SOLUTION: I'm looking for an equation to produce coordinates for a square from two given diagonal vertices that works even with arbitrarily rotated squares. I haven't found a clear answer in
Algebra.Com
Question 479447: I'm looking for an equation to produce coordinates for a square from two given diagonal vertices that works even with arbitrarily rotated squares. I haven't found a clear answer in any of my searches :(
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Place one endpoint of the diagonal at the origin and then the other end point is )
.
Convert to polar coordinates )
:
)
Since the diagonal of a square bisects a vertex angle of 
, each of the sides of the square which can be represented by a vector: )
must be offset in angle from the diagonal by a measure of 
.
Furthermore, if the diagonal of a square is of measure 
, then the side 
.
Hence, the polar coordinates of the side rotated counterclockwise from the diagonal are:
and
Which can then be converted back to rectangular coordinates by:
\ =\ \left(r_1\sin(\varphi_1),\,r_1\cos(\varphi_1)\right))
First look at \ =\ \sin\left(\varphi\ +\ \frac{\pi}{4}\right))
Then using the sum formula for sin:
\ =\ \frac{\sqrt{2}}{2}\left(\sin(\varphi)\ +\ \cos(\varphi)\right))
Next, note that
So:
\ =\ \frac{\sqrt{a^2\ +\ b^2}}{2}\left(sin(\varphi)\ +\ \cos(\varphi)\right))
The sum formula for cos is:
\ =\ \cos(A)\cos(B)\ -\ \sin(A)sin(B))
And from that I'll let you derive that:
\ =\ \frac{\sqrt{a^2\ +\ b^2}}{2}\left(cos(\varphi)\ -\ \sin(\varphi)\right))
And then using the fact that the other side of the square is rotated clockwise from your given diagonal and that all sides of a square are identical in measure, you should be able to see that:
and
I'll let you look up the difference formulas and verify that:
\ =\ \frac{\sqrt{a^2\ +\ b^2}}{2}\left(sin(\varphi)\ -\ \cos(\varphi)\right))
\ =\ \frac{\sqrt{a^2\ +\ b^2}}{2}\left(cos(\varphi)\ +\ \sin(\varphi)\right))
John
My calculator said it, I believe it, that settles it
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