# A “real life” geometry problem.

While designing a system for connecting “educational” cubes together I figured that the holes in the faces had to be positioned very carefully. To achieve what I wanted the holes had to be positioned with  length a equal to length b, and length c had to be twice the length a.

So what is length a, as a fraction of the side of the cube ? There will be eight holes altogether, and all cubes are the same size.

Filed under education, geometry, teaching

### 6 responses to “A “real life” geometry problem.”

1. Elaine Watson

side length = 2a + 2a(tan67.5)
= 2a(1 + tan 67.5)
= 2a(1 + 2.414) approximately
= 2a(3.414) approximately
= 6.828a approximately
Let side length = x
x/6.828 = a
length a is approximately 1/6.828 of x or 0.146 x

2. howardat58

And now without trigonometry . . . . . .

3. Elaine Watson

I’m having a hard time getting my mind around solving it another way. I’ve gotten the side length of the cube to be about [6 times the radius “a” plus 2 lengths that are a little more than 1/2 times “a”]. I’ve played around with phi (1.618…). I’m sure that it’s something simple that I am overlooking!

4. howardat58

Not that simple. I used the “bisector of an angle splits the opposite side in the ratio of the two adjacent sides” theorem. here is the extended picture: 5. Elaine Watson

I can’t figure out how to insert an image of my work here, so I’m going to put it on http://www.watsonmath.com

I worked it using your approach of the angle bisector cutting the opposite side of the triangle into proportional lengths. I also worked it out using the octagon that is formed from all of the 8 holes. In both cases, I came out with the same answer that I derived using trig. If the side length of the square is 1, the length of a is approximately 0.14644.

Fun problem! But not easy!