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Difference between revisions of "2017 AIME I Problems/Problem 4"
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==Solution==
 
==Solution==
 
Let the triangular base be <math>\triangle ABC</math>. Using Simplified Heron's formula for the area of an isosceles triangle gives <math>12\sqrt{32(8)}=192</math>.
 
Let the triangular base be <math>\triangle ABC</math>. Using Simplified Heron's formula for the area of an isosceles triangle gives <math>12\sqrt{32(8)}=192</math>.
 +
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<math>\vdots</math>
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The volume of the pyramid is <math>800\sqrt{3}</math>, so the answer is <math>\boxed{803}</math>

Revision as of 16:12, 8 March 2017

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 $\triangle ABC$. Using Simplified Heron's formula for the area of an isosceles triangle gives $12\sqrt{32(8)}=192$.

$\vdots$

The volume of the pyramid is $800\sqrt{3}$, so the answer is $\boxed{803}$

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