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?
Instead of using pure mathematics to solve this problem, I used
experimentation instead (but there were still some preliminary
mathematics).
Because AP=1 and CP=3, the longest possible diameter AC of square ABCD
is 4 (1+3). Thus, the longest side length is: 2(a^2)=(4^2)
2(a^2)=16
(a^2)=8
a=sqrt(8) or about 2.82.
Using experimentation, one can find the necessary side length so that
arcs with radii of 1 (with an origin at point A), 2 (with an origin at
point B), and 3 (with an origin at point C) meet at a single point P
inside square ABCD is equal to 2.79. Through simple measurement,
DP=2.45 or sqrt(6) and angle APB=137.4 degrees (an estimate).
an answer
