Analytic geometry

[Albert Chan]

Since we know x cos(m) + y sin(m) = 1 is just tangents to the unit circle, we can simplify:
Rotate the angles, so that t1 = 0, line 1 is now simply x*1 + y*0 = x = 1:

Intersect, x=1 and line 2: 1*cos(t2) + y*sin(t2) = 1

(1 - 2 sin(t2/2)^2) + y*sin(t2) = 1
y * (2 sin(t2/2) cos(t2/2)) = 2 sin(t2/2)^2
y = tan(t2/2)

Use symmetry, intersect of x=1 and line 3: y = tan(t3/2)

-> Triangle line 1 side length = | tan((t2-t1)/2) - tan((t3-t1)/2) |

(02-18-2019 09:27 AM)yangyongkang Wrote:  Randomly take three parameters and draw this image

Code:

plotimplicit(x*cos(1)+y*sin(1)=1);
plotimplicit(x*cos(-8)+y*sin(-8)=1);
plotimplicit(x*cos(3)+y*sin(3)=1);
plotimplicit(x^2+y^2=1)

Sure enough, the three tangent lines

Without solving for the vertice coordinates, we can get side length directly.
Example, for above plot:

+1 radian line, slope ≈ -0.642: length = | tan((-8-1)/2 - tan((3-1)/2) |  ≈ 6.1947
−8 radian line, slope ≈ -0.147: length = | tan((1+8)/2 - tan((3+8)/2) | ≈ 5.6329
+3 radian line, slope ≈ 7.015 : length = | tan((1-3)/2 - tan((-8-3)/2) |  ≈ 2.5530

Edit: you can solve the vertices thru "reverse" rotation
Example, the top vertice (line 1 and line 3 intersect):

We know rotated intersect was [1, tan((3-1)/2)], so original intersect
= [[cos(1), -sin(1)], [sin(1), cos(1)]] * [1, tan(1)]
= [cos(1) - sin(1)*tan(1), sin(1) + cos(1)*tan(1)]
= [cos(2)/cos(1), 2*sin(1)]
≈ [-0.7702, 1.6829]