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2004 AMC 10B Problems/Problem 20

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Problem

In \triangle ABC points D and E lie on BC and AC, respectively. If AD and BE intersect at T so that AT/DT=3 and BT/ET=4, what is CD/BD?


\mathrm{(A) \ } \frac{1}{8} \qquad \mathrm{(B) \ } \frac{2}{9} \qquad \mathrm{(C) \ } \frac{3}{10} \qquad \mathrm{(D) \ } \fr...


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--...

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 1 to point A, and a mass of 3 to D. Then, to balance A and D on T, T has a mass of 4.

Now, were we to assign a mass of 1 to B and a mass of 4 to E, we'd have 5T. Scaling this down by 4/5 (to get 4T, which puts B and E in terms of the masses of A and D), we assign a mass of \frac{4}{5} to B and a mass of \frac{16}{5} to E.

Now, to balance A and C on E, we must give C a mass of \frac{16}{5}-1=\frac{11}{5}.

Finally, the ratio of CD to BD is given by the ratio of the mass of B to the mass of C, which is \frac{4}{5}\cdot\frac{5}{11}=\boxed{\textbf{(D)}\ \frac{4}{11}}.

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 \triangle ABC, and we just need to compute it for any single triangle.

We can choose the points A=(-3,0), B=(0,4), and D=(1,0). This way we will have T=(0,0), and E=(0,-1). The situation is shown in the picture below:

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);dra...

The point C is the intersection of the lines BD and AE. The points on the first line have the form (t,4-4t), the points on the second line have the form (t,-1-t/3). Solving for t we get t=15/11, hence C=(15/11,-16/11).

The ratio CD/BD can now be computed simply by observing the x coordinates of B, C, and D:

\frac{CD}{BD} = \frac{15/11 - 1}{1 - 0} = \boxed{\frac 4{11}}

See also

2004 AMC 10B (ProblemsResources)
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
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