2002 AMC 10B Problem 24

The wheel turns at a steady speed, so time depends on the angle turned, not on the height gained. Find the angle.

1. Let $O$ be the center, $A$ the bottom of the wheel, and $C$ the rider's position $10$ feet above the bottom. The radius is $20$, so $OA = OC = 20$.
2. Draw a horizontal line from $C$ to the vertical line $OA$, meeting it at $B$. Then $B$ is $10$ feet above $A$, which is exactly halfway between $A$ and $O$.
3. So $CB$ is a line perpendicular to $OA$ through its midpoint. Every point on such a line is the same distance from $A$ as from $O$. So $CA = CO = 20$.
4. Now triangle $AOC$ has all three sides equal to $20$: it is equilateral, and $\angle AOC = 60^\circ$.
5. The rider has turned $60^\circ$ out of $360^\circ$, which is $\frac16$ of a turn. One full turn takes $60$ seconds, so this takes $\frac{60}{6} = 10$ seconds.

The answer is $\boxed{\textbf{(D)}\ 10}$.

On a wheel turning at a steady rate, equal angles take equal times, but equal heights do not: near the bottom the rider moves mostly sideways. So turn the height into an angle. A drop of half the radius below the center always gives $60^\circ$, a fact worth remembering for any circle problem.

Use the right triangle $OBC$ from step 2. Its long side is $OC = 20$ and the vertical leg is $OB = 20 - 10 = 10$, exactly half. A right triangle whose long side is twice one leg is a $30$-$60$-$90$ triangle, with the $60^\circ$ angle next to the short leg, at $O$. So again the wheel turned $60^\circ$, which takes $10$ seconds.

Let $O$ be the center, $A$ the bottom of the wheel, and $C$ the rider's position when $10$ feet above the bottom. Since $OA = 20$, the point $C$ is $20 - 10 = 10$ feet below the level of $O$.

Drop a perpendicular from $C$ to the vertical line $OA$, meeting it at $B$. In right triangle $OBC$ the leg $OB = 10$ and the hypotenuse $OC = 20$ (a radius). A hypotenuse twice one leg means a $30$-$60$-$90$ triangle with $\angle BOC = 60^\circ$.

The rider has therefore turned through $60^\circ$ out of $360^\circ$, one sixth of a revolution. At one revolution per minute that takes $\frac{60}{6} = 10$ seconds.

The answer is $\boxed{\textbf{(D)}\ 10}$.