2017 AIME I Problem 4

Question

A pyramid has a triangular base with side lengths

20,

20,

and

24.

The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length

25.

The volume of the pyramid is

m\sqrt{n},

where

m

and

n

are positive integers, and

n

is not divisible by the square of any prime. Find

m + n.

Answer

Because all three lateral edges equal

25,

the apex lies directly above the circumcenter of the base triangle

The base has area

192;

compute its circumradius from

R = \frac{abc}{4K},

then get the height by the Pythagorean theorem

Since the apex is equidistant from all three base vertices, its foot is the circumcenter of the base. The base is isosceles with sides

20,

20,

and

24:

its altitude to the side of length

24

is

\sqrt{20^2 - 12^2} = 16,

so its area is

K = \frac{1}{2} \cdot 24 \cdot 16 = 192,

and its circumradius is

R = \frac{abc}{4K} = \frac{20 \cdot 20 \cdot 24}{4 \cdot 192} = \frac{25}{2}.

The height of the pyramid is

\sqrt{25^2 - \left(\frac{25}{2}\right)^2} = \frac{25\sqrt{3}}{2},

so the volume is

\frac{1}{3} \cdot 192 \cdot \frac{25\sqrt{3}}{2} = 800\sqrt{3}.

Then

m + n = 800 + 3 = 803.

Problem 4 in other years