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Some(?) of the Parts

A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle


Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?


M is any point on the line AB. Squares of side length AM and MB are constructed and their circumcircles intersect at P (and M). Prove that the lines AD and BE produced pass through P.

Circle Packing

Age 14 to 16
Challenge Level

James of Christ Church Cathedral School, Oxford and Alexander from Shevah-Mofet School, Israel sent very good solutions to this question.

To find the percentage of the plane covered by the circles in each of the packings we must find, within the original pattern, a shape that tessellates the plane and in each case this can be done in different ways. In one case it could be a tessellation of squares, either surrounding the circles or formed by joining the centres of the circles. In either case the calculation of areas is the same.

Let us say that the radius of the circle is $1$ unit. The area of the circle is $\pi$ and the area of the square is $4$ square units. Therefore the proportion of the plane covered by the circles is $\pi/4 = 0.785398\ldots = 78.5\%$ to 3 significant figures.

In the other case the packing of the plane can be produced by a tessellation of hexagons (like a honeycomb).

The hexagon touches the circle at the midpoints of its sides, the distance between the midpoints of opposite sides is $2$ units, the lengths of the sides of the hexagon are $1/\sqrt{3}$ . The area of the hexagon can be found by splitting it into six equilateral triangles and the total area is $6\times (1/\sqrt{3}) \times 1 = 2\sqrt{3}$ square units.

To get the proportion of the plane covered by the circles we must divide by $pi$ by $2\sqrt{3}$ to get $0.90689\ldots$ or $90.7\%$ to 3 significant figures.

This packing can also be done with a tessellation of rhombuses which have angles of $60$ degrees and $120$ degrees. It can be shown that the area of the rhombus is $2\sqrt{3}$.

Here is another method for Circle Packaging from Suzanne and Nisha of the Mount School, York:

We worked out the percentage of each plane covered by circles, by dividing the two patterns into hexagons and squares (see the diagrams above) We worked out the percentage of each hexagon covered by circles and the percentage of squares covered by circles.


A hexagon is 6 triangles area of 1 triangle: base = 2cm ; height = 1.73cm so area of 1 triangle = 1.73cm2 and the area of the hexagon is 10.38cm2.

There are 3 whole circles in each hexagon. radius of circle = 1cm so area = $\pi$ cm2 and area of 3 circles = 3$\pi$ cm2

% of hexagon covered by circles = (3$\pi$/10.38) x 100 = 90.8% (rounded) Notice how not using exact values for the height of the triangles in the hexagon calculation means this is 0.1 cm2 off - a type of rounding error introduced by rounding off numbers before the answer.


Each square has area = 4cm2

In each square, there is 1 whole circle

area of circle = $\pi$

% of square covered by circles = ($\pi$/4) x 100 = 78.5% (rounded) This means that you could fit more cylindrical cans in a container using the `hexagon' pattern. Advantage of the `square' pattern is that you could maybe fit things in the gaps, which are larger than in the hexagon pattern.