Difference between revisions of "2005 AMC 10A Problems/Problem 25"

Revision as of 21:54, 28 December 2019

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 $BCED$ ?

$\mathrm{(A) \ } \frac{266}{1521}\qquad \mathrm{(B) \ } \frac{19}{75}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{19}{56}\qquad \mathrm{(E) \ } 1$

Solution 1(no trig)

We have that $\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC} = \frac{19}{25} \cdot \frac{14}{42} = \frac{19}{75}.$

$[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label("$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 $\begin{align*} \frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\ &= \frac{1}{[ABC]/[ADE] - 1} \\ &= \frac{1}{75/19 - 1} \\ &= \boxed{\frac{19}{56}\Longrightarrow D}. \end{align*}$

Solution 2(no trig)

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

-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 $\frac{133\sin A}{392\sin A}=\frac{19}{56}\Longrightarrow \mathrm{(D)}$

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

Solution 4

Let $F$ be on $AC$ such that $DE\parallel BF$ then we have $\frac{[ABF]}{[ADE]}=(\frac{AD}{AB})^2=(\frac{19}{25})^2=\frac{361}{625}$ $$ (Error compiling LaTeX. Unknown error_msg)\frac{[ADE]}{[DECB]}= Since $\bigtriangleup ADE\sim\bigtriangleup ABF$ we have $\frac{AD}{AE}=\frac{DB}{EF}\Longrightarrow EF=\frac{84}{19}$ Thus $FC=EC-EF=\frac{448}{19}$ and $\frac{[ABF]}{[BFC]}=\frac{AF}{FC}=\frac{350}{448}$ Finally, $$ (Error compiling LaTeX. Unknown error_msg)\frac{[ADE]}{[DECB]}=\frac{

@@ Line 77: / Line 77: @@
 Let <math>F</math> be on <math>AC</math> such that <math>DE\parallel BF</math> then we have
 <cmath>\frac{[ABF]}{[ADE]}=(\frac{AD}{AB})^2=(\frac{19}{25})^2=\frac{361}{625}</cmath>
-Since <math>\bigtriangleup ADE\sim\bigtriangleup ABF</math>,
+<math></math>\frac{[ADE]}{[DECB]}=
+Since <math>\bigtriangleup ADE\sim\bigtriangleup ABF</math> we have
 <cmath>\frac{AD}{AE}=\frac{DB}{EF}\Longrightarrow EF=\frac{84}{19}</cmath>
+Thus <math>FC=EC-EF=\frac{448}{19}</math> and
+<cmath>\frac{[ABF]}{[BFC]}=\frac{AF}{FC}=\frac{350}{448}</cmath>
+Finally,
+<math></math>\frac{[ADE]}{[DECB]}=\frac{
 ==See also==

2005 AMC 10A (Problems • Answer Key • Resources)
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