document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=479447>Question 479447</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 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 :( </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #328500 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=328500\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "Place one endpoint of the diagonal at the origin and then the other end point is <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large (a,\,b)\">.\r<BR>\n" );
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document.write( "Convert <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large (a,\,b)\"> to polar coordinates <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large (r,\,\varphi)\"> :\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ \sqrt{a^2\ +\ b^2}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \varphi\ =\ \tan^{-1}\left(\frac{b}{a}\right)\">\r<BR>\n" );
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document.write( "Since the diagonal of a square bisects a vertex angle of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \frac{\pi}{2}\">, each of the sides of the square which can be represented by a vector: <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \left(r_i\,\varphi_i\right)\"> must be offset in angle from the diagonal by a measure of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \frac{\pi}{4}\">.\r<BR>\n" );
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document.write( "Furthermore, if the diagonal of a square is of measure <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large d\">, then the side <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large s\ =\ d\frac{\sqrt{2}}{2}\">.\r<BR>\n" );
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document.write( "Hence, the polar coordinates of the side rotated counterclockwise from the diagonal are:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ r_1\ =\ r\frac{\sqrt{2}}{2}\">\r<BR>\n" );
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document.write( "and\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \varphi_1\ =\ \varphi\ +\ \frac{\pi}{4}\">\r<BR>\n" );
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document.write( "Which can then be converted back to rectangular coordinates by:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \left(a_1,\,b_1\right)\ =\ \left(r_1\sin(\varphi_1),\,r_1\cos(\varphi_1)\right)\">\r<BR>\n" );
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document.write( "First look at <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \sin\left(\varphi_1\right)\ =\ \sin\left(\varphi\ +\ \frac{\pi}{4}\right)\">\r<BR>\n" );
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document.write( "Then using the sum formula for sin:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \sin\left(\varphi\ +\ \frac{\pi}{4}\right)\ =\ \frac{\sqrt{2}}{2}\left(\sin(\varphi)\ +\ \cos(\varphi)\right)\">\r<BR>\n" );
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document.write( "Next, note that\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ r_1\ =\ r\frac{\sqrt{2}}{2}\ =\ \sqrt{\frac{a^2\ +\ b^2}{2}}\">\r<BR>\n" );
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document.write( "So:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ r_1\sin\left(\varphi_1\right)\ =\ \frac{\sqrt{a^2\ +\ b^2}}{2}\left(sin(\varphi)\ +\ \cos(\varphi)\right)\">\r<BR>\n" );
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document.write( "The sum formula for cos is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \cos(A\ +\ B)\ =\ \cos(A)\cos(B)\ -\ \sin(A)sin(B)\">\r<BR>\n" );
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document.write( "And from that I'll let you derive that:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ r_1\ =\ r\frac{\sqrt{2}}{2}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ r_1\cos\left(\varphi_1\right)\ =\ \frac{\sqrt{a^2\ +\ b^2}}{2}\left(cos(\varphi)\ -\ \sin(\varphi)\right)\">\r<BR>\n" );
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document.write( "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:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ r_2\ =\ r\frac{\sqrt{2}}{2}\">\r<BR>\n" );
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document.write( "and\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \varphi_2\ =\ \varphi\ -\ \frac{\pi}{4}\">\r<BR>\n" );
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document.write( "I'll let you look up the difference formulas and verify that:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ r_2\sin\left(\varphi_2\right)\ =\ \frac{\sqrt{a^2\ +\ b^2}}{2}\left(sin(\varphi)\ -\ \cos(\varphi)\right)\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ r_2\cos\left(\varphi_2\right)\ =\ \frac{\sqrt{a^2\ +\ b^2}}{2}\left(cos(\varphi)\ +\ \sin(\varphi)\right)\">\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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