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

Revision as of 03:00, 12 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 1

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 $\frac{BC}{AD}=\boxed{\textbf{(B)}~\frac{1}{3}}$ .

Solution 2 (Cheese)

Notice that the question never says what the height of the trapezoid is; the only property we know about it is that $AC=BD$ . Therefore, we can say WLOG that the height of the trapezoid is $0$ and all $5$ points, including $P$ , lie on the same line with $PA=AB=BC=CD=1$ . Notice that this satisfies the problem requirements because $PA=1, PB=2, PC=3,PD=4$ , and $AC=BD=2$ . Now all we have to find is $\frac{BC}{AD} = \frac{1}{3}= \boxed{B}$

~KingRavi

Video Solution

https://youtu.be/hvIOvjjQvIw

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by OmegaLearn

https://youtu.be/jnm2alniaM4

~ pi_is_3.14

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

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

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