Difference between revisions of "2000 AMC 10 Problems/Problem 19"
5849206328x (talk | contribs) (New page: ==Problem== ==Solution== ==See Also== {{AMC10 box|year=2000|num-b=18|num-a=20}}) |
5849206328x (talk | contribs) (→Problem) |
||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| + | |||
| + | Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is <math>m</math> times the area of the square. The ratio of the area of the other small right triangle to the area of the square is | ||
| + | |||
| + | <math>\mathrm{(A)}\ \frac{1}{2m+1} \qquad\mathrm{(B)}\ m \qquad\mathrm{(C)}\ 1-m \qquad\mathrm{(D)}\ \frac{1}{4m} \qquad\mathrm{(E)}\ \frac{1}{8m^2}</math> | ||
==Solution== | ==Solution== | ||
Revision as of 10:20, 11 January 2009
Problem
Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is 
times the area of the square. The ratio of the area of the other small right triangle to the area of the square is
Solution
See Also
| 2000 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 | ||
