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 ?

geometry from real life
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. 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. And now without trigonometry . . . . . .

  3. 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. 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. 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!

  6. Pingback: A real life geometry problem proposed by Howard Phillips | WatsonMath.com

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