Question

Assertion \((A):\) A circle \(C_{1}\) is inscribed in an equilateral triangle \(\mathrm{ABC}\) with side length \(2 .\) Then circle \(\mathrm{C}_{2}\) is inscribed tangent to BC, CA and circle \(\mathrm{C}_{1}\). An infinite sequence of such circles is constructed, each tangent to \(\mathrm{BC}, \mathrm{CA}\) and the previous circle. The sum of areas of all the infinitely many circles is \(\frac{5 \pi}{8}\). Reason ( \(\mathbf{R}\) ) : Radius of \(\mathrm{C}_{1}\) is \(\frac{1}{\sqrt{3}}\), that of \(\mathrm{C}_{2}\) is \(\frac{1}{3 \sqrt{3}}\) and radius of the remaining circle each shrink by a factor \(\frac{1}{3}\).

Short answer

Answer: True

Step-by-step solution

Determine the radius of circle \(C_1\) and circle \(C_2\)

We begin by finding the radius of the incenter of the equilateral triangle \(\mathrm{ABC}\). Let's call the incenter \(I\). Draw the inradii \(IC_1\) from \(I\) to \(BC\) which has length \(r\), and notice that \(IC_1\) splits \(BC\) into two equal segments, each of length 1. Note that \(\triangle ABC\) is equilateral, the altitude drawn from \(A\) is also an angle bisector and median. Let the midpoint of \(BC\) be \(M\). Thus, \(\angle AMI = 90^{\circ}\), and \(\angle IAM = 30^{\circ}\). Using the sine of 30 degrees: \(\sin 30^{\circ} = \frac{1}{2} = \frac{r}{AM}\) Now, we can use the Pythagorean theorem to find that \(AM = \sqrt{3}\), and it follows that \(r = \frac{1}{\sqrt{3}}\). Hence (R) is true. Next, we need to find the radius of circle \(C_2\). Let \(IB\) be the inradius of circle \(C_2\) and \(D\) be the point of tangency of circle \(C_2\) with \(CA\) ( \(BD=IB-r_2 \)) . Then, \(\triangle BDI \sim \triangle IAM\) (They share a common angle and are both right triangles). Thus, \(\frac{r_2}{BD} = \frac{r}{AM-AB} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}-2}\) Multiplying by \(\sqrt{3}\) on the numerator and denominator, and then simplifying: \(r_2 = \frac{1}{3\sqrt{3}}\) Now we have found the radius of circle \(C_1\) and circle \(C_2\).

Determine the sum of the areas of all the infinitely many circles

The sequence of radii of the circles is a geometric series. Given that the radii shrink by a factor of \(\frac{1}{3}\) as stated: \(r_1 = \frac{1}{\sqrt{3}}, r_2 = \frac{1}{3\sqrt{3}}, r_3 = \frac{1}{(3^2)\sqrt{3}}, r_4 = \frac{1}{(3^3)\sqrt{3}}, \ldots\) To find the sum of the areas of all the infinite circles, we take the sum of the areas of the individual circles: \(S_{\text{circle}} = \pi \sum_{k=0}^{\infty} \left(\frac{1}{(3^k)\sqrt{3}}\right)^2\) By the geometric series summation formula: \(S_\text{circle} = \pi \left(\frac{a}{1-r}\right)\) where \(a = \left(\frac{1}{\sqrt{3}}\right)^2\) and \(r = \left(\frac{1}{3}\right)^2\) \(S_\text{circle} = \pi \left(\frac{1/3}{1-\frac{1}{9}}\right) = \pi \left(\frac{1/3}{8/9}\right) = \frac{5\pi}{8}\) Thus, the sum of the areas of all the infinitely many circles is \(\frac{5 \pi}{8}\), and the Assertion (A) is True.