## August 2006

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In my August diary I posed the following brain-teaser, sent in by Boris Zeldovich.

A cylinder is arranged by the "vertical" tangent lines, which are tangent to the Earth's sphere at the equatorial points and are parallel to the axis North Pole — South Pole. The surface of the Earth is projected to that cylinder by the rays, which are "emanated" from the said axis and which are parallel to the equatorial plane. Will the area of projection of France be larger or smaller than the area of France itself?

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*Solution*

The area of France (on the sphere) and the area of its
projection (on the cylinder) are the same.

To see this, consider a point on the sphere, which wolog* I'll suppose to have radius 1, at latitude *A*.
(That's in radians of course:
*A* = 0 at the equator, *A* = π/2 at the North Pole) and longitude *B*
(also in radians, from minus π
to plus
π).

If, with this point as one corner, you mark out a small patch on the sphere by advancing *A* and
*B* by δ*A* and δ*B* respectively, the patch has latitude-height δ*A*,
but longitude-width only
δ*B* cos *A*, because the arc of a latitude-circle subtended by a given longitude angle
diminishes as you approach the
North Pole.

The patch's area is thus, to first order,
δ*A* × δ*B* × cos *A*.

If you now project this wee area on to the cylinder, its image is a patch of the
cylinder with height δ*A* cos *A* (because the patch's projection "loses
height" as you approach the North
Pole) and width δ*B* (no diminution of width now; at any latitude, the *cylinder* still has radius
1).

Area of projected patch:
δ*A* × δ*B* × cos *A*.

The areas are
the same. Stitch together a lot of such patches to make France, and you see that the two areas are equal.

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* **W**ith**o**ut **l**oss
**o**f **g**enerality.