Problem 11 · Symmetric fractions adding up to 2025
Algebraic manipulation: a product hiding the very same sum.
Integer answer, at most 4 digitsThree numbers $a$, $b$, $c$ satisfy $\dfrac{a+b}{c} + \dfrac{b+c}{a} + \dfrac{c+a}{b} = 2025$. What is the value of
$$\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)?$$
Copa Cangur · SCM
Medium
Closed answer
Reasoned solution
Key idea: write each factor over a single denominator and expand: exactly the six quotients from the hypothesis appear.
$$\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right) = \frac{a+b}{a}\cdot\frac{c+a}{c}\cdot\frac{b+c}{b} = \frac{(a+b)(b+c)(c+a)}{abc}.$$
Expanding the numerator:
$$(a+b)(b+c)(c+a) = a^{2}b + a^{2}c + ab^{2} + b^{2}c + ac^{2} + bc^{2} + 2abc.$$
Dividing by $abc$, each term becomes one of the six quotients $\frac{a}{c}, \frac{a}{b}, \frac{b}{c}, \frac{b}{a}, \frac{c}{b}, \frac{c}{a}$, plus the final $2$. But those six quotients are exactly the ones in the condition:
$$\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} = \frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b} = 2025.$$
$$\text{Product} = 2025 + 2 = 2027.$$
Answer: 2027