Difference between revisions of "2017 AIME II Problems/Problem 15"
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==Problem== | ==Problem== | ||
− | 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, | + | 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>. |
− | == | + | ==Solutions== |
===Solution 1=== | ===Solution 1=== | ||
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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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>. | ||
− | |||
− | |||
− | Claim: The gravity center <math>O</math> coincides with the circumcenter. | + | <math>\textrm{Claim } 1 \textrm{:}</math> |
− | Proof | + | |
+ | <math>O</math> is the gravity center <math>\tfrac14(\vec A + \vec B + \vec C + \vec D)</math>. | ||
+ | |||
+ | <math>\textrm{Proof:}</math> | ||
+ | |||
+ | Let <math>M</math> and <math>N</math> denote the midpoints of <math>AB</math> and <math>CD</math>. From <math>\triangle ABD \cong \triangle BAC</math> and <math>\triangle CDA \cong \triangle DCB</math>, we have <math>MC=MD</math>, <math>NA=NB</math> an hence <math>MN</math> is a perpendicular bisector of both segments <math>AB</math> and <math>CD</math>. Then if <math>X</math> is any point inside tetrahedron <math>ABCD</math>, its orthogonal projection onto line <math>MN</math> will have smaller <math>f</math>-value; hence we conclude that <math>O</math> must lie on <math>MN</math>. Similarly, <math>O</math> must lie on the line joining the midpoints of <math>AC</math> and <math>BD</math>. <math>\blacksquare</math> | ||
+ | |||
+ | |||
+ | <math>\textrm{Claim } 2 \textrm{:}</math> | ||
+ | |||
+ | The gravity center <math>O</math> coincides with the circumcenter. | ||
+ | |||
+ | <math>\textrm{Proof:}</math> | ||
+ | |||
+ | Let <math>G_D</math> be the centroid of triangle <math>ABC</math>; then <math>DO = \tfrac 34 DG_D</math> (by vectors). If we define <math>G_A</math>, <math>G_B</math>, <math>G_C</math> similarly, we get <math>AO = \tfrac 34 AG_A</math> and so on. But from symmetry we have <math>AG_A = BG_B = CG_C = DG_D</math>, hence <math>AO = BO = CO = DO</math>. <math>\blacksquare</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>. | 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=== | ===Solution 3=== | ||
− | Isosceles tetrahedron is inscribed in a rectangular box, whose facial diagonals are the edges of the tetrahedron. Minimum | + | 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. |
− | Let the three dimensions of the box be a, b, c. | + | Let the three dimensions of the box be <math>a, b, c</math>. |
− | a^2+b^2=28^2; | + | <math>a^2+b^2=28^2; |
c^2+b^2=52^2; | c^2+b^2=52^2; | ||
− | a^2+c^2=44^2. | + | a^2+c^2=44^2.</math> |
− | Add three equations, d^2=(28^2+52^2+44^2)/2. | + | Add three equations, <math>d^2=(28^2+52^2+44^2)/2</math>. |
− | Hence f(X)= | + | Hence <math>f(X)=4\sqrt {678}</math>. |
− | =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}} | ||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:54, 9 November 2020
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 .
Solutions
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 .
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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