document.write( "Question 1208064: Quadrilateral PQRS is cyclic and side PS = u is a
\n" ); document.write( "diameter of the circle. If PQ = QR = v, RS = w, and u, v,
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Algebra.Com's Answer #846351 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Maybe Ikleyn or Greenestamps can simplify my solution but here it is at last.\r\n" );
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document.write( "Since quadrilaterals are normally lettered counter-clockwise, and PS is a\r\n" );
document.write( "diameter, the quadrilateral must be inscribed in a semi-circle. I will need\r\n" );
document.write( "some right triangles so I will bisect everything including the diameter. The\r\n" );
document.write( "green line segments are the perpendicular bisectors of the upper 3 sides and\r\n" );
document.write( "the angles as well since the triangles are isosceles. So the radius of the\r\n" );
document.write( "circle is u/2, and will be the hypotenuse of all 6 right triangles in this figure:\r\n" );
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document.write( "[eqs. A]  \"sin%28alpha%29=%22v%2F2%22%2F%22u%2F2%22=+v%2Fu\" and \"sin%28beta%29=%22w%2F2%22%2F%22u%2F2%22=+w%2Fu\"\r\n" );
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document.write( "The sum of all 6 angles is 180o\r\n" );
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document.write( "\"4alpha%2B2beta=180%5Eo\"\r\n" );
document.write( "\"2alpha%2Bbeta=90%5Eo\"\r\n" );
document.write( "2α and β are complementary,\r\n" );
document.write( "\"alpha%2B%28alpha%2Bbeta%29=90%5Eo\"\r\n" );
document.write( "α and α+β are also complementary\r\n" );
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document.write( "So \r\n" );
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document.write( "[eq. B]   \"cos%28alpha%29cos%28beta%29=sin%28alpha%29%2Bsin%28alpha%29sin%28beta%29\"\r\n" );
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document.write( "\"2alpha%2Bbeta=90%5Eo\". Taking sines of both sides:\r\n" );
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document.write( "\"sin%282alpha%2Bbeta%29=1\"\r\n" );
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document.write( "\"sin%282alpha%29cos%28beta%29%2Bcos%282alpha%29sin%28beta%29=1\"\r\n" );
document.write( "\"%282sin%28alpha%29cos%28alpha%29%5E%22%22%29cos%28beta%29%2Bcos%282alpha%29sin%28beta%29=1\"\r\n" );
document.write( "Using the associative law and since 2α and β are complementary,\r\n" );
document.write( "\"2sin%28alpha%29%28cos%28alpha%29cos%28beta%29%5E%22%22%29%2Bsin%28beta%29sin%28beta%29=1\"\r\n" );
document.write( "Substituting from [eq. B] above\r\n" );
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document.write( "\"2sin%5E2%28alpha%29%2B2sin%5E2%28alpha%29sin%28beta%29%2Bsin%5E2%28beta%29=1\"\r\n" );
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document.write( "From [eqs. A]\r\n" );
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document.write( "\"2%28v%2Fu%29%5E2%2B2%28v%2Fu%29%5E2%28w%2Fu%29%2B%28w%2Fu%29%5E2=1\"\r\n" );
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document.write( "\"2v%5E2%2Fu%5E2%2B%282v%5E2w%29%2Fu%5E3%2Bw%5E2%2Fu%5E2=1\"\r\n" );
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document.write( "Multiply through by u3\r\n" );
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document.write( "\"2uv%5E2%2B2v%5E2w%2Buw%5E2=u%5E3\"\r\n" );
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document.write( "\"%28u%29w%5E2%2B%282v%5E2%29w%2B%282uv%5E2-u%5E3%29=0\"\r\n" );
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document.write( "Since \"%22%22%2B-u\" are factors of the last term, we try both and\r\n" );
document.write( "find that w=-u is a solution and so we factor it by synthetic \r\n" );
document.write( "division\r\n" );
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document.write( "-u | u      2v2   2uv2-u3\r\n" );
document.write( "   |        -u2  -2uv2+u3\r\n" );
document.write( "     u   2v2-u2         0\r\n" );
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document.write( "\"%28w%2Bu%5E%22%22%5E%22%22%29%28u%2Aw%2B%282v%5E2-u%5E2%29%5E%22%22%29=0\"\r\n" );
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document.write( "\"w%2Bu=0\";  \"u%2Aw%2B%282v%5E2-u%5E2%29=0\"\r\n" );
document.write( "\"w=-u\";    \"u%2Aw=u%5E2-2v%5E2\"\r\n" );
document.write( "              \"w=%28u%5E2-2v%5E2%29%2Fu%5E%22%22\"\r\n" );
document.write( "              \"w=u-2v%5E2%2Fu%5E%22%22\"\r\n" );
document.write( "              \"2v%5E2%2Fu%5E%22%22=u-w\" \r\n" );
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document.write( "So the term \"2v%5E2%2Fu%5E%22%22\" must be an integer, since the terms \r\n" );
document.write( "on the right are integers.\r\n" );
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document.write( "For contradiction, assume u is a prime number.\r\n" );
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document.write( "If u=2, then the radius is 1, and the semicircle is π.\r\n" );
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document.write( "So \"2v%2Bw%3Cpi\", that would make v=1, w=2. then triangle\r\n" );
document.write( "SRO would have sides 1,1,2 which violates the triangle\r\n" );
document.write( "inequality.  So u is not the prime number 2.\r\n" );
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document.write( "Thus u must divide evenly into v2.   But if u is a prime \r\n" );
document.write( "then u must divide evenly into v as well. This cannot be true \r\n" );
document.write( "because v < u, the diameter.\r\n" );
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document.write( "Thus we have a contradiction and u cannot be a prime number. \r\n" );
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document.write( "Edwin
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