Track / Stage 3 / 220 of 260 #220 of 1964

Problem 220

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 is nothing here to pick from. Work it on paper and mark yourself against the solution below.

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.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.