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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.)
Solution Solution | Comment 2 of 5 |
Let M and N be points on Circles 1 and 2 so that /MQP = /NQP = 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 MN 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.

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.

            /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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