1987 AIME Problems/Problem 15

Revision as of 13:57, 17 March 2017 by Warrenwangtennis (talk | contribs) (Solution)

Problem

Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$, as shown in the figures below. Find $AC + CB$ if area $(S_1) = 441$ and area $(S_2) = 440$.

AIME 1987 Problem 15.png

Solution

1987 AIME-15a.png

Because all the triangles in the figure are similar to triangle $ABC$, it's a good idea to use area ratios. In the diagram above, $\frac {T_1}{T_3} = \frac {T_2}{T_4} = \frac {441}{440}.$ Hence, $T_3 = \frac {440}{441}T_1$ and $T_4 = \frac {440}{441}T_2$. Additionally, the area of triangle $ABC$ is equal to both $T_1 + T_2 + 441$ and $T_3 + T_4 + T_5 + 440.$

Setting the equations equal and solving for $T_5$, $T_5 = 1 + T_1 - T_3 + T_2 - T_4 = 1 + \frac {T_1}{441} + \frac {T_2}{441}$. Therefore, $441T_5 = 441 + T_1 + T_2$. However, $441 + T_1 + T_2$ is equal to the area of triangle $ABC$! This means that the ratio between the areas $T_5$ and $ABC$ is $441$, and the ratio between the sides is $\sqrt {441} = 21$. As a result, $AB = 21\sqrt {440} = \sqrt {AC^2 + BC^2}$. We now need $(AC)(BC)$ to find the value of $AC + BC$, because $AB^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2$.

Let $h$ denote the height to the hypotenuse of triangle $ABC$. Notice that $h - \frac {1}{21}h = \sqrt {440}$. (The height of $ABC$ decreased by the corresponding height of $T_5$) Thus, $(AB)(h) = (AC)(BC) = 22\cdot 21^2$. Because $AB^2 + BC^2 +  2(AC)(BC) = (AC + BC)^2 = 21^2\cdot22^2$, $AC + BC = (21)(22) = \boxed{462}$.

See also

1987 AIME (ProblemsAnswer KeyResources)
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