Difference between revisions of "2017 AIME II Problems/Problem 15"

Latest revision as of 01:08, 22 January 2024

Problem

Tetrahedron $ABCD$ has $AD=BC=28$ , $AC=BD=44$ , and $AB=CD=52$ . For any point $X$ in space, suppose $f(X)=AX+BX+CX+DX$ . 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 $m+n$ .

Official Solution (MAA)

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 $f(X)$ .

$\boxed{\textrm{Claim 1: } O \textrm{ is the gravity center } \ \tfrac {1}{4}(\vec A + \vec B + \vec C + \vec D)}$

$\textrm{Proof:}$ 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 $BD$ . $\square$

$\boxed{\textrm{Claim 2: The gravity center } O \textrm{ coincides with the circumcenter.} \phantom{\vec A}}$

$\textrm{Proof:}$ 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 $AO = BO = CO = DO$ . $\square$

Now we use the fact that an isosceles tetrahedron has circumradius $R = \sqrt{\tfrac18(a^2+b^2+c^2)}$ .

Here $R = \sqrt{678}$ so $f(O) = 4R = 4\sqrt{678}$ . Therefore, the answer is $4 + 678 = \boxed{682}$ .

Solution 3

Isosceles tetrahedron $ABCD$ or Disphenoid can be inscribed in a parallelepiped $AB'CD'C'DA'B,$ whose facial diagonals are the pares of equal edges of the tetrahedron $(AC = B'D',$ where $B'D' = BD).$ This parallelepiped is right-angled, therefore it is circumscribed and has equal diagonals. The center O of the circumscribed sphere (coincide with the centroid) has equal distance from each vertex. Tetrachedrons $ABCD$ and $A'B'C'D'$ are congruent, so point of symmetry O is point of minimum $f(X). f(O)= 4R$ , where $R$ is the circumradius of parallelepiped. $8R^2 = 2 CC'^2 = 2CD'^2 + 2D'B^2 + 2BC'^2,$ $2 CC'^2 = (CD'^2 + BC'^2) + (BC'^2 + BD'^2) + (CD'^2 + BD'^2) = AC^2 + AB^2+BC^2,$ $R = OC =\sqrt{\frac {AB^2 + AC^2 + AD^2}{8}}, f(O)= 4R = 4\sqrt {678}.$

vladimir.shelomovskii@gmail.com, vvsss (Reconstruction)

@@ Line 1: / Line 1: @@
 ==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, define <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>.
-==Solution==
-===Solution 1===
+==Official Solution (MAA)==
+[[File:2017 AIME II 15.png|300px|right]]
 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>.
+[[File:2017 AIME II 15a.png|300px|right]]
 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===
+==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>.
+<math>\boxed{\textrm{Claim 1: } O \textrm{ is the gravity center } \ \tfrac {1}{4}(\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>
+<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>\square</math>
-Claim: The gravity center <math>O</math> coincides with the circumcenter.
+<math>\boxed{\textrm{Claim 2: The gravity center } O \textrm{ coincides with the circumcenter.} \phantom{\vec A}}</math>
-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>
+<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>\square</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{\tfrac18(a^2+b^2+c^2)}</math>.
-===Solution 3===
+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>.
-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 <math>a, b, c</math>.
+==Solution 3==
-<math>a^2+b^2=28^2;
+[[File:2017 AIME II 15b.png|300px|right]]
-c^2+b^2=52^2;
+Isosceles tetrahedron <math>ABCD</math> or [https://en.wikipedia.org/wiki/Disphenoid Disphenoid] can be inscribed in a parallelepiped <math>AB'CD'C'DA'B,</math> whose facial diagonals are the pares of equal edges of the tetrahedron <math>(AC = B'D',</math> where <math>B'D' = BD).</math> This parallelepiped is right-angled, therefore it is circumscribed and has equal diagonals. The center O of the circumscribed sphere (coincide with the centroid) has equal distance from each vertex. Tetrachedrons <math>ABCD</math> and <math>A'B'C'D'</math> are congruent, so point of symmetry O is point of minimum <math>f(X). f(O)= 4R</math>, where <math>R</math> is the circumradius of parallelepiped.
-a^2+c^2=44^2.</math>
+<cmath>8R^2 = 2 CC'^2 = 2CD'^2 + 2D'B^2 + 2BC'^2, </cmath>
+<cmath>2 CC'^2 = (CD'^2 + BC'^2) + (BC'^2 + BD'^2) + (CD'^2 + BD'^2) = AC^2 + AB^2+BC^2,</cmath>
+<cmath>R = OC =\sqrt{\frac {AB^2 + AC^2 + AD^2}{8}}, f(O)= 4R = 4\sqrt {678}.</cmath>
-Add three equations, <math>d^2=(28^2+52^2+44^2)/2</math>.
+'''vladimir.shelomovskii@gmail.com, vvsss'''   (Reconstruction)
-Hence <math>f(X)=4\sqrt {678}</math>.
 ==See Also==

2017 AIME II (Problems • Answer Key • Resources)
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