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Geometry Difficulty 5.0 AIME, harder Find the answer United States

Problem:
Let $\ell$ and $m$ be two non-coplanar lines in space, and let $P_{1}$ be a point on $\ell$ . Let $P_{2}$ be the point on $m$ closest to $P_{1}$ , $P_{3}$ be the point on $\ell$ closest to $P_{2}$ , $P_{4}$ be the point on $m$ closest to $P_{3}$ , and $P_{5}$ be the point on $\ell$ closest to $P_{4}$ . Given that $P_{1}P_{2}=5$ , $P_{2}P_{3}=3$ , and $P_{3}P_{4}=2$ , compute $P_{4}P_{5}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:
The figure below shows the situation of the problem when projected appropriately, which will be explained later.

Let $a$ be the answer. By taking the $z$ -axis to be the cross product of these two lines, we can let the lines be on the planes $z=0$ and $z=h$ , respectively. Then, by projecting onto the $xy$ -plane, we get the above diagram. The projected lengths of the first four segments are $\sqrt{25-h^{2}}$ , $\sqrt{9-h^{2}}$ , and $\sqrt{4-h^{2}}$ , and $\sqrt{a^{2}-h^{2}}$ . By similar triangles, these lengths must form a geometric progression. Therefore, $25-h^{2}$ , $9-h^{2}$ , $4-h^{2}$ , $a^{2}-h^{2}$ is a geometric progression. By taking consecutive differences, $16,5,4-a^{2}$ is a geometric progression. Hence, $4-a^{2}=\frac{25}{16} \Longrightarrow a=\frac{\sqrt{39}}{4}$ .

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