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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.)
re(2): Solution | Comment 4 of 5 |
(In reply to re: Solution by Bractals)

Yes, you’re right Bractals, I should have shown that R, S & P are collinear -
inadvertently ‘left to the reader'!
/RPQ and /SPQ are both right angles (subtended in semicircles) should do it.

Since you ask, I first noticed that RS was useful in forming triangle RQS which
is similar to MQN, and now after a little more thought I think this approach
can give a simpler proof that works for all positions of S. Here goes:

Let U be the point of intersection of RP and TQ.

Triangles RQS and MQN are similar since
/SRQ = /NMQ     (angles subtended by PQ in Circle 1)
/RSQ = /MNQ    
            (If P lies between R and S, these are subtended by PQ in Circle 2;
             otherwise, these are the ext. and int. opp. in cyclic quad SPNQ).

Also,     /RQU = /MQP (subtended by equal arcs RT and MP)

So UQ and PQ split triangles RQS and MQN respectively into pairs of similar
triangles (RQU~MQP and SQU~NQP).
Since QU bisects /RQS, it therefore follows that QP bisects /MQN.

Much fun.

  Posted by Harry on 2013-11-17 18:08:26

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