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Difference between revisions of "2015 USAJMO Problems/Problem 5"

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== Solution 2==
 
== Solution 2==
  
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Note that lines <math>AC, AX</math> are isogonal in <math>\triangle ABD</math>, so an inversion centered at <math>A</math> with power <math>r^2=AB\cdot AD</math> composed with a reflection about the angle bisector of <math>\angle DAB</math> swaps the pairs <math>(D,B)</math> and <math>(C,X)</math>. Thus, <cmath>\frac{AD}{XD}\cdot \frac{XD}{CD}=\frac{AC}{BC}\cdot \frac{AB}{CA}\Longrightarrow (A,C;B,D)=-1</cmath>so that <math>ACBD</math> is a harmonic quadrilateral. By symmetry, if <math>Y</math> exists, then <math>(B,D;A,C)=-1</math>. We have shown the two conditions are equivalent, whence both directions follow<math>.\:\blacksquare\:</math>
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All angles are directed. Note that lines <math>AC, AX</math> are isogonal in <math>\triangle ABD</math> and <math>CD, CE</math> are isogonal in <math>\triangle CDB</math>. From the law of sines it follows that
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<cmath>\frac{DX}{XB}\cdot \frac{DE}{ED}=\left(\frac{AD}{DB}\right)^2=\left(\frac{DC}{BC}\right)^2.</cmath>
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Therefore, the ratio equals <math>\frac{AD\cdot DC}{DB\cdot BC}.</math>
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Now let <math>Y</math> be a point of <math>AC</math> such that <math>\angle{ABE}=\angle{CBY}</math>. We apply the above identities for <math>Y</math> to get that <math>\frac{CY}{YC}\cdot \frac{CE}{EA}=\left(\frac{CD}{DA}\right)^2</math>. So <math>\angle{CDY}=\angle{EDA}</math>, the converse follows since all our steps are reversible.
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:24, 13 April 2017

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

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 $\frac{CY}{YC}\cdot \frac{CE}{EA}=\left(\frac{CD}{DA}\right)^2$. So $\angle{CDY}=\angle{EDA}$, the converse follows since all our steps are reversible.

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

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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