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Topic: Problem of the Week (Solution)
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Problem of the Week

Posts: 292
Registered: 12/3/04
Problem of the Week (Solution)
Posted: Aug 3, 1993 2:19 PM
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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.





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