Difference between revisions of "2017 AIME II Problems/Problem 15"
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Tetrahedron <math>ABCD</math> has <math>AD=BC=28</math>, <math>AC=BD=44</math>, and <math>AB=CD=52</math>. For any point <math>X</math> in space, suppose <math>f(X)=AX+BX+CX+DX</math>. The least possible value of <math>f(X)</math> can be expressed as <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. | Tetrahedron <math>ABCD</math> has <math>AD=BC=28</math>, <math>AC=BD=44</math>, and <math>AB=CD=52</math>. For any point <math>X</math> in space, suppose <math>f(X)=AX+BX+CX+DX</math>. The least possible value of <math>f(X)</math> can be expressed as <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. | ||
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− | == | + | ==Official Solution (MAA)=== |
Let <math>M</math> and <math>N</math> be midpoints of <math>\overline{AB}</math> and <math>\overline{CD}</math>. The given conditions imply that <math>\triangle ABD\cong\triangle BAC</math> and <math>\triangle CDA\cong\triangle DCB</math>, and therefore <math>MC=MD</math> and <math>NA=NB</math>. It follows that <math>M</math> and <math>N</math> both lie on the common perpendicular bisector of <math>\overline{AB}</math> and <math>\overline{CD}</math>, and thus line <math>MN</math> is that common perpendicular bisector. Points <math>B</math> and <math>C</math> are symmetric to <math>A</math> and <math>D</math> with respect to line <math>MN</math>. If <math>X</math> is a point in space and <math>X'</math> is the point symmetric to <math>X</math> with respect to line <math>MN</math>, then <math>BX=AX'</math> and <math>CX=DX'</math>, so <math>f(X) = AX+AX'+DX+DX'</math>. | Let <math>M</math> and <math>N</math> be midpoints of <math>\overline{AB}</math> and <math>\overline{CD}</math>. The given conditions imply that <math>\triangle ABD\cong\triangle BAC</math> and <math>\triangle CDA\cong\triangle DCB</math>, and therefore <math>MC=MD</math> and <math>NA=NB</math>. It follows that <math>M</math> and <math>N</math> both lie on the common perpendicular bisector of <math>\overline{AB}</math> and <math>\overline{CD}</math>, and thus line <math>MN</math> is that common perpendicular bisector. Points <math>B</math> and <math>C</math> are symmetric to <math>A</math> and <math>D</math> with respect to line <math>MN</math>. If <math>X</math> is a point in space and <math>X'</math> is the point symmetric to <math>X</math> with respect to line <math>MN</math>, then <math>BX=AX'</math> and <math>CX=DX'</math>, so <math>f(X) = AX+AX'+DX+DX'</math>. | ||
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Because <math>\angle AMN</math> and <math>\angle D'NM</math> are right angles, <cmath>(AD')^2 = (AM+D'N)^2 + MN^2 = (2AM)^2 + MN^2 = 52^2 + 8 = 4\cdot 678.</cmath>It follows that <math>\min f(Q) = 2AD' = 4\sqrt{678}</math>. The requested sum is <math>4+678=\boxed{682}</math>. | Because <math>\angle AMN</math> and <math>\angle D'NM</math> are right angles, <cmath>(AD')^2 = (AM+D'N)^2 + MN^2 = (2AM)^2 + MN^2 = 52^2 + 8 = 4\cdot 678.</cmath>It follows that <math>\min f(Q) = 2AD' = 4\sqrt{678}</math>. The requested sum is <math>4+678=\boxed{682}</math>. | ||
− | + | ==Solution 2== | |
Set <math>a=BC=28</math>, <math>b=CA=44</math>, <math>c=AB=52</math>. Let <math>O</math> be the point which minimizes <math>f(X)</math>. | Set <math>a=BC=28</math>, <math>b=CA=44</math>, <math>c=AB=52</math>. Let <math>O</math> be the point which minimizes <math>f(X)</math>. | ||
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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 <math>f(X)</math> occurs at the center of gravity, and <math>F(x)= 2d</math>, where <math>d</math> is the length of the spatial diagonal of the rectangular box. | Isosceles tetrahedron is inscribed in a rectangular box, whose facial diagonals are the edges of the tetrahedron. Minimum <math>f(X)</math> occurs at the center of gravity, and <math>F(x)= 2d</math>, where <math>d</math> is the length of the spatial diagonal of the rectangular box. | ||
Revision as of 09:36, 4 February 2022
Problem
Tetrahedron has , , and . For any point in space, suppose . 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 .
Official Solution (MAA)=
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 .
is the gravity center .
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 .
The gravity center coincides with the circumcenter.
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 occurs at the center of gravity, and , where is the length of the spatial diagonal of the rectangular box.
Let the three dimensions of the box be .
Add three equations, . Hence .
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 |
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