2004 AMC 10B Problems/Problem 18

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 $\frac{S_{BDF}}{S_{ACE}} = \boxed{\frac 7{16}}$ .

Solution 3 (Coordinate Geometry)

We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9), and let F be (12, 3). You can then use the shoelace theorem to find the area of DBF, which is 42. $\frac {42}{96} = \boxed{\frac 7{16}}$

Solution 4

You can also place a point $X$ on $CE$ such that $CX$ is $12$ , creating trapezoid $CBFX$ . Then, you can find the area of the trapezoid, subtract the area of the two right triangles $DFX$ and $BCD$ , divide by the area of $ABC$ , and get the ratio of $7/16$ .

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

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