concyclic wanted, 2 circles tangent to incircle related

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Source: Hong Kong TST - HKTST 2024 1.6

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30628 posts

#1 h Jul 20, 2024, 11:38 PM • 1 Y

Y by Rounak_iitr

Let $\Omega$ be the incircle of $\Delta ABC$ . There are two smaller circles $\omega_1$ amd $\omega_2$ inside $\Delta ABC$ . The circle $\omega_1$ is tangent to $\Omega$ at $P$ , tangent to $BC$ at $D$ , and also tangent to $AB$ . The circle $\omega_2$ is tangent to $\Omega$ at $Q$ , tangent to $BC$ at $E$ , and also tangent to $AC$ . Prove that $D,E,Q,P$ are concyclic.

This post has been edited 1 time. Last edited by parmenides51, Jul 21, 2024, 7:46 AM
Reason: title typo

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JJeom

#2 h Jul 21, 2024, 3:08 AM

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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.

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277 posts

#3 h Jul 21, 2024, 3:13 AM

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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.

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tugra_ozbey_eratli

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tugra_ozbey_eratli

#4 h Aug 29, 2024, 8:27 PM • 1 Y

Y by ehuseyinyigit

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.

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math-olympiad-clown

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math-olympiad-clown

#5 h Jan 4, 2025, 10:38 AM

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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.

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math-olympiad-clown

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math-olympiad-clown

#6 h Jan 4, 2025, 10:40 AM

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why is Hongkong TST easier then Hongkong Mo?

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834 posts

#7 h Jan 4, 2025, 12:32 PM • 1 Y

Y by Mathematics_enthusiasts

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.

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#8 h Mar 27, 2025, 7:51 AM

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We define $H$ , $K$ as the intersection of the tangent to $\omega_{1}$ at $P$ with $BC$ , and of the tangent to $\omega_{2}$ at $Q$ with $BC$ respectively.
Notice that by radical axis $HD=HZ=HP$ , thus $H$ is the center of $(DPZ)$ . Hence $D$ is the reflection of $Z$ over $H$ .
In the same fashion, we can show $KZ=KE=KQ$ .

Thus we redefine the problem as follows: Let $\triangle PQZ$ have circumcenter $I$ , let $H$ and $K$ be defined as the intersection of the tangents
at $P$ , $Z$ and at $Q$ , $Z$ respectively. Let $D$ be the defined as the reflection of $Z$ over $H$ and let $E$ be defined as the reflection of $Z$ over $K$ .
Show that $PQDE$ is cyclic.

The problem is now trivial by angle chase, complex also works although it is a bit overkill. Notice that
$\angle PDE = \angle PDH = 90 - \frac{\angle PHD}{2} = \frac{\angle PHZ}{2} = 90 - \angle PZH = 90 - \angle PQZ$ However since $ZE$ is a diameter of $(ZQE)$ we know that $\angle ZQE = 90$ , thus
$\angle PQE = 90 + \angle PQZ \Longrightarrow 180 - \angle PQE = 90 - \angle PQZ = \angle PDE$ Thus $PQDE$ is cyclic.

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