1. **State the problem:** We are given three positive real numbers $a,b,c$ such that $$\frac{a}{a+b} = \frac{a+b}{a+b+c} = \frac{c}{b+c}$$ and also given the ratio $\frac{a}{b} = \frac{1 + \sqrt{x}}{y}$. We need to find the value of $xy$. 2. **Set the common value:** Let the common ratio be $k$: $$\frac{a}{a+b} = \frac{a+b}{a+b+c} = \frac{c}{b+c} = k$$ 3. **Write each equation from $k$:** - From $\frac{a}{a+b} = k$, we get: $$a = k(a+b) \implies a = ka + kb \implies a - ka = kb \implies a(1-k) = kb \implies \frac{a}{b} = \frac{k}{1-k}$$ - From $\frac{a+b}{a+b+c} = k$, we have: $$a+b = k(a+b+c) \implies a+b = ka + kb + kc \implies a+b - ka - kb = kc \implies (1-k)(a+b) = kc$$ - From $\frac{c}{b+c} = k$, we have: $$c = k(b+c) \implies c = kb + kc \implies c - kc = kb \implies c(1-k) = kb \implies \frac{c}{b} = \frac{k}{1-k}$$ 4. **Use ratios to express $a$ and $c$ in terms of $b$:** From above, $$\frac{a}{b} = \frac{k}{1-k} \quad\text{and}\quad \frac{c}{b} = \frac{k}{1-k}$$ So, $$a = b \frac{k}{1-k}, \quad c = b \frac{k}{1-k}$$ 5. **Use equation involving $(a+b)$ and $c$:** Recall equation: $$(1-k)(a+b) = kc$$ Rewrite left side: $$(1-k)\Big(a+b\Big) = (1-k)\left(b\frac{k}{1-k} + b\right) = (1-k) b \left(\frac{k}{1-k} + 1\right)$$ Simplify inside parentheses: $$\frac{k}{1-k} + 1 = \frac{k + (1-k)}{1-k} = \frac{1}{1-k}$$ So, $$(1-k)b \cdot \frac{1}{1-k} = b$$ Therefore, $$b = k c = k b \frac{k}{1-k}$$ Dividing both sides by $b$ (since $b>0$): $$1 = k \cdot \frac{k}{1-k} = \frac{k^2}{1-k}$$ 6. **Solve for $k$:** Multiply both sides by $1-k$: $$1-k = k^2 \implies k^2 + k -1 = 0$$ Solve quadratic equation for $k$: $$k = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$$ Since $k$ is a positive ratio (fractions of positive numbers), take the positive root: $$k = \frac{-1 + \sqrt{5}}{2}$$ 7. **Calculate $\frac{a}{b}$:** Recall from step 3: $$\frac{a}{b} = \frac{k}{1-k}$$ Substitute $k$: $$\frac{a}{b} = \frac{\frac{-1 + \sqrt{5}}{2}}{1 - \frac{-1 + \sqrt{5}}{2}} = \frac{\frac{-1 + \sqrt{5}}{2}}{\frac{2 - (-1 + \sqrt{5})}{2}} = \frac{-1 + \sqrt{5}}{2} \cdot \frac{2}{2 +1 - \sqrt{5}} = \frac{-1 + \sqrt{5}}{3 - \sqrt{5}}$$ Rationalize denominator: $$\frac{-1 + \sqrt{5}}{3 - \sqrt{5}} \cdot \frac{3 + \sqrt{5}}{3 + \sqrt{5}} = \frac{(-1 + \sqrt{5})(3 + \sqrt{5})}{(3)^2 - (\sqrt{5})^2} = \frac{(-1)(3) -1(\sqrt{5}) + 3\sqrt{5} + 5}{9 - 5} = \frac{-3 - \sqrt{5} + 3 \sqrt{5} + 5}{4}$$ Simplify numerator: $$-3 + 5 + (-\sqrt{5} + 3 \sqrt{5}) = 2 + 2 \sqrt{5} = 2(1 + \sqrt{5})$$ So, $$\frac{a}{b} = \frac{2(1 + \sqrt{5})}{4} = \frac{1 + \sqrt{5}}{2}$$ 8. **Match given ratio to find $x$ and $y$:** Given: $$\frac{a}{b} = \frac{1 + \sqrt{x}}{y}$$ From above we have: $$\frac{1 + \sqrt{5}}{2} = \frac{1 + \sqrt{x}}{y}$$ Comparing, it suggests: $$x = 5, \quad y = 2$$ 9. **Calculate $xy$:** $$xy = 5 \times 2 = 10$$ **Final answer:** $$\boxed{10}$$