Tetrahedron BRIEF ABSTRACT By folding a triangle, create a tetrahedron and find its volume. This is a short yet challenging problem to enhance 3D visualization. While it could be solved with calculus, a nicer solution relies only on skills learned in HS geometry. GENERAL INFORMATION FileName: Tetra Full title: The Volume of a Tetrahedron Last Update: 5/31/96 Developer: Aaron Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Contact: Aaron Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Klebanoff@rose-hulman.edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. STATEMENT OF PROBLEM The sides of triangle ABC measure 11, 20, and 21 units. We make folds along lines PQ, QR, and RP, where P, Q, and R are the midpoints of its sides, until A, B, and C coincide. What is the volume of the resulting tetrahedron? KEYWORDS Tetrahedron, pyramid, visualization, 3D geometry. TEACHER NOTES ISSUES RELATED TO THE PROBLEM It's quite challenging for many students to understand that a tetrahedron can indeed be formed in the way described. It's recommended to have scissors, paper, and rulers available for students to build accurate models. Prerequisites A course in high school geometry is sufficient. In particular, the students should have been introduced to three dimensional objects. Familiarity with the volume formula for a tetrahedron given height and base would help. Calculus can be used, but is certainly not necessary. Time allotment - time management In class, you could spend between 30 - 50 minutes on this problem. Students would then be expected to complete and write-up solutions as homework with at least 2 nights after the class work to turn it in. Expectations Students have a difficult time visualizing this problem without the help of a model. Students will have a difficult time determining the height of the pyramid -- as opposed to the height of a side of the pyramid (which some may need reviewed.) Be prepared to offer hints. Future payoffs Improves visualization skills. Encourages creative thinking. Extensions Note that the volume of the pyramid was integer valued. Check that the following integer triples (other than (11, 20, 21)) also yield a tetrahedron of integer volume: (33, 65, 72); (69, 91, 100); (21, 99, 100). Can you find other integer triples that yield a tetrahedron of integer volume? Are there infinitely or finitely many such integer triples? (If finite --- can you say how many?) These last two questions are only for the strong willed students who really got involved with the original question and are yearning for more! References and Sources USA Math Talent Search problem: Year 3, Round 3, problem 5. Contact George Berzsenyi, Rose-Hulman Institute of Technology, 5500 Wabash Avenue, Terre Haute, IN 47803. (812) 877-8474. POSSIBLE SOLUTION The key to this solution is to place the final tetrahedron inside a rectangular box (x by y by z) so that its vertices O, P, Q, and R are also vertices of the box and so that half-side lengths of the original triangle are the side lengths of the box. See the figure below. In particular, P, Q, and R are the midpoints of AB, BC, and AC respectively, and O is the coincidence of A, B, and C. The box is partitioned by the desired tetrahedron into five tetrahedra: the one we're interested in, and four other congruent ones each of which has base area xy/2 and height z, and hence volume (1/3)(xy/2) = (1/6)(x y z). This gives us the volume of OPQR as x y z - 4 (1/6)(x y z) = (1/3)(x y z). So, all we have to do is determine x, y, and z and we're done. The side lengths x, y, and z are determined by the system x^2 + y^2 = (21/2)^2 x^2 + z^2 = (10)^2 y^2 + z^2 = (11/2)^2 which yields x^2 = 90, y^2 = 81/4, and z^2 = 10 so that x = 3 sqrt(10), y = 9/2, and z = sqrt(10) so that the volume of the tetrahedron is (1/3) x y z = 45. ISSUES IN SOLUTION If you show the students a picture of the box with the tetrahedron in it (after they have struggled a little first, of course) then most will or should be able to figure out the rest by themselves.