Comments on: Rightagon Inc.- a math/geometry puzzle

By: brianu

brianu — Sat, 20 Jun 2009 21:55:10 +0000

5685.4249492377 sq. ft.

By: .mau.

.mau. — Sat, 20 Jun 2009 17:37:20 +0000

@michaelc: no trigonometry, just rectangles and triangles :-)

By: ignaciordc

ignaciordc — Sat, 20 Jun 2009 11:17:31 +0000

Second oops!
First steps of my solution were correct, since indeed 100/(1+sqr(2))= h = 41.4213.. is the height of the 8 isosceles triangles. However, I messed up the final part. The base of those triangles should be b=2*h/(1+sqr(2)) = 34.314… as you all pointed out, and the octagonal area = 4*h*b = 40000*2/(1+sqr(2))^2,
which is 5685.37… again reached by many.
I liked very much the approach MFox has taken, simple and effective.

By: michaelc

michaelc — Fri, 19 Jun 2009 23:28:00 +0000

Great job guys!

This should have been a great trigonomic work out for most folks, it was for me.

Ok everyone, let’s jazzercise now!

By: RK

RK — Fri, 19 Jun 2009 13:06:02 +0000

.mau., mrman, Shawn, Bizarette18, Mashplum, TheMadEgyptian (#1 Score on Swizzlepop!), MFox, got this one.

The correct answer to MichaelC’s challenge is 5685.4, nice explanations can be found by Mfox and Shawn above.

Ruddwd, Bilbao, Ladyinsomnia and ignaciordc, looks like you were all on the right track

By: Mashplum

Mashplum — Fri, 19 Jun 2009 00:57:04 +0000

I get about 5,685 feet squared. Or 10,100*(40*(2^0.5)-56) to be exact. The octagon should touch all 3 sides of the triangle.

By: TheMadEgyptian

TheMadEgyptian — Thu, 18 Jun 2009 21:14:06 +0000

I get 5685.4 square feet, based on a regular octagon of side 34.3145 ft and an equation for area I pulled off of Wikipedia (Geometry class was a long time ago). I calculated the side of the octagon as Sqrt(20000)/(2 + 3/2 * Sqrt(2)) taking advantage of multiple congruent 45/45/90 triangles. Any more specific explanations would require a diagram, which I’m just too lazy to post.

By: MFox

MFox — Thu, 18 Jun 2009 15:42:59 +0000

I posted an image to help interpret all the verbiage. It can be found at http://www.tulane.edu/~venom/i.....ilding.jpg

By: mimi112

mimi112 — Thu, 18 Jun 2009 15:08:11 +0000

I think the area is 4852.813741 ft(squared)

By: MFox

MFox — Thu, 18 Jun 2009 14:47:12 +0000

I believe the area of the largest octagon they could fit into that triangle (zoning laws and rounding error notwithstanding) is approximately 5685.3704 feet squared.

I drew an equal-sided unit octagon, and then fit a triangle around it. I also extended the sides of the octagons to make squares. Looking at the drawing, I came up with the following relationships:
[C is the length of a side of the octagon.]
[When you extend two sides of the octagon to meet in a right angle, A is the length of that extended segment. Or, in other words, if you make an isosceles right triangle with C as the base, A is the length of one of the sides.]
[B is the length of either side of the plot of land, and D is the hypotenuse of the plot.]

B = 3A + 2C
D = 3C + 4A

I calculated that the sides of our particular plot of land would have to be ~141.42 feet long, and the hypotenuse of the same plot would be 200 feet exactly. So, solving algebraically, C must be 34.32 feet, and A must be 24.26 feet.

With these numbers in hand, I worked out that the area of the octagon must be 5685.3704 square feet.
The area of the big square that contains the octagon is (C + 2A)^2, so 82.84′^2 = 6862.4656′sq.
But the area of the 4 corner triangles must be removed, which adds up to (A^2) * 2, or 1177.0952′sq.

6862.4656 – 1177.0952 = 5685.3704

By: ignaciordc

ignaciordc — Thu, 18 Jun 2009 12:10:33 +0000

Ooops!, that is 6,246.486 square feet, a little over 62% of the land

By: ignaciordc

ignaciordc — Thu, 18 Jun 2009 12:08:36 +0000

This problem is a pure geometric problem
The area is a right rectangle, hence its size is equal to S=100*sqr(2)
The octagon we are seeking is one that has 3 of its sides touching the internal sides of that right triangle.
If we consider the right angle in the origin of coordinates, and the hypotenuse in the positive quadrant, then those sides will be 1,4 and 7, 1 being the bottom horizontal side of the octagon. If the “center” of the octagon is in point (x,y), then x=y; but at the same time, that value is the distance from the center to the hypotenuse, whose equation is x+y-100*sqr(2)=0 -> x = (2*x – 100*sr(2)) / sqr(2)
Solving that, we find that the center of the octagon is at X=Y=100/(1+sqr(2)), and that is the height of all 8 triangles of the octagon.

We can calculate now the proportions between base and height for any right triangle to find out that the maximal area is
40000 * sqrt[sqrt(2)*(2-sqrt(2)] / (3+2*sqrt(2))
which is close to 6,246,486 square feet

By: bizarette18

bizarette18 — Wed, 17 Jun 2009 23:16:53 +0000

5685.4 sq ft?

By: ruddwd

ruddwd — Wed, 17 Jun 2009 13:36:54 +0000

~5700 square feet
CAD makes things a little easier.
But here’s the math I started with.
A right triangle of area 10,000 ft^2
AREA = 1/2 b X h
b = h
AREA = 1/2 b^2
10,000 = 1/2 b^2
20,000 = b^2
b = 141.4213
The hypotenus (sp?) is c^2 = a^2 + b^2
a = b
c^2 = 2b^2
c^2 = 40,000
c = 200

Now you need to find out the maximum size of octagon that can fit in that space. this is where I was lost, so I turned to my friend CAD.

By: bilbao

bilbao — Wed, 17 Jun 2009 07:41:41 +0000

I have tried to solve this puzzle graphically. I have drawn the right triangle with an octagon inside. I have divided the right triangle with a grid, thus forming little squares and triangles. If every little square is area 1 and every little triangle is area 1/2, I get an area of 7 inside the octagon and 5 1/2 outside the octagon. So area of the octagon would be 5.600 sq.feet inside the 10.000 sq.feet right triangle.