Difference between revisions of "2019 USAJMO Problems/Problem 3"

(Solution)
(Solution)
Line 6: Line 6:
  
 
Note that only one point <math>P</math> satisfies the given angle condition. With this in mind, construct <math>P'</math> with the following properties:
 
Note that only one point <math>P</math> satisfies the given angle condition. With this in mind, construct <math>P'</math> with the following properties:
[list]
 
[*] <math>AP' \cdot AB = AD^2</math>
 
[*] <math>BP' \cdot AB = CD^2</math>
 
[/list]
 
  
[b]Claim:[/b]<math>P = P'</math>
+
<math>\dot </math>AP' \cdot AB = AD^2<math>
 +
</math>\dot BP' \cdot AB = CD^2<math>
  
[i]Proof:[/i]
 
  
The conditions imply the similarities <math>ADP \sim ABD</math> and <math>BCP \sim BAC</math> whence <math>\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB</math> as desired. <math>\square</math>
+
Claim:</math>P = P'<math>
  
[b]Claim:[/b] <math>PE</math> is a symmedian in <math>AEB</math>
+
Proof:
  
[i]Proof:[/i]
+
The conditions imply the similarities </math>ADP \sim ABD<math> and </math>BCP \sim BAC<math> whence </math>\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB<math> as desired. </math>\square<math>
 +
 
 +
Claim: </math>PE<math> is a symmedian in </math>AEB<math>
 +
 
 +
Proof:
  
 
We have  
 
We have  
\begin{align*}
 
AP \cdot AB = AD^2 \iff AB^2 \cdot AP &= AD^2 \cdot AB \\
 
\iff \left( \frac{AB}{AD} \right)^2 &= \frac{AB}{AP} \\
 
\iff \left( \frac{AB}{AD} \right)^2 - 1 &= \frac{AB}{AP} - 1 \\
 
\iff \frac{AB^2 - AD^2}{AD^2} &= \frac{BP}{AP} \\
 
\iff \left(\frac{BC}{AD} \right)^2 &= \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP}
 
\end{align*}
 
as desired. <math>\square</math>
 
  
Since <math>P</math> is the isogonal conjugate of <math>N</math>, <math>\measuredangle PEA = \measuredangle MEC = \measuredangle BEN</math>. However <math>\measuredangle MEC = \measuredangle BEN</math> implies that <math>M</math> is the midpoint of <math>CD</math> from similar triangles, so we are done. <math>\square</math>
+
<cmath>AP \cdot AB = AD^2 \iff AB^2 \cdot AP = AD^2 \cdot AB</cmath>
 +
<cmath>\iff \left( \frac{AB}{AD} \right)^2 = \frac{AB}{AP}</cmath>
 +
<cmath>\iff \left( \frac{AB}{AD} \right)^2 - 1 = \frac{AB}{AP} - 1</cmath>
 +
<cmath>\iff \frac{AB^2 - AD^2}{AD^2} = \frac{BP}{AP}</cmath>
 +
<cmath>\iff \left(\frac{BC}{AD} \right)^2 = \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP} </cmath>
 +
 
 +
as desired. </math>\square<math>
 +
 
 +
Since </math>P<math> is the isogonal conjugate of </math>N<math>, </math>\measuredangle PEA = \measuredangle MEC = \measuredangle BEN<math>. However </math>\measuredangle MEC = \measuredangle BEN<math> implies that </math>M<math> is the midpoint of </math>CD<math> from similar triangles, so we are done. </math>\square$
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:34, 25 June 2019

Problem

$(*)$ Let $ABCD$ be a cyclic quadrilateral satisfying $AD^2+BC^2=AB^2$. The diagonals of $ABCD$ intersect at $E$. Let $P$ be a point on side $\overline{AB}$ satisfying $\angle APD=\angle BPC$. Show that line $PE$ bisects $\overline{CD}$.

Solution

Let $PE \cap DC = M$. Also, let $N$ be the midpoint of $AB$.

Note that only one point $P$ satisfies the given angle condition. With this in mind, construct $P'$ with the following properties:

$\dot$ (Error compiling LaTeX. ! Extra }, or forgotten $.)AP' \cdot AB = AD^2$$ (Error compiling LaTeX. ! Missing $ inserted.)\dot BP' \cdot AB = CD^2$Claim:$P = P'$Proof:

The conditions imply the similarities$ (Error compiling LaTeX. ! Missing $ inserted.)ADP \sim ABD$and$BCP \sim BAC$whence$\measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB$as desired.$\square$Claim:$PE$is a symmedian in$AEB$Proof:

We have

<cmath>AP \cdot AB = AD^2 \iff AB^2 \cdot AP = AD^2 \cdot AB</cmath> <cmath>\iff \left( \frac{AB}{AD} \right)^2 = \frac{AB}{AP}</cmath> <cmath>\iff \left( \frac{AB}{AD} \right)^2 - 1 = \frac{AB}{AP} - 1</cmath> <cmath>\iff \frac{AB^2 - AD^2}{AD^2} = \frac{BP}{AP}</cmath> <cmath>\iff \left(\frac{BC}{AD} \right)^2 = \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP} </cmath>

as desired.$ (Error compiling LaTeX. ! Missing $ inserted.)\square$Since$P$is the isogonal conjugate of$N$,$\measuredangle PEA = \measuredangle MEC = \measuredangle BEN$. However$\measuredangle MEC = \measuredangle BEN$implies that$M$is the midpoint of$CD$from similar triangles, so we are done.$\square$

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

See also

2019 USAJMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6
All USAJMO Problems and Solutions
Invalid username
Login to AoPS