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## Question

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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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]

Let's call the 1-coordinate positive to the left and the 2-coordinate positive upward.

Denote a point by its coordinates: X = [x1, x2] and Y = [y1, y2].

Same for A, B, C, D.

Start by moving and rotating the figure to put X at the origin

and Y directly above it: X = [0, 0], Y = [0, L]

A = [-.5, -.5], D = [.5, -.5] or more simply, A = -.5[1, 1], D = -.5[-1, 1]

B = [-1, L+1], C = [1, L+1]

Now rotate and move it back:

[1] Rotate by the known angle a of the line L

Rotating a point z = [z1, z2] thru and angle a makes z' = [z1', z2'] where:

| | = | | | | or,
|z2'| |-sin a cos a| |z2| z2' = -z1 sin a + z2 cos a = -z1*sa + z2*ca - abbreviating the sines and cosines.
`|z1'|   | cos a  sin a| |z1|	 z1' =  z1 cos a + z2 sin a =  z1*ca + z2*sa`

A' = -.5[ca+sa, ca-sa]

D' = -.5[sa-ca, ca+sa]

B' = [-ca+(L+1)*sa, sa+(L+1)*ca]

C' = [ ca+(L+1)*sa, -sa+(L+1)*ca]

[2] Translate by an amount [x1, x2].

A'' = -.5[ca+sa+x1, ca-sa+x2]

D'' = -.5[sa-ca+x1, ca+sa+x2]

B'' = [-ca+(L+1)*sa+x1, sa+(L+1)*ca+x2]

C'' = [ ca+(L+1)*sa+x1, -sa+(L+1)*ca+x2]

Plug in the values of L, sin a, cos a, x1 and x2.

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