Students will learn the sector and arc length formulas for circles, and practice using them. Then give your students this challenge. And after that, this one:

Problem: There is a small circle with center \(Y\) and radius \(r\) rolling around inside a large circle with center \(X\) and radius \(2r.\) Let \(C\) be the point where the two circles are in contact. A pen is fixed somewhere on the circumference of the smaller circle. Let's call the position of the pen \(P.\) Let \(D\) be the point where the ray \(XP\) intersects the large circle. Prove that arc \(CP\) has the same length as \(CD.\) This will show that P always lies on the diameter \(ED.\)

Solution: First, notice that \(\overline{XY} \cong \overline{PY}\) because both are radii of the smaller circle. That makes \(XYP\) an isosceles triangle. Let's call its base angles \(\theta.\) Hence, by the exterior angle theorem, \(\angle PYC = 2\theta.\) Using the arc length formula, we find

$$\begin{align} CP = (2r)(\theta) = 2r\theta \\ CD = (r)(2\theta) = 2r\theta \end{align}$$Here's the source of the problem and solution. I found both lacking, which is why I've altered them slightly.

Conclude by giving your students these challenges:

- In a Spin by NRICH
- 2015 AMC 8, Problem 10

Remainders by NRICH: Give only the challenging extension at the bottom of the page. Here's the solution: If a number is 8 more than a multiple of 9, then it's also 1 less than the next multiple of 9. Likewise, if a number is 7 more than a multiple of 8, then it's 1 less than the next multiple of 8. Continuing in this way, we find the answer is

$$\begin{align} & \text{lcm}(9, 8, 7, 6, 5, 4, 3, 2) - 1 \\[0.3em] =\ & \text{lcm}\left(3^2, 2^3, 7, 2 \cdot 3, 5, 2^2, 3, 2\right) - 1 \\[0.3em] =\ & 2^3 \cdot 3^2 \cdot 5 \cdot 7 - 1 \\[0.3em] =\ & 8 \cdot 9 \cdot 5 \cdot 7 - 1 \\[0.3em] =\ & 40 \cdot 63 \\[0.3em] =\ & 2520 - 1 \\[0.3em] =\ & 2519 \end{align}$$Cube Drilling by NRICH: Here's the solution: Each hole you drill goes through 4 cubes. Each hole your friend drills, goes through 2 new cubes. Thus, the number of cubes having at least one hole, is \(4 \cdot 4 + 4 \cdot 2 = 16 + 8 = 24.\) Therefore, the number of cubes without holes is \(64 - 24 = 40.\) It doesn't matter whether you and your friend drill through the corners or the middle, the reasoning is the same either way.