Difference between revisions of "2015 AIME II Problems/Problem 9"
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Our aim is to find the volume of the part of the cube submerged in the cylinder. | Our aim is to find the volume of the part of the cube submerged in the cylinder. | ||
| − | In the problem, since three edges emanate from each vertex, the boundary of the cylinder touches the cube at three points. Because the space diagonal of the cube is vertical, by the symmetry of the cube, the three points form an equilateral triangle. Because the radius of the circle is 4, by the Law of Cosines, the side length s of the equilateral triangle is | + | In the problem, since three edges emanate from each vertex, the boundary of the cylinder touches the cube at three points. Because the space diagonal of the cube is vertical, by the symmetry of the cube, the three points form an equilateral triangle. Because the radius of the circle is <math>4</math>, by the Law of Cosines, the side length s of the equilateral triangle is |
<cmath>s^2 = 2*(4^2) - 2*(4^2)\cos(120^{\circ}) = 3(4^2)</cmath> | <cmath>s^2 = 2*(4^2) - 2*(4^2)\cos(120^{\circ}) = 3(4^2)</cmath> | ||
| − | so <math>s = 4\sqrt{3}</math>. Again by the symmetry of the cube, the volume we want to find is the volume of a tetrahedron with right angles on all faces at the submerged vertex, so since the lengths of the legs of the tetrahedron are <math>\frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}</math> (the three triangular faces touching the submerged vertex are all 45-45-90 triangles) so | + | so <math>s = 4\sqrt{3}</math>. Again by the symmetry of the cube, the volume we want to find is the volume of a tetrahedron with right angles on all faces at the submerged vertex, so since the lengths of the legs of the tetrahedron are <math>\frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}</math> (the three triangular faces touching the submerged vertex are all <math>45-45-90</math> triangles) so |
<cmath>v = \frac{1}{3}(2\sqrt{6})(\frac{1}{2} \cdot (2\sqrt{6})^2) = \frac{1}{6} \cdot 48\sqrt{6} = 8\sqrt{6}</cmath> | <cmath>v = \frac{1}{3}(2\sqrt{6})(\frac{1}{2} \cdot (2\sqrt{6})^2) = \frac{1}{6} \cdot 48\sqrt{6} = 8\sqrt{6}</cmath> | ||
Revision as of 19:41, 27 March 2015
Problem
A cylindrical barrel with radius 
feet and height 
feet is full of water. A solid cube with side length 
feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is 
cubic feet. Find 
.
![[asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight); triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2)); draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype("2 4")); draw(X+B--X+C+B,linetype("2 4")); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype("2 4")); [/asy]](http://latex.artofproblemsolving.com/9/e/a/9ea530dcf8660904fcdb6c7e02d431954bb03eda.png)
Solution
Our aim is to find the volume of the part of the cube submerged in the cylinder.
In the problem, since three edges emanate from each vertex, the boundary of the cylinder touches the cube at three points. Because the space diagonal of the cube is vertical, by the symmetry of the cube, the three points form an equilateral triangle. Because the radius of the circle is , by the Law of Cosines, the side length s of the equilateral triangle is
![\[s^2 = 2*(4^2) - 2*(4^2)\cos(120^{\circ}) = 3(4^2)\]](http://latex.artofproblemsolving.com/3/0/0/30078cecd5138b66251591614027a86320de513a.png)
so 
. Again by the symmetry of the cube, the volume we want to find is the volume of a tetrahedron with right angles on all faces at the submerged vertex, so since the lengths of the legs of the tetrahedron are 
(the three triangular faces touching the submerged vertex are all 
triangles) so
![\[v = \frac{1}{3}(2\sqrt{6})(\frac{1}{2} \cdot (2\sqrt{6})^2) = \frac{1}{6} \cdot 48\sqrt{6} = 8\sqrt{6}\]](http://latex.artofproblemsolving.com/7/7/8/778318a7868f9abf1dd380a12ff1a1521d3f3201.png)
so
![\[v^2 = 64 \cdot 6 = \boxed{384}\]](http://latex.artofproblemsolving.com/8/1/f/81f0a9c87cc43b3367ae2e659a1f786a8594b8ac.png)
.
See also
| 2015 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
