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2017 AIME II Problems/Problem 15

Revision as of 19:22, 25 March 2017 by Expilncalc (talk | contribs) (→‎Solution 3)

Problem

Tetrahedron $ABCD$ has $AD=BC=28$, $AC=BD=44$, and $AB=CD=52$. For any point $X$ in space, define $f(X)=AX+BX+CX+DX$. The least possible value of $f(X)$ can be expressed as $m\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.

Solution

Solution 1

Let $M$ and $N$ be midpoints of $\overline{AB}$ and $\overline{CD}$. The given conditions imply that $\triangle ABD\cong\triangle BAC$ and $\triangle CDA\cong\triangle DCB$, and therefore $MC=MD$ and $NA=NB$. It follows that $M$ and $N$ both lie on the common perpendicular bisector of $\overline{AB}$ and $\overline{CD}$, and thus line $MN$ is that common perpendicular bisector. Points $B$ and $C$ are symmetric to $A$ and $D$ with respect to line $MN$. If $X$ is a point in space and $X'$ is the point symmetric to $X$ with respect to line $MN$, then $BX=AX'$ and $CX=DX'$, so $f(X) = AX+AX'+DX+DX'$.

Let $Q$ be the intersection of $\overline{XX'}$ and $\overline{MN}$. Then $AX+AX'\geq 2AQ$, from which it follows that $f(X) \geq 2(AQ+DQ) = f(Q)$. It remains to minimize $f(Q)$ as $Q$ moves along $\overline{MN}$.

Allow $D$ to rotate about $\overline{MN}$ to point $D'$ in the plane $AMN$ on the side of $\overline{MN}$ opposite $A$. Because $\angle DNM$ is a right angle, $D'N=DN$. It then follows that $f(Q) = 2(AQ+D'Q)\geq 2AD'$, and equality occurs when $Q$ is the intersection of $\overline{AD'}$ and $\overline{MN}$. Thus $\min f(Q) = 2AD'$. Because $\overline{MD}$ is the median of $\triangle ADB$, the Length of Median Formula shows that $4MD^2 = 2AD^2 + 2BD^2 - AB^2 = 2\cdot 28^2 + 2 \cdot 44^2 - 52^2$ and $MD^2 = 684$. By the Pythagorean Theorem $MN^2 = MD^2 - ND^2 = 8$.

Because $\angle AMN$ and $\angle D'NM$ are right angles, \[(AD')^2 = (AM+D'N)^2 + MN^2 = (2AM)^2 + MN^2 = 52^2 + 8 = 4\cdot 678.\]It follows that $\min f(Q) = 2AD' = 4\sqrt{678}$. The requested sum is $4+678=\boxed{682}$.

Solution 2

Set $a=BC=28$, $b=CA=44$, $c=AB=52$. Let $O$ be the point which minimizes $f(X)$.

Claim: $O$ is the gravity center $\tfrac14(\vec A + \vec B + \vec C + \vec D)$. Proof. Let $M$ and $N$ denote the midpoints of $AB$ and $CD$. From $\triangle ABD \cong \triangle BAC$ and $\triangle CDA \cong \triangle DCB$, we have $MC=MD$, $NA=NB$ an hence $MN$ is a perpendicular bisector of both segments $AB$ and $CD$. Then if $X$ is any point inside tetrahedron $ABCD$, its orthogonal projection onto line $MN$ will have smaller $f$-value; hence we conclude that $O$ must lie on $MN$. Similarly, $O$ must lie on the line joining the midpoints of $AC$ and $BD$. $\blacksquare$

Claim: The gravity center $O$ coincides with the circumcenter. Proof. Let $G_D$ be the centroid of triangle $ABC$; then $DO = \tfrac 34 DG_D$ (by vectors). If we define $G_A$, $G_B$, $G_C$ similarly, we get $AO = \tfrac 34 AG_A$ and so on. But from symmetry we have $AG_A = BG_B = CG_C = DG_D$, hence $AO = BO = CO = DO$. $\blacksquare$

Now we use the fact that an isosceles tetrahedron has circumradius $R = \sqrt{\frac18(a^2+b^2+c^2)}$. Here $R = \sqrt{678}$ so $f(O) = 4R = 4\sqrt{678}$. Therefore, the answer is $4 + 678 = \boxed{682}$.

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 $F(x)= 2d$, 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 $f(X)=4\sqrt 678$.

See Also

2017 AIME II (Problems • Answer Key • Resources)
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Problem 14 		Followed by
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All AIME Problems and Solutions

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