April 14th, 2015, 08:06 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 09:02 AM. |
April 14th, 2015, 09:33 AM | #2 |
Math Team Joined: Jul 2011 From: Texas Posts: 2,805 Thanks: 1449 |
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, 09:56 AM | #3 |
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,898 Thanks: 1093 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, 11:20 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, 01:55 PM | #5 |
Math Team Joined: Jul 2011 From: Texas Posts: 2,805 Thanks: 1449 | you couldn't even copy greg1313's post correctly ... |