AMC 10 · 2014 · #19
Grade 8 geometry-2dTwo concentric circles have radii 1 and 2. Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle?
Pick an answer.
AMC 10 2014 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 circles share the same center, with radii $1$ and $2$. Two points are picked at random, independently and uniformly, on the outer circle of radius $2$. Find the probability that the straight segment joining these two points passes through the inner circle of radius $1$.
Givens: Two concentric circles with radii $1$ (inner) and $2$ (outer).; Two points are chosen independently and uniformly at random on the outer circle.; A chord is drawn joining the two chosen points.
Unknowns: The probability that this chord intersects (cuts through) the inner circle.
Understand
Restated: Two circles share the same center, with radii $1$ and $2$. Two points are picked at random, independently and uniformly, on the outer circle of radius $2$. Find the probability that the straight segment joining these two points passes through the inner circle of radius $1$.
Givens: Two concentric circles with radii $1$ (inner) and $2$ (outer).; Two points are chosen independently and uniformly at random on the outer circle.; A chord is drawn joining the two chosen points.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems
This is a geometry problem about positions and shapes, so tool #1 (Draw a Diagram) carries the work: sketch the two circles and the tangent lines that mark the boundary case. Because everything is symmetric about the center, tool #9 (Solve an Easier Related Problem) lets us freeze the first point wherever we like and only ask where the second point may land. Tool #7 (Identify Subproblems) then splits the job into two clean pieces: first find the boundary angle where a chord just grazes the inner circle, then measure how much of the outer circle gives a chord past that boundary.
Execute — Answer: D
7.SP.C.7 Step 1 Freeze the first point by symmetry
- The picture looks the same no matter how we rotate it about the center, so where the first point sits does not change the probability.
- Fix the first point $A$ at one spot on the outer circle.
- Now the only randomness left is the position of the second point $B$, spread uniformly around the whole outer circle.
- The probability we want equals the fraction of the circle where $B$ can land so that chord $AB$ cuts the inner circle.
💡 By rotational symmetry one point can be nailed down for free, turning a two-point problem into a one-point problem.
8.G.B.7 Step 2 Draw the tangent lines to find the boundary
- A chord from $A$ just barely touches the inner circle when it is tangent to it.
- Draw the tangent line from $A$ to the inner circle and let $T$ be the touch point.
- The radius $OT$ meets the tangent at a right angle, so triangle $OAT$ has a right angle at $T$, hypotenuse $OA = 2$, and leg $OT = 1$.
- A right triangle whose hypotenuse is exactly twice one leg is half of an equilateral triangle, so the angle at $A$ is $30^\circ$ and the angle $AOT$ at the center is $60^\circ$.
💡 A radius always hits its tangent line square, and a 2-to-1 hypotenuse-to-leg right triangle is the tell-tale 30-60-90 shape.
8.G.A.5 Step 3 Turn the tangent into a central-angle rule
- Follow the tangent line past $T$ until it meets the outer circle again at a point $B_0$.
- Because the tangent leaves $A$ at $30^\circ$ from the diameter $AO$, that far point $B_0$ sits $120^\circ$ around the center from $A$.
- The same thing happens on the other side, giving a second boundary point $120^\circ$ away in the other direction.
- A chord $AB$ dips through the inner circle exactly when $B$ lies past these boundaries, that is, when the central angle between $A$ and $B$ is greater than $120^\circ$.
💡 The two tangent lines fence off the danger zone; only points beyond $120^\circ$ swing the chord close enough to the center.
7.RP.A.2 Step 4 Measure the favorable arc and take the ratio
- Going around from $A$, the second point $B$ gives a central angle bigger than $120^\circ$ only while it travels through the far side of the circle, from $120^\circ$ on one side to $120^\circ$ on the other.
- That favorable stretch is the arc from $120^\circ$ to $240^\circ$, a span of $120^\circ$ out of the full $360^\circ$.
- So the probability is $\tfrac{120}{360} = \tfrac13$.
- The answer is (D).
💡 The favorable directions form a single $120^\circ$ wedge, which is one third of all the way around.
7.SP.C.7 The picture looks the same no matter how we rotate it about the center, so where 8.G.B.7 A chord from $A$ just barely touches the inner circle when it is tangent to it. 8.G.A.5 Follow the tangent line past $T$ until it meets the outer circle again at a poin 7.RP.A.2 Going around from $A$, the second point $B$ gives a central angle bigger than $1 Review
Reasonableness: The favorable arc is exactly one third of the outer circle, so $\tfrac13$ is a clean, believable answer that sits sensibly below one half. Sanity check the boundary: a chord tangent to the inner circle is the break-even case, and it corresponds to a $120^\circ$ central angle, matching the $30$-$60$-$90$ triangle from the radius and tangent. Everything closer than that misses the inner circle, everything farther cuts it, so a $120^\circ$-wide favorable slice giving $\tfrac13$ is consistent.
Alternative: Skip the tangent picture and use the distance from the center to a chord. For a chord of the radius-$2$ circle with central angle $\alpha$, that distance is $2\cos\tfrac{\alpha}{2}$. The chord meets the inner circle when this distance is less than $1$: $2\cos\tfrac{\alpha}{2} < 1 \Rightarrow \cos\tfrac{\alpha}{2} < \tfrac12 \Rightarrow \alpha > 120^\circ$. The second point makes $\alpha > 120^\circ$ over a $120^\circ$ arc, again giving $\tfrac{120}{360} = \tfrac13$.
CCSS standards used (min grade 8)
7.SP.C.7Develop probability models and use them to find probabilities of events (Setting up the uniform model so the probability equals the favorable arc for the second point divided by the whole circle.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Analyzing the right triangle formed by the radius and tangent ($OA=2$, $OT=1$) to get the $30$-$60$-$90$ angles.)8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Translating the $30^\circ$ tangent angle into the $120^\circ$ central-angle boundary for the chord.)7.RP.A.2Recognize and represent proportional relationships between quantities (Taking the favorable $120^\circ$ arc as a proportion of the full $360^\circ$ to get the probability $\tfrac13$.)
⭐ Pin one point down, and the chord only reaches the inner circle when the second point lands more than $120^\circ$ away, which is a single one-third slice of the outer circle.
⭐ Pin one point down, and the chord only reaches the inner circle when the second point lands more than $120^\circ$ away, which is a single one-third slice of the outer circle.
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