Parallelogram 15 • Level 4 • 14 Dec 2023Chrismaths - Part 2

Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

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

It’s nearly Christmas, so time for another instalment of your Christmaths challenge paper.

Good luck and happy Chrismaths. Don’t eat too many mince pie charts!

Simon.

PS: I want to say thank you to the UK Mathematics Trust, who own the copyright to these questions.

1.

3 marks

An equilateral triangle, a square and a pentagon all have the same side length.

The triangle is drawn on and above the top edge of the square and the pentagon is drawn on and below the bottom edge of the square.

What is the sum of the interior angles of the resulting polygon?

A) 10 × 180°
B) 9 × 180°
C) 8 × 180°
D) 7 × 180°
E) 6 × 180°
(Not answered)

The sum of the interior angles of the polygon is the sum of the angles in the triangle, the square and the pentagon. The sum of the interior angles of the triangle is 180°, the sum of the angles of the square is 360° = 2 × 180°, and the sum of the angles of the pentagon is 540° = 3 × 180°.

So the sum of the angles is (1 + 2 + 3) × 180° = 6 × 180°.

Note: There is more than one way to see that the sum of the angles of a pentagon is 540°.

Here is one method. Join the vertices of the pentagon to some point, say P, inside the pentagon. This creates 5 triangles whose angles sum to 5 × 180° .The sum of the angles in these triangles is the sum of the angles in a pentagon plus the sum of the angles at P, which is 360° = 2 × 180° . So the sum of the angles in the pentagon is 5 × 180° - 2 × 180° = 3 × 180°.

2.

3 marks

Seb has been challenged to place the numbers 1 to 9 inclusive in the nine regions formed by the Olympic rings so that there is exactly one number in each region and the sum of the numbers in each ring is 11.

The diagram shows part of his solution.

What number goes in the region marked * ?

A) 6
B) 4
C) 3
D) 2
E) 1
(Not answered)

We let u,v,w,x,y and z be the numbers in the regions shown.

Since the sum of the numbers in each ring is 11, we have, from the leftmost ring, that 9+u=11 and so u=2. Then, from the next ring, 2+5+v=11 and so v=4. From the rightmost ring, z+8=11 and so z=3.

We have now used the digits 2, 3, 4, 5, 8 and 9, leaving 1, 6 and 7.

From the middle ring we have that 4+w+x=11, and so w+x=7. From the second ring from the right x+y+3=11, and so x+y=8.

So we need to solve the equations w+x=7 and x+y=8, using 1, 6 and 7.

It is easy to see that the only solution is x=1, y=7 and w=6.

So 6 goes in the region marked * .

3.

3 marks

Alex Erlich and Paneth Farkas shared an opening rally of 2 hours and 12 minutes during their table tennis match at the 1936 World Games. Each player hit around 45 shots per minute.

Which of the following is closest to the total number of shots played in the rally?

A) 200
B) 2,000
C) 8,000
D) 12,000
E) 20,000
(Not answered)

Since they each hit about 45 shots in one minute, between them they hit about 90 shots per minute.

Now 2 hours and 12 minutes is 132 minutes. So the total number of shots in the match is 90 × 132, and 90 × 132 is approximately 100 × 120 = 12,000 .

4.

4 marks

Peri the winkle starts at the origin and slithers anticlockwise around a semicircle with centre (4, 0). Peri then slides anticlockwise around a second semicircle with centre (6, 0), and finally clockwise around a third semicircle with centre (3, 0).

Where does Peri end this expedition?

A) (0, 0)
B) (1, 0)
C) (2, 0)
D) (4, 0)
E) (6, 0)
(Not answered)

As may be seen from the diagram, Peri first moves along the semicircle with centre (4, 0) from the point (0, 0) to the point (8, 0), then along the semicircle with centre (6, 0) to the point (4, 0), and finally along the semicircle with centre (3, 0) to end up at the point (2, 0).

5.

5 marks

In the triangle PQR,PS=2; SR=1; ∠PRQ=45°; T is the foot of the perpendicular from P to QS and ∠PST=60°.

What is the size of ∠QPR?

A) 45°
B) 60°
C) 75°
D) 90°
E) 105°
(Not answered)

In the triangle PST, ∠PTS=90° and ∠PST=60°. Therefore ∠TPS=30° and the triangle PST is half of an equilateral triangle.

It follows that ST=12PS=1. Therefore triangle RST is isosceles, and hence ∠STR=∠SRT.

By the Exterior Angle Theorem, ∠PST=∠STR+∠SRT. Therefore ∠STR=∠SRT=30°. Hence ∠QRT=∠PRQ−∠SRT=45°−30°=15°.

Using the Exterior Angle Theorem again, it follows that ∠STR=∠TQR+∠QRT, and hence ∠TQR=∠STR−∠QRT=45°−30°=15°.

Therefore the base angles of triangle TQR are equal. Hence TQR is an isosceles triangle, and so QT=RT.

We also have that the base angles in triangle TPR are both equal to 30°, and so PT=RT. Therefore QT=RT=PT. So PTQ is an isosceles right-angled triangle.

Therefore ∠QPT=45°. Finally, we deduce that ∠QPR=∠QPT+∠TPS=45°+30°=75°.

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.