Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Find the answer

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 $x.$ )

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

We will use $[...]$ to denote volume (four letters), area (three letters) or length (two letters).
Let $T$ be the top of the tripod, $A,B,C$ are end points of three legs. Let $S$ be the point on $TA$ such that $[TS] = 4$ and $[SA] = 1$ . Let $O$ be the center of the base equilateral triangle $ABC$ . Let $M$ be the midpoint of segment $BC$ . Let $h$ be the distance from $T$ to the triangle $SBC$ ( $h$ is what we want to find).
We have the volume ratio $\frac {[TSBC]}{[TABC]} = \frac {[TS]}{[TA]} = \frac {4}{5}$ .
So $\frac {h\cdot [SBC]}{[TO]\cdot [ABC]} = \frac {4}{5}$ .
We also have the area ratio $\frac {[SBC]}{[ABC]} = \frac {[SM]}{[AM]}$ .
The triangle $TOA$ is a $3-4-5$ right triangle so $[AM] = \frac {3}{2}\cdot[AO] = \frac {9}{2}$ and $\cos{\angle{TAO}} = \frac {3}{5}$ .
Applying Law of Cosines to the triangle $SAM$ with $[SA] = 1$ , $[AM] = \frac {9}{2}$ and $\cos{\angle{SAM}} = \frac {3}{5}$ , we find:

$[SM] = \frac {\sqrt {5\cdot317}}{10}.$
Putting it all together, we find $h = \frac {144}{\sqrt {5\cdot317}}$ .

$\lfloor 144+\sqrt{5 \cdot 317}\rfloor =144+ \lfloor \sqrt{5 \cdot 317}\rfloor =144+\lfloor \sqrt{1585} \rfloor =144+39=\boxed{183}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.