AIME 1987 · 第 15 题

Solution

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 $AC^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2$.

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 $AC^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2 = 21^2\cdot22^2$, $AC + BC = (21)(22) = \boxed{462}$.

Easy Trig Solution

Let $\tan\angle ABC = x$. Now using the 1st square, $AC=21(1+x)$ and $CB=21(1+x^{-1})$. Using the second square, $AB=\sqrt{440}(1+x+x^{-1})$. We have $AC^2+CB^2=AB^2$, or

$$441(x^2+x^{-2}+2x+2x^{-1}+2)=440(x^2+x^{-2}+2x+2x^{-1}+3).$$ Rearranging and letting $u=x+x^{-1} \Rightarrow u^2 - 2 = x^2 + x^{-2}$ gives us $u^2+2u-440=0.$ We take the positive root, so $u=20$, which means $AC+CB=21(2+x+x^{-1})=21(2+u)=\boxed{462}$.

Messy Trig Solution

Let $\theta$ be the smaller angle in the triangle. Then the sum of shorter and longer leg is $\sqrt{441}(2+\tan{\theta}+\cot{\theta})$. We observe that the short leg has length $\sqrt{441}(1+\tan{\theta}) = \sqrt{440}(\sec{\theta}+\sin{\theta})$. Grouping and squaring, we get $\sqrt{\frac{440}{441}} = \frac{\sin{\theta}+\cos{\theta}}{1+\sin{\theta}\cos{\theta}}$. Squaring and using the double angle identity for sine, we get, $110(\sin{2\theta})^2 + \sin{2\theta} - 1 = 0$. Solving, we get $\sin{2\theta} = \frac{1}{10}$. Now to find $\tan{\theta}$, we find $\cos{2\theta}$ using the Pythagorean Identity, and then use the tangent double angle identity. Thus, $\tan{\theta} = 10-3\sqrt{11}$. Substituting into the original sum, we get $\boxed{462}$.

Solution 4 (Algebra)

Label points as above. Let $x=AC$, $y=BC$, $s_1 = 21$ be the side length of $S_1$, and $s_2 = \sqrt{440}$ be the side length of $S_2$.

Since $\triangle ABC\sim\triangle AED$, we have $\frac{x}{y} = \frac{x-s_1}{s_1}$

$\implies xs_1=xy-ys_1$ $\implies xy=s_1(x+y)$

$\implies xy=21(x+y) \qquad \qquad (*)$.

Since $\triangle ABC\sim\triangle AWX\sim\triangle ZBY$, we have $s_2 + \frac{s_2x}{y} + \frac{s_2y}{x}=\sqrt{x^2+y^2}$

$\implies s_2(x^2+xy+y^2)=xy\sqrt{x^2+y^2}$ $\implies s_2^2(x^2+xy+y^2)^2 = x^2y^2(x^2+y^2)$ $\implies 440(x^2+xy+y^2)^2 = x^2y^2(x^2+y^2)$

Let $t=x+y$. Repeatedly applying $(*)$, we get

$$440(t^2-21t)^2 = 441t^2(t^2 - 42t)$$ $$440(t-21)^2 = 441(t^2-42t)$$ $$440t^2 - 42\cdot 440t + 440\cdot 441 = 441t^2 - 441\cdot 42t$$ $$t^2-42t-440\cdot 441=0$$ $$(t-21)^2 = 441^2$$ $$t-21=441$$ $$t=\boxed{462}$$ ~rayfish