Difference between revisions of "2015 AIME II Problems/Problem 9"
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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 | 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 | + | <cmath>s^2 = 2\cdot(4^2) - 2l\cdot(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 |
Revision as of 02:30, 5 December 2019
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 .
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
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
so
In this case, our base was one of the isosceles triangles (not the larger equilateral one). To calculate volume using the latter, note that the height would be .
- Note that in a 30-30-120 triangle, side length ratios are .
- Or, note that the altitude and the centroid of an equilateral triangle are the same point, so since the centroid is 4 units from the vertex (which is the length of the median), the altitude is 6, which gives a hypotenuse of by relationship for 30-60-90 triangles.
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.