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Problem 219

AMC 10/12, early questions

Geometry Difficulty 3.8 Find the answer China Mathematical Competition · China

Given a right triangular prism $A_1B_1C_1 - ABC$ with $\angle BAC = \frac{\pi}{2}$ and $AB = AC = AA_1 = 1$ , let $G, E$ be the midpoints of $A_1B_1$ , $CC_1$ respectively; and $D, F$ be variable points lying on segments $AC, AB$ (not including endpoints) respectively. If $GD \perp EF$ , the range of the length of $DF$ is ( ).

This was a multiple-choice question, but the options didn't survive into the source we have, so there's nothing here to pick from. The solution below still works the problem through.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

We establish a coordinate system with point $A$ as the origin, line $AB$ as the $x$ -axis, $AC$ the $y$ -axis and $AA_1$ the $z$ -axis. Then we have $F(t_1, 0, 0)$ ( $0 < t_1 < 1$ ), $E(0, 1, \frac{1}{2})$ ,
$G(\frac{1}{2}, 0, 1)$ , $D(0, t_2, 0)$ ( $0 < t_2 < 1$ ). Therefore $\vec{EF} = (t_1, -1, -\frac{1}{2})$ , $\vec{GD} = (-\frac{1}{2}, t_2, -1)$ . Since $GD \perp EF$ , we get $t_1 + 2t_2 = 1$ . Then $0 < t_2 < \frac{1}{2}$ . Furthermore, $\vec{DF} = (t_1, -t_2, 0)$ ,
$|\vec{DF}| = \sqrt{t_1^2 + t_2^2} = \sqrt{5t_2^2 - 4t_2 + 1} = \sqrt{5\left(t_2 - \frac{2}{5}\right)^2 + \frac{1}{5}}$
We obtain $\sqrt{\frac{1}{5}} \le |\vec{DF}| < 1$ . Answer: A.

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