We let the incircle be tangent to BC at K. Let N be on the incircle such that KN is the diameter of the incircle. Claim: D, P, N are collinear. Let O₁ be the center of $\omega_1$. ∠DPO₁ = ∠BO₁D / 2 = ∠BIK / 2 = ∠IPN. Similarly, E, Q, N are also collinear. ∠PDE = 90 - ∠PDO₁ = 90 - ∠O₁PD = 90 - ∠IPN = 90 - ∠PNI = ∠NKP = ∠NQP. Thus, ∠PQE + ∠PDE = 180 --> D, E, Q, P are concyclic.

Take an inversion at $D$. Note that $\Omega$ inverts to a line parallel to $BC$, and circles $\omega_1,\omega_2$ invert to two circles between the parallel lines. Thus, the inverted images of $D,E,Q,P$ form a rectangle, and hence are concyclic. Inverting back, we see that $D,E,Q,P$ are concyclic.

Let the incircle be tangent to $BC$ at $X$ and $XY$ is a diameter of the incirle Take the homothety $h$ with centre $P$ which provide $h(\omega_1)=\Omega$ $D$ is the tangency point of $BC$ and $\omega_1$ so $h(D)$ is the tangency point of $\Omega$ and a line which is tangent to $\Omega$, parallel to $BC$ and not $BC$ From this things $h(D)=Y$ so $D,P,Y$ are collinear. Similarly, $E,Q,Y$ are collinear $\angle{YPX}=\angle{YXD}=90^\circ\implies \triangle YPX \sim \triangle YXD\implies YP\cdot YD=YX^2$ Similarly, $YQ\cdot YE=YX^2=YP\cdot YD$ From PoP $D,E,P,Q$ are concyclic. We are done.

just use some very simple angle chasing first let PIQ=90+1/2A and QED=x and PDE=y 540=the total amount of angle in IPDEQ so now we have the relation between A x y and then we can easily find out that PQD=180-y hence proved.

why is Hongkong TST easier then Hongkong Mo?

math-olympiad-clown wrote: why is Hongkong TST easier then Hongkong Mo? That one is a wrong name, it is the Hong Kong (China) Mathematical Olympiad, or CHKMO, which is one of the later tests in the selection cycle. This on the one hand. On the other hand, the TST 1 said here is the first test in the selection cycle, after the computational contest that selects the students for the training and main selection cycle.