Difference between revisions of "2004 AMC 10B Problems/Problem 20"
Pi over two (talk | contribs) m (Show the letter for the answer) |
Pi over two (talk | contribs) m (Use the square bracket notation to denote area) |
||
Line 25: | Line 25: | ||
== Solution (Triangle Areas) == | == Solution (Triangle Areas) == | ||
− | + | We use the square bracket notation <math>[\cdot]</math> to denote area. | |
− | + | Without loss of generality, we can assume <math>[\triangle BTD] = 1</math>. Then <math>[\triangle BTA] = 3</math>, and <math>[\triangle ATE] = 3/4</math>. We have <math>CD/BD = [\triangle ACD]/[\triangle ABD]</math>, so we need to find the area of quadrilateral <math>TDCE</math>. | |
+ | |||
+ | Draw the line segment <math>TC</math> to form the two triangles <math>\triangle TDC</math> and <math>\triangle TEC</math>. Let <math>x = [\triangle TDC]</math>, and <math>y = [\triangle TEC]</math>. By considering triangles <math>\triangle BTC</math> and <math>\triangle ETC</math>, we obtain <math>(1+x)/y=4</math>, and by considering triangles <math>\triangle ATC</math> and <math>\triangle DTC</math>, we obtain <math>(3/4+y)/x=3</math>. Solving, we get <math>x=4/11</math>, <math>y=15/44</math>, so the area of quadrilateral <math>TDEC</math> is <math>x+y=31/44</math>. | ||
Therefore <math>\frac{CD}{BD}=\frac{\frac{3}{4}+\frac{31}{44}}{3+1}=\boxed{\textbf{(D)} \frac{4}{11}}</math> | Therefore <math>\frac{CD}{BD}=\frac{\frac{3}{4}+\frac{31}{44}}{3+1}=\boxed{\textbf{(D)} \frac{4}{11}}</math> |
Revision as of 16:45, 24 January 2011
Contents
Problem
In points and lie on and , respectively. If and intersect at so that and , what is ?
Solution (Triangle Areas)
We use the square bracket notation to denote area.
Without loss of generality, we can assume . Then , and . We have , so we need to find the area of quadrilateral .
Draw the line segment to form the two triangles and . Let , and . 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:
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 |