Difference between revisions of "2005 AMC 10A Problems/Problem 25"
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&= \frac{1}{[ABC]/[ADE] - 1} \\ | &= \frac{1}{[ABC]/[ADE] - 1} \\ | ||
&= \frac{1}{75/19 - 1} \\ | &= \frac{1}{75/19 - 1} \\ | ||
| − | &= \boxed{\frac{19}{56}}. | + | &= \boxed{\frac{19}{56}\LongrightarrowD}. |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
| − | |||
==Solution (trig)== | ==Solution (trig)== | ||
Revision as of 17:40, 21 August 2018
Problem
In 
we have 
, 
, and 
. Points 
and 
are on 
and 
respectively, with 
and 
. What is the ratio of the area of triangle 
to the area of the quadrilateral 
?
Solution 1(no trig)
We have that
![\[\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC} = \frac{19}{25} \cdot \frac{14}{42} = \frac{19}{75}.\]](http://latex.artofproblemsolving.com/c/a/7/ca7f72303ecda76e14992764cd8189ea495884ea.png)
![[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]](http://latex.artofproblemsolving.com/6/e/2/6e2fdc4687b12e3806e8997a889e00dce6eb1d06.png)
But ![$[BCED] = [ABC] - [ADE]$](http://latex.artofproblemsolving.com/5/5/e/55eb2fcfbad02e393d8d58681100faa707370dcb.png)
, so
\begin{align*}
\frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\
&= \frac{1}{[ABC]/[ADE] - 1} \\
&= \frac{1}{75/19 - 1} \\
&= \boxed{\frac{19}{56}\LongrightarrowD}.
\end{align*} (Error compiling LaTeX. Unknown error_msg)
Solution (trig)
.
Using this formula:
![$[ADE]=\frac{1}{2}\cdot19\cdot14\cdot\sin A = 133\sin A$](http://latex.artofproblemsolving.com/4/9/9/4994450bce1d1f4bb79c88144f39f85e954923c3.png)
![$[ABC]=\frac{1}{2}\cdot25\cdot42\cdot\sin A = 525\sin A$](http://latex.artofproblemsolving.com/9/d/5/9d54255e8dd4f8c76b2fe7fb3a02cc6a1a7bd0dc.png)
Since the area of is equal to the area of minus the area of ,
![$[BCED] = 525\sin A - 133\sin A = 392\sin A$](http://latex.artofproblemsolving.com/2/3/8/238b4f47585fadb1facd70b51395692d972ba99d.png)
.
Therefore, the desired ratio is 
Note: 
was not used in this problem
See also
| 2005 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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. 
