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What is the maximum perimeter for an obtuse-angled triangle with each side $\le 100$ cm ? Which property to use?

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  • $\begingroup$ Use the law of cosines, the perimeter $P=a+b+c=a+b+\sqrt{a^2+b^2-2ab\cos C}$, where $\angle C$ is the obtuse angle. There is actually no maximum perimeter but a least upper bound (supremum). $\endgroup$ – Mathis Dec 27 '17 at 16:12
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I think the least upper bound will be just less than $100\sqrt 2 + 100$. This is because the perimeter of a right angled triangle with hypotenuse $100$ is $100\sqrt 2 + 100$.
Not a solid proof but this is the only idea I have and I think it could be right and explainable by more rigid means.

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To explain it a little bit. The triangle is $ABC$, and $c=\overline{AB}$ is the longest edge. Then we can have $c=100$, otherwise we can always scale the triangle so this is met.

Then we imagine a circle with $AB$ as its diameter. To be obtuse, $C$ has to be inside the circle. So for any $C$, we can always find $C'$ on the circle edge, so that $ABC$ is inside $ABC'$, so the perimeter of the former is less than the later. On the other hand, for all $C'$, we can always find $C$ to be close enough to it, so the perimeter of $ABC'$ is an upper limit.

Finally we just need to find the $ABC'$ with the longest perimeter, which is unsurprisingly $100\sqrt{2}+100$

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