USAJMO 2026 (Problem 4)Triangle ABCABCABC has circumcircle ω\omegaω and circumcenter OOO. Lines AOAOAO and BCBCBC meet at point DDD. Let XXX be the AAA-excenter of △ABD\triangle ABD△ABD and let YYY be the AAA-excenter of △ACD\triangle ACD△ACD. Prove that if XXX lies on ω\omegaω, then YYY also lies on ω\omegaω.Related TopicsCoreExcenters and External Angle BisectorsMajorIncenter of a TriangleCircumcenters and Perpendicular BisectorsBasic Rules of Cyclic QuadrilateralsHints (3)Hint 1Hint 2∠AXB=∠ADB2=∠DAC+∠C2=90∘−B+C2.\angle AXB=\frac{\angle ADB}{2}=\frac{\angle DAC+\angle C}{2}=\frac{90^\circ-B+C}{2}.∠AXB=2∠ADB=2∠DAC+∠C=290∘−B+C.XXX lies on ω\omegaω ⇒\Rightarrow⇒ ∠AXB=∠C\angle AXB=\angle C∠AXB=∠C ⇒\Rightarrow⇒90∘−B+C2=C\frac{90^\circ-B+C}{2}=C290∘−B+C=C⇒90∘−B+C=2C\Rightarrow 90^\circ-B+C=2C⇒90∘−B+C=2CB+C=90∘.\boxed{B+C=90^\circ}.B+C=90∘.Hint 3∠AYC=∠ADC2\angle AYC=\frac{\angle ADC}{2}∠AYC=2∠ADC=90∘−C+B2.=\frac{90^\circ-C+B}{2}.=290∘−C+B.YYY lies on ω\omegaω ⟺\Longleftrightarrow⟺ ∠AYC=∠B\angle AYC=\angle B∠AYC=∠B⟺90∘−C+B2=B\Longleftrightarrow\frac{90^\circ-C+B}{2}=B⟺290∘−C+B=B⟺90∘−C+B=2B\Longleftrightarrow 90^\circ-C+B=2B⟺90∘−C+B=2B⟺B+C=90∘.\Longleftrightarrow B+C=90^\circ.⟺B+C=90∘.By Hint 2, B+C=90∘B+C=90^\circB+C=90∘, so YYY lies on ω\omegaω. ✓\checkmark✓Related Problems (3)AMC 12B 2024 (Problem 24)AMC 12A 2024 (Problem 19)AMC 12A 2025 (Problem 8)
