Challenge Problem - 20.5.2013 - The octagon nightmare

[rohan.dd.bhatia]

Is it wrong to assume that a regular Octagon can ALWAYS be divided into 7 square, what is wrong with my approach below?



The area of square ABCD = Area of 4 colored triangles + area of smaller square.
Area of 1 colored triangle = (1/root(2) * 2/root(2))/2 =1/2
Total of 4 such triangles = 2
Area of central square = (1/root(2))^2 = 1/2
Total Area = 2+1/2 = 5/2
(Same thing -- if we use AFB to calculate one side of square)

[RonPurewal]

Your octagon isn't regular; the horizontal and vertical sides are shorter than the diagonal ones.

By the way -- In general, please don't post questions about the current challenge problem. Thanks.

[rohan.dd.bhatia]

Oops missed that.
Okay got it. Thanks Ron

[jlucero]

That's a good thought though. This week's Challenge Problem is definitely a tricky one.

[rohan.dd.bhatia]

Yes, It took some time to notice the pattern, but I got the answer using above approach - I just had to change the octagon into a regular one

[tim]

Let us know if you have any further questions.

[rohan.dd.bhatia]

Thanks Tim,
After following RONS suggestion here is my corrected solution --

The area of square ABCD = Area of 4 colored triangles + area of smaller square(RED) -- Please note the side are now 1 - becuase octagon is regular
Area of 1 colored triangle = ((1+1/root(2)) * 1/root(2))/2 =
(root(2) +1) /4
Total of 4 such triangles = root(2)+1
Area of central square = 1*1 = 1
Total Area = 1+1+root(2) = 2+root(2)
(Same thing -- if we use AFB to calculate one side of square)
AB^2 = (1/root(2) + 1)^2 +( 1/(root(2))^2
AB^2 = 2+root(2)
Area = AB^2 = 2+root(2)

Let me know if this is okay..

[RonPurewal]

Thanks Tim,
After following RONS suggestion here is my corrected solution --

The area of square ABCD = Area of 4 colored triangles + area of smaller square(RED) -- Please note the side are now 1 - becuase octagon is regular
Area of 1 colored triangle = ((1+1/root(2)) * 1/root(2))/2 =
(root(2) +1) /4
Total of 4 such triangles = root(2)+1
Area of central square = 1*1 = 1
Total Area = 1+1+root(2) = 2+root(2)
(Same thing -- if we use AFB to calculate one side of square)
AB^2 = (1/root(2) + 1)^2 +( 1/(root(2))^2
AB^2 = 2+root(2)
Area = AB^2 = 2+root(2)

Let me know if this is okay..

that looks legitimate.