Given a square ABCD, let P be such that AP=1, BP=2 and CP=3.
* What is the length of DP?
* What is the angle APB?
I used Excel to solve for the side of the square, using the law of cosines.
cos(APB) = (1+4S^2)/4
cos(BPC) = (9+4S^2)/12
cos(APC) = (1+92*S^2)/6
The factor of 2 times the square of the side in the last is due to the side being not the side of the square but the diagonal of the square.
This was solved so that The total of the three angles is 2*pi.
S comes out to 2.797933, and angle APB is 2.356194 radians, or 135 degrees.
This assumes that the point P is inside the square. The same Pythagorean relationships make the distance to D the same even if point P is outside the square, but the side is smaller, and I haven't worked out what the angle would be for that, but it would be acute (having drawn it on graph paper, with the locus of points at a 1:2 ratio from A vs. B,being a circle and the locus of points at a 2:3 ratio from B vs.C; therefore these are two intersecting circles with the points of intersection being the two possible locations for P, with the side shrinking in absolute size when the outer point is chosen).
Edited on March 17, 2004, 10:04 pm

Posted by Charlie
on 20040317 22:03:40 