[Math] Find the biggest area of an isosceles triangle in an unit circle

calculusgeometry

An isosceles triangle is inscribed in the unit circle. Determine the largest area of ​​the triangle can assume.

My solution attempt:

An isosceles triangle has two sides of equal size, and the sides will be at some point on the circumference of the unit circle during suggestion the $x$-axis. The height will be $1 + h \le 2$ and width $b + 1 \le 2$

Another way would be to make the vectors of the three points that will be on the periphery (assuming that it can provide the largest area on the periphery)

I really do not know how to solve it.

Best Answer

Hint

The triangle is $\Delta ABC$: $AB=AC=b$ and $BC=a$.

Also $\angle A=2x \rightarrow \angle B=\frac{\pi - 2x}{2}=\frac{\pi}{2}-x$, so:

$$S(ABC)=\frac{b^2\sin 2x}{2}$$

And by sine rule:

$$\frac{b}{\sin \angle B}=2\cdot 1 \rightarrow b=2\cdot \cos x$$

and then:

$$S(ABC)=2\cdot \cos^2x\cdot\sin 2x=(1+\cos 2x)\cdot \sin 2x$$

Can you finish?

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