Difference between revisions of "2001 AIME II Problems/Problem 15"
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Set the coordinate system so that vertex <math>E</math>, where the drilling starts, is at <math>(8,8,8)</math>. Using a little visualization (involving some [[similar triangles]], because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining <math>(1,0,0)</math> to <math>(2,2,0)</math>, and <math>(0,1,0)</math> to <math>(2,2,0)</math>, and similarly for the other three faces meeting at the origin (by symmetry). So one face of the tunnel is the polygon with vertices (in that order), <math>S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0)</math>, and the other two faces of the tunnel are congruent to this shape. | Set the coordinate system so that vertex <math>E</math>, where the drilling starts, is at <math>(8,8,8)</math>. Using a little visualization (involving some [[similar triangles]], because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining <math>(1,0,0)</math> to <math>(2,2,0)</math>, and <math>(0,1,0)</math> to <math>(2,2,0)</math>, and similarly for the other three faces meeting at the origin (by symmetry). So one face of the tunnel is the polygon with vertices (in that order), <math>S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0)</math>, and the other two faces of the tunnel are congruent to this shape. | ||
− | Observe that this shape is made up of two congruent [[trapezoid]]s each with height <math>\sqrt {2}</math> and bases <math>7\sqrt {3}</math> and <math>6\sqrt {3}</math>. Together they make up an area of <math>\sqrt {2}(7\sqrt {3} + 6\sqrt {3}) = 13\sqrt {6}</math>. The total area of the tunnel is then <math>3\cdot13\sqrt {6} = 39\sqrt {6}</math>. Around the | + | Observe that this shape is made up of two congruent [[trapezoid]]s each with height <math>\sqrt {2}</math> and bases <math>7\sqrt {3}</math> and <math>6\sqrt {3}</math>. Together they make up an area of <math>\sqrt {2}(7\sqrt {3} + 6\sqrt {3}) = 13\sqrt {6}</math>. The total area of the tunnel is then <math>3\cdot13\sqrt {6} = 39\sqrt {6}</math>. Around the corner <math>E</math> and the corner opposite <math>E</math> we're missing an area of <math>6</math>. So the outside area is <math>6\cdot 64 - 2\cdot 6 = 372</math>. Thus the the total surface area is <math>372 + 39\sqrt {6}</math>, and the answer is <math>372 + 39 + 6 = \boxed{417}</math>. |
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Revision as of 14:18, 6 February 2016
Problem
Let ,
, and
be three adjacent square faces of a cube, for which
, and let
be the eighth vertex of the cube. Let
,
, and
, be the points on
,
, and
, respectively, so that
. A solid
is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to
, and containing the edges,
,
, and
. The surface area of
, including the walls of the tunnel, is
, where
,
, and
are positive integers and
is not divisible by the square of any prime. Find
.
Solution
![[asy] import three; currentprojection = perspective(5,-40,12); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8)--(0,0,8)--(0,0,1)); draw((1,0,0)--(8,0,0)--(8,8,0)--(0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,8,8)--(0,0,8)--(0,0,1)); draw((8,8,0)--(8,8,6),l); draw((8,0,8)--(8,6,8)); draw((0,8,8)--(6,8,8)); draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle); [/asy]](http://latex.artofproblemsolving.com/4/1/4/4149e9121b6bad141b8f2860b3c45b17cd77c9a1.png)
![[asy] import three; currentprojection = perspective(5,40,12); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8),l); draw((8,0,8)--(0,0,8)); draw((0,0,8)--(0,0,1),l); draw((8,0,0)--(8,8,0)); draw((8,8,0)--(0,8,0)); draw((0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,0,8)--(0,0,1),l); draw((8,8,0)--(8,8,6)); draw((8,0,8)--(8,6,8)); draw((0,0,8)--(0,8,8)--(6,8,8)); draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle); [/asy]](http://latex.artofproblemsolving.com/c/6/f/c6f7e4ede0c4d7cc77c040c58a09164295de9135.png)
Set the coordinate system so that vertex , where the drilling starts, is at
. Using a little visualization (involving some similar triangles, because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining
to
, and
to
, and similarly for the other three faces meeting at the origin (by symmetry). So one face of the tunnel is the polygon with vertices (in that order),
, and the other two faces of the tunnel are congruent to this shape.
Observe that this shape is made up of two congruent trapezoids each with height and bases
and
. Together they make up an area of
. The total area of the tunnel is then
. Around the corner
and the corner opposite
we're missing an area of
. So the outside area is
. Thus the the total surface area is
, and the answer is
.
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See also
2001 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.