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Difference between revisions of "2015 AIME II Problems/Problem 9"

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<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 <math>45-45-90</math> 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})\left(\frac{1}{2} \cdot (2\sqrt{6})^2\right) = \frac{1}{6} \cdot 48\sqrt{6} = 8\sqrt{6}</cmath>
 
<cmath>v = \frac{1}{3}(2\sqrt{6})\left(\frac{1}{2} \cdot (2\sqrt{6})^2\right) = \frac{1}{6} \cdot 48\sqrt{6} = 8\sqrt{6}</cmath>
Line 37: Line 37:
  
 
<cmath>v^2 = 64 \cdot 6 = \boxed{384}</cmath>.
 
<cmath>v^2 = 64 \cdot 6 = \boxed{384}</cmath>.
 +
 +
*Note that in a 30-30-120 triangle, side length ratios are 1:1:<math>\sqrt{3}</math>.
  
 
==See also==
 
==See also==

Revision as of 20:42, 28 August 2017

Problem

A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$.

[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]

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 $4$, by the Law of Cosines, the side length s of the equilateral triangle is

\[s^2 = 2*(4^2) - 2*(4^2)\cos(120^{\circ}) = 3(4^2)\]

so $s = 4\sqrt{3}$.* 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 $\frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}$ (the three triangular faces touching the submerged vertex are all $45-45-90$ triangles) so

\[v = \frac{1}{3}(2\sqrt{6})\left(\frac{1}{2} \cdot (2\sqrt{6})^2\right) = \frac{1}{6} \cdot 48\sqrt{6} = 8\sqrt{6}\]

so

\[v^2 = 64 \cdot 6 = \boxed{384}\].

  • Note that in a 30-30-120 triangle, side length ratios are 1:1:$\sqrt{3}$.

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

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