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Topic: POW Solution, Jan. 10-14
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Problem of the Week

Posts: 292
Registered: 12/3/04
POW Solution, Jan. 10-14
Posted: Jan 17, 1994 10:26 AM
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Problem of the Week, January 10-14

A regular HEXAGON and an equilateral triangle are both inscribed in the
same circle so that the hexagon and the triangle share three vertices. The
RADIUS of the circle is 8 units. What is the area of the region between the
two polygons?


We received only one solution to the problem of the week this time around!
The winner is:

Dan Hirschhorn, Illinois State University

Please note the first line of Dan's submission!


From: (Dan Hirschhorn)

I assume you'll get lots of good stuff from the students--here is my own

If you draw in the radii to the vertices of the equilateral triangle, you
get 6 triangles all congruent by SAS (8-120 deg-8). 3 make up the
equilateral triangle, 3 are outside of it, so the area of region desired
is 3 * area of 1 of these 8-120 deg-8 triangles or 1/2 (area of hexagon)
or the area of the equilateral triangle. I choose 1/2 hexagon to make it
interesting. Draw all the radii to the vertices of the hexagon. You form
6 equilateral triangles each with side of 8. 3 of these triangles equal
1/2 the hexagon. Using formula "side squared root 3 over 4" for the area
of an equilateral triangle , you get 48 sqt (3) as desired area.

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