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Indian National Mathematical Olympiad

Geometry Difficulty 6.7 National Olympiad Prove it India

Problem:

Let Γ1\Gamma_{1} and Γ2\Gamma_{2} be two circles of unequal radii, with centres O1O_{1} and O2O_{2} respectively, in the plane intersecting in two distinct points AA and BB. Assume that the centre of each of the circles Γ1\Gamma_{1} and Γ2\Gamma_{2} is outside the other. The tangent to Γ1\Gamma_{1} at BB intersects Γ2\Gamma_{2} again in CC, different from BB; the tangent to Γ2\Gamma_{2} at BB intersects Γ1\Gamma_{1} again in DD, different from BB. The bisectors of ∠DAB\angle D A B and ∠CAB\angle C A B meet Γ1\Gamma_{1} and Γ2\Gamma_{2} again in XX and YY, respectively, different from AA. Let PP and QQ be the circumcentres of triangles ACDA C D and XAYX A Y, respectively. Prove that PQP Q is the perpendicular bisector of the line segment O1O2O_{1} O_{2}.

Solution

Solution:

Figure 1
Let ∠CBA=α\angle C B A=\alpha and ∠DBA=β\angle D B A=\beta. Then ∠BDA=α\angle B D A=\alpha and ∠BCA=β\angle B C A=\beta. We also observe that ∠AO1O2=(∠AO1B/2)=α\angle A O_{1} O_{2}=\left(\angle A O_{1} B / 2\right)=\alpha and, simiarly, ∠AO2O1=β\angle A O_{2} O_{1}=\beta. Hence
∠O1AO2=180∘−(α+β) \angle O_{1} A O_{2}=180^{\circ}-(\alpha+\beta)
We also have
∠PO1A=∠DO1A2=2∠DBA2=∠DBA=β \angle P O_{1} A=\frac{\angle D O_{1} A}{2}=\frac{2 \angle D B A}{2}=\angle D B A=\beta
Hence ∠PO1O2=∠PO1A+∠AO1O2=β+α\angle P O_{1} O_{2}=\angle P O_{1} A+\angle A O_{1} O_{2}=\beta+\alpha. Similarly, we can get ∠PO2O1=α+β\angle P O_{2} O_{1}=\alpha+\beta. It follows that PP lies on the perpendicular bisector of O1O2O_{1} O_{2}.

Now we observe that
∠XQY=360∘−2∠XAY=360∘−2(180∘−α−β)=2(α+β) \angle X Q Y=360^{\circ}-2 \angle X A Y=360^{\circ}-2\left(180^{\circ}-\alpha-\beta\right)=2(\alpha+\beta)
This gives
∠O1QO2=12(∠XQA+∠YQA)=∠XQY2=α+β \angle O_{1} Q O_{2}=\frac{1}{2}(\angle X Q A+\angle Y Q A)=\frac{\angle X Q Y}{2}=\alpha+\beta
This shows that A,O1,O2,QA, O_{1}, O_{2}, Q are concyclic. We also have
∠ABX=∠ABD+∠DBX=β+∠DAX=β+∠DAB2∠ABY=∠ABC+∠CBY=α+∠CAY=α+∠BAC2 \begin{aligned} & \angle A B X=\angle A B D+\angle D B X=\beta+\angle D A X=\beta+\frac{\angle D A B}{2} \\ & \angle A B Y=\angle A B C+\angle C B Y=\alpha+\angle C A Y=\alpha+\frac{\angle B A C}{2} \end{aligned}
Adding we obtain
∠ABX+∠ABY=α+β+12(∠DAB+∠BAC)=α+β+(180∘−α−β)=180∘ \angle A B X+\angle A B Y=\alpha+\beta+\frac{1}{2}(\angle D A B+\angle B A C)=\alpha+\beta+\left(180^{\circ}-\alpha-\beta\right)=180^{\circ}
Hence X,B,YX, B, Y are collinear. Now
∠QAX=12(180∘−∠AQX)=90∘−β∠XAO1=12(180∘−∠XO1A)=90∘−12(360∘−2∠ABX)=∠ABX−90∘ \begin{gathered} \angle Q A X=\frac{1}{2}\left(180^{\circ}-\angle A Q X\right)=90^{\circ}-\beta \\ \angle X A O_{1}=\frac{1}{2}\left(180^{\circ}-\angle X O_{1} A\right)=90^{\circ}-\frac{1}{2}\left(360^{\circ}-2 \angle A B X\right)=\angle A B X-90^{\circ} \end{gathered}
Hence
∠QAO1=90∘−β+∠ABX−90∘=∠ABX−β=∠DAB2=∠O1AO22 \angle Q A O_{1}=90^{\circ}-\beta+\angle A B X-90^{\circ}=\angle A B X-\beta=\frac{\angle D A B}{2}=\frac{\angle O_{1} A O_{2}}{2}
This shows that AQA Q bisects ∠O1AO2\angle O_{1} A O_{2} and therefore the chords QO1Q O_{1} and QO2Q O_{2} subtend equal angles on the circumference of the circle passing through QO2AO1Q O_{2} A O_{1}. Hence QO2=QO1Q O_{2}=Q O_{1}. This means QQ lies on the perpendicular bisector of O1O2O_{1} O_{2}.

Combining, we get that PQP Q is the perpendicular bisector of O1O2O_{1} O_{2}.

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