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2006 AIME I Problems/Problem 14

Revision as of 20:23, 28 November 2007 by Azjps (talk | contribs) (solution by fuzzy_logic)

Problem

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 $x.$)

Solution

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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 $h = \frac {144}{\sqrt {5\cdot317}}$.

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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