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

Revision as of 21:40, 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 $DP = BP\cdot\frac{PC}{PA} = 2BP$ $EP = BP\cdot\frac{PA}{PC} = \frac12 BP$ 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 $\frac34x = \boxed{\textbf{(D) }12\sqrt2}$

@@ Line 12: / Line 12: @@
 ===Solution 2===
 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} = 2BP</cmath>
-<cmath>EP = BP\cdot\frac{PA}{PC} = 2BP</cmath>
+<cmath>EP = BP\cdot\frac{PA}{PC} = \frac12 BP</cmath>
 So
 <cmath>DE = \frac{3}{2}BP = \boxed{\textbf{(D) }12\sqrt2}</cmath>
+===Solution 3===
+Let <math>x</math> be the length <math>PE</math>. From the similar triangles <math>BPA\sim DPC</math> and <math>BPC\sim EPA</math> we have
+<cmath>BP = \frac{PA}{PC}x = \frac12 x</cmath>
+<cmath>PD = \frac{PA}{PC}BP = \frac14 x</cmath>
+Therefore <math>BD = DE = \frac{3}{4}x</math>. Now extend line <math>CD</math> to the point <math>Z</math> on <math>AE</math>, forming parallelogram <math>ZABC</math>. As <math>BD = DE</math> we also have <math>EZ = ZC = 13</math> so <math>EC = 26</math>.
+We now use the Law of Cosines to find <math>x</math> (the length of <math>PE</math>):
+<cmath>x^2 = EC^2 + PC^2 - 2(EC)(PC)\cos{(PCE)} = 26^2 + 10^2 - 2\cdot 26\cdot 10\cos(\angle PCE)</cmath>
+As <math>\angle PCE = \angle BAC</math>, we have (by Law of Cosines on triangle <math>BAC</math>)
+<cmath>\cos(\angle PCE) = \frac{13^2 + 15^2 - 14^2}{2\cdot 13\cdot 15}.</cmath>
+Therefore
+<cmath>\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*}</cmath>
+And <math>x = 16\sqrt2</math>. The answer is then <math>\frac34x = \boxed{\textbf{(D) }12\sqrt2}</math>
 ==See Also==
 {{AMC12 box|year=2021|ab=B|num-b=10|num-a=12}}
 {{MAA Notice}}

2021 AMC 12B (Problems • Answer Key • Resources)
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