Question

A circle is inscribed in an equilateral triangle with side lengths 6 unit. Another circle is drawn inside the triangle (but outside the first circle), tangent to the first circle and two of the sides of the triangle. The radius of the smaller circle is:

• A. $\frac{1}{\sqrt{3}}$
• B. 2/3
• C. 1/2
• D. 1

Answer 1

Answer: (A)

$\frac{1}{\sqrt{3}}$

Answer 2

1. Properties of the Equilateral Triangle:

   • Side length $a = 6$.
   • Height $h = \frac{\sqrt{3}}{2}a = \frac{\sqrt{3}}{2}(6) = 3\sqrt{3}$.

2. Properties of the Inscribed Circle (First Circle):

   • The center $O_1$ of the inscribed circle is the incenter, which is also the centroid of the equilateral triangle.
   • The radius of the inscribed circle (inradius) is $r_1 = \frac{1}{3}h = \frac{1}{3}(3\sqrt{3}) = \sqrt{3}$.
   • The distance from any vertex (say $A$) to the incenter $O_1$ is $AO_1 = \frac{2}{3}h = \frac{2}{3}(3\sqrt{3}) = 2\sqrt{3}$.

3. Properties of the Smaller Circle (Second Circle):

   • Let its radius be $r_2$ and its center be $O_2$.
   • The circle is tangent to two sides of the triangle. Let these sides be $AB$ and $AC$.
   • The center $O_2$ must lie on the angle bisector of the angle between these two sides, which is the angle bisector of $\angle A$. In an equilateral triangle, this is also the altitude from $A$.
   • Consider the right-angled triangle formed by vertex $A$, center $O_2$, and the point of tangency $P$ on side $AB$. The angle $\angle O_2AP = \frac{1}{2}\angle BAC = \frac{1}{2}(60^\circ) = 30^\circ$.
   • In $\triangle APO_2$, $\sin(30^\circ) = \frac{O_2P}{AO_2} = \frac{r_2}{AO_2}$.
   • Therefore, the distance from vertex $A$ to the center $O_2$ is $AO_2 = \frac{r_2}{\sin(30^\circ)} = \frac{r_2}{1/2} = 2r_2$.

4. Tangency Condition:

   • The two circles are tangent to each other. Since the smaller circle is "outside the first circle" and in a corner, they are tangent externally.
   • The distance between their centers is $O_1O_2 = r_1 + r_2 = \sqrt{3} + r_2$.
   • Both $O_1$ and $O_2$ lie on the altitude from $A$. Since the smaller circle is in the corner near $A$, $O_2$ is closer to $A$ than $O_1$.
   • Thus, the points are collinear in the order $A, O_2, O_1$. The distances add up: $AO_1 = AO_2 + O_1O_2$.

5. Solving for $r_2$:

   • Substitute the expressions for $AO_1$, $AO_2$, and $O_1O_2$ into the equation: $2\sqrt{3} = 2r_2 + (\sqrt{3} + r_2)$
   • Simplify the equation: $2\sqrt{3} = 3r_2 + \sqrt{3}$
   • Isolate $r_2$: $3r_2 = 2\sqrt{3} - \sqrt{3}$ $3r_2 = \sqrt{3}$ $r_2 = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.