Problem 5

An arbitrary point

M

is selected in the interior of segment

AB

. Squares

AMCD

and

MBEF

are constructed on the same side of

AB

, with circumcenters

P

and

Q

. Their circumcircles meet again at

N

. Let

N'

be the intersection of

AF

and

BC

. (a) Prove

N=N'

. (b) Prove that

MN

passes through a fixed point independent of

M

. (c) Find the locus of the midpoint of

PQ

M

varies.

Step 2 of 5: Identify the circle with diameter

AB

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\angle AMN=\angle BMN=90^\circ\quad\Longrightarrow\quad A,M,N,B\text{ are concyclic}

Detailed analysis

Since $N$ lies on both square circumcircles, the right-angle relations imply $\angle ANB=90^\circ$ . Hence $N$ lies on the circle with diameter $AB$ . The same conclusion for $N'$ follows from $\angle AN'B=90^\circ$ .