Points A and B lie on a circle with radius 1, and arc AB has length pi/3. What fraction of...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
Points \(\mathrm{A}\) and \(\mathrm{B}\) lie on a circle with radius \(1\), and arc \(\mathrm{AB}\) has length \(\frac{\pi}{3}\). What fraction of the circumference of the circle is the length of arc \(\mathrm{AB}\)?
1. TRANSLATE the problem information
- Given information:
- Circle has radius 1
- Arc AB has length \(\mathrm{\pi/3}\)
- What we need to find: The fraction of the circumference that arc AB represents
2. INFER the solution strategy
- To find what fraction one quantity is of another, we divide: part ÷ whole
- We have the arc length (part = \(\mathrm{\pi/3}\)), but we need the total circumference (whole)
- Strategy: Find circumference first, then divide arc length by circumference
3. Calculate the total circumference
- Using the circumference formula: \(\mathrm{C = 2\pi r}\)
- With radius = 1: \(\mathrm{C = 2\pi(1) = 2\pi}\)
4. SIMPLIFY the fraction calculation
- Fraction = (arc length) ÷ (circumference)
- Fraction = \(\mathrm{(\pi/3) \div 2\pi}\)
- Convert division to multiplication: \(\mathrm{(\pi/3) \times (1/2\pi)}\)
- Multiply: \(\mathrm{\pi/(3 \times 2\pi) = \pi/6\pi}\)
- Cancel π terms: \(\mathrm{1/6}\)
Answer: \(\mathrm{1/6}\) (also acceptable as 0.166, 0.167, .1666, .1667)
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students don't recognize that finding 'what fraction' means dividing part by whole, or they try to work directly with the arc length without finding the total circumference first.
Without the circumference, they get stuck trying to make sense of \(\mathrm{\pi/3}\) alone, leading to confusion and guessing.
Second Most Common Error:
Poor SIMPLIFY execution: Students set up the division correctly but make algebraic errors when simplifying \(\mathrm{(\pi/3) \div 2\pi}\), particularly in handling the π terms or converting division to multiplication.
Common mistakes include getting \(\mathrm{2/3}\) (forgetting the π entirely) or \(\mathrm{\pi/6}\) (failing to cancel the π terms), leading them to select incorrect answer choices.
The Bottom Line:
This problem tests whether students can connect the abstract concept of 'fraction of circumference' to the concrete procedure of dividing arc length by total circumference, then execute the algebraic simplification correctly.