2004 AMC 10B Problems/Problem 18
Contents
Problem
In the right triangle 
, we have 
, 
, and 
. Points 
, 
, and 
are located on 
, 
, and 
, respectively, so that 
, 
, and 
. What is the ratio of the area of 
to that of ?
![[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]](http://latex.artofproblemsolving.com/8/b/9/8b90a1d12a3618d41a089a2890533013e611b558.png)
Solution 1
Let ![$x = [DBF]$](http://latex.artofproblemsolving.com/2/6/3/263dc6cad183b002bbc16658c481d8935e2373a7.png)
. Because is divided into four triangles, ![$[ACE] = [BCD] + [ABF] + [DEF] + x$](http://latex.artofproblemsolving.com/3/2/6/3265f63783f8f764b76e9f3825300b6fe5687e17.png)
.
Because the area of $\triangleXYZ = \frac12 \cdot XY \cdot XZ \cdot \sin(\angle X), \frac12 \cdot 12 \cdot 16 = \frac12 \cdot 9 \cdot 4 + \frac12 \cdot 3 \cdot 15 \cdot \sin(\angle A) + \frac12 \cdot 5 \cdot 12 \cdot \sin(\angle E) + x$ (Error compiling LaTeX. Unknown error_msg).
and 
, so 
.
, so ![$\frac{[DBF]}{[ACE]} = \frac{42}{96} = \boxed{\textbf{(E)}\frac 7{16}}$](http://latex.artofproblemsolving.com/f/7/8/f7896c8dff2f70ba2153cd44766cc4f5507ec7c5.png)
.
Solution 2
First of all, note that 
, and therefore
.
Draw the height from onto 
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]](http://latex.artofproblemsolving.com/e/3/a/e3a991ca861b53810c15c17e5a93234eddefa082.png)
Now consider the area of 
. Clearly the triangles 
and 
are similar, as they have all angles equal. Their ratio is 
, hence 
.
Now the area 
of can be computed as 
= 
.
Similarly we can find that 
as well.
Hence 
, and the answer is 
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
