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 $DP = BP\cdot\frac{PC}{PA} = \frac{1}{2} BP$ $EP = BP\cdot\frac{PA}{PC} = 2BP$ So $DE = \frac{3}{2}BP = \boxed{\textbf{(D) }12\sqrt2}$

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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2021 AMC 12B Problems/Problem 11

Contents

Problem

Solutions

Solution 1 (cheese)

Solution 2

See Also