AIME 1986 · 第 14 题

Solution

In the above diagram, we focus on the line that appears closest and is parallel to $BC$. All the blue lines are perpendicular lines to $BC$ and their other points are on $AB$, the main diagonal. The green lines are projections of the blue lines onto the bottom face; all of the green lines originate in the corner and reach out to $AC$, and have the same lengths as their corresponding blue lines. So we want to find the shortest distance between $AC$ and that corner, which is $\frac {wl}{\sqrt {w^2 + l^2}}$.

So we have:

$$\frac {lw}{\sqrt {l^2 + w^2}} = \frac {10}{\sqrt {5}}$$ $$\frac {hw}{\sqrt {h^2 + w^2}} = \frac {30}{\sqrt {13}}$$ $$\frac {hl}{\sqrt {h^2 + l^2}} = \frac {15}{\sqrt {10}}$$ Notice the familiar roots: $\sqrt {5}$, $\sqrt {13}$, $\sqrt {10}$, which are $\sqrt {1^2 + 2^2}$, $\sqrt {2^2 + 3^2}$, $\sqrt {1^2 + 3^2}$, respectively. (This would give us the guess that the sides are of the ratio 1:2:3, but let's provide the complete solution.)

$$\frac {l^2w^2}{l^2 + w^2} = \frac {1}{\frac {1}{l^2} + \frac {1}{w^2}} = 20$$ $$\frac {h^2w^2}{h^2 + w^2} = \frac {1}{\frac {1}{h^2} + \frac {1}{w^2}} = \frac {900}{13}$$ $$\frac {h^2l^2}{h^2 + l^2} = \frac {1}{\frac {1}{h^2} + \frac {1}{l^2}} = \frac {45}{2}$$ We invert the above equations to get a system of linear equations in $\frac {1}{h^2}$, $\frac {1}{l^2}$, and $\frac {1}{w^2}$:

$$\frac {1}{l^2} + \frac {1}{w^2} = \frac {45}{900}$$ $$\frac {1}{h^2} + \frac {1}{w^2} = \frac {13}{900}$$ $$\frac {1}{h^2} + \frac {1}{l^2} = \frac {40}{900}$$ We see that $h = 15$, $l = 5$, $w = 10$. Therefore $V = 5 \cdot 10 \cdot 15 = \boxed{750}$