Difference between revisions of "2017 AIME I Problems/Problem 4"
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<cmath>(x-12)^2+(y+16)^2+z^2=625</cmath> | <cmath>(x-12)^2+(y+16)^2+z^2=625</cmath> | ||
− | Adding the last two equations and substituting in the first equation, we get that <math>y=-\frac{25}{2}</math>. If you drew a good diagram, it should be obvious that <math>x=0</math>. Now, solving for <math>z</math>, we get that <math>z=\frac{25\sqrt{3}}{2}</math>. So, the height of the pyramid is <math>\frac{25\sqrt{3}}{2}</math>. The base is equal to the area of the triangle, which is <math>\frac{1}{2} \cdot 24 \cdot 16 = 192</math>. The volume is | + | Adding the last two equations and substituting in the first equation, we get that <math>y=-\frac{25}{2}</math>. If you drew a good diagram, it should be obvious that <math>x=0</math>. Now, solving for <math>z</math>, we get that <math>z=\frac{25\sqrt{3}}{2}</math>. So, the height of the pyramid is <math>\frac{25\sqrt{3}}{2}</math>. The base is equal to the area of the triangle, which is <math>\frac{1}{2} \cdot 24 \cdot 16 = 192</math>. The volume is <math>\frac{1}{3} \cdot 192 \cdot \frac{25\sqrt{3}}{2} = 800\sqrt{3}</math>. Thus, the answer is <math>800+3 = \boxed{803}</math>. |
'''-RootThreeOverTwo''' | '''-RootThreeOverTwo''' |
Revision as of 22:34, 14 May 2018
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
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.