# 2005 AMC 10A Problems/Problem 25

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## Problem

In $ABC$ we have $AB = 25$, $BC = 39$, and $AC=42$. Points $D$ and $E$ are on $AB$ and $AC$ respectively, with $AD = 19$ and $AE = 14$. What is the ratio of the area of triangle $ADE$ to the area of the quadrilateral $BCED$? $\textbf{(A) } \frac{266}{1521}\qquad \textbf{(B) } \frac{19}{75}\qquad \textbf{(C) }\frac{1}{3}\qquad \textbf{(D) } \frac{19}{56}\qquad \textbf{(E) } 1$

## Solution 1

We have $$\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC} = \frac{19}{25} \cdot \frac{14}{42} = \frac{19}{75}.$$ $[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label("A", A, N); label("B", B, SW); label("C", C, SE); label("D", D, W); label("E", E, NE); label("19", (A + D)/2, W); label("6", (B + D)/2, W); label("14", (A + E)/2, NE); label("28", (C + E)/2, NE); [/asy]$

But $[BCED] = [ABC] - [ADE]$, so \begin{align*} \frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\ &= \frac{1}{\frac{ABC}{ADE} - 1} \\ &= \frac{1}{\frac{75}{19} - 1} \\ &= \boxed{\textbf{(D) }\frac{19}{56}}. \end{align*}

Note: If it is hard to understand why $$\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC}$$, you can use the fact that the area of a triangle equals $\frac{1}{2} \cdot ab \cdot \sin(C)$. If angle $DAE = Z$, we have that $$\frac{[ADE]}{[ABC]} = \frac{\frac{1}{2} \cdot 19 \cdot 14 \cdot \sin(Z)}{\frac{1}{2} \cdot 25 \cdot 42 \cdot \sin(Z)} = \frac{19 \cdot 14}{25 \cdot 42} = \frac{ab}{cd}$$. - SuperJJ

## Video Solution

CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00

## Solution 2 (no trig) $[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label("A", A, N); label("B", B, SW); label("C", C, SE); label("D", D, W); label("E", E, NE); label("19", (A + D)/2, W); label("6", (B + D)/2, W); label("14", (A + E)/2, NE); label("28", (C + E)/2, NE); [/asy]$

We can let $[ADE]=x$. Since $EC=2 \cdot EA$, $[DEC]=2x$. So, $[ADC]=3x$. This means that $[BDC]=\frac{6}{19}\cdot3x=\frac{18x}{19}$. Thus, $\frac{[ADE]}{[BCED]} = \frac{x}{\frac{18x}{19}+2x}= \boxed{\textbf{(D) }\frac{19}{56}}.$

-Conantwiz2023

## Solution 3 (trig)

The area of a triangle is $\frac{1}{2} bc\sin A$.

Using this formula: $[ADE]=\frac{1}{2}\cdot19\cdot14\cdot\sin A = 133\sin A$ $[ABC]=\frac{1}{2}\cdot25\cdot42\cdot\sin A = 525\sin A$

Since the area of $BCED$ is equal to the area of $ABC$ minus the area of $ADE$, $[BCED] = 525\sin A - 133\sin A = 392\sin A$.

Therefore, the desired ratio is $\frac{133\sin A}{392\sin A}=\boxed{\textbf{(D) }\frac{19}{56}}$

Note: $BC=39$ was not used in this problem.

## Solution 4

Let $F$ be on $AC$ such that $DE\parallel BF$ then we have $$\frac{[ADE]}{[ABF]}=\left(\frac{AD}{AB}\right)^2=\left(\frac{19}{25}\right)^2=\frac{361}{625}$$ $$\frac{[ADE]}{[DEFB]}=\frac{361}{625-361}=\frac{361}{364}$$ Since $\bigtriangleup ADE\sim\bigtriangleup ABF$ we have $$\frac{AD}{AE}=\frac{DB}{EF}\Longrightarrow EF=\frac{84}{19}$$ Thus $FC=EC-EF=\frac{448}{19}$ and $$\frac{[ABF]}{[BFC]}=\frac{AF}{FC}=\frac{350}{448}$$ $$\frac{[ADE]}{[BFC]}=\left(\frac{[ADE]}{[ABF]}\right)\left(\frac{[ABF]}{[BFC]}\right)=\left(\frac{361}{625}\right)\left(\frac{350}{448}\right)=\frac{126350}{280000}$$ Finally, after some calculations, $$\frac{[ADE]}{[DECB]}=\frac{[ADE]}{[BFC]+[DECB]}=\boxed{\textbf{(D) \ } \frac{19}{56}}$$.

~ Nafer

~ LaTeX changes by tkfun

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