The sides of a triangle are $n$, $n-1$, and $n+1$, for natural $n$, and angles $A$, $2A$, $\pi-3A$. Find $n$.

Question

Problem:

$\triangle ABC$ is such that $AB=n$, $AC=n-1$, $BC=n+1$ (with $n$ a natural number) and angles $A$, $2A$, $\pi-3A$. Find $n$.

The answer is $n=5$.

My try:

By the Law of Cosines we have:

$$\begin{align} n^{2} &=(n-1)^{2}+(n+1)^{2}-2(n^{2}-1)\cos (\pi-3A) \\ (n-1)^{2}&=n^{2}+(n+1)^{2}-2n(n+1)\cos A \\ (n+1)^{2}&=n^{2}+(n-1)^{2}\cos (2A) \end{align}$$

I know that $$\cos(\pi-3A)=-\cos(3A)=3\cos A-4\cos^{3}A \quad\text{and}\quad \cos(2A)=2cos^{2}A-1$$

From the second equation we have

$$\cos A=\frac{n+4}{2(n+1)}$$

But when I applied I can't get correct answer.

I'm searching a simple way to find $n$.

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And we can generalized for sides $4k$, $5k$, $6k$ ($k$ a natural number).

Answer 1

By law of sines: $$\frac{\sin\alpha}{n-1}=\frac{\sin2\alpha}{n+1},$$ which gives $$\cos\alpha=\frac{n+1}{2(n-1)}.$$ Also, $$\frac{\sin\alpha}{n-1}=\frac{\sin3\alpha}{n}$$ or $$\frac{n}{n-1}=3-4\sin^2\alpha$$ or $$\frac{n}{n-1}=3-4+\frac{(n+1)^2}{(n-1)^2}$$ or $$n^2-n=-n^2+2n-1+n^2+2n+1.$$ Can you end it now?

You made a mistake for $\cos2\alpha$. $$\cos2\alpha=\frac{n^2+(n-1)^2-(n+1)^2}{2n(n-1)},$$ which gives $$\cos2\alpha=\frac{n-4}{2(n-1)}.$$

Answer 2

By the sine theorem, we have: $\frac{n-1}{sinA}=\frac{n+1}{sin2A}=\frac{n}{sin(\pi-3A)}$ Using the equality of the first and second relations, as well as the formula for the sine of the double argument, we find $cosA=\frac{n+1}{2(n-1)}$ (*); You have already found $cosA=\frac{n+4}{2(n+1)}$ (**); Equating the right-hand sides of equalities (*) and (**) we obtain an equation that gives a single root $n=5$

Answer 3

If $\angle A=A, \angle B=2A, \angle C=\pi -3A$, there is no solution.

If $\angle A=2A, \angle B=A, \angle C=\pi -3A$, then by the Sine theorem: $$\frac{n-1}{\sin A}=\frac{n+1}{\sin 2A}=\frac{n}{\sin (\pi -3A)}$$ From the first: $$\frac{\sin 2A}{\sin A}=\frac{n+1}{n-1} \Rightarrow \cos A=\frac{n+1}{2n-2}$$ From the second: $$\frac{\sin(\pi -3A)}{\sin A}=\frac{n}{n-1} \Rightarrow 3-4\sin^2A=\frac{n}{n-1} \Rightarrow \\ 3-4\left(1-\left(\frac{n+1}{2n-2}\right)^2\right)=\frac{n}{n-1} \Rightarrow \\ 12(n-1)^2-4[4(n-1)^2-(n+1)^2]=4n(n-1) \Rightarrow \\ 4n^2-20n=0 \Rightarrow n=5.$$