Bertrand's paradox. A chord of a circle is chosen at random.…

Bertrand's paradox. A chord of a circle is chosen at random. Find the probability that its length is greater than the side of the inscribed equilateral triangle.

Step 1: The critical length

For a circle of radius $R$, the side of the inscribed equilateral triangle is $R\sqrt3$. A chord at distance $x$ from the centre has length $2\sqrt{R^2-x^2}$, which is greater than $R\sqrt3$ iff $x\lt\frac R2$. Equivalently, the chord's central angle exceeds $120^\circ$.

Step 2: Model 1 — random endpoints

Fix one endpoint $P$ (by symmetry) and choose the other uniformly on the circle. The chord is longer than $R\sqrt3$ iff the second endpoint lies on the arc of $120^\circ$ opposite $P$ (between the other two vertices of the inscribed triangle with vertex $P$):

\[ P_1 = \frac{120^\circ}{360^\circ} = \frac13. \]

Step 3: Model 2 — random distance from the centre

Choose a direction, then the distance $x$ of the chord from the centre uniformly on $[0,R]$ (as in Problem 76). The chord is long iff $x\lt\frac R2$:

\[ P_2 = \frac{R/2}{R} = \frac12. \]

Step 4: Model 3 — random midpoint

Choose the midpoint of the chord uniformly in the disc (every point of the disc, except the centre, is the midpoint of exactly one chord). The chord is long iff its midpoint is within $\frac R2$ of the centre:

\[ P_3 = \frac{\pi(R/2)^2}{\pi R^2} = \frac14. \]

Step 5: Why the answers differ

The three procedures produce different distributions of chords: model 3 favours chords near the edge (short chords), because most of the disc's area is near the boundary; model 2 spreads distances evenly. There is no contradiction — "random chord" is simply an incomplete description. The model must match the physical experiment (for example, model 2 describes a long straw dropped at random onto a circle, as in Problem 76).