Let’s first draw a diagram given all the information:
(let G denote the intersection of \overline{AP} with (ABC) and F the intersection of \overline{BQ} and (ABC))

Now we have to prove that the quadrilateral P_1Q_1QP is cyclic.
Let’s draw a circle around them and see if anything interesting happens.

It looks like F and G are also on this circle!
Maybe proving that will help.
\angle GAB = \angle GPQ (lines \overline{PQ} and \overline{AB} are parallel)
But \angle GAB = \angle GFB as well because they subtend the same arc in (ABC)!

Now we can conclude that the four points G, F, P, and Q are concylic.

From here we would like to prove that \angle GQ_1Q = \angle GFQ (we can apply the same technique to P_1)

Time for some wishful thinking!
Wishful Thinking
Since \angle QFG = \angle A_1CG (they subtend arc GB), if we can prove that \angle A_1CG = \angle A_1Q_1G then we’d be done!
But this means that points G, Q_1, C and A_1 are concylic. Could we prove that?
Yes we can!

We know that \angle CBA = \angle CGA.
But \angle CBA= \angle CQ_1A_1 as well (looking at (CQ_1B))!

Thus points G, Q_1, C and A_1 are concylic and the same applies for P_1, C, B_1 and F.


(Here’s the final diagram)


Absolutely unreal! Can I ask how you make the diagrams?

It’s by using Geogebra. (on an actual contest it would be using a straightedge and a compass)

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