# 2006 AIME I Problems/Problem 14

## 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 1

$[asy] size(200); import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65); currentprojection = perspective(1,-10,3.3); triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0); triple M=(B+C)/2,S=(4*A+T)/5; draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,ddash);draw(T--O--M,ddash); label("T",T,N);label("A",A,SW);label("B",B,SE);label("C",C,NE);label("S",S,NW);label("O",O,SW);label("M",M,NE); label("4",(S+T)/2,NW);label("1",(S+A)/2,NW);label("5",(B+T)/2,NE);label("4",(O+T)/2,W); dot(S);dot(O); [/asy]$

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

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

## Solution 2

We note that $AO=3$. From this we can derive that the side length of the equilateral is $3\sqrt{3}$. We now use 3D coordinate geometry. $$A = (0,0,0)$$ $$B = (3\sqrt{3},0,0)$$ $$C = (\frac{3\sqrt{3}}{2}, \frac{9}{2}, 0)$$ $$T = (\frac{3\sqrt{3}}{2}, \frac{3}{2}, 4)$$ $$S= (\frac{3\sqrt{3}}{10}, \frac{3}{10}, \frac{4}{5})$$ We know three points of plane $SCB$ hence we can write out the equation for the plane. Plane $SCB$ can be expressed as $$4\sqrt{3}x+4y+39z-36=0.$$

Applying the distance between a point and a plane formula.

$$\frac{ax+by+cz+d}{\sqrt{a^{2}+b^{2}+c^{2}}} = \frac{4\sqrt{3} \cdot \frac{3\sqrt{3}}{2} + 4\cdot \frac{3}{2} + 39 \cdot 4 -36}{\sqrt{(4\sqrt{3})^2+4^2+39^2}} = \frac{144}{\sqrt{1585}}$$

$$\lfloor m+\sqrt{n}\rfloor = \lfloor 144+\sqrt{1585}\rfloor = 183$$

Solution by SimonSun

## Solution 3 (Cosine Law & Pythagorean Bash)

Diagram borrowed from Solution 1

$[asy] size(200); import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65); currentprojection = perspective(1,-10,3.3); triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0); triple M=(B+C)/2,S=(4*A+T)/5; draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,ddash);draw(T--O--M,ddash); label("T",T,N);label("A",A,SW);label("B",B,SE);label("C",C,NE);label("S",S,NW);label("O",O,SW);label("M",M,NE); label("4",(S+T)/2,NW);label("1",(S+A)/2,NW);label("5",(B+T)/2,NE);label("4",(O+T)/2,W); dot(S);dot(O); [/asy]$

Apply Pythagorean Theorem on $\bigtriangleup TOB$ yields $$BO=\sqrt{TB^2-TO^2}=3$$ Since $\bigtriangleup ABC$ is equilateral, we have $\angle MOB=60^{\circ}$ and $$BC=2BM=2(OB\sin MOB)=3\sqrt{3}$$ Apply Pythagorean Theorem on $\bigtriangleup TMB$ yields $$TM=\sqrt{TB^2-BM^2}=\sqrt{5^2-(\frac{3\sqrt{3}}{2})^2}=\frac{\sqrt{73}}{2}$$ Apply Law of Cosines on $\bigtriangleup TBC$ we have $$BC^2=TB^2+TC^2-2(TB)(TC)\cos BTC$$ $$(3\sqrt{3})^2=5^2+5^2-2(5)^2\cos BTC$$ $$\cos BTC=\frac{23}{50}$$ Apply Law of Cosines on $\bigtriangleup STB$ using the fact that $\angle STB=\angle BTC$ we have $$SB^2=ST^2+BT^2-2(ST)(BT)\cos STB$$ $$SB=\sqrt{4^2+5^2-2(4)(5)\cos BTC}=\frac{\sqrt{565}}{5}$$ Apply Pythagorean Theorem on $\bigtriangleup BSM$ yields $$SM=\sqrt{SB^2-BM^2}=\frac{\sqrt{1585}}{10}$$ Let the perpendicular from $T$ hits $SBC$ at $P$. Let $SP=x$ and $PM=\frac{\sqrt{1585}}{10}-x$. Apply Pythagorean Theorem on $TSP$ and $TMP$ we have $$TP^2=TS^2-SP^2=TM^2-PM^2$$ $$4^2-x^2=(\frac{\sqrt{73}}{2})^2-(\frac{\sqrt{1585}}{10}-x)^2$$ Cancelling out the $x^2$ term and solving gets $x=\frac{181}{2\sqrt{1585}}$.

Finally, by Pythagorean Theorem, $$TP=\sqrt{TS^2-SP^2}=\frac{144}{\sqrt{1585}}$$ so $\lfloor m+\sqrt{n}\rfloor=\boxed{183}$

~ Nafer