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I was given $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=9$ and $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=20$ and was asked one of the possible values for $\frac{a}{b}+\frac{c}{d}$.

So far, what I did is isolate $\frac{a}{b}+\frac{c}{d}=9-\left(\frac{b}{c}+\frac{d}{a}\right)=9-\left(\frac{ab+cd}{ac}\right)$ and then changed the second equation so which I have $\frac{a^2+c^2}{ac}+\frac{b^{2\:}+d^{2\:}}{bd}=20$, but I had no idea how to continue. The choices are 4, 3, 2, 1, and 0.

aco
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1 Answers1

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Hint: $\left( \frac{a}{b} + \frac{c}{d} \right) \left( \frac{b}{c} + \frac{d}{a} \right) = ?? $

Expanding, we get $ \frac{a}{c} + \frac{ d}{b} + \frac{b}{d} + \frac{c}{a}$

Let $ x = \frac{a}{b} + \frac{c}{d} , y = \frac{b}{c} + \frac{d}{a}$, what do the equations transform into?

$x+y = 9, xy = 20$.

Hence, $ x \in \{4, 5 \}$.

Note: Solutions exist like by setting $ a = 1, c = 2$ and solving the diophantine equations

Calvin Lin
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