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Collinear and Equal Angles (Posted on 2013-11-09) Difficulty: 5 of 5
Let Γ1 and Γ2 be arbitrary circles that intersect at points P and Q.

Prove or disprove that there exist points M and N such that

(1) M ∈ Γ1\{P,Q},
(2) N ∈ Γ2\{P,Q},
(3) M, N, and P are collinear, and
(4) ∠MQP = ∠NQP.

If they exist, prove or disprove that they can be constructed with
straightedge and compass.

Here is a link to Wolfram MathWorld:
Definition of Set Difference

See The Solution Submitted by Bractals    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts Specific case. Almost full solution | Comment 1 of 5
It's easy enough to coordinatize the situation:  Let the circles have their centers on the x-axis at points (a1,0) and (a2,0)
with radii a1^2 + 1 and a2^2 + 1
so the points p and q are (0,1) and (0,-1) respectively.
The line has equation y = mx+1

To save on algebra let a1=-3 and a2=2 [I just wanted to see if this could be done.][Most algebra and simplifying not shown.]

The circles now have equations (x+3)^2 + y^2 =10 and (x-2)^2 + y^2 = 5
The line intersects these circles at
M= (-(2m+6)/(1+m^2),(-m^2-6m+1)/(1+m^2))
N= (-(2m-4)/(1+m^2),(-m^2+4m+1)/(1+m^2))

Call the angles sought θ1 and θ2
cos(θ1)=(-6m+2)/√(40(1+m^2))
cos(θ2)=(4m+2)/√(20(1+m^2))
so
(-6m+2)/√(40(1+m^2)) = (4m+2)/√(20(1+m^2))
m = 7-5√2
and
θ1=θ2=67.5

Seeing how things went it seems reasonable that had I not chosen specific values for a1 and a2 this still would have worked.  Furthermore, because the equations and solutions involved nothing but quadratic equations the problem is solvable by straightedge and compass.

  Posted by Jer on 2013-11-12 09:53:12
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