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April 14th, 2015, 07:06 AM   #1
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A parallelogram




A parallelogram is cut into 178 pieces of equilateral triangles with sides of 1 unit.

Find the maximum value of the perimeter of the parallelogram.



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Last edited by skipjack; April 14th, 2015 at 08:02 AM.
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April 14th, 2015, 08:33 AM   #2
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178 equilateral triangles can be put together to form 89 individual rhombi of side length 1. Since the rhombi will tessellate the larger parallelogram, they may only be arranged in a 1 x 89 row.

max/min ... doesn't matter, the only perimeter can be 180 units.
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April 14th, 2015, 08:56 AM   #3
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The area of the parallelogram is

$\displaystyle a\cdot\sin(60^\circ)\cdot b= 178\cdot\frac{\sqrt3}{4}\text{ units}$

hence

$\displaystyle a\cdot b=89\text{ units}$

As 89 is prime and the sides of the parallelogram must be positive integers, the only possible configuration has a perimeter of 180 units (with sides 1 unit and 89 units).
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April 15th, 2015, 10:20 AM   #4
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The area of the parallelogram is a⋅sin(60∘)⋅b=178⋅3√4


there fore

a⋅b=89

As 89 is prime and the sides of the parallelogram must be positive integers, the only possible configuration has a perimeter of 180 units (with sides 1 unit and 89 units).
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April 15th, 2015, 12:55 PM   #5
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Quote:
Originally Posted by aaliyah View Post
The area of the parallelogram is a⋅sin(60∘)⋅b=178⋅3√4


there fore

a⋅b=89

As 89 is prime and the sides of the parallelogram must be positive integers, the only possible configuration has a perimeter of 180 units (with sides 1 unit and 89 units).
you couldn't even copy greg1313's post correctly ...
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