Difference between revisions of "2017 AIME II Problems/Problem 15"
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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>. | ||
Revision as of 02:42, 16 June 2022
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
.
Official Solution (MAA)
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
.
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
.
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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