Parallelogram 47 • Level 4 • 24 Jul 2025Inflating a balloon

Let the vertices of the square be labelled K, L, M, N and those of the regular pentagon P, Q, R, S, T, as shown in the diagram.

We use the fact that the interior angles of a regular pentagon are each 108°. It follows that ∠RST=∠STP=108°. Because it is the angle of a square, ∠SNT=90°. By the Exterior Angle Theorem, ∠RST=∠SNT+∠NTS. Therefore, 108°=90°+∠NTS. It follows that ∠NTS=108°−90°=18°.

Finally, because the angles at T on the line KN have sum 180°,

∠NTS+∠STP+∠PTK=180°,

that is,

18°+108°+x°=180°.

We conclude that

x=180−18−108=54.

The two-digit squares are 16, 25, 36, 49, 64 and 81. Of these, just 49 = 50 − 1 and 81 = 80 + 1 differ by 1 from a multiple of 10. So there are 2 such two-digit squares.

Each of the 12 pentagons has 5 edges, and each of the 20 hexagons has 6 edges. So, altogether, the panels have 12×5+20×6=180 edges. Each join connects 2 edges, So there are 12180=90 joins.

We form a complete grid inside the large square by extending the sides of the small squares, as shown in the figure. In this way the large square is divided up into 61 small squares, 20 triangles along the edges whose areas are equal to half that of the small squares, and 4 triangles in the corners whose areas are equal to one quarter that of the small squares.

So the area of the large square is equal to that of

61×1+20×12+4×14=72

of the small squares. The shaded area is made up of 24 small squares. Therefore the fraction of the large square that is shaded is

2472=13.

[Note that we could save some work by taking advantage of the symmetry of the figure, and just work out which fraction of, say, the top left hand corner of the square is shaded.]