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It was easy to see that for an equilateral triangle maximum value of $\angle B$ can be $\angle B=60^\circ$ as for other triangles $\angle C$ or $\angle A$ would have the largest angle, depending on the common difference of the sides.

But how do I prove it using trigonometry, geometry or even calculus

I tried taking sides as $a-d,\; a,\; a+d$ and applying $Law \;of\; sines\;$ but couldn't get the result.

1 Answers 1

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Let the sides be $b-d, b, b+d$. To form a triangle, one needs $d

Then apply the law of cosines:

$$b^2=(b+d)^2+(b-d)^2-2(b+d)(b-d)\cos B$$

Long story short, you will get (unless I made some mistakes):

$$\cos B= \frac{3}{2(1-(d/b)^2)}-1$$

$B$ is maximum when $\cos B$ is minimum, which happens when $\frac{d}{b}=0$, so $d=0$ and the triangle is equilateral.

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    In the law of cosines, there is a minus sign in front of the $2.$ Also more steps needed to answer OP's question about getting the max of angle $B.$ – 2017-02-04
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    Thanks, corrected. Also added more steps. – 2017-02-04
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    @David K. Yes, this is is what I was trying to write, but I kept making mistakes in the formula for $\cos B$. Hope it's correct now. – 2017-02-04
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    If $x=b/d$ I'm getting $\cos B=(x^2+2)/[2(x^2-1)].$ This function is decreasing on the relevant interval $[2,\infty)$ so max at $2$ where $\cos B=1.$ I must have some error... – 2017-02-04
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    @coffeemath you want the minimum of $\cos B$, not the maximum. And this happens when $b/d\to\infty$, so $d=0$ – 2017-02-04
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    Thank you all, I believe $\cos B=(1+2x^2)/[2(1-x^2)] $ – 2017-02-04
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    @Momo Good point. So then from $d=0$ get equilateral triangle as OP thought. Nice, and +1. – 2017-02-04
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    OK, I see it's finished. Looks good! – 2017-02-04