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

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==Problem==
 
==Problem==
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<math>(*)</math>  Let <math>ABCD</math> be a cyclic quadrilateral satisfying <math>AD^2+BC^2=AB^2</math>. The diagonals of <math>ABCD</math> intersect at <math>E</math>. Let <math>P</math> be a point on side <math>\overline{AB}</math> satisfying <math>\angle APD=\angle BPC</math>. Show that line <math>PE</math> bisects <math>\overline{CD}</math>.
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Let <math>ABCD</math> be a cyclic quadrilateral satisfying <math>AD^2+BC^2=AB^2</math>. The diagonals of <math>ABCD</math> intersect at <math>E</math>. Let <math>P</math> be a point on side <math>\overline{AB}</math> satisfying <math>\angle APD=\angle BPC</math>. Show that line <math>PE</math> bisects <math>\overline{CD}</math>.
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==Solution 1==
  
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==Solution==
 
 
Let <math>PE \cap DC = M</math>. Also, let <math>N</math> be the midpoint of <math>AB</math>.
 
Let <math>PE \cap DC = M</math>. Also, let <math>N</math> be the midpoint of <math>AB</math>.
  
 
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:
  
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1) <math>AP' \cdot AB = AD^2</math>
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<math>AP' \cdot AB = AD^2 \quad \text{and} \quad BP' \cdot AB = CD^2</math>
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2) <math>BP' \cdot AB = CD^2</math>
 
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Claim:<math>P = P'</math>
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Claim: <math>P = P'</math>
  
 
Proof:
 
Proof:
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<cmath>\iff \left(\frac{BC}{AD} \right)^2 = \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP} </cmath>
 
<cmath>\iff \left(\frac{BC}{AD} \right)^2 = \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP} </cmath>
  
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as desired. <math>\square</math>
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as desired. <math>\blacksquare</math>
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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>\blacksquare</math>
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~sriraamster
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==Solution 2==
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By monotonicity, we can see that the point <math>P</math> is unique. Therefore, if we find another point <math>P'</math> with all the same properties as <math>P</math>, then <math>P = P'</math>
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Part 1) Let <math>N</math> be a point on <math>\overline{AB}</math> such that <math>AN\cdot AB = AD^2</math>, and <math>BN \cdot AB = BC^2</math>. Obviously <math>N</math> exists because adding the two equations gives <math>AN\cdot AB + BN \cdot AB = AD^2 + BC^2 = AB^2</math>, which is the problem statement. Notice that converse PoP gives<cmath>AN\cdot AB = AD^2 \implies \bigtriangleup ADN \sim \bigtriangleup ABD</cmath><cmath>BN\cdot AB = AD^2 \implies \bigtriangleup CBN \sim \bigtriangleup ABC</cmath>Therefore, <math>\angle AND = \angle ADB = \angle ACB = \angle CNB</math>, so <math>N</math> does indeed satisfy all the conditions <math>P</math> does, so <math>N = P</math>. Hence, <math>\bigtriangleup ADP \sim \bigtriangleup ABD</math> and <math>\bigtriangleup CBP \sim\bigtriangleup ABC</math>.
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Part 2) Define <math>G</math> as the midpoint of <math>CD</math>. Furthermore, create a point <math>X</math> such that <math>DX || EC</math> and <math>CX || ED</math>. Obviously <math>XCED</math> must be a parallelogram. Now we set up for Jacobi's. The problem already gives us that <math>\angle APD = \angle CPB</math>, which is good for starters. Furthermore, <math>\bigtriangleup ADP \sim \bigtriangleup ABD</math> tells us that<cmath>\angle ADP = \angle ABD = \angle ACD = \angle XDC</cmath>This gives us our second needed angle equivalence. Lastly, <math>\bigtriangleup CBP \sim\bigtriangleup ABC</math> will give<cmath>\angle BCP = \angle BAC = \angle BDC = \angle XCD</cmath>which is our last necessary angle equivalence to apply Jacobi's. Finally, applying Jacobi's tells us that <math>AC</math>, <math>BD</math>, and <math>XP</math> are concurrent <math>\implies</math> <math>X</math>, <math>E</math>, <math>P</math> collinear. Additionally, since parallelogram diagonals bisect each other, <math>X</math>, <math>G</math>, and <math>E</math> are collinear, so finally we obtain that <math>P</math>, <math>E</math>, and <math>G</math> are collinear, as desired.
  
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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>
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-jj_ca888
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:13, 12 April 2021

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 1

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:

$AP' \cdot AB = AD^2 \quad \text{and} \quad BP' \cdot AB = CD^2$

Claim: $P = P'$

Proof:

The conditions imply the similarities $ADP \sim ABD$ 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

\[AP \cdot AB = AD^2 \iff AB^2 \cdot AP = AD^2 \cdot AB\] \[\iff \left( \frac{AB}{AD} \right)^2 = \frac{AB}{AP}\] \[\iff \left( \frac{AB}{AD} \right)^2 - 1 = \frac{AB}{AP} - 1\] \[\iff \frac{AB^2 - AD^2}{AD^2} = \frac{BP}{AP}\] \[\iff \left(\frac{BC}{AD} \right)^2 = \left(\frac{BE}{AE} \right)^2 = \frac{BP}{AP}\]

as desired. $\blacksquare$

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. $\blacksquare$

~sriraamster

Solution 2

By monotonicity, we can see that the point $P$ is unique. Therefore, if we find another point $P'$ with all the same properties as $P$, then $P = P'$

Part 1) Let $N$ be a point on $\overline{AB}$ such that $AN\cdot AB = AD^2$, and $BN \cdot AB = BC^2$. Obviously $N$ exists because adding the two equations gives $AN\cdot AB + BN \cdot AB = AD^2 + BC^2 = AB^2$, which is the problem statement. Notice that converse PoP gives\[AN\cdot AB = AD^2 \implies \bigtriangleup ADN \sim \bigtriangleup ABD\]\[BN\cdot AB = AD^2 \implies \bigtriangleup CBN \sim \bigtriangleup ABC\]Therefore, $\angle AND = \angle ADB = \angle ACB = \angle CNB$, so $N$ does indeed satisfy all the conditions $P$ does, so $N = P$. Hence, $\bigtriangleup ADP \sim \bigtriangleup ABD$ and $\bigtriangleup CBP \sim\bigtriangleup ABC$.

Part 2) Define $G$ as the midpoint of $CD$. Furthermore, create a point $X$ such that $DX || EC$ and $CX || ED$. Obviously $XCED$ must be a parallelogram. Now we set up for Jacobi's. The problem already gives us that $\angle APD = \angle CPB$, which is good for starters. Furthermore, $\bigtriangleup ADP \sim \bigtriangleup ABD$ tells us that\[\angle ADP = \angle ABD = \angle ACD = \angle XDC\]This gives us our second needed angle equivalence. Lastly, $\bigtriangleup CBP \sim\bigtriangleup ABC$ will give\[\angle BCP = \angle BAC = \angle BDC = \angle XCD\]which is our last necessary angle equivalence to apply Jacobi's. Finally, applying Jacobi's tells us that $AC$, $BD$, and $XP$ are concurrent $\implies$ $X$, $E$, $P$ collinear. Additionally, since parallelogram diagonals bisect each other, $X$, $G$, and $E$ are collinear, so finally we obtain that $P$, $E$, and $G$ are collinear, as desired.

-jj_ca888

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

See also

2019 USAJMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions
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