Parallelogram 46 • Level 4 • 17 Jul 2025Freaky dot patterns

Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

a portmanteau word combining parallel and telegram. A message sent each week by the Parallel Project to bright young mathematicians.

Tackle each Parallelogram in one go. Don’t get distracted.
Finish by midnight on Sunday if your whole class is doing parallelograms.
Your score & answer sheet will appear immediately after you hit SUBMIT.
Don’t worry if you score less than 50%, because it means you will learn something new when you check the solutions.

1.

2 marks

1.1 What is a half of a third, plus a third of a quarter, plus a quarter of a fifth?

11440
338
130
13
310
(Not answered)

A half of a third is 12×13=16, a third of a quarter is 13×14=112, and a quarter of a fifth is 14×15=120.

It follows that the required answer is

16+112+120=10+5+360=1860=310.

2.

3 marks

2.1. A tetrahedron is a solid figure which has four faces, all of which are triangles.

What is the product of the number of edges and the number of vertices of the tetrahedron?

8
10
12
18
24
(Not answered)

We can think of a tetrahedron as a pyramid with a triangular base.

The base has 3 vertices and there is 1 more vertex, called the apex, at the peak of the pyramid. So a tetrahedron has 4 vertices.

The base has 3 edges, and there are 3 more edges joining the vertices of the base to the apex. Therefore a tetrahedron has 3 + 3 = 6 edges.

Finally, we observe that 4 × 6 = 24.

3.

3 marks

3.1 Three different positive integers have a mean of 7.

What is the largest positive integer that could be one of them?

15
16
17
18
19
(Not answered)

Because the three integers have mean 7, their sum is 3×7=21. For the largest of the three positive integers to be as large as possible the other two need to be as small as possible. Since they are different positive integers, the smallest they can be is 1 and 2. The largest of the three integers is then 21−1+2=18. So 18 is the largest any of them could be.

4. Numberphile – Freaky dot patterns

(If you have problems watching the video, right click to open it in a new window)

4 marks

4.1 Tadashi slightly moves the pattern on the transparency over the identical pattern on the paper and sees a new pattern of circles. If the transparency is moved left/right, which direction does the centre of the circles move?

Diagonally
Does not move
Left/right
Up/down
In a loop
(Not answered)

5.

5 marks

5.1 A rectangle is placed obliquely on top of an identical rectangle, as shown.

The area X of the overlapping region (shaded more darkly) is one eighth of the total shaded area.

What fraction of the area of one rectangle is X?

13
27
14
29
15
(Not answered)

The rectangles are identical. Therefore they have the same area. Let this area be Y.

It follows that the area of each rectangle that is not part of the overlap is Y−X. Hence the total shaded area, made up of the two areas of the rectangles which are not part of the overlap, together with the overlap, is

2Y−X+X=2Y−X.

We are given that the area of the overlap is one eighth of the total shaded area. Therefore

X2Y−X=18.

This equation may be rearranged to give

8X=2Y−X,

from which it follows that

9X=2Y,

and therefore that

XY=29.

Before you hit the SUBMIT button, here are some quick reminders:

You will receive your score immediately, and collect your reward points.
You might earn a new badge... if not, then maybe next week.
Make sure you go through the solution sheet – it is massively important.
A score of less than 50% is ok – it means you can learn lots from your mistakes.
The next Parallelogram is next week, at 3pm on Thursday.
Finally, if you missed any earlier Parallelograms, make sure you go back and complete them. You can still earn reward points and badges by completing missed Parallelograms.

Cheerio, Simon.