1986 AIME Problem 14

Attempt Problem 14 of the 1986 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1986 AIME solutions, or check the answer key.

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14.

The shortest distances between an interior diagonal of a rectangular parallelepiped PP and the edges it does not meet are 25,2\sqrt5, 3013,\frac{30}{\sqrt{13}}, and 1510.\frac{15}{\sqrt{10}}. Determine the volume of P.P.

Answer: 750
Concepts:rectangular prismsystem of equationsvector
Difficulty rating: 3060
Small Hint:

Let the side lengths be a,a, b,b, and cc and use a vector formula for the distance between skew lines

Big Hint:

Taking reciprocals of the squared distances makes the equations linear in 1a2,\frac{1}{a^2}, 1b2,\frac{1}{b^2}, and 1c2\frac{1}{c^2}

Solution:

Let the side lengths be a,a, b,b, and c.c. For example, the distance from the space diagonal with direction (a,b,c)(a,b,c) to a nonintersecting edge parallel to the aa-direction is bcb2+c2. \frac{bc}{\sqrt{b^2+c^2}}. The other two distances are obtained cyclically. We may assign the three given distances to these three directions in the listed order, since permuting them only permutes the side lengths.

Put X=1a2,X=\frac{1}{a^2}, Y=1b2,Y=\frac{1}{b^2}, and Z=1c2.Z=\frac{1}{c^2}. Taking reciprocal squares gives Y+Z=120,X+Z=13900,X+Y=245. \begin{aligned} Y+Z&=\frac1{20},\\ X+Z&=\frac{13}{900},\\ X+Y&=\frac2{45}. \end{aligned} Solving, X=1225,Y=125,Z=1100. \begin{aligned} X&=\frac1{225},\\ Y&=\frac1{25},\\ Z&=\frac1{100}. \end{aligned} Hence the side lengths are 15,15, 5,5, 10,10, and the volume is 15510=750.15\cdot5\cdot10=750.

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