Question

A tripod has three legs each of length $5$ feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is $4$ feet from the ground. In setting up the tripod, the lower $1$ foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\frac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\lfloor m + \sqrt{n} \rfloor.$ (The notation $\lfloor x \rfloor$ denotes the greatest integer that is less than or equal to $x.$ )

答案

Solution:

Place the top at $T = (0, 0, 4).$ Each leg has length $5,$ so each foot is $3$ from the origin: $A = (3, 0, 0),$ $B = \left(-\frac{3}{2}, \frac{3\sqrt{3}}{2}, 0\right),$ $C = \left(-\frac{3}{2}, -\frac{3\sqrt{3}}{2}, 0\right).$ The broken leg has length $4,$ so its tip is $A' = \frac{4}{5}(3, 0, 0) + \frac{1}{5}(0, 0, 4) = \left(\frac{12}{5}, 0, \frac{4}{5}\right),$ and the tripod now stands on the plane $A'BC,$ with $h$ equal to the distance from $T$ to that plane.

The plane $A'BC$ contains $\overline{BC},$ which is parallel to the $y$ -axis and passes through the midpoint $M = \left(-\frac{3}{2}, 0, 0\right),$ so its distance from $T$ can be measured in the $xz$ -plane: it is the distance from $(0, 4)$ to the line through $\left(-\frac{3}{2}, 0\right)$ and $\left(\frac{12}{5}, \frac{4}{5}\right),$ whose equation is $8x - 39z + 12 = 0.$ Therefore $h = \frac{|8 \cdot 0 - 39 \cdot 4 + 12|}{\sqrt{8^2 + 39^2}} = \frac{144}{\sqrt{1585}}.$

Here $m = 144$ and $n = 1585 = 5 \cdot 317$ is squarefree. Since $39^2 = 1521 \lt 1585 \lt 1600,$ we get $\lfloor 144 + \sqrt{1585} \rfloor = 144 + 39 = 183.$

2006 AIME I Problem 14

Solution:

Problem 14 in Other Years