AIME II 2026 (Problem 8)Plane GeometryPolynomialsHardIsosceles triangle △ABC\triangle ABC△ABC has AB=BCAB=BCAB=BC. Let III be the incenter of △ABC\triangle ABC△ABC. The perimeters of △ABC\triangle ABC△ABC and △AIC\triangle AIC△AIC are in the ratio 125:6125:6125:6, and all the sides of both triangles have integer lengths. Find the minimum possible value of ABABAB.Related TopicsCoreRational Root TheoremHomogeneity and Ratio SubstitutionTrigonometric IdentitiesCheck AnswerYour answer:CheckHints (7)Hint 1Let AB=BC=aAB=BC=aAB=BC=a, AC=bAC=bAC=b, and AI=CI=xAI=CI=xAI=CI=x.Hint 2cosA2=b2x=b2x\cos\frac{A}{2}=\frac{\frac b2}{x}=\frac{b}{2x}cos2A=x2b=2xbcosA=b2a=b2a\cos A=\frac{\frac b2}{a}=\frac{b}{2a}cosA=a2b=2abHint 3cosA=2cos2A2−1\cos A=2\cos^2\frac A2-1cosA=2cos22A−1b2a=2(b2x)2−1\frac b{2a}=2\left(\frac b{2x}\right)^2-12ab=2(2xb)2−1b2a=b2−2x22x2\frac b{2a}=\frac{b^2-2x^2}{2x^2}2ab=2x2b2−2x2⇒bx2=b2a−2x2a\Rightarrow bx^2=b^2a-2x^2a⇒bx2=b2a−2x2aHint 4PABCPAIC=1256\frac{P_{ABC}}{P_{AIC}}=\frac{125}{6}PAICPABC=61252a+b2x+b=1256\frac{2a+b}{2x+b}=\frac{125}{6}2x+b2a+b=612512a+6b=250x+125b12a+6b=250x+125b12a+6b=250x+125b12a=250x+119b12a=250x+119b12a=250x+119bHint 5By Hint 3 and Homogeneity and Ratio Substitutionbx2=b2a−2x2a÷x3⇒bx=b2x2⋅ax−2ax bx^2=b^2a-2x^2a \div x^3 \Rightarrow \frac bx=\frac{b^2}{x^2}\cdot\frac ax-2\frac ax bx2=b2a−2x2a÷x3⇒xb=x2b2⋅xa−2xa12a=250x+119b÷x⇒12ax=250+119bx 12a=250x+119b \div x \Rightarrow 12\frac ax=250+119\frac bx 12a=250x+119b÷x⇒12xa=250+119xbLet r=ax,t=bx \text{Let } r=\frac ax,\qquad t=\frac bx Let r=xa,t=xb12r=250+119t⇒r=250+119t12 12r=250+119t \Rightarrow r=\frac{250+119t}{12} 12r=250+119t⇒r=12250+119tt=t2r−2r⇒t=r(t2−2) t=t^2r-2r \Rightarrow t=r(t^2-2) t=t2r−2r⇒t=r(t2−2)t=(250+119t12)(t2−2) t=\left(\frac{250+119t}{12}\right)(t^2-2) t=(12250+119t)(t2−2)12t=(250+119t)(t2−2)=119t3+250t2−238t−500 12t=(250+119t)(t^2-2) =119t^3+250t^2-238t-500 12t=(250+119t)(t2−2)=119t3+250t2−238t−500119t3+250t2−250t−500=0 119t^3+250t^2-250t-500=0 119t3+250t2−250t−500=0Hint 6By Finding Rational Roots0=119t3+250t2−250t−5000=119t^3+250t^2-250t-5000=119t3+250t2−250t−500=(7t−10)(17t2+60t+50)=(7t-10)(17t^2+60t+50)=(7t−10)(17t2+60t+50)b,x>0⇒t=bx>0b,x>0\Rightarrow t=\frac{b}{x}>0b,x>0⇒t=xb>017t2+60t+50>017t^2+60t+50>017t2+60t+50>0⇒7t−10=0⇒t=107\Rightarrow 7t-10=0\Rightarrow t=\frac{10}{7}⇒7t−10=0⇒t=710Hint 7t=107=bx⇒b=10k, x=7kt=\frac{10}{7}=\frac{b}{x}\Rightarrow b=10k,\ x=7kt=710=xb⇒b=10k, x=7kBy Hint 3bx2=b2a−2x2ab x^2=b^2a-2x^2abx2=b2a−2x2a(10k)(7k)2=(10k)2a−2(7k)2a(10k)(7k)^2=(10k)^2a-2(7k)^2a(10k)(7k)2=(10k)2a−2(7k)2a490k3=100k2a−98k2a=2k2a 490k^3=100k^2a-98k^2a=2k^2a 490k3=100k2a−98k2a=2k2aa=245k a=245k a=245kIf k=1k=1k=1, thenmin{a}=245,b=10k=10,x=7k=7\min\{a\}=245,\quad b=10k=10,\quad x=7k=7min{a}=245,b=10k=10,x=7k=7and they satisfy all the conditions⇒Ans=min{a}=245\Rightarrow Ans=\min\{a\}=245⇒Ans=min{a}=245Final Answer245Related Problems (6)AMC 12B 2025 (Problem 16)2024 AMC 12A Problem 102024 AMC 12A Problem 23AMC 12B 2024 (Problem 21)AMC 12B 2024 (Problem 19)AMC 12B 2024 (Problem 22)
