2021 AMC 12B Problems/Problem 11

Problem

Triangle $ABC$ has $AB=13,BC=14$ and $AC=15$ . Let $P$ be the point on $\overline{AC}$ such that $PC=10$ . There are exactly two points $D$ and $E$ on line $BP$ such that quadrilaterals $ABCD$ and $ABCE$ are trapezoids. What is the distance $DE?$

$\textbf{(A) }\frac{42}5 \qquad \textbf{(B) }6\sqrt2 \qquad \textbf{(C) }\frac{84}5\qquad \textbf{(D) }12\sqrt2 \qquad \textbf{(E) }18$

Diagram

$[asy] size(8cm); pair A = (5,12); pair B = (0,0); pair C = (14,0); pair P = 2/3*A+1/3*C; pair D = 3/2*P; pair E = 3*P; draw(A--B--C--A); draw(A--D); draw(C--E--B); dot("$A$",A,N); dot("$B$",B,W); dot("$C$",C,ESE); dot("$D$",D,N); dot("$P$",P,W); dot("$E$",E,N); defaultpen(fontsize(9pt)); label("$13$", (A+B)/2, NW); label("$14$", (B+C)/2, S); label("$5$",(A+P)/2, NE); label("$10$", (C+P)/2, NE); [/asy]$

Solution 1 (analytic geometry)

Toss on the Cartesian plane with $A=(5, 12), B=(0, 0),$ and $C=(14, 0)$ . Then $\overline{AD}\parallel\overline{BC}, \overline{AB}\parallel\overline{CE}$ by the trapezoid condition, where $D, E\in\overline{BP}$ . Since $PC=10$ , point $P$ is $\tfrac{10}{15}=\tfrac{2}{3}$ of the way from $C$ to $A$ and is located at $(8, 8)$ . Thus line $BP$ has equation $y=x$ . Since $\overline{AD}\parallel\overline{BC}$ and $\overline{BC}$ is parallel to the ground, we know $D$ has the same $y$ -coordinate as $A$ , except it'll also lie on the line $y=x$ . Therefore, $D=(12, 12). \, \blacksquare$

To find the location of point $E$ , we need to find the intersection of $y=x$ with a line parallel to $\overline{AB}$ passing through $C$ . The slope of this line is the same as the slope of $\overline{AB}$ , or $\tfrac{12}{5}$ , and has equation $y=\tfrac{12}{5}x-\tfrac{168}{5}$ . The intersection of this line with $y=x$ is $(24, 24)$ . Therefore point $E$ is located at $(24, 24). \, \blacksquare$

The distance $DE$ is equal to the distance between $(12, 12)$ and $(24, 24)$ , which is $\boxed{\textbf{(D)} ~12\sqrt{2}}$

Solution 2

Using Stewart's Theorem we find $BP = 8\sqrt{2}$ . From the similar triangles $BPA\sim DPC$ and $BPC\sim EPA$ we have $DP = BP\cdot\frac{PC}{PA} = 2BP$ $EP = BP\cdot\frac{PA}{PC} = \frac12 BP$ So $DE = \frac{3}{2}BP = \boxed{\textbf{(D) }12\sqrt2}$

Solution 3

Let $x$ be the length $PE$ . From the similar triangles $BPA\sim DPC$ and $BPC\sim EPA$ we have $BP = \frac{PA}{PC}x = \frac12 x$ $PD = \frac{PA}{PC}BP = \frac14 x$ Therefore $BD = DE = \frac{3}{4}x$ . Now extend line $CD$ to the point $Z$ on $AE$ , forming parallelogram $ZABC$ . As $BD = DE$ we also have $EZ = ZC = 13$ so $EC = 26$ .

We now use the Law of Cosines to find $x$ (the length of $PE$ ): $x^2 = EC^2 + PC^2 - 2(EC)(PC)\cos{(PCE)} = 26^2 + 10^2 - 2\cdot 26\cdot 10\cos(\angle PCE)$ As $\angle PCE = \angle BAC$ , we have (by Law of Cosines on triangle $BAC$ ) $\cos(\angle PCE) = \frac{13^2 + 15^2 - 14^2}{2\cdot 13\cdot 15}.$ Therefore $\begin{align*} x^2 &= 26^2 + 10^2 - 2\cdot 26\cdot 10\cdot\frac{198}{2\cdot 13\cdot 15}\\ &= 776 - 264\\ &= 512 \end{align*}$ And $x = 16\sqrt2$ . The answer is then $\frac34x = \boxed{\textbf{(D) }12\sqrt2}$

Video Solution by Punxsutawney Phil

https://YouTube.com/watch?v=yxt8-rUUosI&t=450s

Video Solution by OmegaLearn (Using properties of 13-14-15 triangle)

https://youtu.be/mTcdKf5-FWg

~ pi_is_3.14

Video Solution by Hawk Math

https://www.youtube.com/watch?v=p4iCAZRUESs

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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All AMC 12 Problems and Solutions

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