5

We want to find the maximum area of a triangle inscribed in a circle with radius r and with constant difference of two of its angles.

If $a, b, c$ are the angles of the triangle, if we set, wlog that $a>b$, we need to have:

$a-b = k$ (constant) and

$a+b+c=180$, so $a+b = 180-c$

I know that in general, without the restriction of the $2$ angles fixed difference, the largest triangle is the equilateral. Any assistance is much appreciated (by the way this is not homework or anything; just challenge between friends).

Or Shahar
  • 1,766

1 Answers1

1

Hint: use this formula"

$S=\frac{abc}{4R}$

Now we want S be maximum for a constant R. We apply sine rule:

$\frac a{ \sin \alpha}=\frac b { \sin \beta}=\frac c{\sin \gamma}=2R$

$\alpha +\beta=180-\gamma$

$\alpha - \beta =k$

These two relations give:

$\alpha = \frac {\pi}2+\frac {k-\gamma}2$

$\beta = \frac {\pi}2 -\frac{k +\gamma}2$

Plug these in S you get:

$S=2R^2 \sin{\left( \frac {\pi}2+\frac {k-\gamma}2 \right)} \sin { \left(\frac {\pi}2 -\frac{k +\gamma}2 \right)} \sin (\gamma)$

Now take derivative of S ,$\gamma$ is the variable, set ir to 0 find $\gamma$ and put it in S and find the value of S. It can be seen (for example plotting S in wolfram) that for any k , S is maximum when $\gamma=90^o$, that is triangle must be right triangle to have maximum ares for any k.

Mailbox
  • 896
sirous
  • 10,751