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 Collinear and Equal Angles (Posted on 2013-11-09)
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)

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 Solution | Comment 2 of 5 |
Existence
Let M’ and N’ be points on Circles 1 and 2 so that /M’QP = /N’QP = x.
As x decreases from 90 degrees to 0, the lengths of QM’ and QN’ first
increase then decrease as M’ and N’ approach P.
Accordingly, the point P’, where M’N’ crosses PQ, also moves so that QP’
increases and then decreases as P’ approaches P, showing that, at some
point, P’ passed through P, confirming that M’, N’ and P can be collinear.

Construction
If the circles are equal, MPN is perpendicular to PQ.
Otherwise, wlog., label them so that Circle 1 is the larger.
Draw the diameters, QR and QS, of Circles 1 and 2.
Construct the bisector of /RQS and let T be its intersection with Circle 1.
Draw a line through T, parallel to PR. M is the point where it crosses Circle 1.
Draw MP produced to cross Circle 2 at N.

Proof
/PQT     = /MQR  subtended by equal arcs MR and TP)       (1)
= /MPR  (subtended by arc MR in Circle 1)
= /NPS  (vert.opp.)
= /NQS  (subtended by arc NS in Circle 2)            (2)

Thus     /MQP     = /RQT   using (1)
= /SQT  (construction)
= /NQP  using(2)

 Posted by Harry on 2013-11-12 18:19:35

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