Difference between revisions of "2017 AIME I Problems/Problem 4"
m (→Solution) |
(→Solution) |
||
Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
− | Let the triangular base be <math>\triangle ABC</math>, with <math>\overline {AB} = 24</math>. | + | Let the triangular base be <math>\triangle ABC</math>, with <math>\overline {AB} = 24</math>. We find that the altitude to side <math>AB</math> is <math>16</math>, so the area of <math>\triangle ABC</math> is <math>(24*16)/2 = 192</math>. |
− | Let the fourth vertex of the tetrahedron be <math>P</math>, and let the midpoint of <math>\overline {AB}</math> be <math>M</math>. Since <math>P</math> is equidistant from <math>A</math>, <math>B</math>, and <math>C</math>, the line through <math>P</math> perpendicular to the plane of <math>\triangle ABC</math> will pass through the circumcenter of <math>\triangle ABC</math>, which we will call <math>O</math>. Note that <math>O</math> is equidistant from each of <math>A</math>, <math>B</math>, and <math>C | + | Let the fourth vertex of the tetrahedron be <math>P</math>, and let the midpoint of <math>\overline {AB}</math> be <math>M</math>. Since <math>P</math> is equidistant from <math>A</math>, <math>B</math>, and <math>C</math>, the line through <math>P</math> perpendicular to the plane of <math>\triangle ABC</math> will pass through the circumcenter of <math>\triangle ABC</math>, which we will call <math>O</math>. Note that <math>O</math> is equidistant from each of <math>A</math>, <math>B</math>, and <math>C</math>. Then, |
<cmath>\overline {OM} + \overline {OC} = \overline {CM} = 16</cmath> | <cmath>\overline {OM} + \overline {OC} = \overline {CM} = 16</cmath> |
Revision as of 20:10, 20 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 , 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 .
Solution by Zeroman
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.