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

Revision as of 19:54, 12 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 $DE?$

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

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 (fakesolve)

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

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

@@ Line 74: / Line 74: @@
 ==See Also==
 {{AMC12 box|year=2021|ab=B|num-b=10|num-a=12}}
+[[Category:Intermediate Geometry Problems]]
 {{MAA Notice}}

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