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

(Solution 2)
m (Solution 2: I believe the added statement about how directed angles are not standard is true.)
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== Solution 2==
 
== Solution 2==
  
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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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  
  
 
<cmath>\frac{DX}{XB}\cdot \frac{DE}{ED}=\left(\frac{AD}{DB}\right)^2=\left(\frac{DC}{BC}\right)^2.</cmath>
 
<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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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}{YA}\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.
 
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}{YA}\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.
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*Beware that directed angles, or angles <math>mod 180</math>, 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.]
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:51, 5 September 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 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 (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6
All USAJMO Problems and Solutions
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