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 $DE?$

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

Solutions

Solution 1 (cheese)

Using Stewart's Theorem of $man+dad=bmb+cnc$ calculate the cevian to be $8\sqrt{2}$ . It then follows that the answer must also have a factor of the $\sqrt{2}$ . Having eliminated 3 answer choices, we then proceed to draw a rudimentary semiaccurate diagram of this figure. Drawing that, we realize that $6\sqrt2$ is too small making out answer $\boxed{\textbf{(D) }12\sqrt2}$ ~Lopkiloinm

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 $\frac34x = \boxed{\textbf{(D) }12\sqrt2}$

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

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