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

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==Problem==
 
==Problem==
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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, 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>.
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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>.
  
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==Solution==
+
 
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===Solution 1===
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==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>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>.
 
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>.
 
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>.
 
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>.
  
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===Solution 2===
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==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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Claim:  <math>O</math> is the gravity center <math>\tfrac14(\vec A + \vec B + \vec C + \vec D)</math>.
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<math>\boxed{\textrm{Claim 1: } O \textrm{ is the gravity center } \ \tfrac {1}{4}(\vec A + \vec B + \vec C + \vec D)}</math>
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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>
 +
 
 +
<math>\boxed{\textrm{Claim 2: The gravity center } O \textrm{ coincides with the circumcenter.} \phantom{\vec A}}</math>
  
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Claim: The gravity center <math>O</math> coincides with the circumcenter.
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<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>
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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>
 
  
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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>.
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Now we use the fact that an isosceles tetrahedron has circumradius <math>R = \sqrt{\tfrac18(a^2+b^2+c^2)}</math>.  
  
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===Solution 3===
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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>.
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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.
 
  
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Let the three dimensions of the box be <math>a, b, c</math>.
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==Solution 3==
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<math>a^2+b^2=28^2; 
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[[File:2017 AIME II 15b.png|300px|right]]
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c^2+b^2=52^2;  
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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.
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a^2+c^2=44^2.</math>
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<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>
  
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Add three equations, <math>d^2=(28^2+52^2+44^2)/2</math>.
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'''vladimir.shelomovskii@gmail.com, vvsss'''  (Reconstruction)
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Hence <math>f(X)=4\sqrt {678}</math>.
 
  
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=See Also=
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==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 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)

2017 AIME II 15.png

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 $\overline{MN}$.

2017 AIME II 15a.png

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

2017 AIME II 15b.png

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)

See Also

2017 AIME II (Problems • Answer Key • Resources)
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