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

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

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. $A = (0,0,0)$ $B = (3\sqrt{3},0,0)$ $C = (\frac{3\sqrt{3}}{2}, \frac{9}{2}, 0)$ $T = (\frac{3\sqrt{3}}{2}, \frac{3}{2}, 4)$

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.

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

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

Solution by SimonSun

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
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2006 AIME I Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

See also