Doesn't seem like Geometry!

Geometry Level 5

Consider $3$ circles $\omega_{1}, \omega_{2}$ and $\omega_{3}$ on the Euclidean plane. Given that there exists a point $M$ such that $\omega_{1} \cap \omega_{2} \cap \omega_{3} = \{M\}$ , and also $\omega_{1} \cap \omega_{2} = \{A,M\}$ , $\omega_{1} \cap \omega_{3}=\{B,M\}$ and $\omega_{2} \cap \omega_{3}=\{C,M\}$ , consider the following statements :

1) There exist points $D, E$ and $F$ on $\omega_{1}, \omega_{2}$ and $\omega_{3}$ respectively such that the points $(D,A,E), (E,C,F)$ and $(D,B,F)$ are collinear and points $(D,A,E,C,F,B,D)$ joined in the given order form a triangle.

2) There exist points $X, Y$ and $Z$ on $\omega_{1}, \omega_{2}$ and $\omega_{3}$ respectively such that $\angle XBM = \angle YAM = \angle ZCM$ .

3) Points $(D,E,F)$ satisfy the properties of points $(X,Y,Z)$ mentioned in 2).

4) $M$ can lie only in the interior of $\triangle DEF$ .

5) $M$ can lie only in the interior of $\triangle ABC$ .

6) $\angle AMB = \angle AMC$ .

Which of the above statements do you think always hold true ?

Enter the sum of the serial numbers of such statements as your answer.

NOTE :

Suppose you think that statements $4), 5)$ and $6)$ always hold true, you must enter the answer as $15$ .
Tuples are ordered lists, so be careful.

This is an original problem.

The answer is 6.

1 solution

Venkata Karthik Bandaru
Jul 17, 2015

Well, the whole problem was designed to shed some light on Miquel's Theorem and its converse . The statements that are true are $1), 2)$ and $3)$ . Do try to prove these, some of them come along your way of proving the theorem true , and some of them are direct corollaries of the theorem.

Moderator note:

The problem statement is unnecessarily complicated and confusing, which would make this a problem that few would be interested in working through.

You could simplify the first paragraph greatly, by starting off with "consider 3 circles which intersect at a point M". Also, it is not clear if
1. These circles are allowed to be identical
2. These circles are allowed to pass through the same points (IE A=B=C)
3. If we allow for A/B/C = M.

Given the ambiguity in the question, it is not immediately apparent why 1, 2, 3 must hold.

In response to the Challenge Master Note:

I do not feel that the statements are unnecessarily complicated. I used standard notation, instead of long sentences that would be tough to read. Without notation, it would be tough to convey what I mean. Your "simplification" of the first paragraph does not convey everything about the intersections of pairs of individual circles.As for your ambiguity cases 1., 2., and 3., The circles are allowed to be identical, The circles are not allowed to pass through the same points ( read the first paragraph, intersections of pairs of individual circles are just two points ) and we can allow for A/B/C = M.

Assuming that the challenge master is you, @Calvin Lin .

@Brian Charlesworth @Nihar Mahajan @Michael Mendrin

Venkata Karthik Bandaru - 5 years, 11 months ago

Doesn't seem like Geometry!

This is an original problem.

The answer is 6.

1 solution

Moderator note:

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