EGMO 2017 1


Nothing like a geo problem to end a hiatus.


To show that MNAC is a cyclic quadrilateral we will prove that \angle CMA = \angle CNA.

Now let’s angle chase!
First, set \angle BAC = y, \angle BCA = x, and \angle PQB = z.

Now we know that M is the circumcenter of \triangle ABD. So, by the inscribed angle theorem, \angle BMA = 2 \cdot \angle BCA = 2x and similarly, \angle CMB = 2 \cdot \angle BAC = 2y.

Thus \angle CMA = \angle CMB + \angle BMA = 2x+2y.

Now because we have cyclic quadrilateral ABCD, \angle BDA = x and \angle BDC = y.

Since \angle DAB = \angle BCD = 90, \angle ABD = 90-x and \angle CBD = 90-y.
This means that \angle PBQ = 180-(90-y)-(90-x)=x+y. And because the interior angles of a triangle add up to 180^{\circ}, we know that \angle BPQ = 180 - (x+y) - z (by looking at \triangle PBQ).

By a similar angle chasing argument one can also get that \angle RDS = 180-x-y.

Now, because \triangle PAS is a right triangle, \angle PSA = 90 - \angle APS = 90 - (180-x-y-z) = x+y+z-90.

And because angles in a triangle add up to 180^{\circ}, \angle DRS = 180 - (x+y+z-90) - (180-x-y) = 90 - z.

By vertical angles, \angle CRQ = 90-z as well and now one can notice that in right triangle CRQ, N is the midpoint of the hypotenuse and thus the circumcenter. And thus we can see that \angle CNQ = 2 \cdot \angle CRQ = 180 - 2z.

Now for the final crucial observation, one can see that N is also the midpoint of the hypotenuse of \triangle PAS as PQ = RS. So by an analogous argument \angle PNA = 2 \cdot \angle PSA = 2(x+y+z-90) = 2x+2y+2z-180.

From here we add \angle QNA (\angle PNA = \angle QNA) and \angle CNQ to get \angle CNA = 2x+2y+2z-180+180-2z = 2x+2y.

This is exactly the same as \angle CMA and so we are done. \blacksquare

Remark: This solution only required simple angle chasing and important observations about N and M (especially N) Neat problem!