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2015 USAJMO Problems/Problem 5

Revision as of 15:51, 5 September 2017 by Expilncalc2 (talk | contribs) (→‎Solution 2: I believe the added statement about how directed angles are not standard is true.)

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 1

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

All angles are directed*. Note that lines $AC, AX$ are isogonal in $\triangle ABD$ and $CD, CE$ are isogonal in $\triangle CDB$. From the law of sines it follows that

\[\frac{DX}{XB}\cdot \frac{DE}{ED}=\left(\frac{AD}{DB}\right)^2=\left(\frac{DC}{BC}\right)^2.\]

Therefore, the ratio equals $\frac{AD\cdot DC}{DB\cdot BC}.$

Now let $Y$ be a point of $AC$ such that $\angle{ABE}=\angle{CBY}$. We apply the above identities for $Y$ to get that $\frac{CY}{YA}\cdot \frac{CE}{EA}=\left(\frac{CD}{DA}\right)^2$. So $\angle{CDY}=\angle{EDA}$, the converse follows since all our steps are reversible.

  • Beware that directed angles, or angles $mod 180$, are not standard olympiad material. If you use them, make sure to define and indicate what they mean. [Mentioned in Evan Chen's Geometry Olympiad book.]

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See Also

2015 USAJMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions
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