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Problem of the Week (Solution)
Posted:
Aug 3, 1993 2:19 PM


Only one solution was received this week. (I can corroborate the answer, though, since it is the same one I got and the same one other people I gave the problem to have found.)
From PDALEY@fair2.fairfield.edu (Pat Daley):
Before attempting this proof, I constructed the problem on the Geometer's Sketchpad. So many relationships can be investigated by dragging the triangle's points and measuring. Construct an altitude DG from D to AB and an altitude CH from C to AB. Let the height of triangle ABD, DG, be h. The height of triangle ABE is also h since D and E lie on a line parallel to AB. The area of the lower region of triangle ABC, that is the entire region excluding quadrilateral CDFE, is then equal to .5bh + .5bh  4 = bh  4. The area of triangle ABC, therefore, is bh. Triangles ADG and ACH are similar triangles. Since the ratio of DG to CH is 1:2, the ratio of AD to AC is also 1:2. BD, therefore, is a median. Similarly, AE is also a median. Medians intersect in a triangle so that the ratio of BF to BD is 2:3. Construct an altitude FJ from F to AB. Triangles BFJ and BDG are similar triangles. Since the ratio of BF to BD is 2:3, the ratio of FJ to DG is 2:3. The ratio, therefore, of FJ to CH is 1:3. Thus, the area of triangle ABC is 3 times the area of triangle ABF. The area of triangle ABC is 12 square units.



