1999 AIME 第 15 题

The three folded corners meet at a single apex, whose distance to each midpoint equals half of the corresponding original side

Keep the midpoint triangle in the plane $z = 0$ and solve three distance equations for the apex $(x, y, z)$

The midpoints are $M_1 = (17, 0),$ $M_2 = (25, 12),$ and $M_3 = (8, 12).$ Folding the three corner triangles up along the sides of the midpoint triangle brings the corners together at one apex $Q$ (each pair of glued half-sides has equal length). The apex keeps its folded distances: $QM_1 = 17$ (half of the side of length $34$ that $M_1$ bisects), $QM_2 = 15$ (half of $30$), and $QM_3 = 4\sqrt{13}$ (half of $\sqrt{16^2 + 24^2} = 8\sqrt{13}$).

Keep the midpoint triangle in the plane $z = 0$ and let $Q = (x, y, z).$ Subtracting $|Q - M_3|^2 = 208$ from $|Q - M_2|^2 = 225$ gives $(x - 25)^2 - (x - 8)^2 = 17,$ so $x = 16;$ subtracting $|Q - M_2|^2 = 225$ from $|Q - M_1|^2 = 289$ gives $2x + 3y = 68,$ so $y = 12.$ Then $z^2$ $= 289 - (16 - 17)^2$ $- 12^2$ $= 144,$ so the apex is at height $z = 12.$

The base is the midpoint triangle, with area one quarter of the original triangle’s $\frac{1}{2} \cdot 34 \cdot 24 = 408,$ i.e. $102.$ The volume is $$\frac{1}{3} \cdot 102 \cdot 12 = 408.$$