2015 USAJMO Problems/Problem 5

Problem

Let $ABCD$ be a cyclic quadrilateral. Prove that there exists a point $X$ on segment $\overline{BD}$ such that $\angle BAC=\angle XAD$ and $\angle BCA=\angle XCD$ if and only if there exists a point $Y$ on segment $\overline{AC}$ such that $\angle CBD=\angle YBA$ and $\angle CDB=\angle YDA$ .

Solution

Note that lines $AC, AX$ are isogonal in $\triangle ABD$ , so an inversion centered at $A$ with power $r^2=AB\cdot AD$ composed with a reflection about the angle bisector of $\angle DAB$ swaps the pairs $(D,B)$ and $(C,X)$ . Thus, $\frac{AD}{XD}\cdot \frac{XD}{CD}=\frac{AC}{BC}\cdot \frac{AB}{CA}\Longrightarrow (A,C;B,D)=-1$ so that $ACBD$ is a harmonic quadrilateral. By symmetry, if $Y$ exists, then $(B,D;A,C)=-1$ . We have shown the two conditions are equivalent, whence both directions follow $.\:\blacksquare\:$

Solution 2

Note that lines $AC, AX$ are isogonal in $\triangle ABD$ , so an inversion centered at $A$ with power $r^2=AB\cdot AD$ composed with a reflection about the angle bisector of $\angle DAB$ swaps the pairs $(D,B)$ and $(C,X)$ . Thus, $\frac{AD}{XD}\cdot \frac{XD}{CD}=\frac{AC}{BC}\cdot \frac{AB}{CA}\Longrightarrow (A,C;B,D)=-1$ so that $ACBD$ is a harmonic quadrilateral. By symmetry, if $Y$ exists, then $(B,D;A,C)=-1$ . We have shown the two conditions are equivalent, whence both directions follow $.\:\blacksquare\:$

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2015 USAJMO (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
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2015 USAJMO Problems/Problem 5

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Problem

Solution

Solution 2

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