Running Track

Geometry Level 3

Runners run the track above from start line A B \mathrm{AB} to end line C D \mathrm{CD} . The curve part of the track is a half circle which shares O , O ′ \mathrm{O, O^\prime} as the center. What is the value of θ \theta to the nearest degree?. (All runners should run equal distances)


The answer is 72.

Junghwan Han by Junghwan Han , 19, South Korea

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1 solution

Chew-Seong Cheong
Jan 22, 2019

Let the external radius of the circular part of the track O E = R OE=R and the internal radius O A = r OA=r . To be a fair track, we have:

A C = B D 2 O O ′ + π r = 2 O O ′ − E B + π R π r = − E B + π R = − A E tan ⁡ θ + π R = − ( R − r ) tan ⁡ θ + π R ⟹ ( R − r ) tan ⁡ θ = π ( R − r ) tan ⁡ θ = π ⟹ θ ≈ 72 ∘ \begin{aligned} AC & = BD \\ 2OO'+\pi r & = 2OO' - EB + \pi R \\ \pi r & = - EB + \pi R \\ & = - AE \tan \theta + \pi R \\ & = - (R-r) \tan \theta + \pi R \\ \implies (R-r) \tan \theta & = \pi (R-r) \\ \tan \theta & = \pi \\ \implies \theta & \approx \boxed {72}^\circ \end{aligned}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Cool problem. Entering t a n − 1 π tan^{-1}\pi feels so wrong. It looks like the type of error students make when learning trigonometry -- confusing the angle with the tangent ratio. The dissonance compounds when we use π \pi which looks like an angle in radians but then we answer in degrees.

Jeremy Galvagni - 2 years, 4 months ago

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