In square ABCD, we select a point P such that AP=a, BP=2a, CP=3a. Find ∠APB?
Let ABCD be (0,0), (1,0), (1,1), (0,1)
The locus of points that is a certain ratio from 2 points is a circle.
A and B:
2 points that satisfy are (1/3,0) and (-1,0). The circle has
center (-1/3,0)
radius 2/3
Equation: (x + 1/3)^2 + y^2 = 4/9
A and C:
2 points that satisfy are (1/4, 1/4) and (-1/2, -1/2). The circle has
center (-1/8, -1/8)
radius 3√2/8
Equation: (x + 1/8)^2 + (y + 1/8)^2 = 9/32
With a little help from Wolfram Alpha, the 2 equations intersect at
x = 1/17 + 3√2/17 = 0.3083906286540756
y = -4/17 + 5√2/17 = 0.1806510477567927
BAP is 180*atan(y/x)/pi = 30.36119340482172 degrees
ABP is 180*atan(y/(1-x))/pi = 14.63880659517829 degrees
APB is 180 - ABP - BAP = 135 degrees
https://www.desmos.com/calculator/6flagrjlgu
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Posted by Larry
on 2024-06-21 12:55:45 |
Solution and Desmos chart
