1965 IMO Problems/Problem 3
Problem
Given the tetrahedron whose edges and have lengths and respectively. The distance between the skew lines and is , and the angle between them is . Tetrahedron is divided into two solids by plane , parallel to lines and . The ratio of the distances of from and is equal to . Compute the ratio of the volumes of the two solids obtained.
Solution
Let the plane meet at , at , at and at . Take a plane parallel to through and let it meet in .
Since the distance of from is times the distance of , we have that and hence Similarly is parallel to , so also
vol vol vol is similar to the tetrahedron The sides are times smaller, so vol vol
The base of the prism is which is similar to with sides times smaller and hence area times smaller. Its height is times the height of above so vol prism vol
Thus vol vol
We get the volume of the other piece as vol vol and hence the ratio is (after a little manipulation)
Remark (added by pf02, November 2024)
Note that the problem is untypically sloppy or misleading. It mentions the sizes as if they are needed. In fact, as the solution above shows, they are not needed either in expressing the result, or in solving the problem. A thorough problem solver might worry about not having used all the data in the problem.
However, one can imagine other solutions, where these quantities would be used in the process of solving the problem. For example one could break up the doubly truncated triangular prism into pyramids . Computing the volume of each of these pyramids would require all the data in the problem. The end result should of course be the same, but the thorough problem solver would not have the uneasy feeling of not having used all the data in the problem.
See Also
1965 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |