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 April 14th, 2015, 07:06 AM #1 Member     Joined: Jun 2014 From: Math Forum Posts: 67 Thanks: 4 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. MR.Brhum Last edited by skipjack; April 14th, 2015 at 08:02 AM.
 April 14th, 2015, 08:33 AM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 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.
 April 14th, 2015, 08:56 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond 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).
 April 15th, 2015, 10:20 AM #4 Newbie   Joined: Jun 2014 From: United state Posts: 14 Thanks: 0 Math Focus: algebra, geometry, trigonometry, calculus 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).
April 15th, 2015, 12:55 PM   #5
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Quote:
 Originally Posted by aaliyah 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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