Difference between revisions of "2017 AIME I Problems/Problem 4"
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<cmath>V = (192)(25\sqrt{3}/2)/3</cmath> | <cmath>V = (192)(25\sqrt{3}/2)/3</cmath> | ||
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>. | ||
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+ | NOTE : If you don’t know or remember the formula for the volume of a triangular pyramid, you can derive it using calculus as follows : | ||
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+ | Take a small triangular element in the pyramid. We know that it’s area is proportional to the height from the vertex to the base. Hence, we know that <math>\frac{A_{small element}}{A} = \frac{h^2}{H^2} \implies A_{small element} = \frac{Ah^2}{H^2}</math>. Now integrate it taking the limits <math>0</math> to <math>H</math> | ||
==Shortcut== | ==Shortcut== |
Latest revision as of 06:59, 28 December 2020
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 .
NOTE : If you don’t know or remember the formula for the volume of a triangular pyramid, you can derive it using calculus as follows :
Take a small triangular element in the pyramid. We know that it’s area is proportional to the height from the vertex to the base. Hence, we know that . Now integrate it taking the limits to
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
Video Solution
https://youtu.be/Mk-MCeVjSGc ~Shreyas S
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.