Difference between revisions of "2015 AIME II Problems/Problem 9"
Expilncalc (talk | contribs) m (→Solution) |
Soyamyboya (talk | contribs) (→Solution) |
||
Line 41: | Line 41: | ||
*Note that in a 30-30-120 triangle, side length ratios are <math>1:1:\sqrt{3}</math>. | *Note that in a 30-30-120 triangle, side length ratios are <math>1:1:\sqrt{3}</math>. | ||
+ | *Or, note that the altitude and the centroid of an equilateral triangle are concurrent, so since the centroid is 4 units from the vertex, the altitude is 6, which gives a hypotenuse of <math>\frac{12}{\sqrt{3}}=4\sqrt{3}</math> | ||
==See also== | ==See also== |
Revision as of 22:41, 18 January 2018
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 concurrent, so since the centroid is 4 units from the vertex, the altitude is 6, which gives a hypotenuse of
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.