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

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== Solution ==
 
== Solution ==
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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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== 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>
 
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>

Revision as of 15:16, 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

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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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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