Difference between revisions of "2004 AMC 10B Problems/Problem 20"
(New page: == Problem == In <math>\triangle ABC</math> points <math>D</math> and <math>E</math> lie on <math>BC</math> and <math>AC</math>, respectively. If <math>AD</math> and <math>BE</math> inter...) |
|||
Line 23: | Line 23: | ||
</asy> | </asy> | ||
− | == Solution == | + | == 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 <math>1</math> to point <math>A</math>, and a mass of <math>3</math> to <math>D</math>. Then, to balance <math>A</math> and <math>D</math> on <math>T</math>, <math>T</math> has a mass of <math>4</math>. | ||
+ | |||
+ | Now, were we to assign a mass of <math>1</math> to <math>B</math> and a mass of <math>4</math> to <math>E</math>, we'd have <math>5T</math>. Scaling this down by <math>4/5</math> (to get <math>4T</math>, which puts <math>B</math> and <math>E</math> in terms of the masses of <math>A</math> and <math>D</math>), we assign a mass of <math>\frac{4}{5}</math> to <math>B</math> and a mass of <math>\frac{16}{5}</math> to <math>E</math>. | ||
+ | |||
+ | Now, to balance <math>A</math> and <math>C</math> on <math>E</math>, we must give <math>C</math> a mass of <math>\frac{16}{5}-1=\frac{11}{5}</math>. | ||
+ | |||
+ | Finally, the ratio of <math>CD</math> to <math>BD</math> is given by the ratio of the mass of <math>B</math> to the mass of <math>C</math>, which is <math>\frac{4}{5}\cdot\frac{5}{11}=\boxed{\textbf{(D)}\ \frac{4}{11}}</math>. | ||
+ | |||
+ | == 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 <math>\triangle ABC</math>, and we just need to compute it for any single triangle. | 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 <math>\triangle ABC</math>, and we just need to compute it for any single triangle. |
Revision as of 18:19, 30 March 2009
Problem
In points and lie on and , respectively. If and intersect at so that and , what is ?
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:
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 :
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 |