AMC 10 · 2011 · #20
Grade 7 geometry-2dTwo points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?
Pick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Two points are placed on a circle of radius r, each chosen at random and independently. From each point, draw a chord of length r going clockwise. Find the probability that these two chords cross.
Givens: A circle of radius r.; Two points chosen independently and uniformly at random on the circle.; From each point a chord of length r is drawn in the clockwise direction.
Unknowns: The probability that the two chords intersect.
Understand
Restated: Two points are placed on a circle of radius r, each chosen at random and independently. From each point, draw a chord of length r going clockwise. Find the probability that these two chords cross.
Givens: A circle of radius r.; Two points chosen independently and uniformly at random on the circle.; From each point a chord of length r is drawn in the clockwise direction.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #9 Solve an Easier Related Problem, #4 Introduce a Variable
A picture is the key: drawing the circle shows that a chord of length r cuts off a 60-degree arc, and it makes the crossing condition visible. Fixing the first point by symmetry turns a two-random-point problem into one uniform random point, and a variable for the second point's angle lets us measure exactly when the chords cross.
Execute — Answer: D
4.G.A.2 Step 1 Chord equal to radius cuts a 60 degree arc
- Draw a chord of length r and join both of its endpoints to the center.
- The two joins are radii of length r, and the chord is also length r, so the triangle has three equal sides and is equilateral.
- Every angle of an equilateral triangle is 60 degrees, so the central angle at the center is 60 degrees.
- That means each chord of length r cuts off a 60-degree arc.
💡 Three equal sides forces all angles equal, and equal angles in a triangle are each 60 degrees.
7.SP.C.7 Step 2 Fix the first point by symmetry
- Both points are chosen the same random way, so we may rotate the whole picture until the first point sits at a fixed spot.
- Its clockwise chord then covers the 60-degree arc just clockwise of it.
- Only the position of the second point relative to this fixed picture matters, and that position is still uniform all the way around the circle.
💡 Rotating the circle changes nothing about crossing, so we can pin down one point for free.
4.MD.C.6 Step 3 Give the endpoints angle labels
- Put the first point at 0 degrees.
- Its clockwise chord ends 60 degrees clockwise, at 300 degrees, so chord A has endpoints at 0 and 300 degrees.
- Let the second point sit at angle b, measured the same way around the circle.
- Its clockwise chord has endpoints at b and at b minus 60 degrees.
💡 Naming each endpoint by its degree position turns the geometry into something we can measure.
4.MD.C.7 Step 4 Chords cross when endpoints alternate
- Two chords of a circle cross inside it exactly when their four endpoints alternate around the circle, meaning one endpoint of chord B falls in the 60-degree gap of chord A (the arc from 300 to 360 degrees) and the other endpoint falls outside it.
- Endpoint b lands in that gap when b is between 300 and 360 degrees; endpoint b minus 60 lands in it when b is between 0 and 60 degrees.
- Each case is a 60-degree window, and the two windows do not overlap, giving 120 degrees of favorable positions.
💡 A crossing needs exactly one of B's ends tucked inside A's short 60-degree gap.
6.RP.A.3 Step 5 Turn favorable arc into probability
- The second point is uniform around the full 360 degrees, so the probability is the favorable measure divided by the whole circle.
- That is 120 degrees out of 360 degrees, which reduces to one third.
- So the probability that the two chords intersect is 1/3, which is answer (D).
💡 For a uniform point, probability is just the favorable arc as a fraction of the whole circle.
4.G.A.2 Draw a chord of length r and join both of its endpoints to the center. The two j 7.SP.C.7 Both points are chosen the same random way, so we may rotate the whole picture u 4.MD.C.6 Put the first point at 0 degrees. Its clockwise chord ends 60 degrees clockwise, 4.MD.C.7 Two chords of a circle cross inside it exactly when their four endpoints alterna 6.RP.A.3 The second point is uniform around the full 360 degrees, so the probability is t Review
Reasonableness: The result 1/3 lies between 0 and 1, as any probability must. It also makes sense in scale: each chord only blocks a short 60-degree arc, so a crossing is possible but not likely, and two such 60-degree windows out of 360 degrees giving one third fits that expectation. This matches choice (D).
Alternative: Instead of labeling angles, use the fixed-point hexagon picture directly: from the first point's chord, the second point produces a crossing only if it lands within a 60-degree arc on either flanking side, a total of 120 degrees. Dividing 120 by 360 again gives 1/3, confirming the answer.
CCSS standards used (min grade 7)
4.G.A.2Classify two-dimensional figures based on their properties (Recognizing that three equal sides make an equilateral triangle with 60-degree angles.)7.SP.C.7Develop a uniform probability model and use it to find probabilities (Reducing two random points to one uniform random point using the circle's symmetry.)4.MD.C.6Measure angles in whole-number degrees (Assigning degree positions to each point and chord endpoint.)4.MD.C.7Recognize angle measure as additive (Adding the two 60-degree favorable windows to get 120 degrees.)6.RP.A.3Use ratio and rate reasoning to solve problems (Converting the favorable 120-degree arc into the probability 120/360 = 1/3.)
⭐ A chord as long as the radius always cuts a 60-degree arc, so the two chords cross only when the second point falls in one of two 60-degree windows: 120 out of 360 degrees, or one third of the time.
⭐ A chord as long as the radius always cuts a 60-degree arc, so the two chords cross only when the second point falls in one of two 60-degree windows: 120 out of 360 degrees, or one third of the time.
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