Inscribed circle sums.

Let the total area be $A$ . It's easy to see that $A=\dfrac{1}{3}A+\dfrac{1}{3}\pi$ Solving gives $A=\dfrac{\pi}{2}$

The red circle is the incircle of the large triangle. If A represents the area of the triangle, S is half the perimeter and r is the radius then: A=rs, so r=A/s. This gives r= 1/ $\sqrt{3}$ , so it follows that area of circle = pi/3. The circles areas decrease by 1/9 each time, but the number of those circles increases by 3 each time. This gives us the infinite sum: Total area of circles = (1/3)pi(1+1/3 + 1/9 +1/27+...) = (1/3) (pi)/(1-1/3) = (1/3) (3pi/2) = pi/2 = 1.57 (2d.p).

The area of triangle is actually the sum of the areas of regular hexagons. $A_t = Σ A_h$

Since the ratio of hexagon to circle is always constant, the sum of the areas of circles can be calculated by taking the ratio of one regular hexagon to a circle and apply it to the triangle. $A_t : Σ A_c = Σ A_h : Σ A_c = A_h : A_c$

Here, area of triangle is: $A_t = \frac {1}{2} \times 2^2 \times \sin 60º = \sqrt{3}$

Assume that the radius of one of these circle is r.

Area of circle: $A_c = πr^2$

Area of regular hexagon is treated as 12 similar right angled triangle with length r on one side which subtends an angle of 30º: $A_h = 12 \times \frac {1}{2} r (r\tan 30º)$

Hence, $A_h = 2\sqrt{3}r^2$

Returning to the original problem, $A_t : Σ A_c = A_h : A_c$

Hence, $\sqrt{3} : Σ A_c = 2\sqrt{3}r^2:πr^2$

and

$Σ A_c = \frac {π}{2}$

Let $r_{0}$ ne the radius of the bigger inscribed circle which is: $r_{0}=\frac {a*\tan(30)}{2}=\frac{1}{\sqrt(3)}$ where $a$ is side length of the triangle wich is in This case 2. We now find the radius $r_{1}$ of the next three circles in terms of $r_{0}$ : Using the above formula: $r_{1}=\frac {a^{'}*\tan(30)}{2}=\frac{a^{'}}{2\sqrt(3)}$ Where $a^{'}$ is the side length of the three triangles wich equals: $a^{'}=2r_{0}\tan(30)=\frac{2r_{0}}{\sqrt(3)}$ And subtituting the value of $a^{'}$ in the above equation we get: $r_{1}=\frac{r_{0}}{3}$ , so we Can easily conclude the following recurrence relation : $r_{n+1}=\frac{r_{n}}{3}$ , wich also gives: $r_{n}=r_{0}/3^{n}$ for each n>=0. Now we Can compute the area of circles: $A_{n}=\pi*r_{n}^{2}=\frac{\pi*r_{0}^{2}}{3^{2n}}$ . Now observe That the number of triangles gets multiplied by 3 each time, so the total area Will be: $A_{t}=\sum_{0}^{\infty} A(n)3^{n}=\frac{\pi}{2}$