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2004 AMC 10B Problems/Problem 18

Revision as of 21:37, 24 January 2015 by Warrenwangtennis (talk | contribs) (→‎Solution 1)

Problem

In the right triangle $\triangle ACE$, we have $AC=12$, $CE=16$, and $EA=20$. Points $B$, $D$, and $F$ are located on $AC$, $CE$, and $EA$, respectively, so that $AB=3$, $CD=4$, and $EF=5$. What is the ratio of the area of $\triangle DBF$ to that of $\triangle ACE$?

$\mathrm{(A) \ } \frac{1}{4} \qquad \mathrm{(B) \ } \frac{9}{25} \qquad \mathrm{(C) \ } \frac{3}{8} \qquad \mathrm{(D) \ } \frac{11}{25} \qquad \mathrm{(E) \ } \frac{7}{16}$


[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label("$A$",A,N); label("$B$",B,W); label("$C$",C,SW); label("$D$",D,S); label("$E$",E,SE); label("$F$",F,NE); label("$3$",A--B,W); label("$9$",C--B,W); label("$4$",C--D,S); label("$12$",D--E,S); label("$5$",E--F,NE); label("$15$",F--A,NE); [/asy]

Solution 1

Let $x = [DBF]$. Because $\triangle ACE$ is divided into four triangles, $[ACE] = [BCD] + [ABF] + [DEF] + x$.

Because of $SAS$ triangle area, $\frac12 \cdot 12 \cdot 16 = \frac12 \cdot 9 \cdot 4 + \frac12 \cdot 3 \cdot 15 \cdot \sin(\angle A) + \frac12 \cdot 5 \cdot 12 \cdot \sin(\angle E) + x$.

$\sin(\angle A) = \frac{16}{20}$ and $\sin(\angle E) = \frac{12}{20}$, so $96 = 18 + 18 + 18 + x$.

$x = 42$, so $\frac{[DBF]}{[ACE]} = \frac{42}{96} = \boxed{\textbf{(E)}\frac 7{16}}$.

Solution 2

First of all, note that $\frac{AB}{AC} = \frac{CD}{CE} = \frac{EF}{EA} = \frac 14$, and therefore $\frac{BC}{AC} = \frac{DE}{CE} = \frac{FA}{EA} = \frac 34$.

Draw the height from $F$ onto $AB$ as in the picture below:

[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label("$A$",A,N); label("$B$",B,W); label("$C$",C,SW); label("$D$",D,S); label("$E$",E,SE); label("$F$",F,NE); label("$3$",A--B,W); label("$9$",0.5*C + 0.5*B,4*W); label("$4$",C--D,S); label("$12$",D--E,S); label("$5$",E--F,NE); label("$15$",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label("$G$",G,W); [/asy]

Now consider the area of $\triangle ABF$. Clearly the triangles $\triangle AFG$ and $\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\frac {AF}{AE} = \frac 34$, hence $FG = \frac 34 \cdot CE$. Now the area $S_{ABF}$ of $\triangle ABF$ can be computed as $S_{ABF} = \frac 12 \cdot AB \cdot FG$ = $\frac 12 \cdot \left( \frac 14 \cdot AC \right) \cdot \left( \frac 34 \cdot EC \right) = \frac 14 \cdot \frac 34 \cdot S_{ACE}$.

Similarly we can find that $S_{BCD} = S_{DEF} = \frac 3{16}\cdot S_{ACE}$ as well.

Hence $S_{BDF} = S_{ACE} - 3\cdot\left( \frac 3{16} \cdot S_{ACE} \right) = \frac 7{16} \cdot S_{ACE}$, and the answer is $\frac{S_{BDF}}{S_{ACE}} = \boxed{\frac 7{16}}$.

See also

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 10 Problems and Solutions

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