# 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 $DE?$

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

## 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 (Analytic Geometry)

Toss on the Cartesian plane with $A=(5, 12), B=(0, 0),$ and $C=(14, 0)$. Then $\overline{AD}\parallel\overline{BC}, \overline{AB}\parallel\overline{CE}$ by the trapezoid condition, where $D, E\in\overline{BP}$. Since $PC=10$, point $P$ is $\tfrac{10}{15}=\tfrac{2}{3}$ of the way from $C$ to $A$ and is located at $(8, 8)$. Thus line $BP$ has equation $y=x$. Since $\overline{AD}\parallel\overline{BC}$ and $\overline{BC}$ is parallel to the ground, we know $D$ has the same $y$-coordinate as $A$, except it'll also lie on the line $y=x$. Therefore, $D=(12, 12). \, \blacksquare$

To find the location of point $E$, we need to find the intersection of $y=x$ with a line parallel to $\overline{AB}$ passing through $C$. The slope of this line is the same as the slope of $\overline{AB}$, or $\tfrac{12}{5}$, and has equation $y=\tfrac{12}{5}x-\tfrac{168}{5}$. The intersection of this line with $y=x$ is $(24, 24)$. Therefore point $E$ is located at $(24, 24). \, \blacksquare$

The distance $DE$ is equal to the distance between $(12, 12)$ and $(24, 24)$, which is $\boxed{\textbf{(D)} ~12\sqrt{2}}$

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

## Solution 4 (Heron's Formula, Pythagorean Theorem, Similar Triangles)

Let the brackets denote areas. By Heron's Formula, we have \begin{align*} [ABC]&=\sqrt{\frac{13+14+15}{2}\left(\frac{13+14+15}{2}-13\right)\left(\frac{13+14+15}{2}-14\right)\left(\frac{13+14+15}{2}-15\right)} \\ &=\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)} \\ &=\sqrt{21\left(8\right)\left(7\right)\left(6\right)} \\ &=\sqrt{\left(3\cdot7\right)\left(2^3\right)\left(7\right)\left(2\cdot3\right)} \\ &=2^2\cdot3\cdot7 \\ &=84. \end{align*} It follows that the height of $ABCD$ is $\frac{2[ABC]}{14}=12.$

Next, we drop the altitudes $\overline{AF}$ and $\overline{DG}$ of $ABCD.$ By the Pythagorean Theorem on right $\triangle AFB,$ we get $BF=5.$ By the AA Similarity, we have $\triangle ADP\sim\triangle CBP,$ with the ratio of similitude $1:2.$ It follows that $AD=7.$ Since $ADGF$ is a rectangle, we get $FG=AD=7.$ By the Pythagorean Theorem on right $\triangle DGB,$ we get $BD=12\sqrt2.$

By $\triangle ADP\sim\triangle CBP$ again, we get $BP=8\sqrt2$ and $DP=4\sqrt2.$ Also, by the AA Similarity, we have $\triangle ABP\sim\triangle CEP,$ with the ratio of similitude $1:2.$ It follows that $EP=16\sqrt2.$

Finally, we obtain $DE=EP-DP=\boxed{\textbf{(D) }12\sqrt2}.$

Remark

If you memorize that the area of a $13\text{-}14\text{-}15$ triangle is $84,$ then the Heron's Formula part will be done instantly.

~MRENTHUSIASM

## Solution 5 (Barycentric Coordinates)

(For those unfamiliar with barycentric coordinates, consider reading the barycentric coordinates article written by Evan Chen and Max Schindler here: https://artofproblemsolving.com/wiki/index.php/Resources_for_mathematics_competitions#Articles)

We can find $P$ in barycentric coordinates as $\Bigr(\frac{2}{3},0,\frac{1}{3}\Bigr)$. We can then write $\overline{BP}$ as $x-2z=0$, where $(x,y,z)$ defines a point in barycentric coordinates. We have $\overline{AD}\parallel \overline{BC}$ as $y+z=0$ and $\overline{CE}\parallel \overline{AB}$ as $x+y=0$. We can then compute $D$ and $E$ by intersecting lines:

$$\begin{cases} x-2z=0\\ y+z=0\\ x+y+z=1 \end{cases}$$

Which gives us $D=(1, -\frac{1}{2}, \frac{1}{2})$. We can get $E$ with:

$$\begin{cases} x-2z=0\\ x+y=0\\ x+y+z=1 \end{cases}$$

Which gives us $E=(2, -2, 1)$. Then, finding the displacement vector, we have $\overrightarrow{ED}=(1,-\frac{3}{2},\frac{1}{2})$. Using the barycentric distance formula:

$$\text{dist}(D,E)=\sqrt{-a^2yx-b^2zx-c^2xy}$$ $$=\sqrt{-14^2\Bigr(-\frac{3}{2}\Bigr)\Bigr(\frac{1}{2}\Bigr)-15^2\Bigr(1\Bigr)\Bigr(\frac{1}{2}\Bigr)-13^2\Bigr(-\frac{3}{2}\Bigr)\Bigr(\frac{1}{2}\Bigr)}$$ We get $\boxed{\textbf{(D)}12\sqrt{2}}$

~ pi_is_3.14

## Video Solution by Hawk Math

 2021 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 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