Difference between revisions of "1987 AIME Problems/Problem 15"
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Because all the [[triangle]]s in the figure are [[similar]] to triangle <math>ABC</math>, it's a good idea to use [[area ratios]]. In the diagram above, <math>\frac {T_1}{T_3} = \frac {T_2}{T_4} = \frac {441}{440}.</math> Hence, <math>T_3 = \frac {440}{441}T_1</math> and <math>T_4 = \frac {440}{441}T_2</math>. Additionally, the area of triangle <math>ABC</math> is equal to both <math>T_1 + T_2 + 441</math> and <math>T_3 + T_4 + T_5 + 440.</math> | Because all the [[triangle]]s in the figure are [[similar]] to triangle <math>ABC</math>, it's a good idea to use [[area ratios]]. In the diagram above, <math>\frac {T_1}{T_3} = \frac {T_2}{T_4} = \frac {441}{440}.</math> Hence, <math>T_3 = \frac {440}{441}T_1</math> and <math>T_4 = \frac {440}{441}T_2</math>. Additionally, the area of triangle <math>ABC</math> is equal to both <math>T_1 + T_2 + 441</math> and <math>T_3 + T_4 + T_5 + 440.</math> | ||
− | Setting the equations equal and solving for <math>T_5</math>, <math>T_5 = 1 + T_1 - T_3 + T_2 - T_4 = 1 + \frac {T_1}{441} + \frac {T_2}{441}</math>. Therefore, <math>441T_5 = 441 + T_1 + T_2</math>. However, <math>441 + T_1 + T_2</math> is equal to the area of triangle <math>ABC</math>! This means that the ratio between the areas <math>T_5</math> and <math>ABC</math> is <math>441</math>, and the ratio between the sides is <math>\sqrt {441} = 21</math>. As a result, <math>AB = 21\sqrt {440} = \sqrt {AC^2 + BC^2}</math>. We now need <math>(AC)(BC)</math> to find the value of <math>AC + BC</math>, because <math>AB^2 + 2(AC)(BC) = (AC + BC)^2</math>. | + | Setting the equations equal and solving for <math>T_5</math>, <math>T_5 = 1 + T_1 - T_3 + T_2 - T_4 = 1 + \frac {T_1}{441} + \frac {T_2}{441}</math>. Therefore, <math>441T_5 = 441 + T_1 + T_2</math>. However, <math>441 + T_1 + T_2</math> is equal to the area of triangle <math>ABC</math>! This means that the ratio between the areas <math>T_5</math> and <math>ABC</math> is <math>441</math>, and the ratio between the sides is <math>\sqrt {441} = 21</math>. As a result, <math>AB = 21\sqrt {440} = \sqrt {AC^2 + BC^2}</math>. We now need <math>(AC)(BC)</math> to find the value of <math>AC + BC</math>, because <math>AB^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2</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 + 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>. |
== See also == | == See also == |
Revision as of 13:57, 17 March 2017
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 , .
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 |
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