Difference between revisions of "2017 AIME II Problems/Problem 15"
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Now we use the fact that an isosceles tetrahedron has circumradius <math>R = \sqrt{\frac18(a^2+b^2+c^2)}</math>. Here <math>R = \sqrt{678}</math> so <math>f(O) = 4R = 4\sqrt{678}</math>. Therefore, the answer is <math>4 + 678 = \boxed{682}</math>. | Now we use the fact that an isosceles tetrahedron has circumradius <math>R = \sqrt{\frac18(a^2+b^2+c^2)}</math>. Here <math>R = \sqrt{678}</math> so <math>f(O) = 4R = 4\sqrt{678}</math>. Therefore, the answer is <math>4 + 678 = \boxed{682}</math>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Isosceles tetrahedron is inscribed in a rectangular box, whose facial diagonals are the edges of the tetrahedron. Minimum F(X) occurs at the center of gravity, and = 2d, where d is the length of the spatial diagonal of the rectangular box. | ||
+ | |||
+ | Let the three dimensions of the box be a, b, c. | ||
+ | a^2+b^2=28^2; | ||
+ | c^2+b^2=52^2; | ||
+ | a^2+c^2=44^2. | ||
+ | |||
+ | Add three equations, d^2=(28^2+52^2+44^2)/2. | ||
+ | Hence f(X)=4sqrt(678). | ||
=See Also= | =See Also= | ||
{{AIME box|year=2017|n=II|num-b=14|after=Last Question}} | {{AIME box|year=2017|n=II|num-b=14|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:48, 25 March 2017
Problem
Tetrahedron has , , and . For any point in space, define . The least possible value of can be expressed as , where and are positive integers, and is not divisible by the square of any prime. Find .
Solution
Solution 1
Let and be midpoints of and . The given conditions imply that and , and therefore and . It follows that and both lie on the common perpendicular bisector of and , and thus line is that common perpendicular bisector. Points and are symmetric to and with respect to line . If is a point in space and is the point symmetric to with respect to line , then and , so .
Let be the intersection of and . Then , from which it follows that . It remains to minimize as moves along .
Allow to rotate about to point in the plane on the side of opposite . Because is a right angle, . It then follows that , and equality occurs when is the intersection of and . Thus . Because is the median of , the Length of Median Formula shows that and . By the Pythagorean Theorem .
Because and are right angles, It follows that . The requested sum is .
Solution 2
Set , , . Let be the point which minimizes .
Claim: is the gravity center . Proof. Let and denote the midpoints of and . From and , we have , an hence is a perpendicular bisector of both segments and . Then if is any point inside tetrahedron , its orthogonal projection onto line will have smaller -value; hence we conclude that must lie on . Similarly, must lie on the line joining the midpoints of and .
Claim: The gravity center coincides with the circumcenter. Proof. Let be the centroid of triangle ; then (by vectors). If we define , , similarly, we get and so on. But from symmetry we have , hence .
Now we use the fact that an isosceles tetrahedron has circumradius . Here so . Therefore, the answer is .
Solution 3
Isosceles tetrahedron is inscribed in a rectangular box, whose facial diagonals are the edges of the tetrahedron. Minimum F(X) occurs at the center of gravity, and = 2d, where d is the length of the spatial diagonal of the rectangular box.
Let the three dimensions of the box be a, b, c. a^2+b^2=28^2; c^2+b^2=52^2; a^2+c^2=44^2.
Add three equations, d^2=(28^2+52^2+44^2)/2. Hence f(X)=4sqrt(678).
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.