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Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 3 / 12E / Solution

In polar coordinates the metric is

d s^{2} = \frac{4 ( d r ^{2} + r ^{2} d θ ^{2} )}{( 1 - r ^{2} ) ^{2}} .

Along a radius, the hyperbolic distance from the origin to Euclidean radius

r

is therefore

R = \int_{0}^{r} \frac{2 d t}{1 - t ^{2}} = lo g \frac{1 + r}{1 - r},

so

r = tanh (R /2)

. The Riemannian area element is

d A = \frac{4 r}{( 1 - r ^{2} ) ^{2}} d r d θ .

Consequently a hyperbolic disc of radius

R

has area

A (R) = \int_{0}^{2 π} \int_{0}^{t a n h (R /2)} \frac{4 r}{( 1 - r ^{2} ) ^{2}} d r d θ = 4 π sinh^{2} (R /2) = 2 π (cosh R - 1) .

This proves the area of a hyperbolic disc formula from the metric.

For area

π /2

we get

cosh R = 5/4

. Hence

sinh R = 3/4

and

e^{R} = 2

, so

R = lo g 2

. Tangent equal discs have centers at distance

2 R

. Since the centers lie successively on the same radial geodesic,

d (O, c_{n}) = 2 n R = n lo g 4.

If

r_{n}

is the Euclidean coordinate of

c_{n}

, the radial distance formula gives

\frac{1 + r _{n}}{1 - r _{n}} = 4^{n} .

Therefore the radial chain of equal hyperbolic discs has

r_{n} = \frac{4 ^{n} - 1}{4 ^{n} + 1} .

No such isometry to the stated upper-half-plane configuration exists for

n \geq 3

. The centers

c_{0}, \dots, c_{n}

lie on one hyperbolic geodesic, so their images under an isometry would also lie on one geodesic. In the upper-half-plane model, geodesics are vertical lines or semicircles orthogonal to the real axis. Neither type can contain three distinct points of the horizontal line

y = 1

, while the centers of the distinct discs

D_{0}^{'}, \dots, D_{n}^{'}

all lie on that line. This is the geodesic obstruction for a horizontal chain of hyperbolic discs.

Solved by gpt-5.6-sol high.

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