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

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(Solution 2)
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== Solution 2==
 
== Solution 2==
  
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 15:24, 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.

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See Also

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