HMMT 2025 — Algebra & Number Theory Round (Problem 7)There exists a unique triple (a,b,c)(a, b, c)(a,b,c) of positive real numbers that satisfies the equations 2(a2+1)=3(b2+1)=4(c2+1)2(a^2 + 1) = 3(b^2 + 1) = 4(c^2 + 1)2(a2+1)=3(b2+1)=4(c2+1) and ab+bc+ca=1ab + bc + ca = 1ab+bc+ca=1. Compute a+b+ca + b + ca+b+c.Related TopicsToolkit 27 — Expansion (or factorization) of two binomialsHints (11)Hint 1Use 27. Expansion (or factorization) of two binomials.Hint 2a2+1=a2+ab+ac+ca=(a+b)(a+c)a^2 + 1 = a^2 + ab + ac + ca = (a + b)(a + c)a2+1=a2+ab+ac+ca=(a+b)(a+c)Hint 32(a+b)(a+c)=3(b+a)(b+c)=4(c+a)(c+b)2(a + b)(a + c) = 3(b + a)(b + c) = 4(c + a)(c + b)2(a+b)(a+c)=3(b+a)(b+c)=4(c+a)(c+b)Hint 4Let a+b=x, a+c=y, b+c=za + b = x,\ a + c = y,\ b + c = za+b=x, a+c=y, b+c=z.Hint 52xy=3yz=4xz2xy = 3yz = 4xz2xy=3yz=4xzHint 6y=3x4,z=x2y = \tfrac{3x}{4},\quad z = \tfrac{x}{2}y=43x,z=2xHint 7a+b=x, a+c=3x4, b+c=x2a + b = x,\ a + c = \tfrac{3x}{4},\ b + c = \tfrac{x}{2}a+b=x, a+c=43x, b+c=2xHint 8a=5x8,b=3x8,c=x8a = \tfrac{5x}{8},\quad b = \tfrac{3x}{8},\quad c = \tfrac{x}{8}a=85x,b=83x,c=8xHint 9ab+ac+bc=1ab + ac + bc = 1ab+ac+bc=1Hint 10x=823x = \tfrac{8}{\sqrt{23}}x=238Hint 11a+b+c=x+y+z2a + b + c = \tfrac{x + y + z}{2}a+b+c=2x+y+zFinal Answer923=92323\tfrac{9}{\sqrt{23}} = \tfrac{9\sqrt{23}}{23}239=23923