A Mathematical Enrichment: Analysis of a Two-horse Race

TO THE TEACHER: This note can be a project assigned to small groups or it can be a student assignment to be given at math club. Written to stir up interest in mathematics and to become a foundation on which to develop group cohesion, the teacher's role is limited to strategic questioning to keep participants actively engaged. For example: What do you think would happen if the lane confinement requirement were changed? How important is the requirement of equal speed throughout the race? And "Can you think of another application for these ideas?"

To master the material, the student must know elementary facts about circles, how to compute the circumference (length) of a circle, the definition of parallel line and the distance between parallel lines.

THE PROBLEM: Two horses race on a standard racing track. They are confined to specific racing lanes and they run the same speeds throughout the race.

THE STANDARD RACING TRACK: The center of the inner lane of a racing track can be viewed as a closed figure of two parallel line segments of equal length capped on each end by a semicircle. The length of the diameters of the semicircles is equal to the separation between the line segments. The centers of the second and third lanes are concentric paths. All lanes are separated by a distance ∆ from the adjacent lane.

THE ANALYSIS - TWO HORSES

Throughout the race the horses are confined to the inner and second racing lanes.

Let D be the length of the second lane minus the length of the inner lane; r , the length of radius of the inner semicircle; ∆, the distance separating the centers of adjacent lanes and l is the length of the linear segments.

D = { 2π (r + ∆) + 2l } - { 2πr + 2l}

=2π ∆

This number is positive. Thus the second lane is longer.

Since horses run the same speed throughout the race, the horse in the inner lane must cross the finish line first, thereby winning the race.

Let's take 18 inches as the measurement between the centers of the inner and second lanes.

So ∆ = 18/12 feet.

Hence D = 2π ∆ > 6 •18/12 feet.

D > 9 feet.

The horse in the inner lane wins by at least 9 feet if the horses run at the same speed and stay in prescribed lanes throughout the race.

Hardly a photo finish!

Suppose the two horses are confined to the first and third racing lanes. Where will the horse in the third racing lane be when the horse in the inner lane touches the finish line?

Discuss strategies for the second horse to win the race.

CONCLUDING REMARK: As an assigned study to a small group, the interactions and communication among the members have the potential to strengthen group bonding thereby lessening student fears to communicate mathematics. As an added benefit, students are lightly to put other situations to mathematical scrutiny.