# Difference between revisions of "2017 AIME I Problems/Problem 4"

## Problem 4

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 $m+n$.

## Solution

Let the triangular base be $\triangle ABC$, with $\overline {AB} = 24$. Using Simplified Heron's formula for the area of an isosceles triangle gives $12\sqrt{32(8)}=192$.

Let the fourth vertex of the tetrahedron be $P$, and let the midpoint of $\overline {AB}$ be $M$. Since $P$ is equidistant from $A$, $B$, and $C$, the line through $P$ perpendicular to the plane of $\triangle ABC$ will pass through the circumcenter of $\triangle ABC$, which we will call $O$. Note that $O$ is equidistant from each of $A$, $B$, and $C$. We find that $\overline {CM} = 16$. Then, $\overline {OM} + \overline {OC} = \overline {CM} = 16$ $d + \sqrt {d^2 + 144} = 16$ (1)

Squaring both sides, we have $d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256$ $2d^2 + 2d\sqrt {d^2+144} = 112$ $2d(d + \sqrt {d^2+144}) = 112$

Substituting with equation (1): $2d(16) = 112$ $d = 7/2$.

We now find that $\sqrt{d^2 + 144} = 25/2$.

Let the distance $\overline {OP} = h$. Using the Pythagorean Theorem on triangle $AOP$, $BOP$, or $COP$ (all three are congruent by SSS): $25^2 = h^2 + (\sqrt {d^2 + 144})^2$ $625 = h^2 + 625/4$ $1875/4 = h^2$ $25\sqrt {3} / 2 = h$.

Finally, by the formula for volume of a pyramid, $V = Bh/3$ $V = (192)(25\sqrt{3}/2)/3$. This simplifies to $V = 800\sqrt {3}$, so $m+n = \boxed {803}$.