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For $\vartriangle$ABC it is given that $$\frac1{b+c}+\frac1{a+c}=\frac3{a+b+c}$$ Find the measure of angle $C$.

This is a "challenge problem" in my precalculus book that I was assigned. How do I find an angle from side lengths like this? I have tried everything I can. I think I may need to employ the law of cosines or sines. Thanks.

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  • $\begingroup$ What are a,b,c? are they sides of $\Delta$ $\endgroup$
    – Ganesh
    Feb 15, 2016 at 15:24
  • $\begingroup$ On a closely related note (as the answers below illustrate), see Eisenstein triple. $\endgroup$
    – PM 2Ring
    Feb 16, 2016 at 13:46

3 Answers 3

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Multiply both sides by $a+b+c$ to get $$\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}=3$$

We get

$$\frac{a}{b+c}+\frac{b}{a+c}=1$$

$$a(a+c)+b(b+c)=(a+c)(b+c)$$

$$a^2+ac+b^2+bc=ab+ac+bc+c^2$$

$$a^2+b^2=ab+c^2$$

$$a^2-ab+b^2=c^2$$

By the cosine law, we have $a^2-2\cos(C)ab+b^2=c^2$. Hence $\cos(C)=\frac12$, so $C=60^\circ$.

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Short answer:

According to the problem, the solution is unique, so any triple of values that satisfies the equation provides a solution. We immediately see that $a=b=c=1$ is a solution, hence the angle is 60 degrees.

Medium answer.

What count are the ratios between sides. So we can assume that $c=1$. The equation is symmetric in $a$ and $b$, and we know it's unique, so as $a$ varies, $b$ varies. We can try to see if this infinite family of solutions has an intersection with isoscele triangles, so we put $b=a$ and we solve $$\frac{1}{a+1}+\frac{1}{a+1}=\frac{3}{2a+1}$$ finding $a=b=c=1$. So the angle is 60 deg.

Notice that the fun thing is that $k\cdot (1,1,1)$ is not the unique solution. They are indeed infinite. Example, $a=15$, $b=8$, $c=13$.

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Hint:

Simplify the equation to $$a^2 + b^2 - ab = c^2$$ And use cosine law.

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