Difference between revisions of "2004 AMC 10A Problems/Problem 20"
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==Solution 2 (Non-trig) == | ==Solution 2 (Non-trig) == | ||
− | Without loss of generality, let the side length of <math>ABCD</math> be 1. Let <math>DE = x</math> | + | Without loss of generality, let the side length of <math>ABCD</math> be 1. Let <math>DE = x</math>. It suffices that <math>AE = 1 - x</math>. Then triangles <math>ABE</math> and <math>CBF</math> are congruent by HL, so <math>CF = AE</math> and <math>DE = DF</math>. We find that <math>BE = EF = x \sqrt{2}</math>, and so, by the Pythagorean Theorem, we have |
<math>(1 - x)^2 + 1 = 2x^2.</math> This yields <math>x^2 + 2x = 2</math>, so <math>x^2 = 2 - 2x</math>. Thus, the desired ratio of areas is | <math>(1 - x)^2 + 1 = 2x^2.</math> This yields <math>x^2 + 2x = 2</math>, so <math>x^2 = 2 - 2x</math>. Thus, the desired ratio of areas is | ||
− | <cmath>\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = 2.</cmath> | + | <cmath>\frac{\frac{x^2}{2}}{\frac{1-x}{2}} = \frac{x^2}{1 - x} = \boxed{\text{D) }2}.</cmath> |
==See also== | ==See also== |
Revision as of 18:16, 14 August 2017
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 . It suffices that . Then triangles and are congruent by HL, so and . 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.