Cure for the Common Chord

Let $\theta$ be the angle subtended by the smaller arc, at the center of the circle. Then the length of the smaller arc is also $\theta$ , and the length of the larger arc is $2 \pi - \theta$ . Let $d$ be length of the chord. (We can compute that $d = 2 \sin \frac{\theta}{2}$ , but this fact is not necessary.) From the given condition, $\frac{2 \pi - 2 \theta}{d} = 2,$ which simplifies to $\theta + d = \pi$ .

Then the ratio of the perimeter of the smaller segment to the perimeter of the circle is $\frac{\theta + d}{2 \pi} = \frac{\pi}{2 \pi} = \frac{1}{2}.$

Let $\theta_1$ and $\theta_2$ denote the angles of the sectors, such that

• $k > 1$ is a constant
• $\theta_1 = k\theta_2$

Let $d$ denote the length of the chord. Here are the formulas used for the solutions:

• Circumference of the sector : $C_{\theta} = \theta r$ where $r$ is the radius
• Circumference of the segment: $C_s = \theta + d$
• Law of Cosine : $c^2 = a^2 + b^2 - 2ab\cos(\alpha)$ where $a,b,c$ are side lengths of the triangle and $\alpha$ is the angle that "faces" the side of length $c$

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Since $C_1 + C_2 = 2\pi$ , and the radius of both sectors is $1$ , $C_1 = \theta_1 = k\theta_2$ and $C_2 = \theta_2$ , which implies that $\begin{array}{rl} \theta_2 (k + 1) &= 2\pi\\ \theta_2 &= \dfrac{2\pi}{k + 1} \end{array}$ From the given, we can determine the form of the chord length in terms of $k$ : $\begin{array}{rl} C_1 - C_2:\text{chord} &= 2\\ \dfrac{\theta_2 (k - 1)}{d} &= 2\\ \theta_2 (k - 1) &= 2d\\ d &= \dfrac{\theta_2 (k - 1)}{2}\\ &= \dfrac{\pi (k - 1)}{k + 1} \end{array}$ For the triangular region of the circle, we can apply the Law of Cosine to determine the value of $k$ and $d$ . For $c^2 = a^2 + b^2 - 2ab\cos(\alpha)$ set $c = d$ , $a = b = 1$ (since the radius of the sector is $1$ ) and $\alpha = \theta_2$ . Then, we have $\begin{array}{rl} d^2 &= a^2 + b^2 - 2ab\cos(\theta_2)\\ \left(\dfrac{\pi (k - 1)}{k + 1}\right)^2 &= 2 - 2\cos\left(\dfrac{2\pi}{k + 1}\right) \end{array}$ Since $k > 1$ , $k \approx 2.77726$ , so $d \approx 1.47817$ . Thus, $\dfrac{\text{minor segment circumference}}{\text{unit circle circumference}} = \dfrac{\theta_2 + d}{2\pi} \approx \boxed{0.50}$