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Difference between revisions of "2022 AMC 12A Problems/Problem 20"

(Created page with "==Problem== Isosceles trapezoid <math>ABCD</math> has parallel sides <math>\overline{AD}</math> and <math>\overline{BC},</math> with <math>BC < AD</math> and <math>AB = CD.</m...")
 
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==Solution==
 
==Solution==
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Consider the reflection <math>P^{\prime}</math> of <math>P</math> over the perpendicular bisector of <math>\overline{BC}</math>, creating two new isosceles trapezoids <math>DAPP^{\prime}</math> and <math>CBPP^{\prime}</math>. Under this reflection, <math>P^{\prime}A=PD=4</math>, <math>P^{\prime}D=PA=1</math>, <math>P^{\prime}C=PB=2</math>, and <math>P^{\prime}B=PC=3</math>. By Ptolmey's theorem <cmath>\begin{align*} PP^{\prime}\cdot AD+1=16 \\ PP^{\prime}\cdot BC+4=9\end{align*}</cmath> Thus <math>PP^{\prime}\cdot AD=15</math> and <math>PP^{\prime}\cdot BC=5</math>; dividing these two equations yields <math>\frac{BC}{AD}=\boxed{\textbf{(B)}~\frac{1}{3}}</math>.
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Consider the reflection <math>P^{\prime}</math> of <math>P</math> over the perpendicular bisector of <math>\overline{BC}</math>, creating two new isosceles trapezoids <math>DAPP^{\prime}</math> and <math>CBPP^{\prime}</math>. Under this reflection, <math>P^{\prime}A=PD=4</math>, <math>P^{\prime}D=PA=1</math>, <math>P^{\prime}C=PB=2</math>, and <math>P^{\prime}B=PC=3</math>. By Ptolmey's theorem <cmath>\begin{align*} PP^{\prime}\cdot AD+1=16 \\ PP^{\prime}\cdot BC+4=9\end{align*}</cmath> Thus <math>PP^{\prime}\cdot AD=15</math> and <math>PP^{\prime}\cdot BC=5</math>; dividing these two equations and taking the reciprocal yields <math>\frac{BC}{AD}=\boxed{\textbf{(B)}~\frac{1}{3}}</math>.
  
 
==See also==
 
==See also==

Revision as of 20:02, 11 November 2022

Problem

Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC},$ with $BC < AD$ and $AB = CD.$ There is a point $P$ in the plane such that $PA=1, PB=2, PC=3,$ and $PD=4.$ What is $\tfrac{BC}{AD}?$

$\textbf{(A) }\frac{1}{4}\qquad\textbf{(B) }\frac{1}{3}\qquad\textbf{(C) }\frac{1}{2}\qquad\textbf{(D) }\frac{2}{3}\qquad\textbf{(E) }\frac{3}{4}$

Solution

Consider the reflection $P^{\prime}$ of $P$ over the perpendicular bisector of $\overline{BC}$, creating two new isosceles trapezoids $DAPP^{\prime}$ and $CBPP^{\prime}$. Under this reflection, $P^{\prime}A=PD=4$, $P^{\prime}D=PA=1$, $P^{\prime}C=PB=2$, and $P^{\prime}B=PC=3$. By Ptolmey's theorem \begin{align*} PP^{\prime}\cdot AD+1=16 \\ PP^{\prime}\cdot BC+4=9\end{align*} Thus $PP^{\prime}\cdot AD=15$ and $PP^{\prime}\cdot BC=5$; dividing these two equations and taking the reciprocal yields $\frac{BC}{AD}=\boxed{\textbf{(B)}~\frac{1}{3}}$.

See also

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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