2004 AMC 10A Problems/Problem 20
Contents
Problem
Points and are located on square so that is equilateral. What is the ratio of the area of to that of ?
Solution 1
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)
WLOG, 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
Solution 3
is equilateral, so , and so they must each be . Then let , which gives and . The area of is then . is an isosceles right triangle with hypotenuse 1, so and therefore its area is . The ratio of areas is then
Solution 4(system of equations)
Assume AB=1 then FC is x ED is then we see that using HL FCB is congruent is EAB. Using Pythagoras of triangles FCB and FDE we get . Expanding we get . Simplifying gives solving using completing the square(or other methods) gives 2 answers and because then then using the areas we get the answer to be D
Solution 5
First, since is equilateral and is a square, by the Hypothenuse Leg Theorem, is congruent to . Then, assume length and length , then . is equilateral, so and , it is given that is a square and and are right triangles. Then we use the Pythagorean theorem to prove that and since we know that and , which means . Now we plug in the variables and the equation becomes , expand and simplify and you get . We want the ratio of area of to . Expressed in our variables, the ratio of the area is and we know , so the ratio must be 2. Choice D
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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 |
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