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Taiwan IMO Selection Camp, 2022

Combinatorics Difficulty 5.6 AIME, harder Prove it Taiwan

Let $N$ and $s$ be positive integers with $N > s$ . In the Taipower Park there are several buildings, of which exactly $N$ are power plants, and one other is the headquarters. Between certain pairs of buildings there are wires that can only transmit power in one direction, satisfying:

(i) All wires connected to a power plant only send power out of the power plant.

(ii) For every building other than the headquarters, there exists a unique sequence of wires forming a circuit from that building to the headquarters.

A building is called $s$ -level powered if and only if, when we remove any one wire in the park, the building can still receive power from at least $s$ power plants. Find the maximum possible number of $s$ -level powered buildings.

Solution

Solution.
In fact, the order of $ABCDE$ on $Ω$ doesn't affect the result. Let $U, V$ be the second intersections of $⊙(ABC)$ and $DS, DT$ , respectively, and $R$ be the point on $ST$ such that $∠BAD = ∠RAC$ . Since $△ABD \sim \triangle ART$ , we have $∠ART = ∠ABD = ∠AVD = ∠AVT$ , implying that $A, R, T, V$ are concyclic. Hence we obtain that $B, R, V$ are collinear since $∠BVD = ∠BAD = ∠RAC = ∠RVT$ . Similarly, $C, R, U$ are collinear. Applying Pascal's theorem to $BECUDV$ , we obtain that $X = BZ ∩ UD$ , $Y = EC ∩ DR$ , and $R = CU ∩ BV$ are collinear.
By the assumption that $AD$ is tangent to $⊙(DXY)$ , we obtain
$∠SXR = ∠ADV = ∠ABV = ∠SBR,$

implying that $B, S, R, X$ are concyclic. Therefore,
$\angle AEY = \angle AEC = \angle ABC = \angle BSR = \angle EXY,$
implying that $AE$ is tangent to $\odot(EXY)$ , as desired.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.