Parallelogram 15 • Level 5 • 12 Dec 2024Chrismaths - 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

Four of these numbers can make two pairs so that each pair adds up to 98,765.

Which number is the odd one out?

37,373
45,678
53,087
61,392
70,082
(Not answered)

37,373 + 61,392 = 98,765, and 45,678 + 53,087 = 98,765, so 70,082 is the odd one out.

2.

4 marks

In March 2003, Welshman Tony Evans dropped a ball from an aircraft a mile above the Mojave desert to see if it would bounce. The ball was made from 6 million rubber bands, had a circumference of 14ft 8in, weighed 2600 pounds and took Mr Evans five years to build.

On average, roughly how many rubber bands did he add each day while building the ball?

3
33
330
3300
33,000
(Not answered)

The average number of rubber bands added each day was approximately 60000005×365≈60000001800≈3300

3.

4 marks

Four of these points lie on a single straight line. Which is the odd one out?

(−3, −3)
(−2, −1)
(2, 5)
(4, 11)
(5, 13)
(Not answered)

Points (−3, −3), (−2, −1), (4, 11) and (5, 13) all lie on the line whose equation is y=2x+3, but (2, 5) does not lie on this line.

4.

4 marks

In the triangle PQR, there is a right angle at Q and angle QPR is 60°.

The bisector of the angle QPR meets QR and S, as shown.

What is the ratio of QS : SR?

1 : 1
1 : 2
1 : (3 - 3)
1 : 3
1 : 2
(Not answered)

Angle PRS = 30°, so triangle PRS is isosceles with SP = SR.

Hence QSSR=QSSP=12 as PQS is half of an equilateral triangle.

Alternatively, we can use the angle bisector theorem: QSSR=PQPR=12as PQR is also half of an equilateral triangle.

5.

5 marks

In the diagram, the letter S is made from two arcs, KL and MN, which are each five-eights of the circumference of a circle of radius 1, and the line segment LM, which is tangent to both circles.

At points K and N, common tangents to the circles touch one of the circles.

What is the length LM?

32
3−2
2
322
1+2
(Not answered)

Let the centres of the circles be J and O and let NI be the common tangent shown.

Let P be the point of intersection of JO and LM.

As arc KL is 5/8 of the circumference of the top circle, ∠IJL = 45°.

Consider quadrilateral IJON: sides IJ and NO and are radii, so they are both of unit length and they are both perpendicular to tangent NI. So IJON is a rectangle.

Hence ∠IJO=90° and ∠LJP=90°−45°=45°. In triangle JLP, ∠JPL=180°−90°+45°=45°.

So triangle JLP is isosceles and LP=LJ= 1 unit.

By a similar argument, it may be shown that PM is also of length 1 unit, so LM=LP+PM= 2 units.

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.