2002 AMC 10B Problem 24

P21-25~3 minPrint

Circles Triangles: Area & Pythagorean Add construction lines/points (drop altitudes, extend segments, connect centers)Set up the equation/formula and compute; no special trick needed

Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $20$ feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point $10$ vertical feet above the bottom?

Show solution without answering

Key insight

Ten feet up is halfway from the bottom to the center, so bottom, center and rider form an equilateral triangle: a 60 degree turn, one sixth of a minute.

Solution

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

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 $OA = OC = 20$ .
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 $O$ .
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 $CA = CO = 20$ .
Now triangle $AOC$ has all three sides equal to $20$ : it is equilateral, and $\angle AOC = 60^\circ$ .
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 $\boxed{\textbf{(D)}\ 10}$ .

Why this works

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.

Alternative approach

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)
Introduction to Geometry, Chapter 12: Circles and Angles