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

Revision as of 20:00, 13 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 Ptolemy'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 (Coordinate Bashing)

Since we're given distances and nothing else, we can represent each point as a coordinate and use the distance formula to set up a series of systems and equations. Let the height of the trapezoid be $h$ , and let the coordinates of $A$ and $D$ be at $(-a,0)$ and $(a,0)$ , respectively. Then let $B$ and $C$ be at $(-b,h)$ and $(b,h)$ , respectively. This follows the rules that this is an isosceles trapezoid since the origin is centered on the middle of $AD$ . Finally, let $P$ be located at point $(c,d)$ .

The distance from $P$ to $A$ is $1$ , so by the distance formula: $\sqrt{(c+a)^2+(d-h)^2} = 1 \implies (c+a)^2+(d-h)^2 = 1$ The distance from $P$ to $D$ is $4$ , so $\sqrt{(c-a)^2+(d-h)^2} = 1 \implies (c-a)^2+(d-h)^2 = 16$

Looking at these two equations alone, notice that the second term is the same for both equations, so we can subtract the equations. This yields $-4ac = 15$

Next, the distance from $P$ to $B$ is $2$ , so $\sqrt{(c+b)^2+(d-h)^2} = 2 \implies (c+b)^2+(d-h)^2 = 4$ The distance from $P$ to $C$ is $3$ , so $\sqrt{(c-b)^2+(d-h)^2} = 3 \implies (c-b)^2+(d-h)^2 = 9$

Again, we can subtract these equations, yielding $-4bc = 5$

We can now divide the equations to eliminate $c$ , yielding $\frac{b}{a} = \frac{5}{15} = \frac{1}{3}$

We wanted to find $\frac{BC}{AD}$ . But since $b$ is half of $BC$ and $a$ is half of $AD$ , this ratio is equal to the ratio we want.

Therefore $\frac{BC}{AD} = \frac{1}{3} = \boxed{B}$

~KingRavi

Solution 3 (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

Solution 4 (Coordinate Bashing 2)

Let the point $P$ be at the origin, and draw four concentric circles around $P$ each with radius $1$ , $2$ , $3$ , and $4$ , respectively. The vertices of the trapezoid would be then on each of the four concentric circles. WLOG, let $BC$ and $AD$ be parallel to the x-axis. Assigning coordinates to each point, we have: $A=(x_1,y_1)$ $B=(x_2,y_2)$ $C=(x_3,y_2)$ $D=(x_4,y_1)$ which satisfy the following: $x_1^2 + y_1^2 = 1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (1)$ $x_2^2 + y_2^2 = 4 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (2)$ $x_3^2 + y_2^2 = 9 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (3)$ $x_4^2 + y_1^2 = 16 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (4)$ In addition, because the trapezoid is isosceles ( $AB=CD$ ), the midpoints of the two bases would then have the same x-coordinate, giving us $x_1 + x_4 = x_2 + x_3 \;\;\;\;\;\;\;\;\;\;\;\;\; (5)$ Subtracting Equation $2$ from Equation $3$ , and Equation $1$ from Equation $4$ , we have

$x_3^2 - x_2^2 = 5 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (6)$ $x_4^2 - x_1^2 = 15 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (7)$

Dividing Equation $6$ by Equation $7$ , we have $\frac{x_3^2-x_2^2}{x_4^2-x_1^2}=\frac{1}{3}$ $\frac{(x_3-x_2)(x_3+x_2)}{(x_4-x_1)(x_4+x_1)}=\frac{1}{3}$

Cancelling $(x_3+x_2)$ and $(x_4+x_1)$ with Equation $5$ , we get

$\frac{(x_3-x_2)}{(x_4-x_1)}=\frac{1}{3}$

In other words,

$\frac{BC}{AD}=\frac{1}{3}=\boxed{B}$

~G63566

Video Solution By ThePuzzlr

https://youtu.be/KqtpaHy-eoU

~ MathIsChess

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=uYXtEzX4fb0

Video Solution

https://youtu.be/hvIOvjjQvIw

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

Video Solution by OmegaLearn using Pythagorean Theorem

https://youtu.be/jnm2alniaM4

~ pi_is_3.14

@@ Line 70: / Line 70: @@
 ~ MathIsChess
+==Video Solution by Punxsutawney Phil==
+https://youtube.com/watch?v=uYXtEzX4fb0
 ==Video Solution==

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AMC 12 Problems and Solutions

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Preceded by Problem 22	Followed by Problem 24
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