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2019 USAMO Problems/Problem 2

Revision as of 01:03, 27 February 2020 by Ciceronii (talk | contribs)

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

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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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