Difference between revisions of "2021 AMC 12B Problems/Problem 11"

(Solutions)
(Solution 2)
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===Solution 2===
 
===Solution 2===
Using Stewart's Theorem we find <math>BP = 8\sqrt{2}</math>. From the similar triangles <math>BPA~DPC</math> and <math>BPC~EPA</math> we have
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Using Stewart's Theorem we find <math>BP = 8\sqrt{2}</math>. From the similar triangles <math>BPA\sim DPC</math> and <math>BPC\sim EPA</math> we have
 
<cmath>DP = BP\cdot\frac{PC}{PA} = \frac{1}{2} BP</cmath>
 
<cmath>DP = BP\cdot\frac{PC}{PA} = \frac{1}{2} BP</cmath>
 
<cmath>EP = BP\cdot\frac{PA}{PC} = 2BP</cmath>
 
<cmath>EP = BP\cdot\frac{PA}{PC} = 2BP</cmath>

Revision as of 20:17, 11 February 2021

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 \[DE = \frac{3}{2}BP = \boxed{\textbf{(D) }12\sqrt2}\]

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
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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