2017 AIME I Problems/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 .
Let the triangular base be , with . We find that the altitude to side is , so the area of is .
Let the fourth vertex of the tetrahedron be , and let the midpoint of be . Since is equidistant from , , and , the line through perpendicular to the plane of will pass through the circumcenter of , which we will call . Note that is equidistant from each of , , and . Then,
Let . Equation :
Squaring both sides, we have
Substituting with equation :
We now find that .
Let the distance . Using the Pythagorean Theorem on triangle , , or (all three are congruent by SSS):
Finally, by the formula for volume of a pyramid,
This simplifies to , so .
Here is a shortcut for finding the radius of the circumcenter of .
As before, we find that the foot of the altitude from lands on the circumcenter of . Let , , and . Then we write the area of in two ways:
Plugging in , , and for , , and respectively, and solving for , we obtain .
Then continue as before to use the Pythagorean Theorem on , find , and find the volume of the pyramid.
Solution 2 (Coordinates)
We can place a three dimensional coordinate system on this pyramid. WLOG assume the vertex across from the line that has length is at the origin, or . Then, the two other vertices can be and . Let the fourth vertex have coordinates of . We have the following equations from the distance formula.
Adding the last two equations and substituting in the first equation, we get that . If you drew a good diagram, it should be obvious that . Now, solving for , we get that . So, the height of the pyramid is . The base is equal to the area of the triangle, which is . The volume is therefore . Thus, the answer is .
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