2004 AMC 10A Problems/Problem 20
Problem
Points and are located on square so that is equilateral. What is the ratio of the area of to that of ?
Solution
Since triangle is equilateral, , and and are congruent. Thus, triangle is an isosceles right triangle. So we let . Thus . If we go angle chasing, we find out that , thus . . Thus , or . Thus , and , and . Thus the ratio of the areas is
Solution 2 (Non-trig)
Without loss of generality let the side length of be 1. Let and . Then , so , and ; i.e. that is 45-45-90. We find that , and so, by the Pythagorean Theorem, we have This yields , so . Thus, the desired ratio of areas is
See also
2004 AMC 10A (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.