Difference between revisions of "1987 AIME Problems/Problem 15"
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== Solution == | == Solution == | ||
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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> |
Revision as of 19:23, 23 October 2007
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 |