# 2021 Fall AMC 12A Problems/Problem 21

## Problem

Let $ABCD$ be an isosceles trapezoid with $\overline{BC}\parallel \overline{AD}$ and $AB=CD$. Points $X$ and $Y$ lie on diagonal $\overline{AC}$ with $X$ between $A$ and $Y$, as shown in the figure. Suppose $\angle AXD = \angle BYC = 90^\circ$, $AX = 3$, $XY = 1$, and $YC = 2$. What is the area of $ABCD?$

$[asy] size(10cm); usepackage("mathptmx"); import geometry; void perp(picture pic=currentpicture, pair O, pair M, pair B, real size=5, pen p=currentpen, filltype filltype = NoFill){ perpendicularmark(pic, M,unit(unit(O-M)+unit(B-M)),size,p,filltype); } pen p=black+linewidth(1),q=black+linewidth(5); pair C=(0,0),Y=(2,0),X=(3,0),A=(6,0),B=(2,sqrt(5.6)),D=(3,-sqrt(12.6)); draw(A--B--C--D--cycle,p); draw(A--C,p); draw(B--Y,p); draw(D--X,p); dot(A,q); dot(B,q); dot(C,q); dot(D,q); dot(X,q); dot(Y,q); label("2",C--Y,S); label("1",Y--X,S); label("3",X--A,S); label("A",A,2*E); label("B",B,2*N); label("C",C,2*W); label("D",D,2*S); label("Y",Y,2*sqrt(2)*NE); label("X",X,2*N); perp(B,Y,C,8,p); perp(A,X,D,8,p); [/asy]$ $\textbf{(A)}\: 15\qquad\textbf{(B)} \: 5\sqrt{11}\qquad\textbf{(C)} \: 3\sqrt{35}\qquad\textbf{(D)} \: 18\qquad\textbf{(E)} \: 7\sqrt{7}$

## Solution 1

First realize that $\triangle BCY \sim \triangle DAX.$ Thus, because $CY: XA = 2:3,$ we can say that $BY = 2s$ and $DX = 3s.$ From the Pythagorean Theorem, we have $AB^2 =(2s)^2 + 4^2 = 4s^2 + 16$ and $CD^2 = (3s)^2 + 3^2 = 9s^2 + 9.$ Because $AB = CD,$ from the problem statement, we have that $$4s^2 + 16 = 9s^2 + 9.$$ Solving, gives $s = \frac{\sqrt{7}}{\sqrt{5}}.$ To find the area of the trapezoid, we can compute the area of $\triangle ABC$ and add it to the area of $\triangle ACD.$ Thus, the area of the trapezoid is $\frac{1}{2} \cdot 2 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 + \left(\frac{1}{2} \cdot 3 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 \right) = \frac{15\sqrt{7}}{\sqrt{5}} = 3\sqrt{35}.$ Thus, the answer is $\boxed{\textbf{(C)} \: 3\sqrt{35}}.$

~NH14

## Solution 2

We put the graph to a coordinate plane. We put $X$ to the origin, $AC$ to the $x$-axis, and $DX$ to the $y$-axis.

We have the coordinates of the following points: $X = \left( 0 , 0 \right)$, $A = \left( 3 , 0 \right)$, $Y = \left( - 1 , 0 \right)$, $C = \left( - 3 , 0 \right)$.

Denote $BY = u$, $DX = v$. Hence, $B = \left( - 1 , u \right)$, $D = \left( 0 , - v \right)$.

Hence, the slope of $BC$ is $m_{BC} = \frac{u}{2}$. The slope of $AD$ is $m_{AD} = \frac{v}{3}$.

Because $BC \parallel AD$, $m_{BC} = m_{AD}$. Hence, $$\frac{u}{2} = \frac{v}{3} . \hspace{1cm} (1)$$

In $\triangle AYB$, following from the Pythagorean theorem, we have $AB^2 = AY^2 + BY^2 = 4^2 + u^2$.

In $\triangle CXD$, following from the Pythagorean theorem, we have $CD^2 = CX^2 + XD^2 = 3^2 + v^2$.

Because $AB = CD$, $$4^2 + u^2 = 3^2 + v^2 . \hspace{1cm} (2)$$

Solving Equations (1) and (2), we get $u = \frac{2 \sqrt{35}}{5}$, $V = \frac{3 \sqrt{35}}{5}$.

Therefore, \begin{align*} {\rm Area} \ ABCD & = {\rm Area} \ \triangle ABC + {\rm Area} \ \triangle ADC \\ & = \frac{1}{2} AC \cdot BY + \frac{1}{2} AC \cdot DX \\ & = \frac{1}{2} AC \left( BY + DY \right) \\ & = 3 \sqrt{35} . \end{align*}

Therefore, the answer is $\boxed{\textbf{(C)} \: 3\sqrt{35}}$.

~Steven Chen (www.professorchenedu.com)

## Solution 3

Since both $\overline{BY}$ and $\overline{DX}$ are perpendicular to $\overline{AC}$, we have $BD^2=XY^2+(BY+DX)^2$.

Since the trapezoid is equilateral, $BD=AC=6$. Hence, $BY+DX=\sqrt{6^2-1^2}=\sqrt{35}$.

Notice that both triangles $ABC$ and $ADC$ share a common base $\overline{AC}$.

Therefore, $$[ABCD]=[ABC]+[ADC]=\frac{1}{2}AC\cdot(BY+DX)=\frac{1}{2}\cdot6\cdot\sqrt{35}=\boxed{\textbf{(C)} \: 3\sqrt{35}}.$$ ~RunyangWang

## Video Solution by The Power of Logic

~math2718281828459

~IceMatrix

~ pi_is_3.14