, quadrilateral $AKLB$ is also cyclic, because \[\angle'AKB = \angle'CKB = \angle'CP B\] and $\angle'ALB = \angle'APD$, and these are equal.

As the quadrilaterals are cyclic, we have that $\angle'KCD = \angle'ABD = \angle'ABL = \angle'AKL = \angle'CKL$, meaning that $CD \parallel KL$. Thus $CDKL$ is a trapezoid whose legs intersect at $P$ and whose diagonals intersect at $E$. Therefore, line $PE$ bisects the bases $CD$ and $KL$, as wished. ~ciceronii

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

See also

2019 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
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