Parallelogram 13 Year 9 12 Dec 2019Chrismaths Y9

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Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

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

It’s nearly Christmas. Not very many people realise that the original name for this holiday was Chrismaths. For centuries, children traditionally completed mathematics problems during the festive period. The amount of effort that children put into their mathematics problems allowed Santa to decide who had been naughty and who had been nice.

With this in mind, this week’s Parallelogram contains a WHOLE maths challenge paper – 25 glorious questions to stretch your brain over Chrismaths.

The time limit on maths challenges is 60 minutes, but it might take you up to 90 minutes to complete all the questions. However, you can tackle the challeges in one, two or three or more sessions. Just complete all the questions and hit the submit button before you go back to school in January. Bear in mind, the last 10 questions are tough. Also, in the real maths challenge paper, there is negative marking for tough questions, which means you lose marks for incorrect answers – don’t worry, in this Parallelogram you will not lose marks for wrong answers.

As this Parallelogram is longer than a typical Parallelogram, you can win a maximum of 400 (rather than 100) rewards points – this is a real opportunity to win an extra badge (or two).

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

How many of the following four numbers are prime?

3
33
333
3333

  • A) 0
  • B) 1
  • C) 2
  • D) 3
  • E) 4

The number 3 is prime, but the other numbers listed are not prime:

  • 33 = 3 × 11
  • 333 = 3 × 111
  • 3333 = 3 × 1111.

2.

3 marks

Three positive integers are all different. Their sum is 7.

What is their product?

  • A) 12
  • B) 10
  • C) 9
  • D) 8
  • E) 5

It can be seen that 1 + 2 + 4 = 7 and 1 × 2 × 4 = 8, so assuming that there is just one solution, the answer must be 8.

In the context of the IMC, that is enough, but if you are asked to give a full solution, you need to give an argument to show there are no other possibilities. This is not difficult.

For example, suppose a,b and c are three different positive integers with sum 7, and that a<b<c.

If a2, then b3 and c4, and so a+b+c9.

So we must have that a=1. It follows that b+c=6.

If b3 then c4 and hence b+c3+4=7. So b=2. Since a=1 and b=2, it follows that c=4.

3.

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°

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°.

4.

3 marks

All four digits of two 2-digit numbers are different.

What is the largest possible sum of two such numbers?

  • A) 169
  • B) 174
  • C) 183
  • D) 190
  • E) 197

To get the largest possible sum we need to take 9 and 8 as the tens digits, and 7 and 6 as the units digits. For example, 97 + 86 = 183.

5.

3 marks

How many minutes will elapse between 20:12 today and 21:02 tomorrow?

  • A) 50
  • B) 770
  • C) 1250
  • D) 1490
  • E) 2450

From 20:12 today until 20.12 tomorrow is 24 hours, that is 24 × 60 = 1440 minutes. There are 50 minutes from 20:12 tomorrow to 21:02 tomorrow. This gives a total of 1440 + 50 = 1490 minutes.

6.

3 marks

Triangle QRS is isosceles and right-angled.

Beatrice reflects the P-shape in the side QR to get an image.

She reflects the first image in the side QS to get a second image.

Finally, she reflects the second image in the side RS to get a third image.

What does the third image look like?

  • A)
  • B)
  • C)
  • D)
  • E)

The effect of the successive reflections is shown in the diagram.

7.

3 marks

The prime numbers p and q are the smallest primes that differ by 6.

What is the sum of p and q?

  • A) 12
  • B) 14
  • C) 16
  • D) 20
  • E) 28

Suppose p<q. Then q=p+6. The prime numbers are 2, 3, 5, 7,….

With p=2,q=8, which is not prime.

Similarly if p=3,q=9, which is also not prime.

However, when p=5,q=11, which is prime.

So, p=5, q=11 gives the smallest primes that differ by 6. Then p+q=5+11=16.

8.

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

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 * .

9.

3 marks

Auntie Fi’s dog Itchy has a million fleas. His anti-flea shampoo claims to leave no more than 1% of the original number of fleas after use.

What is the least number of fleas that will be eradicated by the treatment?

  • A) 900000
  • B) 990000
  • C) 999000
  • D) 999990
  • E) 999999

Since no more than 1% of the fleas will remain, at least 99% of them will be eradicated.

Now 99% of a million is 99100×1000000=99×10000=9900.

10.

3 marks

An ‘abundant’ number is a positive integer N, such that the sum of the factors of N (excluding N itself) is greater than N.

What is the smallest abundant number?

  • A) 5
  • B) 6
  • C) 10
  • D) 12
  • E) 15

In the IMC, it is only necessary to check the factors of the numbers given as the options.

However, to be sure that the smallest of these which is abundant, is the overall smallest abundant number, we would need to check the factors of all the positive integers in turn, until we find an abundant number.

The following table gives the sum of the factors of N (excluding N itself), for 1N12.

From this table we see that 12 is the smallest abundant number.

11.

3 marks

In the diagram, PQRS is a parallelogram; QRS=50°; SPT=62° and PQ=PT.

What is the size of TQR?

  • A) 84°
  • B) 90°
  • C) 96°
  • D) 112°
  • E) 124°

Because PQRS is a parallelogram, SPQ=QRS=50°. Therefore TPQ=62+50°=112°.

Therefore, as the angles in a triangle add up to 180°, PQT+PTQ=180°1120=68°.

Because PQ=PT, the triangle QPT is isosceles, and so PQT=PTQ. Therefore PQT=PTQ=34°.

Because PQRS is a parallelogram, PQR+QRS=180°, and therefore PQR=180°50°=130°.

Therefore, TQR=PQRPQT=130°34°=96°.

12.

3 marks

Which of the following has a different value from the others?

  • A) 18% of £30
  • B) 12% of £50
  • C) 6% of £90
  • D) 4% of £135
  • E) 2% of £270

We have that 18% of £30 = £18100×30 = £5.40.

Similarly, 12% of £50 is £6.00, and 6% of £90 is £5.40.

We already see that option B must be the odd one out.

It is easy to check that 4% of £135 and 2% of £270 are also both £5.40.

13.

3 marks

Alex Erlich and Paneth Farcas 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

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 .

14.

3 marks

What value of x makes the mean of the first three numbers in this list equal to the mean of the last four?

15 ; 5 ; x ; 7 ; 9 ; 17

  • A) 19
  • B) 21
  • C) 24
  • D) 25
  • E) 27

The mean of the first three numbers in the list is 15+5+x3 and the mean of the last four is x+7+9+174.

Now:
15+5+x3=x+7+9+174
415+5+x=3x+7+9+17
80+4x=3x+99
x=19.

An alternative method in the context of the IMC would be just to try the given options in turn. This runs the risk of involving a lot of arithmetic, but here, as the first option is the correct answer, the gamble would pay off.

15.

3 marks

Which of the following has a value that is closest to 0?

  • A) 12+13×14
  • B) 12+13÷14
  • C) 12×13÷14
  • D) 1213÷14
  • E) 1213×14

When working out the values of these expressions it is important to remember the convention (sometimes known as BODMAS or BIDMAS) that tells us that Divisions and Multiplications are carried out before Additions and Subtractions.

Some work can be saved by noting that the expressions A and B have values greater than 12, whereas the value of expression E lies between 0 and 12. So it must be C, D or E that has the value closest to 0.

Now, noting that 13÷14=13×41=43,
we obtain that the value of C is 12×13÷14=12×43=23;
that of D is 1213÷14=1243=56;
and that of E is 1213×14=12112=512.

From these calculations we see that E gives the value closest to 0.

16.

4 marks

The diagram shows a large equilateral triangle divided by three straight lines into seven regions.

The three grey regions are equilateral triangles with sides of length 5 cm and the central black region is an equilateral triangle with sides of length 2 cm.

What is the side length of the original large triangle?

  • A) 18 cm
  • B) 19 cm
  • C) 20 cm
  • D) 21 cm
  • E) 22 cm

Let P,Q,R,S,T,U and V be the points shown. All the angles in all the triangles are 60°. So QRT=PSU and hence RT is parallel to SU.

Similarly, as RSV=TUV, RS is parallel to TU. Therefore RSUT is a parallelogram.

Therefore RS has the same length as TU, namely, 2 + 5 = 7 cm. Similarly PQ has length 7 cm.

So the length of PS which is the sum of the lengths of PQ, QS and RS is 7 + 5 + 7 = 19 cm.

17.

4 marks

The first term in a sequence of positive integers is 6. The other terms in the sequence follow these rules:

  • if a term is even then divide it by 2 to obtain the next term;
  • if a term is odd then multiply it by 5 and subtract 1 to obtain the next term.

For which values of n is the n th term equal to n?

  • A) 10 only
  • B) 13 only
  • C) 16 only
  • D) 10 and 13 only
  • E) 13 and 16 only

Since the options refer only to the 10th, 13th and 16th terms of the sequence, as far as this IMC question is concerned it is only necessary to check the first 16 terms in the sequence. These are as shown in the table below:

From this we see that the 13th term is 13, and the 16th term is 16, and that these are the only cases where the n th term is equal to n.

However, a complete answer requires a proof that for all n > 16, the nth term is not equal to n. It can be seen that after the 16th term the sequence continues 8, 4, 2, 1, 4, 2, 1… with the cycle 4, 2, 1 now repeating for ever. It follows that, for n ≥ 17, the only values taken by the nth term are 8, 4, 2 and 1. We deduce that for n ≥ 16, the nth term is not equal to n.

18.

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)

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).

19.

4 marks

The shaded region shown in the diagram is bounded by four arcs, each of the same radius as that of the surrounding circle.

What fraction of the surrounding circle is shaded?

  • A) 4π1
  • B) 1π4
  • C) 12
  • D) 13
  • E) It depends on the radius of the circle

Suppose that the surrounding circle has radius r. In the diagram we have drawn the square with side length 2r which touches the circle at the points where it meets the arcs.

The square has area 2r2=4r2. The unshaded area inside the square is made up of four quarter circles with radius r, and thus has area πr2. Hence the shaded area is 4r2πr2=4πr2.

The circle has area πr2. So the fraction of the circle that is shaded is:

4πr2πr2=4ππ=4π1.

20.

4 marks

A rectangle with area 125 cm2 has sides in the ratio 4:5. What is the perimeter of the rectangle?

  • A) 18 cm
  • B) 22.5 cm
  • C) 36 cm
  • D) 45 cm
  • E) 54 cm

Since the side lengths of the rectangle are in the ratio 4:5, they are 4a cm and 5a cm, for some positive number a.

This means that the rectangle has area 4a×5a=20a2cm2.

Hence 20a2=125. So a2 = 12520=254, and hence a=52.

Hence the rectangle has perimeter 24a+5a=18a=18×52=45 cm.

21.

5 marks

The parallelogram PQRS is formed by joining together four equilateral triangles of side 1 unit, as shown.

What is the length of the diagonal SQ?

  • A) 7
  • B) 8
  • C) 3
  • D) 6
  • E) 5


Let T be the foot of the perpendicular from Q to the line SR extended. Now RQT is half of an equilateral triangle with side length 1. Hence the length of RT is 1/2 and hence ST has length 1+1+12=52.

By Pythagoras’ Theorem applied to the right angled triangle RQT,12=122+QT2.

Therefore QT2=12122=114=34

Hence, by Pythagoras’ Theorem applied to the right angled triangle SQT,SQ2=ST2+QT2=522+34=254+34=7.

Therefore, SQ=7.

22.

5 marks

What is the maximum possible value of the median number of cups of coffee bought per customer on a day when Sundollars Coffee Shop sells 477 cups of coffee to 190 customers, and every customer buys at least one cup of coffee?

  • A) 1.5
  • B) 2
  • C) 2.5
  • D) 3
  • E) 3.5

Put the set of numbers of cups of coffee drunk by the individual customers into numerical order with the smallest first. This gives an increasing sequence of positive integers with sum 477. Because 190 is even, the median of these numbers is the mean of the 95th and 96th numbers in this list. Suppose these are a and b, respectively. Then the median is 12a+b.

We note that 1ab. Also, each of the first 94 numbers in the list is between 1 and a, and each of the last 94 numbers is at least b. So if we replace the first 94 numbers by 1, and the last 94 numbers by b, we obtain the sequence of numbers

whose sum does not exceed 477, the sum of the original sequence. Therefore

As 1a, it follows that 95+95b94+a+95b477, hence 95b47795=382 and therefore b38295.

Therefore, since b is an integer, b4.

When b=4, it follows from (2) that 94+a+380477, giving a3.

This shows that the maximum possible values for a and b are 3 and 4, respectively. We can see that these values are possible, as, if we substitute these values in (1), we obtain a sequence of numbers with sum 94×1+3+95×4=477.

So 3.5 is the maximum possible value of the median.

23.

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°


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=PRQSRT=45°30°=15°.

Using the Exterior Angle Theorem again, it follows that STR=TQR+QRT, and hence TQR=STRQRT=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°.

24.

5 marks


All the positive integers are written in the cells of a square grid.

Starting from 1, the numbers spiral anticlockwise. The first part of the spiral is shown in the diagram.

Which number is immediately below 2012?

  • A) 1837
  • B) 2011
  • C) 2013
  • D) 2195
  • E) 2210

The key to the solution is to note that the squares of the odd numbers occur on the diagonal leading downwards and to the right from the cell which contains the number 1, and the squares of the even numbers occur on the diagonal which leads upwards and to the left of the cell which contains the number 4.

The squares of the even numbers have the form (2n)2, that is, 4n2. We see that the number 4n2 + 1 occurs to the left of the cell containing 4n2. Below 4n2 + 1 there occur the numbers 4n2 + 2, 4n2 + 3, .., 4n2 + 2n + 1, and then in the cells to the right of the cell containing 4n2 + 2n + 1, there occur the numbers 4n2 + 2n + 2, 4n2 + 2n + 3,..., 4n2 + 4n + 1 = (2n + 1)2.

Now 442 = 1936 and 452 = 2025. Thus 2011 is in the same row as 2025 and to the left of it, in the sequence 1981, … ,2012, … ,2025, and below these occur the numbers 2163, … , 2209 = 472, with 2208 below 2025 as shown below.

It follows that 2195 is the number below 2012.

In the diagram below the square numbers are shown in bold.

25.

5 marks


The diagram shows a ceramic design by the Catalan architect Antoni Gaudi.

It is formed by drawing eight lines connecting points which divide the edges of the outer regular octagon into three equal parts, as shown.

What fraction of the octagon is shaded?

  • A) 15
  • B) 29
  • C) 14
  • D) 310
  • E) 516

We consider the triangular segment of the octagon formed by joining two adjacent vertices, P and Q to the centre, O. For convenience, we show this segment, drawn on a larger scale, on the right, where we have added the lines RW,ST,TW and UV.

These lines are parallel to the edges of the triangle POQ, as shown and together with the lines RU and SV they divide the triangle OPQ into 9 congruent triangles, of which 2 are shaded.

Thus 29 of the segment is shaded. The same holds for all the other congruent segments of the octagon.

So 29 of the whole octagon is shaded.

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 will appear on Thursday 10 January, at 3pm.
  • 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.