2002 AMC 10B Problem 24

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 $\boxed{\textbf{(D)}\ 10}$.

Constant angular speed means time is proportional to the angle swept, not to the height gained, so the problem is really "find the central angle." Height above the bottom equals $r - r\cos\theta$, and a target height of $r/2$ forces $\cos\theta = \frac12$. Dropping the perpendicular to the vertical diameter always exposes this right triangle.

Trig directly: after $t$ seconds the angle is $6t$ degrees and the height is $20 - 20\cos(6t^\circ)$. Setting this to $10$ gives $\cos(6t^\circ) = \frac12$, so $6t = 60$ and $t = 10$.

• Assuming height grows in proportion to time (10 of 40 feet is a quarter turn, 15 seconds), when height is a cosine of the angle turned.
• Placing $C$ at $10$ feet above the center instead of above the bottom, which gives a $120^\circ$ turn and $20$ seconds (not a choice, a useful red flag).