Problem 3 of Asian Pacific Mathematical Olympiad 2002

Let ABC be an equilateral triangle. Let P be a point on the side AC and Q be a point on the side AB so that both triangles ABP and ACQ are acute. Let R be the orthocentre of triangle ABP and S be the orthocentre of triangle ACQ. Let T be the point common to the segments BP and CQ. Find all possible values of ∠CBP and ∠BCQ such that triangle TRS is equilateral.

Solution by Vo Duc Dien (dedicated to former colleague David Bengel):

Let a be the side length of equilateral triangle ABC and

	∠CBP = α
	∠BCQ = β
	∠SBR = δ
	∠ABS = σ
	∠TBR = ψ

We have:

BT / sinβ = a /sin (α+β) or BT = a sinβ /sin (α+β)

BP / sin 60° = PC / sinα = a / sin (α+60°)

but α+60° = 90° - ψ therefore sin (α+60°) = cosψ

BP = sin(60°) a/cosψ and PC = a sinα /cosψ

AP = NP /sinψ or

NP = sinψ AP = sinψ (a – PC) = a sinψ (1 – sinα /cosψ)

TN = BP – BT – NP

TN = sin(60°) a /cosψ - a sinβ /sin (α+β) - a sinψ (1 – sinα/cosψ)

Now for RN:

RN / sinψ = BN /cosψ or RN = BN tanψ = (BP – NP) tanψ

RN = [sin(60°) a /cosψ - a sinψ ( 1 – sinα /cosψ )] tanψ

TR² = TN² + RN² = [sin(60°) a/cosψ - a sinβ/sin (α+β) - a sinψ (1 – sinα /cosψ )] ² + [sin(60°)a /cosψ - a sinψ (1 – sinα /cosψ )] ² tan²ψ

Using the same process to find TS, we have

TS² = TM² + SM² = [sin(60°) a/cosσ - a sinα/sin (α+β) - a sinσ (1 – sinβ /cosσ )] ² + [sin(60°) a /cosσ - a sinσ (1 – sinβ /cosσ )] ² tan²σ

So for TR = TS one obvious solution is that α = β, ψ = σ to make the corresponding terms of TR² and TR² above equal, and when α = β the points P and Q are symmetrical across AI where I is the foot of A to BC.

Since SA = SB and CG ⊥ AB and CQ ⊥ AM, we then also have

∠ABS = ∠SAB = ∠TCS = σ

Assume a solution has been attained and that ∠CBP = α₁ and ∠BCQ = β₁

We will prove that for every unique value α₁ for ∠CBP there is one and only one corresponding value β₁ for ∠BCQ to satisfy the problem.

Indeed, let’s keep angle α₁ and increase ∠BCQ. As we do so point T moves to T’ closer to N and RT’ < RT, or RT decreases.

We also know that ∠MAN = α + β. So ∠MAN increases by the same amount of the increase of ∠BCQ, and ∠SAG also simulatneously decreases by the same amount. Therefore, as we increase ∠BCQ, point S moves to S’ closer to point G and RS’ > RS, or RS increases.

The same but opposite effect occurs if we decrease ∠BCQ.

Therefore TR will no longer equal SR if ∠BCQ ≠ β₁. So for every angle α there is only one unique angle β to satisfy the condition for triangle TRS to be equilateral.

We earlier proved that ∠BCQ = ∠CBP is a condition for ST = RT. Based on the above reasoning, there is no other value for ∠CBP to equal ∠BCQ except that their values are the same, or α = β and ψ = σ, and point T has to always be on AI. Now let’s find ∠α.

Since triangle TRS is equilateral, and R is on the bisector BH of ∠ABC and T is on bisector AI of ∠BAC, we have SR || BC, ST || AC and RT || AB, or BH is the bisector of ∠SBT or δ = ψ. We have

σ = 30° - δ = 30° - ψ = α

We also have ∠BCG = σ + β = 30°

Therefore, α = β = δ = σ = ψ = 15°

or ∠CBP = ∠BCQ = 15°