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,

$\overline {OM} + \overline {OC} = \overline {CM} = 16$

Let $\overline {OM} = d$ . Equation $(1)$ : $d + \sqrt {d^2 + 144} = 16$

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 $m+n = \boxed {803}$ .

Solution by Zeroman

2017 AIME I (Problems • Answer Key • Resources)
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2017 AIME I Problems/Problem 4

Problem 4

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