2017 AIME I Problems/Problem 4: Difference between revisions
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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>. | ||
<math>\vdots</math> | |||
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 
, , and 
. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length 
. The volume of the pyramid is 
, where 
and 
are positive integers, and is not divisible by the square of any prime. Find 
.
Solution
Let the triangular base be 
. Using Simplified Heron's formula for the area of an isosceles triangle gives 
.
The volume of the pyramid is 
, so the answer is 
