1. **State the problem:** We have positive real numbers $a$, $b$, $c$ such that $$\frac{a}{a+b} = \frac{a+b}{a+b+c} = \frac{c}{b+c}$$ and we know $$\frac{a}{b} = \frac{1 + \sqrt{x}}{y}$$ We want to find the value of $xy$. 2. **Set a common ratio:** Since all three fractions are equal, let this common value be $k$, so $$\frac{a}{a+b} = \frac{a+b}{a+b+c} = \frac{c}{b+c} = k$$ 3. **Express variables in terms of $k$: ** From the first equality, $$\frac{a}{a+b} = k \implies a = k(a+b) \implies a = ka + kb \implies a - ka = kb \implies a(1-k) = kb \implies \frac{a}{b} = \frac{k}{1-k}$$ 4. **From the third equality,** $$\frac{c}{b+c} = k \implies c = k(b+c) \implies c = kb + kc \implies c - kc = kb \implies c(1 - k) = kb \implies c = \frac{kb}{1-k}$$ 5. **From the second equality,** $$\frac{a+b}{a+b+c} = k \implies a + b = k(a+b+c)$$ Substitute $a$ and $c$ with expressions in terms of $b$ and $k$: Recall from Step 3: $a = \frac{k}{1-k}b$, then $a + b = \frac{k}{1-k}b + b = \left( \frac{k}{1-k} + 1 \right) b = \frac{k + 1 - k}{1 - k}b = \frac{1}{1-k}b$ Similarly, $a + b + c = a + b + c = \frac{k}{1-k}b + b + \frac{kb}{1-k} = \left( \frac{k}{1-k} + 1 + \frac{k}{1-k} \right) b = \left( 1 + \frac{2k}{1-k} \right) b = \frac{1 - k + 2k}{1-k}b = \frac{1 + k}{1 - k} b$ Now, the second ratio equality: $$\frac{a+b}{a+b+c} = \frac{ \frac{1}{1-k} b }{ \frac{1+k}{1-k} b } = \frac{1}{1+k}$$ But we defined this as $k$, so $$k = \frac{1}{1+k} \implies k(1+k) = 1 \implies k + k^2 = 1 \implies k^2 + k -1=0$$ 6. **Solve the quadratic for $k$: ** $$k = \frac{-1 \pm \sqrt{1+4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$$ Since $k$ represents a ratio of positive quantities, $0 < k < 1$, so take $$k = \frac{-1 + \sqrt{5}}{2}$$ 7. **Recall from step 3:** $$\frac{a}{b} = \frac{k}{1-k}$$ Calculate this: $$1-k = 1 - \frac{-1 + \sqrt{5}}{2} = \frac{2 - (-1 + \sqrt{5})}{2} = \frac{3 - \sqrt{5}}{2}$$ Therefore, $$\frac{a}{b} = \frac{ \frac{-1 + \sqrt{5}}{2} }{ \frac{3 - \sqrt{5}}{2} } = \frac{-1 + \sqrt{5}}{3 - \sqrt{5}}$$ Multiply numerator and denominator by conjugate $(3 + \sqrt{5})$: $$\frac{-1 + \sqrt{5}}{3 - \sqrt{5}} \times \frac{3 + \sqrt{5}}{3 + \sqrt{5}} = \frac{(-1 + \sqrt{5})(3 + \sqrt{5})}{9 - 5} = \frac{(-1)(3) + (-1)(\sqrt{5}) + 3\sqrt{5} + \sqrt{5}\sqrt{5}}{4}$$ Simplify numerator: $$-3 - \sqrt{5} + 3\sqrt{5} + 5 = (-3 + 5) + (-\sqrt{5} + 3\sqrt{5}) = 2 + 2\sqrt{5}$$ Therefore, $$\frac{a}{b} = \frac{2 + 2\sqrt{5}}{4} = \frac{2(1 + \sqrt{5})}{4} = \frac{1 + \sqrt{5}}{2}$$ 8. **Match with given form:** Given $$\frac{a}{b} = \frac{1 + \sqrt{x}}{y}$$ We found $$\frac{a}{b} = \frac{1 + \sqrt{5}}{2}$$ So, $$x = 5, \quad y = 2$$ 9. **Calculate $xy$: ** $$xy = 5 \times 2 = 10$$ **Final answer:** $$\boxed{10}$$