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

(→‎Solution 4)
(→‎Solution 4)
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Let <math>F</math> be on <math>AC</math> such that <math>DE\parallel BF</math> then we have
 
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>
 
<cmath>\frac{[ABF]}{[ADE]}=(\frac{AD}{AB})^2=(\frac{19}{25})^2=\frac{361}{625}</cmath>
−
Since <math>\bigtriangleup ADE\sim\Bigtriangleup ABF</math>,  
+
Since <math>\bigtriangleup ADE\sim\bigtriangleup ABF</math>,  
 
<cmath>\frac{AD}{AE}=\frac{DB}{EF}\Longrightarrow EF=\frac{84}{19}</cmath>
 
<cmath>\frac{AD}{AE}=\frac{DB}{EF}\Longrightarrow EF=\frac{84}{19}</cmath>
  

Revision as of 21:49, 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 \[\frac{[ABF]}{[ADE]}=(\frac{AD}{AB})^2=(\frac{19}{25})^2=\frac{361}{625}\] Since $\bigtriangleup ADE\sim\bigtriangleup ABF$, \[\frac{AD}{AE}=\frac{DB}{EF}\Longrightarrow EF=\frac{84}{19}\]

See also

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