Difference between revisions of "1987 AIME Problems/Problem 15"
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Let <math>h</math> denote the height to the [[hypotenuse]] of triangle <math>ABC</math>. Notice that <math>h - \frac {1}{21}h = \sqrt {440}</math>. (The height of <math>ABC</math> decreased by the corresponding height of <math>T_5</math>) Thus, <math>(AB)(h) = (AC)(BC) = 22\cdot 21^2</math>. Because <math>AB^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2 = 21^2\cdot22^2</math>, <math>AC + BC = (21)(22) = \boxed{462}</math>. | Let <math>h</math> denote the height to the [[hypotenuse]] of triangle <math>ABC</math>. Notice that <math>h - \frac {1}{21}h = \sqrt {440}</math>. (The height of <math>ABC</math> decreased by the corresponding height of <math>T_5</math>) Thus, <math>(AB)(h) = (AC)(BC) = 22\cdot 21^2</math>. Because <math>AB^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2 = 21^2\cdot22^2</math>, <math>AC + BC = (21)(22) = \boxed{462}</math>. | ||
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+ | == Easy Trig Solution == | ||
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+ | Let <math>\tan\angle ABC = x</math>. Now using the 1st square, <math>AC=21(1+x)</math> and <math>CB=21(1+x^{-1})</math>. Using the second square, <math>AB=\sqrt{440}(1+x+x^{-1})</math>. We have <math>AC^2+CB^2=AB^2</math>, or <cmath>441(x^2+x^{-2}+2x+2x^{-1}+2)=440(x^2+x^{-2}+2x+2x^{-1}+3).</cmath> Rearranging and letting <math>u=x+x^{-1} \Rightarrow u^2 - 2 = x^2 + x^{-2}</math> gives <math>u^2+2u-440=0.</math> Taking the positive root gives <math>u=20</math>, so <math>AC+CB=21(2+x+x^{-1})=21(2+u)=\boxed{462}</math>. | ||
== See also == | == See also == |
Revision as of 05:04, 26 February 2018
Contents
[hide]Problem
Squares and are inscribed in right triangle , as shown in the figures below. Find if area and area .
Solution
Because all the triangles in the figure are similar to triangle , it's a good idea to use area ratios. In the diagram above, Hence, and . Additionally, the area of triangle is equal to both and
Setting the equations equal and solving for , . Therefore, . However, is equal to the area of triangle ! This means that the ratio between the areas and is , and the ratio between the sides is . As a result, . We now need to find the value of , because .
Let denote the height to the hypotenuse of triangle . Notice that . (The height of decreased by the corresponding height of ) Thus, . Because , .
Easy Trig Solution
Let . Now using the 1st square, and . Using the second square, . We have , or Rearranging and letting gives Taking the positive root gives , so .
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.