Let G be a circle of radius R 0. Let G₁, G₂, ..., Gₙ be n circles of equal radius r 0. Suppose e

Q: Wait, why is the central angle $\frac{2\pi}{n}$? Shouldn't it be $\frac{2\pi}{n+1}$ because there are $n$ circles?

That is a classic trap! Think about the gaps. If you have $n$ items in a circle, you have $n$ gaps between them. Imagine a clock face with 12 numbers; there are 12 gaps between the numbers, not 11. Similarly, with $n$ circles, there are $n$ central angles, each of size $\frac{2\pi}{n}$.

Q: Why do we use $\csc\left(\frac{\pi}{n}\right)$? Can't we just leave it as $\sin\left(\frac{\pi}{n}\right)$?

You absolutely can! Using $\csc$ is just an algebraic preference to isolate $R$ more cleanly. If you prefer $\sin$, you get $\frac{r}{R+r} = \sin\left(\frac{\pi}{n}\right)$. Solving for $R$ will lead you to the exact same conclusion. Never feel forced to use a specific function if another one makes more sense to your intuition.

Q: How do I know if the inequality is strict or not?

Always look at the boundary condition. In this problem, we found $R = r(\csc(\frac{\pi}{n}) - 1)$. If the option asks for an inequality, check if the value of $R$ is strictly greater or less than the expression provided. For $n=4$, $R$ is exactly equal to $r(\sqrt{2}-1)$, so an inequality like $R < r(\sqrt{2}-1)$ would be false.

Question

Let $G$ be a circle of radius $R > 0$. Let $G_1, G_2, \dots, G_n$ be $n$ circles of equal radius $r > 0$. Suppose each of the $n$ circles $G_1, G_2, \dots, G_n$ touches the circle $G$ externally. Also, for $i = 1, 2, \dots, n-1$, the circle $G_i$ touches $G_{i+1}$ externally, and $G_n$ touches $G_1$ externally. Then, which of the following statements is/are TRUE? Options: (A) If $n = 4$, then $(\sqrt{2} - 1)r < R$ (B) If $n = 5$, then $r < R$ (C) If $n = 8$, then $(\sqrt{2} - 1)r < R$ (D) If $n = 12$, then $\sqrt{2}(\sqrt{3} + 1)r > R$

Sigma Percentile · Accepted Answer

### The Geometric Vision: A Necklace of Pearls Imagine you are standing in a vast, empty space, and before you, a master jeweler is arranging a set of precious gems. At the center lies a magnificent, large circle $G$ with radius $R$. Surrounding this central gem, the jeweler places $n$ smaller, identical circles, each with radius $r$. These circles are not just placed randomly; they are locked in a perfect, symmetric embrace. Each small circle touches the central circle $G$ externally, and each small circle touches its neighbors on either side. It is a necklace of pearls, perfectly packed. To solve this, we must stop looking at the circles as mere shapes and start seeing the skeleton of lines connecting their centers. Let $O$ be the center of the central circle $G$, and let $O_1$ and $O_2$ be the centers of two adjacent small circles. When we connect $O$, $O_1$, and $O_2$, we form a triangle that holds the secret to the entire arrangement. ### The Bridge of Trigonometry Now, let us analyze this triangle $\Delta OO_1O_2$. Because the circles touch externally, the distance between their centers is the sum of their radii. Thus, the distance $OO_1$ is $R+r$, and $OO_2$ is also $R+r$. The distance between the two adjacent small circles, $O_1O_2$, is simply $r+r = 2r$. We have an isosceles triangle with sides $(R+r)$, $(R+r)$, and $2r$. Since there are $n$ circles arranged symmetrically around the center, the total angle of $2\pi$ radians is divided equally among them. Therefore, the angle $\angle O_1OO_2$ is exactly $\frac{2\pi}{n}$. To make this manageable, let us drop a perpendicular from $O$ to the base $O_1O_2$. This line bisects the base and the angle. We now have a right-angled triangle with a hypotenuse of $R+r$, an opposite side of $r$, and an angle of $\frac{\pi}{n}$. The relationship is elegant and simple: $$\sin\left(\frac{\pi}{n}\right) = \frac{r}{R+r}$$ ### The Analytical Siege With this equation, we have conquered the geometry. Now, we must manipulate it to isolate $R$. Rearranging $\sin\left(\frac{\pi}{n}\right) = \frac{r}{R+r}$, we find: $$\frac{R+r}{r} = \frac{1}{\sin(\pi/n)} = \csc\left(\frac{\pi}{n}\right)$$ Thus, $\frac{R}{r} + 1 = \csc\left(\frac{\pi}{n}\right)$, leading us to our master formula: $$R = r\left(\csc\left(\frac{\pi}{n}\right) - 1\right)$$ ### Verification and Conclusion This formula is our key. Let us test it against the options provided. For $n=4$, we have $R = r(\csc(\pi/4) - 1) = r(\sqrt{2}-1)$. Option A claims $r(\sqrt{2}-1) < R$, but since they are equal, this is false. For $n=5$, we know $\sin(36^\circ) > 0.5$, so $\csc(36^\circ) < 2$. This implies $R < r$, making Option B false. For $n=8$, we compare $\csc(\pi/8)$ with $\sqrt{2}$. Since $\pi/8 < \pi/4$, $\csc(\pi/8) > \csc(\pi/4) = \sqrt{2}$. Thus, $R > r(\sqrt{2}-1)$, and Option C is true. Finally, for $n=12$, we calculate $\csc(15^\circ)$. Using the identity for $\sin(15^\circ) = \frac{\sqrt{3}-1}{2\sqrt{2}}$, we find: $$\csc(15^\circ) = \sqrt{2}(\sqrt{3}+1)$$ Substituting this, $R = r(\sqrt{2}(\sqrt{3}+1) - 1)$. This value is clearly less than $r\sqrt{2}(\sqrt{3}+1)$, confirming Option D is true. We have navigated the geometry, mastered the trigonometry, and verified the logic. The beauty of this problem lies in how a complex physical arrangement collapses into a single, elegant trigonometric identity.

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Visualized Solution

The Circle Arrangement

Connecting the Centers

Sides of $Δ O O_{1} O_{2}$

Central Angle of $Δ O O_{1} O_{2}$

Dropping a Perpendicular

Trigonometric Relation

Isolating $R$

Checking Option A ( $n = 4$ )

Checking Option B ( $n = 5$ )

Checking Option C ( $n = 8$ )

Checking Option D ( $n = 12$ )

Concluding Option D

Final Conclusion

The Sigma Insight: Multiple and Sub-multiple Angles

The Geometric Vision

The Bridge of Trigonometry

The Analytical Siege

Verification and Conclusion

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Visualized Solution

The Circle Arrangement

Connecting the Centers

Sides of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔOO1​O2​

Central Angle of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔOO1​O2​

Dropping a Perpendicular

Trigonometric Relation

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Checking Option A ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=4)

Checking Option B ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=5)

Checking Option C ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=8)

Checking Option D ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=12)