Parallelogram 30 • Level 2 • 28 Mar 2024Manipulating cubes

There are two approaches.

(A) Draw a net of the cube, and straight away the middle diagram allows us to fill in the large letters. If U is in the middle, then the square at the bottom of the net is opposite U. Then, we can look at the first diagram, which gives us the possibilities in small letters, because I and M must be next to K. The third diagram demands that M is next to O, so M must be in the bottom square of the net, so M is opposite U.

(B) We see from the left hand diagram that K is not opposite either I or M, and from the middle diagram that it is not opposite either O or U. Therefore K must be opposite P. So neither K nor P is opposite U. From the middle diagram O is also not opposite U. So U is opposite either I or M. But if U is opposite I, then O must be opposite M, and this possibility is ruled out by the right hand diagram. So U must be opposite to M (and, also, I is opposite O).

9002 - 2002 = 7000 so 9002 - 2009 = 7000 - 7 = 6993.

Method 1: In the figure we have labelled some of the vertices so that we may refer to them.

Suppose that there is a flag whose pole is in the direction of GP and pointing as shown. Consider the effect of carrying out the following operations. First rotate the flag anti-clockwise about G through ∠PGQ so that now its pole lies along GQ. Next slide the flag without rotation so that its pole lies along HR. Next rotate the pole about H through ∠RHS so that it lies along HS, and so on, until the flag returns to its original position.

The total angle that the flag has turned through is the sum of the seven marked angles. But in returning to its original position the flag has completed a full rotation of 360°.

Therefore the sum of the seven marked angles is 360°.

Method 2: We use the same labelling of the vertices that we used in Method 1. We let the sum of the marked angles be X°.

The sum of the angles at the vertex G is 360°. Now ∠FGPand∠QGH are angles of a square, and so are each 90°. It follows that ∠PGQ+∠HGF=360°−90°−90°=180°. Therefore

∠PGQ=180°−∠HGF.

A similar equation holds for each of the seven marked angles. It follows, by adding the seven equations obtained in this way, that

X°=7×180° − the sum of the interior angles of the heptagon.

The sum of the interior angles of a polygon with n vertices is n−2×180°. Therefore the sum of the interior angles of a heptagon is 5×180°. Therefore

X°=7×180°−5×180°=2×180°=360°.

In the context of the JMC we can assume that just one of the given options is correct, so we can find which it is by eliminating the ones that are wrong. We can do this by just considering the last digit (the units digit) of the given options.

The last digit of 56 is 5. Since 1+56−1×4=1+56−4 its last digit is the same as that of 1 + 5 − 4, that is, 2. So the first option is not the correct answer.

In a similar way, it follows that the last digit of 1+56−1×5 is 1, the last digit of 1+56−1×6 is 0, and the last digit of 1+56−1×7 is 9.

So this leaves 15618=1+56−1×8 as the only possible correct option.

In this way there is no need to evaluate 56. However, to give a complete answer we would need to check our answer is correct. This is straightforward:

1+56−1×8=1+15625−8=15626−8=15618.

If we draw in the additional lines shown in the diagram on the right, the hexagon is divided into 54 small congruent equilateral triangles of which 12 are shaded. So the fraction that is shaded is 1254 = 29.

In fact, it is easier to note that the hexagon can be divided into 6 congruent equilateral triangles, like the one shown with the bold edges.

Each of these is made up of 9 of the small equilateral triangles of which 2 are shaded.