AIME 2006 I · 第 14 题

Solution 1

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

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