# What is the average area of all triangles that can be inscribed in a unit circle?

There is only one circle that passes through any three given points. Hence by suitable scaling, we can inscribe every triangle inside a unit circle of radius $$1$$. We define distinct triangles as triangles which have different sides regardless of of the order. Hence a triangle with sides $$(a,b,c)$$ and a triangle with sides $$(b,c,a)$$ are not distinct.

Question 1: What is the average area of all distinct triangles that can be inscribed in a unit circle?

Question 2: What is the average perimeter of all distinct triangles that can be inscribed in a unit circle?

Motivation: This question was motivated by this related question where it was proved that the average perimeter of all right triangles inscribed in a semi-circle of unit diameter is $$1+4/\pi$$.

• Would make sense to write the areas with respect to the angles of the triangles, then. What have you tried? – Andrew Chin Oct 8 at 21:47
• If you don't specify your distribution a bit more, I suspect you will run into problems very similar to the Bertrand paradox – Arthur Oct 8 at 21:51
• Perhaps you'd like to edit into the post what method you have been trying so we can build on it. – Andrew Chin Oct 8 at 21:51
• With respect to what measure? For example: Are you choosing three points uniformly from the unit circle? Or are you choosing three points from some distribution on the whole plane, then scaling them to be inscribed in the unit circle? – Chris Culter Oct 8 at 21:52
• The sides of a triangle with angles $A$, $B$, $C$ are $2\sin A$, $2\sin B$, $2\sin C$. In this context, you're asking for the averages of $2\sin A\sin B\sin C$ and $2(\sin A+\sin B+\sin C)$ over $A$, $B$, $C$ with $A+B+C=\pi$ and, say, $A\geq B\geq C > 0$. – Blue Oct 8 at 21:55

Comment:

As can be seen in figure the area and perimeter of right angled triangle in half circle is maximum when the height is maximum , i.e. $$h=r$$ where h is height of triangle and r is the radius of circle. The area and perimeter are minimum when $$h → 0$$. So we can write:

$$a/2 = r= 1$$

$$S_{max}=(\frac{(\sqrt 2)^2}{2})=1$$

$$S_{min}= 0$$

$$S_{ave.}=\frac{1+2\times 0}{3}=\frac{1}{3}$$

and perimeter:

$$P_{max}=2+2\sqrt2$$

$$P_{min}=0$$

$$P_{ave}=\frac{2(1+\sqrt2)+2\times 0}{3} ≈ 2.27 ≈ 1+\frac{4}{\pi}$$

Now suppose we want to find the average area and perimeter of isosceles triangles which can be inscribed in circle with radius unit. In triangle $$OHC_1$$ we have:

$$(\frac{a}{2})^2+(dr)^2=r^2=1$$

$$S_{AB_1C_1}=\frac{a}{2}\times (r+dr)=\frac{a}{2}(1+dr)$$

Eliminating $$dr$$ and letting $$\frac{a}{2}=x$$ we get:

$$S=x(1+\sqrt{1-x^2})$$

$$S'=\frac{\sqrt{1-x^2}+1-2x^2}{\sqrt{1-x^2}}=0$$

$$x=0$$ and $$x=\frac{\sqrt 3}{2}$$

$$a= \sqrt 3$$

$$S_{max}=\frac{3}{2}\times\frac{\sqrt3}{2}=\frac{3\sqrt3}{4}$$

That is S is maximum when $$a=b=c=\sqrt 3$$, i.e. when the triangle is equilateral.The minimum value is when $$h →0$$ or $$a →0$$ , so the average of area can be:

$$S_{ave.}=\frac{\frac{3\sqrt3}{4}+2\times 0}{3}=\frac{\sqrt 3}{4}$$

Similarly you can find the average of perimeter:

$$P=a+2\sqrt{2+2\sqrt{1-(a/2)^2}}$$

Now take derivative and so on.