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Difference between revisions of "2021 AMC 12B Problems/Problem 11"
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==Solutions==
 
==Solutions==
  
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===Solution 1===
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===Solution 1 (cheese)===
  
 
Using Stewart's Theorem of <math>man+dad=bmb+cnc</math> calculate the cevian to be <math>8\sqrt{2}</math>. It then follows that the answer must also have a factor of the <math>\sqrt{2}</math>. Having eliminated 3 answer choices, we then proceed to draw a rudimentary semiaccurate diagram of this figure. Drawing that, we realize that <math>6\sqrt2</math> is too small making out answer <math>\boxed{\textbf{(D) }12\sqrt2}</math> ~Lopkiloinm
 
Using Stewart's Theorem of <math>man+dad=bmb+cnc</math> calculate the cevian to be <math>8\sqrt{2}</math>. It then follows that the answer must also have a factor of the <math>\sqrt{2}</math>. Having eliminated 3 answer choices, we then proceed to draw a rudimentary semiaccurate diagram of this figure. Drawing that, we realize that <math>6\sqrt2</math> is too small making out answer <math>\boxed{\textbf{(D) }12\sqrt2}</math> ~Lopkiloinm
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 +
===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
 +
<cmath>DP = BP\cdot\frac{PC}{PA} = \frac{1}{2} BP</cmath>
 +
<cmath>EP = BP\cdot\frac{PA}{PC} = 2BP</cmath>
 +
So
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<cmath>DE = \frac{3}{2}BP = \boxed{\textbf{(D) }12\sqrt2}</cmath>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021|ab=B|num-b=10|num-a=12}}
 
{{AMC12 box|year=2021|ab=B|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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 $BPA~DPC$ and $BPC~EPA$ 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 (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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