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Geometry Difficulty 8.4 Shortlist Find the answer

Can there be drawn on a circle of radius 11 a number of 19751975 distinct points, so that the distance (measured on the chord) between any two points (from the considered points) is a rational number?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

We are asked whether it is possible to draw 19751975 distinct points on a circle of radius 11 such that the chord distance between any two points is a rational number.

### Key Observations

1. Chord Distance Formula: For a circle of radius 11, the chord distance dd between two points subtending an angle θ\theta at the center is given by:
d=2sin(θ2). d = 2 \sin\left(\frac{\theta}{2}\right).
We need this distance to be rational for any pair of chosen points.

2. **Rational sin\sin values**: The value sin(θ2) \sin(\frac{\theta}{2}) must be a rational number divided by 2 for the chord distance to be rational. This occurs when θ\theta is such that sinθ\sin \theta is rational.

3. Vertices of Regular Polygons: On a unit circle, regular polygons can help maintain rational sine values. Specifically, if the circle is divided such that the central angle θ=2πn\theta = \frac{2\pi}{n} for integer nn where sinθ\sin \theta is rational, then vertices of such regular polygons can be potential points.

### Chebyshev Polynomials to Ensure Rational Sine

Chebyshev polynomials, Tn(x)T_n(x), preserve the property:
- Tk(cosθ)=cos(kθ) T_k(\cos \theta) = \cos(k \theta) .

Thus, for an integer kk, if cosθ\cos \theta is rational (hence sinθ\sin \theta can be calculated from it), then Tk(cosθ)T_k(\cos \theta) remains rational, maintaining the rationality of all involved sine values for angles that are integer multiples of θ\theta.

### Selecting Points

To ensure every pair of the 19751975 points has a chord distance that is rational, the points can be aligned with the vertices of a regular polygon that obeys the rational sine condition.

- nineteenth roots of unity: The angles of 2πk1975\frac{2\pi k}{1975} for k=0,1,2,,1974k = 0, 1, 2, \ldots, 1974 on a circle generate points of a regular 1975-gon on the circle.

Since these points are symmetrically distributed and use regular divisions based on integer multiples of mainly ensured rational sine values, all resulting chord lengths for these combinations of points fulfill the condition of being a rational distance.

### Conclusion

Yes, it is indeed possible to draw 19751975 distinct points on a circle of radius 11, such that the chord distance between any two points is a rational number. This arrangement leverages the geometric and algebraic properties of the unit circle and sine rationality.

Thus, the final answer is:
yes \boxed{\text{yes}}

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.