This is not so much a puzzle as it is a request for some help. My uncle told me, "Smart men are those who can walk into a room and know that they are the dumb ones." So, with that in mind, I need some help with my geometry. (This isn't homework, I'm too old to be in school.)

Here are the criteria:

You are given two points, X and Y, and the direction in degrees, L, between X and Y, assuming up is 0 or 360. X is the center of a circle with a radius of .5 and the center of a square with sides of 1; Y is the center of a circle with a radius of 1 and the center of a square with sides of 2. The squares are tilted so that their sides are parallel with each other and a line connecting X and Y will bisect one side of each square. With this information, can you determine a formula to find the outside corners of the two squares, A B C and D. All points will lie in the upper-right quadrant of a two dimensional graph (all numbers will be positive). I've attached a picture for clarification.

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Guest Message by DevFuse

Started by Jkyle1980, Jan 31 2008 10:22 AM

1 reply to this topic

### #1

Posted 31 January 2008 - 10:22 AM

### #2

Posted 31 January 2008 - 11:46 AM

Assuming you know the coordinates of X, the length L and its angle a, here are the points.

If instead you're given X and Y, then simply calculate L and a from them.

Don't bet anything significant on this, there could be a typo or two [physical or mental].

A = -.5[cos a+sin a+x1, cos a-sin a+x2]

D = -.5[sin a-cos a+x1, cos a+sin a+x2]

B = [-cos a+(L+1)*sin a+x1, sin a+(L+1)*cos a+x2]

C = [ cos a+(L+1)*sin a+x1, -sin a+(L+1)*cos a+x2]

If instead you're given X and Y, then simply calculate L and a from them.

Don't bet anything significant on this, there could be a typo or two [physical or mental].

A = -.5[cos a+sin a+x1, cos a-sin a+x2]

D = -.5[sin a-cos a+x1, cos a+sin a+x2]

B = [-cos a+(L+1)*sin a+x1, sin a+(L+1)*cos a+x2]

C = [ cos a+(L+1)*sin a+x1, -sin a+(L+1)*cos a+x2]

Spoiler for The calculation...

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

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