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 Post subject: Challenge Problem #10Posted: Fri, 15 Aug 2003 03:26:03 UTC
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Joined: Mon, 19 May 2003 19:55:19 UTC
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Location: Lexington, MA
Challenge Problem #10

An equilateral triangle and a regular hexagon have equal perimeters.
Find the ratio of their areas.

[Time limit: 1 minute]

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 Post subject: Posted: Fri, 15 Aug 2003 05:23:06 UTC
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Joined: Fri, 2 May 2003 16:33:24 UTC
Posts: 781
Location: Hong Kong
add 3 equilateral triangle to the hexagon and make it a larger equilateral triangle.
original equilateral triangle area : new equilateral triangle area = 1:9
original equilateral triangle area : hexagon 1:(9-3) = 1:6
Have same perimeter = sides of equilateral triangle : sides of hexagon = 2:1
So the final ans is 4:6=2:3

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 Post subject: Posted: Fri, 15 Aug 2003 09:25:22 UTC
 S.O.S. Oldtimer

Joined: Sat, 12 Jul 2003 09:40:00 UTC
Posts: 216
An alternative approach :-

Divide the hexagon (H) into 6 equilateral triangles. Call one of these triangles T.

Area(H) = 6Area(T) so Area(T) = Area(H)/6

T has same side length as H but only 3 sides instead of 6, so
Perimeter(T)=Perimeter(H)/2

To make a triangle with same perimeter as H, we must double the length of the sides of T - call this triangle U.

Then Area(U) = 4Area(T) = 4Area(H)/6 = 2Area(H)/3

So Area(H):Area(U) = 3:2

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 Post subject: Posted: Fri, 15 Aug 2003 14:54:02 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Mon, 19 May 2003 19:55:19 UTC
Posts: 7949
Location: Lexington, MA
Hello, Soarer and Simon!

Well done! Your derivations are correct, of course.

But the "30 second limit" was only half in jest.

There IS a "Quickie" solution.

Solution to Challenge Problem #10

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Divide the equilateral triangle into four equilateral triangles.
The hexagon divides into six identical equilateral triangles.
Code:
/|\
/  |  \
/\                           |  \|/  |
/__\                          |  /|\  |
/\  /\                          \  |  /
/__\/__\                           \|/

Therefore, the ratio of the areas is:

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