Difference between revisions of "2017 AIME II Problems/Problem 15"
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==Solution== | ==Solution== | ||
− | <math>\boxed{682}</math> | + | ===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>Q</math> be the intersection of <math>\overline{XX'}</math> and <math>\overline{MN}</math>. Then <math>AX+AX'\geq 2AQ</math>, from which it follows that <math>f(X) \geq 2(AQ+DQ) = f(Q)</math>. It remains to minimize <math>f(Q)</math> as <math>Q</math> moves along <math>\overline{MN}</math>. | ||
+ | |||
+ | Allow <math>D</math> to rotate about <math>\overline{MN}</math> to point <math>D'</math> in the plane <math>AMN</math> on the side of <math>\overline{MN}</math> opposite <math>A</math>. Because <math>\angle DNM</math> is a right angle, <math>D'N=DN</math>. It then follows that <math>f(Q) = 2(AQ+D'Q)\geq 2AD'</math>, and equality occurs when <math>Q</math> is the intersection of <math>\overline{AD'}</math> and <math>\overline{MN}</math>. Thus <math>\min f(Q) = 2AD'</math>. Because <math>\overline{MD}</math> is the median of <math>\triangle ADB</math>, the Length of Median Formula shows that <math>4MD^2 = 2AD^2 + 2BD^2 - AB^2 = 2\cdot 28^2 + 2 \cdot 44^2 - 52^2</math> and <math>MD^2 = 684</math>. By the Pythagorean Theorem <math>MN^2 = MD^2 - ND^2 = 8</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>. | ||
+ | |||
+ | Claim: <math>O</math> is the gravity center <math>\tfrac14(\vec A + \vec B + \vec C + \vec D)</math>. | ||
+ | Proof. 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> | ||
+ | |||
+ | Claim: The gravity center <math>O</math> coincides with the circumcenter. | ||
+ | Proof. 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>. | ||
=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 18:55, 23 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 .
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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