2006 AIME I Problem 14

Question

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.

)

Answer

Answer: 183

Concepts:3D geometry coordinate geometry distance formula symmetry

Difficulty rating: 3270

Small hint:

Set coordinates: top at

(0, 0, 4)

and feet

3

from the origin; the broken leg now ends

\frac{4}{5}

of the way from top to old foot

Big hint:

By symmetry

h

is a 2D distance in the

xz

-plane: from

(0, 4)

to the line through the broken tip and the midpoint of the other two feet

Solution:

Place the top at

T = (0, 0, 4).

Each leg has length

5,

so each foot is

3

from the origin:

A = (3, 0, 0),

B = \left(-\frac{3}{2}, \frac{3\sqrt{3}}{2}, 0\right),

C = \left(-\frac{3}{2}, -\frac{3\sqrt{3}}{2}, 0\right).

The broken leg has length

4,

so its tip is

A' = \frac{4}{5}(3, 0, 0)

+ \frac{1}{5}(0, 0, 4)

= \left(\frac{12}{5}, 0, \frac{4}{5}\right),

and the tripod now stands on the plane

A'BC,

with

h

equal to the distance from

T

to that plane.

The plane

A'BC

contains

\overline{BC},

which is parallel to the

y

-axis and passes through the midpoint

M = \left(-\frac{3}{2}, 0, 0\right),

so its distance from

T

can be measured in the

xz

-plane: it is the distance from

(0, 4)

to the line through

\left(-\frac{3}{2}, 0\right)

and

\left(\frac{12}{5}, \frac{4}{5}\right),

whose equation is

8x - 39z + 12 = 0.

Therefore

\begin{aligned} h &= \frac{|8 \cdot 0 - 39 \cdot 4 + 12|}{\sqrt{8^2 + 39^2}} \\ &= \frac{144}{\sqrt{1585}}. \end{aligned}

Here

m = 144

and

n = 1585 = 5 \cdot 317

is squarefree. Since

39^2 = 1521 \lt 1585 \lt 1600,

we get

\lfloor 144 + \sqrt{1585} \rfloor

= 144 + 39 = 183.