Question 14

If $$\frac{1}{b+c} + \frac{1}{c+a} = \frac{2}{a+b}$$, then the value of $$\frac{a^2 + b^2}{c^2}$$ is

Solution

Combining the left side gives $$\frac{a + b + 2c}{(b+c)(c+a)} = \frac{2}{a+b}$$, so $$(a+b)(a+b+2c) = 2(b+c)(c+a)$$. Expanding both sides gives $$a^2 + 2ab + b^2 + 2ac + 2bc = 2ab + 2bc + 2ac + 2c^2$$, which reduces to $$a^2 + b^2 = 2c^2$$. Hence $$\frac{a^2 + b^2}{c^2} = 2$$.

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