Difference between revisions of "2017 AIME I Problems/Problem 4"
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From this, we can determine the height of both <math>\triangle CPM</math> and tetrahedron <math>ABCP</math> to be <math>\frac{100\sqrt{3}}{8}</math>; therefore, the volume of the tetrahedron equals <math>\frac{100\sqrt{3}}{8} \cdot 192=800\sqrt{3}</math>; thus, <math>m+n=800+3=\boxed{803}.</math> | From this, we can determine the height of both <math>\triangle CPM</math> and tetrahedron <math>ABCP</math> to be <math>\frac{100\sqrt{3}}{8}</math>; therefore, the volume of the tetrahedron equals <math>\frac{100\sqrt{3}}{8} \cdot 192=800\sqrt{3}</math>; thus, <math>m+n=800+3=\boxed{803}.</math> | ||
+ | -dzhou100 | ||
==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 14:10, 6 July 2019
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 ,
, 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
. Thus, the answer is
.
-RootThreeOverTwo
Solution 3 (Heron's Formula)
Label the four vertices of the tetrahedron and the midpoint of , and notice that the area of the base of the tetrahedron,
, equals
, according to Solution 1.
Notice that the altitude of from
to point
is the height of the tetrahedron. Side
is can be found using the Pythagorean Theorem on
, giving us
Using Heron's Formula, the area of can be written as
Notice that both and
can be rewritten as differences of squares; thus, the expression can be written as
From this, we can determine the height of both and tetrahedron
to be
; therefore, the volume of the tetrahedron equals
; thus,
-dzhou100
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.