2017 AIME I Problems/Problem 4

Revision as of 16:41, 8 March 2017 by Zeroman (talk | contribs) (Solution)

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

Solution by Zeroman

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