Difference between revisions of "2017 AIME I Problems/Problem 4"
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This simplifies to <math>V = 800\sqrt {3}</math>, so <math>m+n = \boxed {803}</math>. | This simplifies to <math>V = 800\sqrt {3}</math>, so <math>m+n = \boxed {803}</math>. | ||
− | == | + | ==Shortcut== |
− | + | Here is a shortcut for finding the radius <math>R</math> of the circumcenter of <math>\triangle ABC</math>. | |
+ | |||
+ | As before, we find that the foot of the altitude from <math>P</math> lands on the circumcenter of <math>\triangle ABC</math>. Let <math>BC=a</math>, <math>AC=b</math>, and <math>AB=c</math>. | ||
Then we write the area of <math>\triangle ABC</math> in two ways: | Then we write the area of <math>\triangle ABC</math> in two ways: | ||
− | <cmath>[ABC] = </cmath> | + | <cmath>[ABC]= \frac{1}{2} \cdot 24 \cdot 16 = \frac{abc}{4R}</cmath> |
+ | |||
+ | Plugging in the values of <math>20</math>, <math>20</math>, and <math>24</math> for <math>a</math>, <math>b</math>, and <math>c</math> respectively, and solving for <math>R</math>, we obtain <math>R= \frac{25}{2}=OA=OB=OC</math>. | ||
+ | |||
+ | Then continue as before to use the Pythagorean Theorem on <math>\triangle AOP</math>, find <math>h</math>, and find the volume of the pyramid. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2017|n=I|num-b=3|num-a=5}} | {{AIME box|year=2017|n=I|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:58, 5 January 2018
Contents
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
.
Shortcut
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 the values of ,
, 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.
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.