PG 35 4 Jul 2019Summer Sums

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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.
  • Tackle each Parallelogram in one go. Don’t get distracted.
  • 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.

It’s summer and the school year is over (or maybe it’s nearly over), but that doesn’t mean that it’s the end of mathematics. If you are taking mathematics seriously and if you are having fun with it (being serious and having fun can go together), then it is important that you keep the numbers and geometry parts of your brain at least a little bit busy over the summer. I am hoping that your teacher will have set you some holiday work, but on top of that here is a longer than usual Parallelogram.

Most of the questions are from past Intermediate Maths Challenge papers, but I have also added a sprinkling of other things to make the Parallelogram (even) more interesting.

It will take you an hour or two to complete this Parallelogram, so maybe tackle it across three or four sessions … but don’t forget to complete it before the new term starts.

And... if you have missed any earlier Parallelograms then the summer is a great time to do some catching up.

1.

2 marks

1.1 What is the value of 1 − 0.2 + 0.03 − 0.004?

  • 0.826
  • 0.834
  • 0.926
  • 1.226
  • 1.234

We have 1 − 0.2 = 0.8 and so 1 − 0.2 + 0.03 = 0.8 + 0.03 = 0.83, and therefore 1 − 0.2 + 0.03 − 0.004 = 0.83 − 0.004 = 0.826.

Note that in the context of the IMC where we just need to decide which of the options is correct, it isn’t really necessary to do this exact arithmetic. Looking at the terms making up the sum it is evident that the answer is just greater than 0.8, and the presence of the term −0.004 shows that the digit in the thousandths column is 6. So 0.826 is the correct option.

2.

2 marks

2.1 In 2014, Australian Suzy Walsham won the annual women’s race up the 1576 steps of the Empire State Building in New York for a record fifth time. Her winning time was 11 minutes 57 seconds.

Approximately how many steps did she climb per minute?

  • 13
  • 20
  • 80
  • 100
  • 130

We see that 11 minutes 53 seconds is just under 12 minutes, and 1576 is just under 1600. So the number of steps per minute is approximately

160012=4003133.

We conclude that, of the given options, 130 is the best approximation.

3.

2 marks

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

  • 11440
  • 338
  • 130
  • 13
  • 310

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.

4.

2 marks

4.1 The diagram shows a regular pentagon inside a square.

What is the value of x?

  • 48
  • 51
  • 54
  • 60
  • 72
Show Hint (–1 mark)
–1 mark

The interior angles of a regular pentagon are each 108°.

Let the vertices of the square be labelled K, L, M, N and those of the regular pentagon P, Q, R, S, T, as shown in the diagram.

We use the fact that the interior angles of a regular pentagon are each 108°. It follows that RST=STP=108°. Because it is the angle of a square, SNT=90°. By the Exterior Angle Theorem, RST=SNT+NTS. Therefore, 108°=90°+NTS. It follows that NTS=108°90°=18°.

Finally, because the angles at T on the line KN have sum 180°,

NTS+STP+PTK=180°,

that is,

18°+108°+x°=180°.

We conclude that

x=18018108=54.

5. The amazing Tadashi and perplexing paperclips

Tadashi Tokieda is one of my favourite mathematicians. He is one of the most curious people in the world, finding wonder and joy in the most ordinary of objects. I have sprinkled some Tadashi videos through this Parallelogram, and to start with here is a video about paperclips.

2 marks

5.1 According to Tadashi, in terms of learning and advancing science, “it’s always important to...”

  • smile
  • guess
  • be polite
  • eat your greens
  • stick your chewing gum on your bedpost

6.

3 marks

6.1. Which of the following numbers is not a square?

  • 16
  • 25
  • 34
  • 43
  • 52

We have 25=32 which is not a square.

In the context of the IMC we can assume that only one option is correct and that it is sufficient just to find it. However, for a complete solution we need to check that none of the other options is a square. Note that we can show this without evaluating any of them, as follows. We have 16=132, 34=322, 43=223=26=232, and so these are all squares, and it is immediate that 52 is a square.

7.

3 marks

7.1. The equilateral triangle and regular hexagon shown have perimeters of the same length.

What is the ratio of the area of the triangle to the area of the hexagon?

  • 5:6
  • 4:5
  • 3:4
  • 2:3
  • 1:1

We can assume that we have chosen units so that each side of the regular hexagon has length 1. It follows that the length of the perimeter of this hexagon is 6. Therefore the length of the perimeter of the equilateral triangle is 6. Hence each of its sides has length 2.

It follows that we can divide the equilateral triangle into 4 smaller equilateral triangles each with side length 1 as shown. The hexagon may be divided into 6 equilateral triangles with side length 1 as shown.

It follows that the ratio of their areas is 4:6, that is, 2:3.

8.

3 marks

8.1. 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

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.

9.

3 marks

9.1. How many two-digit squares differ by 1 from a multiple of 10?

  • 1
  • 2
  • 3
  • 4
  • 5

The two-digit squares are 16, 25, 36, 49, 64 and 81. Of these, just 49 = 50 − 1 and 81 = 80 + 1 differ by 1 from a multiple of 10. So there are 2 such two-digit squares.

10. Money catching

2 marks

10.1 Tadashi explained that an object falls a distance equal to (12×g×t2), which is (12×10×t2), which is 5×t2.

How many meters will an object fall in 1 second?

Correct Solution: 5 metres

3 marks

10.2 How many seconds will it take for an object to fall 80m?

Correct Solution: 4 seconds

If distance = 5 × (time)2, then we know that 80 = 5 × (time)2.

That means that 16 = time2.

So, the time required is 4 seconds.

11.

3 marks

11.1. What is the value of p+q+r+s+t+u+v+w+x+y in the diagram?

  • 540
  • 720
  • 900
  • 1080
  • 1440
Show Hint (–1 mark)
–1 mark

Remember, for every polygon the sum of its exterior angles is 360°.

We use the fact that for every polygon the sum of its exterior angles is 360°.

The shaded angles in the diagram on the right form one set of external angles. Therefore

q+s+u+w+y=360.

The other marked angles form another set of external angles and therefore

r+t+v+x+p=360.

[Alternatively, we could argue that the angles marked q° and r° are vertically opposite and therefore q=r, and, similarly, s=t, u=v, w=x and p=y, and therefore r+t+v+x+p=q+s+u+w+y=360.]

It follows that

p+q+r+s+t+u+v+w+x+y+z=q+s+u+w+y+r+t+v+x+p
=360+360
=720.

12.

3 marks

12.1. What is the remainder when 22×33×55×77 is divided by 8?

  • 2
  • 3
  • 4
  • 5
  • 7
Show Hint (–1 mark)
–1 mark

Notice that 33×55×77 is a positive odd integer greater than 1 and so is equal to 2n+1, for some positive integer n.

First, we note that 33×55×77 is a positive odd integer greater than 1 and so is equal to 2n+1, for some positive integer n.

It follows that

22×33×55×77=4×33×55×77
=42n+1
=8n+4.

Therefore when 22×33×55×77 is divided by 8, the quotient is n and the remainder is 4.

13.

3 marks

13.1 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

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 211+2=18. So 18 is the largest any of them could be.

14.

3 marks

14.1 An ant is on the square marked with a black dot.

The ant moves across an edge from one square to an adjacent square four times and then stops.

How many of the possible finishing squares are black?

  • 0
  • 2
  • 4
  • 6
  • 8
Show Hint (–1 mark)
–1 mark

The crucial point with this question is that we are not dealing with a flat board, but rather a board with a step, and this means it is possible move from a white square to a white square. So you can reach a black square by moving from white to black to white to white to black.

The figure shows the part of the layout that is relevant. Some of the squares have been labelled so that we may refer to them. A movement of the ant will be indicated by the sequence of labels of the four squares, in order, that the ant travels to after leaving the square labelled P. For example, QTUV is one such sequence. In this example the ant ends up on the black square V.

The ant starts on a white square. If at each move the colour of the square it is on changes, after four moves it will end up on a white square. So the only way the ant can end up on a black square is if on one of its moves the colour of the square does not change. That is, during its sequence of moves it either moves from T to U, or from U to T, or from W to X, or from X to W.

We therefore see that the only movements which result in the ant ending up on a black square are

QTUR, QTUV, QTUX, QTWX, RUTQ, RUTS, RUTW and RUXW.

From the final letters of these sequences we see that there are exactly 6 black squares that the ant can end up on after four moves, namely, Q, R, S, V, W and X.

15. An unexpected way to inflate a balloon

3 marks

15.1 The video shows a graph explaining how “spherical” the balloon is as time passes. What would be a good way to describe the line in the graph?

  • Spiral
  • Saw tooth
  • Flat
  • Rollercoaster

16.

4 marks

16.1 What is the area of the shaded region in the rectangle?

  • 21 cm2
  • 22 cm2
  • 23 cm2
  • 24 cm2
  • more information needed
Show Hint (–1 mark)
–1 mark

This question is not that hard to solve algebraically or by drawing some lines and seeing some symmetries. However, I decided to take an extreme version of squeezing one of the triangles and expanding the other, so that we have the following, which has an area which is easy to calculate. This means that the answer is either the shaded area or “more information needed”.

We divide the rectangle into four smaller rectangles by the vertical dashed lines as shown in the figure. It is then evident that the shaded area is the sum of exactly half the area of each of the smaller rectangles.

Therefore the total shaded area is half the total area of the original rectangle. So the shaded area is 123×14 cm2 = 21 cm2.

We could also solve this question algebraically. We let the top edges of the two shaded triangles have lengths a cm and b cm, respectively, so that a+b=14. The two shaded triangles have areas 12a×3 cm2 =32a cm2 and 12b×3 cm2 =32b cm2, respectively. Hence their total area is

32a+32b cm2 =32a+b cm2 =32×14 cm2 = 21 cm2.

17.

4 marks

17.1 In a sequence, each term after the first two terms is the mean of all the terms which come before that term. The first term is 8 and the tenth term is 26.

What is the second term?

  • 17
  • 18
  • 44
  • 52
  • 68

Here it is actually a little easier to see what is going on if we think about the general case of a sequence in which every term after the second is the mean of all the terms that come before it.

Consider a sequence generated by this rule whose first two terms are a and b. The mean of these terms is a+b2, and so the first three terms are

a,b,a+b2.

The sum of these three terms is the same as the sum of the three terms

a+b2,a+b2,a+b2

and it follows that the mean of the first three terms is also (a + b)/2. So the first four terms of the sequence are

a,b,a+b2,a+b2.

These have the same sum as the four terms

a+b2,a+b2,a+b2,a+b2.

It follows that the mean of the first four terms is also a+b2 and that therefore this is the fifth term of the sequence, and so on. We thus see that each term of the sequence after the first two is equal to the mean of the first two terms.

In the sequence we are given in the question the first term is 8 and the tenth term is 26. It follows that 26 is the mean of the first two terms. Hence their sum is 2 × 26 = 52. Since the first term is 8, we deduce that the second term is 52 − 8 = 44.

18.

4 marks

18.1. A flag is in the shape of a right-angled triangle, as shown, with the horizontal and vertical sides being of length 72 cm and 24 cm respectively.

The flag is divided into 6 vertical stripes of equal width.

What, in cm2, is the difference between the areas of any two adjacent stripes?

  • 96
  • 72
  • 48
  • 32
  • 24
Show Hint (–1 mark)
–1 mark

I did not follow the approach of the solution you will after you have completed the Parallelogram, but here is something to get you started. If the difference in area is the same for any two stripes, then just consider the two smallest stripes. They make up a sub-triangle that is only a third the size of the whole triangle. And the shaded mini-triangle is half the length, so a quarter of the area of the outlined sub-triangle.

The lines we have added in the figure on the right divide the flag into 15 congruent rectangles and 6 congruent triangles. Let the area of each of the 15 rectangles be x cm2. The area of each of the 6 congruent triangles is half that of the rectangles, that is, 12x cm2.

The difference between the area of the adjacent vertical stripes of the flag is the area of one of the rectangles.

From its dimensions, we see that the area of the flag is 1272×24 cm2 =36×24 cm2. It is made up of 15 rectangles, each with area x cm2, and 6 triangles, each with area 12x cm2.

Therefore

15x+612x=36×24,

that is,

18x=36×24,

and it follows that

x=36×2418=2×24=48.

Hence the difference between the area of adjacent stripes of the flag is 48 cm2.

19.

4 marks

19.1 You are asked to choose two positive integers, m and n with m>n, so that as many as possible of the expressions m+n, mn, m×n and m÷n have values that are prime.

When you do this correctly, how many of these four expressions have values that are prime?

  • 0
  • 1
  • 2
  • 3
  • 4
Show Hint (–1 mark)
–1 mark

The answer is not equal to 4.

With m=2 and n=1, we have m+n=3, mn=1, m×n=2 and m÷n=2. So it is possible for three of the expressions to have values that are prime. [Note that 1 is not a prime, so in this case mn is not prime].

We now show that this is the best that can be achieved by showing that it is not possible for all four of the expressions to have values that are primes.

Since m and n are positive integers with m>n, we have m>1. Therefore m×n is prime if, and only if, m is prime and n=1. Then m+n=m+1 and mn=m1. Therefore, if all four expressions are primes, m1, m and m+1 would be three consecutive primes. However a trio of consecutive positive integers that are all primes does not exist. We conclude that not all four of the expressions can have values that are primes.

20. Numberphile – Freaky dot patterns

4 marks

20.1 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

21.

4 marks

21.1 The football shown is made by sewing together 12 black pentagonal panels and 20 white hexagonal panels.

There is a join wherever two panels meet along an edge.

How many joins are there?

  • 20
  • 32
  • 60
  • 90
  • 180
Show Hint (–1 mark)
–1 mark

Try harder.

Each of the 12 pentagons has 5 edges, and each of the 20 hexagons has 6 edges. So, altogether, the panels have 12×5+20×6=180 edges. Each join connects 2 edges, So there are 12180=90 joins.

22.

4 marks

22.1. The total weight of a box, 20 plates and 30 cups is 4.8 kg. The total weight of the box, 40 plates and 50 cups is 8.4 kg.

What is the total weight of the box, 10 plates and 20 cups?

  • 3 kg
  • 3.2 kg
  • 3.6 kg
  • 4 kg
  • 4.2 kg

The difference between the box, 20 plates, 30 cups and the box, 40 plates, 50 cups, is 20 plates and 20 cups. From the information in the question we see that the extra weight of 20 plates and 20 cups is 8.4 kg − 4.8 kg = 3.6 kg.

Therefore the weight of 10 plates and 10 cups is 123.6 kg = 1.8 kg.

If we remove 10 plates and 10 cups from the box, 20 plates and 30 cups we end up with the box, 10 plates and 20 cups. Removing the 10 plates and 10 cups reduces the original weight of 4.8 kg by 1.8 kg. Hence the weight of the box, 10 plates and 20 cups is 4.8 kg − 1.8 kg = 3 kg.

23.

4 marks

23.1 The figure shows four smaller squares in the corners of a large square.

The smaller squares have sides of length 1 cm, 2 cm, 3 cm and 4 cm (in anticlockwise order) and the sides of the large square have length 11 cm.

What is the area of the shaded quadrilateral?

  • 35 cm2
  • 36 cm2
  • 37 cm2
  • 38 cm2
  • 39 cm2

We work out the area that is not shaded and subtract this area from that of the large square.

The area that is not shaded is made up of the four corner squares and four trapeziums.

The total area of the squares is, in cm2,

12+22+32+42=1+4+9+16=30.

We now use the fact that the area of a trapezium is given by the formula

12a+bh,

where a and b are the lengths of the parallel sides, and h is their distance apart.

The values of a, b and h for the trapeziums P, Q, R and S, as labelled in the diagram, can be calculated from the side lengths of the squares, and are shown in the diagram. We therefore have that

area of P=121+2×8=12,
area of Q=122+3×6=15,
area of R=123+4×4=14, and
area of S=124+1×6=15.

It follows that the total area of the trapeziums is

12+15+14+15=56.

The large square has area 112 cm2 = 121 cm2. Therefore the area of the shaded quadrilateral is, in cm2, 1213056=35.

24.

4 marks

24.1 A voucher code is made up of four characters. The first is a letter: V, X or P. The second and third are different digits. The fourth is the units digit of the sum of the second and third digits.

How many different voucher codes like this are there?

  • 180
  • 243
  • 270
  • 300
  • 2700

There are 3 choices for the first character, namely V, X or P. There are 10 choices for the second character, namely one of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9. Because the second and third characters are different digits, the third character can be any one of the 10 digits, other than the second character. Therefore there are 9 choices for the third character. There is just 1 choice for the fourth character as this is the units digit of the sum of the second and third digits.

It follows that the total number of voucher codes is 3×10×9×1=270.

25. Catching Kendama

3 marks

25.1 In order to make one of the toughest Kendama catches, Tadashi has to take advantage of a principle from physics known as…

  • The conservation of energy
  • The conservation of angular momentum
  • The conservation of mass
  • The conservation of atoms
  • The conservation of curiosity

Finally, it’s up to you, but you might enjoy watching the rest of Tadashi’s video about paperclips

26.

5 marks

26.1 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

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 YX. 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

2YX+X=2YX.

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

X2YX=18.

This equation may be rearranged to give

8X=2YX,

from which it follows that

9X=2Y,

and therefore that

XY=29.

27.

5 marks

27.1. The diagram shows a shaded region inside a large square.

The shaded region is divided into small squares.

What fraction of the area of the large square is shaded?

  • 310
  • 13
  • 38
  • 25
  • 37
Show Hint (–1 mark)
–1 mark

We form a complete grid inside the large square by extending the sides of the small squares, as shown in the figure. In this way the large square is divided up into 61 small squares, 20 triangles along the edges whose areas are equal to half that of the small squares, and 4 triangles in the corners whose areas are equal to one quarter that of the small squares.

We form a complete grid inside the large square by extending the sides of the small squares, as shown in the figure. In this way the large square is divided up into 61 small squares, 20 triangles along the edges whose areas are equal to half that of the small squares, and 4 triangles in the corners whose areas are equal to one quarter that of the small squares.

So the area of the large square is equal to that of

61×1+20×12+4×14=72

of the small squares. The shaded area is made up of 24 small squares. Therefore the fraction of the large square that is shaded is

2472=13.

[Note that we could save some work by taking advantage of the symmetry of the figure, and just work out which fraction of, say, the top left hand corner of the square is shaded.]

28.

5 marks

28.1 There are 120 different ways of arranging the letters, U, K, M, I and C. All of these arrangements are listed in dictionary order, starting with CIKMU.

Which position in the list does UKIMC occupy?

  • 110th
  • 112th
  • 114th
  • 116th
  • 118th

Because the sequence UKIMC comes towards the bottom of the list in dictionary order of all 120 arrangements of the letters U, K, M, I, C, we can most easily find its position by listing the arrangements in reverse dictionary order, until we reach UKIMC, as follows

120 UMKIC
119 UMKCI
118 UMIKC
117 UMICK
116 UMCKI
115 UMCIK
114 UKMIC
113 UKMCI
112 UKIMC

We see from this that UKIMC is the 112th arrangement in the dictionary order list.

29.

5 marks

29.1 In square RSTU a quarter-circle arc with centre S is drawn from T to R.

A point P on this arc is 1 unit from TU and 8 units from RU.

What is the length of the side of square RSTU?

Show Hint (–1 mark)
–1 mark

This is a tough geometry question. So here is an unusual hint – the answer is a vowel.

  • 9
  • 10
  • 11
  • 12
  • 13

Suppose that x is the length of the sides of the square RSTU. We need to find the value of x.

We let K be the point where the perpendicular from P to RS meets RS, and L be the point where the perpendicular from P to ST meets ST, as shown in the figure. Then PK has length x1, and so also does LS. Also, PL has length x8.

Because PS is a radius of the quarter circle, its length is the same as that of the sides of the square, namely, x.

Therefore SLP is a right-angled triangle in which the hypotenuse has length x, and the other two sides have lengths x1 and x8. Hence, by Pythagoras’ Theorem,

x12+x82=x2.

Multiplying out the squares, we get

x22x+1+x216x+64=x2,

which we can rearrange as

x218x+65=0.

We can factorize the quadratic on the left-hand side of this equation to give the equivalent equation

x5x13=0,

which has the solutions x=5 and x=13. Since x8 is the length of PL and therefore cannot be negative, x5. Therefore x=13.

30.

5 marks

30.1 A point is marked one quarter of the way along each side of a triangle, as shown.

What fraction of the area of the triangle is shaded?

  • 716
  • 12
  • 916
  • 58
  • 1116

We use the notation XYZ for the area of a triangle XYZ.

We let P, Q and R be the vertices of the triangle, and let S, T and U be the points one quarter of the way along each side, as shown in the figure.

We let K and L be the points where the perpendiculars from P and T, respectively, meet QR.

The area of a triangle is half the base multiplied by the height of the triangle. Therefore

PQR=12QR×PK.

We now obtain an expression for TSR.

Because S is one quarter of the way along QR, SR=34QR.

In the triangles PKR and TLR, the angles at K and L are both right angles, and the angle at R is common to both triangles. It follows that these triangles are similar. Therefore

TLPK=TRPR=14,

and hence

TL=14PK.

It follows that

STR=12SR×TL=1234QR×14PK=31612QR×PK=316PQR.

Similarly,

TPU=316PQR.

It follows that the fraction of the area of triangle PQR that is shaded is

1316316=58.

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.
  • 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.