2004 AMC 10B Problems/Problem 20
Contents
Problem
In 
points 
and 
lie on 
and 
, respectively. If 
and 
intersect at 
so that 
and 
, what is 
?
![[asy] unitsize(1cm); defaultpen(0.8); pair A=(0,0), B=5*dir(60), C=5*(1,0), D=B + (11/15)*(C-B), E = A + (11/16)*(C-A); draw(A--B--C--cycle); draw(A--D); draw(B--E); pair T=intersectionpoint(A--D,B--E); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,S); label("$T$",T,2*WNW); [/asy]](http://latex.artofproblemsolving.com/0/5/b/05b2f751804dc3a7866b1e5c5e9c3f214a502ad1.png)
Solution (Triangle Areas)
We use the square bracket notation ![$[\cdot]$](http://latex.artofproblemsolving.com/5/7/9/579a0ec9e030cc17f3b2c9cd1510c9493b1813b2.png)
to denote area.
Without loss of generality, we can assume ![$[\triangle BTD] = 1$](http://latex.artofproblemsolving.com/8/d/0/8d018803340bfaddea348768ffc2ee0afba70e94.png)
. Then ![$[\triangle BTA] = 3$](http://latex.artofproblemsolving.com/5/3/6/5368e579e079501f1cdc63487d84ef2096b861a4.png)
, and ![$[\triangle ATE] = 3/4$](http://latex.artofproblemsolving.com/5/f/c/5fca20e19ee98031a4040c8426601ee6ea6e8fa2.png)
. We have ![$CD/BD = [\triangle ACD]/[\triangle ABD]$](http://latex.artofproblemsolving.com/0/3/a/03a7d84d4281bd486ab6e39d8f26a7dc11817030.png)
, so we need to find the area of quadrilateral 
.
Draw the line segment 
to form the two triangles 
and 
. Let ![$x = [\triangle TDC]$](http://latex.artofproblemsolving.com/6/f/9/6f9790d3ce4ba9b1c7b4a0b306f321d56acd516c.png)
, and ![$y = [\triangle TEC]$](http://latex.artofproblemsolving.com/e/d/4/ed4e725364fd4d07e4c4dffc2e673367c2758e78.png)
. By considering triangles 
and 
, we obtain 
, and by considering triangles 
and 
, we obtain 
. Solving, we get 
, 
, so the area of quadrilateral 
is 
.
Therefore 
Solution (Mass points)
The presence of only ratios in the problem essentially cries out for mass points.
As per the problem, we assign a mass of 
to point 
, and a mass of 
to . Then, to balance and on , has a mass of 
.
Now, were we to assign a mass of to 
and a mass of to , we'd have 
. Scaling this down by 
(to get 
, which puts and in terms of the masses of and ), we assign a mass of 
to and a mass of 
to .
Now, to balance and 
on , we must give a mass of 
.
Finally, the ratio of 
to 
is given by the ratio of the mass of to the mass of , which is 
.
Solution (Coordinates)
Affine transformations preserve ratios of distances, and for any pair of triangles there is an affine transformation that maps the first one onto the second one. This is why the answer is the same for any , and we just need to compute it for any single triangle.
We can choose the points 
, 
, and 
. This way we will have 
, and 
. The situation is
shown in the picture below:
![[asy] unitsize(1cm); defaultpen(0.8); pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1); draw(A--B--C--cycle); draw(A--D); draw(B--E); pair T=intersectionpoint(A--D,B--E); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,S); label("$T$",T,NW); label("$3$",A--T,N); label("$4$",B--T,W); label("$1$",D--T,N); label("$1$",E--T,W); [/asy]](http://latex.artofproblemsolving.com/a/3/3/a33ad88d1e8b0bec0d13245fe244c3096f00077e.png)
The point is the intersection of the lines and 
. The points on the first line have the form 
, the points on the second line have the form 
. Solving for 
we get 
, hence 
.
The ratio can now be computed simply by observing the 
coordinates of , , and :
![\[\frac{CD}{BD} = \frac{15/11 - 1}{1 - 0} = \boxed{\frac 4{11}}\]](http://latex.artofproblemsolving.com/1/9/5/1957a477c93b2f66ad7b9d706b732a4940a6173f.png)
See also
| 2004 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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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