Cayley 2003 2009 PDF [PDF]

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Cayley Olympiad Past Papers and Solutions 2003-2009

The United Kingdom Mathematics Trust

Intermediate Mathematical Olympiad and Kangaroo (IMOK) Olympiad Cayley Paper Thursday 20th March 2003 All candidates must be in School Year 9 or below (England and Wales), S2 or below (Scotland), or School Year 10 or below (Northern Ireland).

READ THESE INSTRUCTIONS CAREFULLY BEFORE STARTING 1.

Time allowed: 2 hours.

2.

The use of calculators, protractors and squared paper is forbidden. Rulers and compasses may be used.

3.

For questions in Section A only the answer is required. Enter each answer neatly in the relevant box on the Cover Sheet. For questions in Section B start each question on a fresh A4 sheet and give full written solutions, including clear mathematical explanations as to why your method is correct. Solutions must be written neatly on A4 paper. Sheets must be STAPLED together in the top left corner with the Cover Sheet on top. Do not hand in rough work.

4.

Questions A1-A5 are relatively short questions. Try to complete Section A within the first 20 minutes so as to allow sufficient time for Section B.

5.

Questions B1-B5 are longer questions requiring full written solutions. This means that each answer must be accompanied by clear explanations and proofs. Work in rough first, then set out your final solution with clear explanations of each step.

6.

These problems are meant to be challenging! Do not hurry. Try the earlier questions in each section first (they tend to be easier). Try to finish whole questions even if you can't do many. A good candidate will have done most of Section A and given solutions to at least two questions in Section B.

7.

Numerical answers must be FULLY SIMPLIFIED, and EXACT using symbols like π, fractions, or square roots if appropriate, but NOT decimal approximations.

DO NOT OPEN THE PAPER UNTIL INSTRUCTED BY THE INVIGILATOR TO DO SO! The United Kingdom Mathematics Trust is a Registered Charity.

Section A Write your answers in the boxes provided on the Cover Sheet. Do not hand in your working. You should aim to spend no more than 20 minutes on this section.

A1

There are 17 trees along the road from Basil’s home to a swimming pool. On his way to and from a swim Basil marks some trees with a red stripe. On his way to the pool he marks the first tree, the third tree, the fifth tree and so on. On his way back again, he marks the first tree he comes to, the fourth tree, the seventh tree and so on, missing out two trees each time. By the time he gets home, how many trees have no mark?

A2

A straight line is drawn across a 4 × 4 grid (like a chessboard). What is the greatest number of 1 × 1 squares which can be divided into two by the line?

A3

There used to be 5 parrots in my cage. Their average value was ¤6000. One day while I was cleaning out the cage the most beautiful parrot flew away. The average value of the remaining four parrots was ¤5000. What was the value of the parrot that escaped?

A4

Start with a positive integer with 2 digits. Crossing out the units digit gives a new single digit number. If you multiply this new number by an integer x you get the original number back. What is the greatest possible value of x?

A5

Mike has 42 identical cubes, each with edges of length 1 cm. He used all the cubes to construct a cuboid. The perimeter of the base of that cuboid was 18 cm. What was its height?

Section B Answer each question on a separate sheet of A4 paper. Do not hand in rough working. Try to finish whole questions even if you cannot do many: few candidates will do all five questions. You should give full solutions, including clear mathematical explanations, and express all calculations and answers as exact numbers such as 4π, 2 + 7. Just stating an answer, even a correct one, will earn you very few marks; also, incomplete or poorly presented solutions will not receive full marks.

B1

p q × r q

In this multiplication sum, p, q, r and s stand for different digits. Find the digit which each letter represents, explaining how you know that you have found all possible solutions.

s s s B2

In the diagram (which is not to scale), AQ bisects ∠PQR, AR bisects ∠KRP and ∠RPQ = 80°.

A P 80°

What is the size of ∠RAQ?

Q

R

K

B3

At McBride Academy there are 300 children each of whom represents the school in both summer and winter sports. In summer, 60% of these play tennis and the other 40% play badminton. In winter they play hockey or swim but not both. 56% of the hockey players play tennis in summer and 30% of the tennis players swim. How many both swim and play badminton?

B4

In the diagram, P, Q, R and S are four points on a line such that PQ = RS. Semicircles are drawn above the line with diameters PQ, RS and PS, and another semicircle with diameter QR is drawn below the line. The line MN is the line of symmetry of the figure. Prove that the shaded area is equal to the area of the circle with diameter MN.

M

P

Q

R

N

B5

The numbers 1 to 9 are to be placed in the circles in such a way that the sum of the four numbers along each side of the triangle has the same value, T say. (a) Prove that 17 ≤ T ≤ 23. (b) Find a suitable arrangement of the numbers when T = 23. (c) Show that when T = 20 then there are at most 8 different choices for the collection of three numbers which should be placed at the vertices of the triangle.

S

The United Kingdom Mathematics Trust

Intermediate Mathematical Olympiad and Kangaroo (IMOK) Olympiad Cayley Paper Thursday 18th March 2004 All candidates must be in School Year 9 or below (England and Wales), S2 or below (Scotland), or School Year 10 or below (Northern Ireland).

READ THESE INSTRUCTIONS CAREFULLY BEFORE STARTING 1.

Time allowed: 2 hours.

2.

The use of calculators, protractors and squared paper is forbidden. Rulers and compasses may be used.

3.

Solutions must be written neatly on A4 paper. Sheets must be STAPLED together in the top left corner with the Cover Sheet on top.

4.

Start each question on a fresh A4 sheet. You may wish to work in rough first, then set out your final solution with clear explanations and proofs. Do not hand in rough work.

5.

Answers must be FULLY SIMPLIFIED, and EXACT using symbols like π, fractions, or square roots if appropriate, but NOT decimal approximations.

6.

Give full written solutions, including mathematical reasons as to why your method is correct. Just stating an answer, even a correct one, will earn you very few marks; also, incomplete or poorly presented solutions will not receive full marks.

7.

These problems are meant to be challenging! The earlier questions tend to be easier; the last two questions are the most demanding. Do not hurry, but spend time working carefully on one question before attempting another. Try to finish whole questions even if you cannot do many: you will have done well if you hand in full solutions to two or more questions.

DO NOT OPEN THE PAPER UNTIL INSTRUCTED BY THE INVIGILATOR TO DO SO! The United Kingdom Mathematics Trust is a Registered Charity. Enquiries should be sent to: Maths Challenges Office, School of Mathematics, University of Leeds, Leeds, LS2 9JT. (Tel. 0113 343 2339)

http://www.ukmt.org.uk

1.

The “star” octagon shown in the diagram is beautifully symmetrical and the centre of the star is at the centre of the circle. If angle NAE = 110°, how big is angle DNA ?

N D

A

W

E C

B S

2.

Mars, his wife Venus and grandson Pluto have a combined age of 192. The ages of Mars and Pluto together total 30 years more than Venus's age. The ages of Venus and Pluto together total 4 years more than Mars's age. Find their three ages.

3.

A quadrilateral ABCD has sides AB, BC, CD, DA of length x, y, z and t , respectively. The diagonals AC and BD cross at right angles. Prove that x2 + z2 = y2 + t 2.

4.

The first two terms of a sequence are the numbers 1, 2. From then on, each term is obtained by dividing the previous term by the term before that. Thus the third term is obtained by dividing the second term, 2, by the first term, 1. (a) Write down the first five terms. (b) Calculate the fiftieth term. (c) What happens if other non-zero numbers are chosen for the first two terms, but the rule for calculating the next term remains the same?

5.

Four football teams—the Apes, the Baboons, the Chimps and the Gorillas—play each other once in a season. After some of the matches have been played the table of results, with some entries missing, looks like this: Team

Played

A

Won 0

B C G

Lost

Drawn

Goals for

Goals against

0

2

3

0 2

1 4

0

1

5

Complete the table, explaining how each entry is worked out, and find the score in each match played so far.

6.

How many different solutions are there to this word sum, where each letter stands for a different non-zero digit?

M A T H

S

+ M A T H

S

C

Y

A

Y L E

The United Kingdom Mathematics Trust

Intermediate Mathematical Olympiad and Kangaroo (IMOK) Olympiad Cayley Paper Thursday 17th March 2005 All candidates must be in School Year 9 or below (England and Wales), S2 or below (Scotland), or School Year 10 or below (Northern Ireland).

READ THESE INSTRUCTIONS CAREFULLY BEFORE STARTING 1.

Time allowed: 2 hours.

2.

The use of calculators, protractors and squared paper is forbidden. Rulers and compasses may be used.

3.

Solutions must be written neatly on A4 paper. Sheets must be STAPLED together in the top left corner with the Cover Sheet on top.

4.

Start each question on a fresh A4 sheet. You may wish to work in rough first, then set out your final solution with clear explanations and proofs. Do not hand in rough work.

5.

Answers must be FULLY SIMPLIFIED, and EXACT using symbols like π, fractions, or square roots if appropriate, but NOT decimal approximations.

6.

Give full written solutions, including mathematical reasons as to why your method is correct. Just stating an answer, even a correct one, will earn you very few marks; also, incomplete or poorly presented solutions will not receive full marks.

7.

These problems are meant to be challenging! The earlier questions tend to be easier; the last two questions are the most demanding. Do not hurry, but spend time working carefully on one question before attempting another. Try to finish whole questions even if you cannot do many: you will have done well if you hand in full solutions to two or more questions.

DO NOT OPEN THE PAPER UNTIL INSTRUCTED BY THE INVIGILATOR TO DO SO! The United Kingdom Mathematics Trust is a Registered Charity. Enquiries should be sent to: Maths Challenges Office, School of Mathematics, University of Leeds, Leeds, LS2 9JT. (Tel. 0113 343 2339)

http://www.ukmt.org.uk

1.

The edge length, in centimetres, of a solid wooden cube is a whole number greater than two. The outside of the cube is painted blue and the cube is then cut into small cubes whose edge length is 1 cm. The number of small cubes with exactly one blue face is ten times the number of small cubes with exactly two blue faces. Find the edge length of the original cube.

2.

When two congruent isosceles triangles are joined to form a parallelogram, as shown in the first diagram, the perimeter of the parallelogram is 3 cm longer than the perimeter of one of the triangles. When the same two triangles are joined to form a rhombus, as shown in the second diagram, the perimeter of the rhombus is 7 cm longer than the perimeter of one of the triangles. What is the perimeter of one of the triangles?

3.

In triangle ABC, angle B is a right angle and X is the point on BC so that BX : XC = 5 : 4. Also, the length of AB is three times the length of CX and the area of triangle CXA is 54 cm2. Calculate the length of the perimeter of triangle CXA.

C X B

A

4.

The five-digit number ‘a679b’, where a and b are digits, is divisible by 36. Find all possible such five-digit numbers.

5.

In the diagram, O is the centre of the circle and the straight lines AOBP and RQP meet at P. The length of PQ is equal to the radius of the circle. Prove that ∠AOR = 3 × ∠BOQ.

6.

R Q A

O

B

P

If you have an endless supply of 3 × 2 rectangular tiles, you can place 100 tiles end to end to tile a 300 × 2 rectangle. Similarly, you can put k tiles side by side to tile a 3k × 2 rectangle. Find the values of the integers k and m for which it is possible to tile a 6k × m rectangle with 3 × 2 tiles.

The United Kingdom Mathematics Trust

Intermediate Mathematical Olympiad and Kangaroo (IMOK) Olympiad Cayley Paper Thursday 16th March 2006 All candidates must be in School Year 9 or below (England and Wales), S2 or below (Scotland), or School Year 10 or below (Northern Ireland).

READ THESE INSTRUCTIONS CAREFULLY BEFORE STARTING 1.

Time allowed: 2 hours.

2.

The use of calculators, protractors and squared paper is forbidden. Rulers and compasses may be used.

3.

Solutions must be written neatly on A4 paper. Sheets must be STAPLED together in the top left corner with the Cover Sheet on top.

4.

Start each question on a fresh A4 sheet. You may wish to work in rough first, then set out your final solution with clear explanations and proofs. Do not hand in rough work.

5.

Answers must be FULLY SIMPLIFIED, and EXACT using symbols like π, fractions, or square roots if appropriate, but NOT decimal approximations.

6.

Give full written solutions, including mathematical reasons as to why your method is correct. Just stating an answer, even a correct one, will earn you very few marks; also, incomplete or poorly presented solutions will not receive full marks.

7.

These problems are meant to be challenging! The earlier questions tend to be easier; the last two questions are the most demanding. Do not hurry, but spend time working carefully on one question before attempting another. Try to finish whole questions even if you cannot do many: you will have done well if you hand in full solutions to two or more questions.

DO NOT OPEN THE PAPER UNTIL INSTRUCTED BY THE INVIGILATOR TO DO SO! The United Kingdom Mathematics Trust is a Registered Charity. Enquiries should be sent to: Maths Challenges Office, School of Mathematics, University of Leeds, Leeds, LS2 9JT. (Tel. 0113 343 2339)

http://www.ukmt.org.uk

1.

A rectangular piece of paper is cut into two pieces by a straight line passing through one corner, as shown. Given that area X : area Y = 2 : 7, what is the value of the ratio a : b?

a

X Y

2.

Show that there are no solutions to this “letter sum”. [Each letter stands for one of the digits 0–9; different letters stand for different digits; no number begins with the digit 0.]

3.

In the diagram, rectangles ABCD and AZYX are congruent, and angle ADB = 70°. Find angle BMX.

b

SEVEN + ONE EIGHT X

A

70°

M

D

Z

B

C

Y

4.

Find the positive integer whose value is increased by 518059 when the digit 5 is placed at each end of the number.

5.

Mij the magician has a large bag of red balls and a large bag of green balls. Mij wanders round the audience selecting volunteers, asking each volunteer to remove two balls, one from each bag, until 25 of the red balls and 37 of the green balls have been removed. The balls remaining in the bags are then emptied into a bucket. What fraction of all the balls does the bucket contain?

6.

A mathematician has a full one-litre bottle of concentrated orange squash, a large container and a tap. He first pours half of the bottle of orange squash into the container. Then he fills the bottle from the tap, shakes well, and pours half of the resulting mixture into the container. He then repeats this step over and over again: filling the bottle from the tap each time, shaking the mixture well, and then pouring half of the contents into the container. Suppose that on the final occasion he fills the bottle from the tap and empties it completely into the container. How many times has he filled the bottle from the tap if the final mixture consists of 10% orange squash concentrate?

The United Kingdom Mathematics Trust

Intermediate Mathematical Olympiad and Kangaroo (IMOK) Olympiad Cayley Paper Thursday 15th March 2007 All candidates must be in School Year 9 or below (England and Wales), S2 or below (Scotland), or School Year 10 or below (Northern Ireland).

READ THESE INSTRUCTIONS CAREFULLY BEFORE STARTING 1.

Time allowed: 2 hours.

2.

The use of calculators, protractors and squared paper is forbidden. Rulers and compasses may be used.

3.

Solutions must be written neatly on A4 paper. Sheets must be STAPLED together in the top left corner with the Cover Sheet on top.

4.

Start each question on a fresh A4 sheet. You may wish to work in rough first, then set out your final solution with clear explanations and proofs. Do not hand in rough work.

5.

Answers must be FULLY SIMPLIFIED, and EXACT using symbols like π, fractions, or square roots if appropriate, but NOT decimal approximations.

6.

Give full written solutions, including mathematical reasons as to why your method is correct. Just stating an answer, even a correct one, will earn you very few marks; also, incomplete or poorly presented solutions will not receive full marks.

7.

These problems are meant to be challenging! The earlier questions tend to be easier; the last two questions are the most demanding. Do not hurry, but spend time working carefully on one question before attempting another. Try to finish whole questions even if you cannot do many: you will have done well if you hand in full solutions to two or more questions.

DO NOT OPEN THE PAPER UNTIL INSTRUCTED BY THE INVIGILATOR TO DO SO! The United Kingdom Mathematics Trust is a Registered Charity. Enquiries should be sent to: Maths Challenges Office, School of Mathematics, University of Leeds, Leeds, LS2 9JT. (Tel. 0113 343 2339)

http://www.ukmt.org.uk

1.

Four copies of the polygon shown are fitted together (without gaps or overlaps) to form a rectangle. How many different rectangles are possible?

2.

Before the last of a series of tests, Sam calculated that a mark of 17 would enable her to average 80 over the series, but that a mark of 92 would raise her average mark over the series to 85. How many tests were there in the series?

3.

The diagram shows a square ABCD of side 10 units. Line segments AP, AQ, AR and AS divide the square into five regions of equal area, as shown. Calculate the length of QR.

B

A

S R D

P

Q

C

4.

How many right-angled triangles can be made by joining three vertices of a cube?

5.

In a quadrilateral ABCD, AB = BC, ∠BAC = 60°, ∠CAD = 40°, AC and BD cross at X and ∠BXC = 100°. Calculate ∠BDC.

6.

(a) You are told that one of the integers in a list of distinct positive integers is 97 and that their average value is 47. If the sum of all the integers in the list is 329, what is the largest possible value for a number in the list? (b) Suppose the sum of all the numbers in the list can take any value. What would the largest possible number in the list be then?

The United Kingdom Mathematics Trust

Intermediate Mathematical Olympiad and Kangaroo (IMOK) Olympiad Cayley Paper Thursday 13th March 2008 All candidates must be in School Year 9 or below (England and Wales), S2 or below (Scotland), or School Year 10 or below (Northern Ireland).

READ THESE INSTRUCTIONS CAREFULLY BEFORE STARTING 1.

Time allowed: 2 hours.

2.

The use of calculators, protractors and squared paper is forbidden. Rulers and compasses may be used.

3.

Solutions must be written neatly on A4 paper. Sheets must be STAPLED together in the top left corner with the Cover Sheet on top.

4.

Start each question on a fresh A4 sheet. You may wish to work in rough first, then set out your final solution with clear explanations and proofs. Do not hand in rough work.

5.

Answers must be FULLY SIMPLIFIED, and EXACT using symbols like π, fractions, or square roots if appropriate, but NOT decimal approximations.

6.

Give full written solutions, including mathematical reasons as to why your method is correct. Just stating an answer, even a correct one, will earn you very few marks; also, incomplete or poorly presented solutions will not receive full marks.

7.

These problems are meant to be challenging! The earlier questions tend to be easier; the last two questions are the most demanding. Do not hurry, but spend time working carefully on one question before attempting another. Try to finish whole questions even if you cannot do many: you will have done well if you hand in full solutions to two or more questions.

DO NOT OPEN THE PAPER UNTIL INSTRUCTED BY THE INVIGILATOR TO DO SO! The United Kingdom Mathematics Trust is a Registered Charity. Enquiries should be sent to: Maths Challenges Office, School of Mathematics, University of Leeds, Leeds, LS2 9JT. (Tel. 0113 343 2339)

http://www.ukmt.org.uk

1.

How many four-digit multiples of 9 consist of four different odd digits?

2.

A hexagon is made by cutting a small equilateral triangle from each corner of a larger equilateral triangle. The sides of the smaller triangles have lengths 1, 2 and 3 units. The lengths of the perimeters of the hexagon and the original triangle are in the ratio 5 : 7. What fraction of the area of the original triangle remains?

3.

In the rectangle ABCD the midpoint of AB is M and AB : AD = 2 : 1. The point X is such that triangle MDX is equilateral, with X and A lying on opposite sides of the line MD. Find the value of ∠XCD.

4.

The number N is the product of the first 99 positive integers. The number M is the product of the first 99 positive integers after each has been reversed. That is, for example, the reverse of 8 is 8; of 17 is 71; and of 20 is 02. Find the exact value of N ÷ M.

5.

A kite has sides AB and AD of length 25 cm and sides CB and CD of length 39 cm. The perpendicular distance from B to AD is 24 cm. The perpendicular distance from B to CD is h cm. Find the value of h.

25

cm A

m

C

h cm

m 39 c

c 24

6.

B

39 c m

25

cm

D

A regular tetrahedron ABCD has edges of length 2 units. The midpoint of the edge AB is M and the midpoint of the edge CD is N . Find the exact length of the segment MN .

The United Kingdom Mathematics Trust

Intermediate Mathematical Olympiad and Kangaroo (IMOK) Olympiad Cayley Paper Thursday 19th March 2009 All candidates must be in School Year 9 or below (England and Wales), S2 or below (Scotland), or School Year 10 or below (Northern Ireland).

READ THESE INSTRUCTIONS CAREFULLY BEFORE STARTING 1.

Time allowed: 2 hours.

2.

The use of calculators, protractors and squared paper is forbidden. Rulers and compasses may be used.

3.

Solutions must be written neatly on A4 paper. Sheets must be STAPLED together in the top left corner with the Cover Sheet on top.

4.

Start each question on a fresh A4 sheet. You may wish to work in rough first, then set out your final solution with clear explanations and proofs. Do not hand in rough work.

5.

Answers must be FULLY SIMPLIFIED, and EXACT. They may contain symbols such as π, fractions, or square roots, if appropriate, but NOT decimal approximations.

6.

Give full written solutions, including mathematical reasons as to why your method is correct. Just stating an answer, even a correct one, will earn you very few marks; also, incomplete or poorly presented solutions will not receive full marks.

7.

These problems are meant to be challenging! The earlier questions tend to be easier; the last two questions are the most demanding. Do not hurry, but spend time working carefully on one question before attempting another. Try to finish whole questions even if you cannot do many: you will have done well if you hand in full solutions to two or more questions.

DO NOT OPEN THE PAPER UNTIL INSTRUCTED BY THE INVIGILATOR TO DO SO! The United Kingdom Mathematics Trust is a Registered Charity. Enquiries should be sent to: Maths Challenges Office, School of Mathematics, University of Leeds, Leeds, LS2 9JT. (Tel. 0113 343 2339)

http://www.ukmt.org.uk

1.

An aquarium contains 280 tropical fish of various kinds. If 60 more clownfish were added to the aquarium, the proportion of clownfish would be doubled. How many clownfish are in the aquarium?

2.

The boundary of the shaded figure consists of four semicircular arcs whose radii are all different. The centre of each arc lies on the line AB, which is 10 cm long. What is the length of the perimeter of the figure?

3.

B

A

Two different rectangles are placed together, edge-to-edge, to form a large rectangle. The length of the perimeter of the large rectangle is 23 of the total perimeter of the original two rectangles. Prove that the final rectangle is in fact a square.

4.

In the rectangle ABCD, the side AB has length 2 and the side AD has length 1. Let the circle with centre B and passing through C meet AB at X. Find ∠ADX (in degrees).

5.

Two candles are the same height. The first takes 10 hours to burn completely whilst the second takes 8 hours to burn completely. Both candles are lit at midday. At what time is the height of the first candle twice the height of the second candle?

6.

Teams A, B, C and D competed against each other once. The results table was as follows: Team

Win

Draw

Loss

Goals for

Goals against

A

3

0

0

5

1

B

1

1

1

2

2

C

0

2

1

5

6

D

0

1

2

3

6

(a) Find (with proof) which team won in each of the six matches. (b) Find (with proof) the scores in each of the six matches.

60

Solutions to the Olympiad Cayley Paper Section A

Al Numbering the trees 1 to 17, Basil marks 1, 3, 5, 7, 9, 11, 13, 15 and 17 on his way out, and 17, 14, 11, 8, 5 and 2 on his way back. So the five trees 4, 6, 10, 12 and 16 have no mark.

A2 The diagram to the right shows that you can cut seven squares. You cannot cut more because once the straight line has entered the grid, and so cut into one small square, it can enter other squares only by crossing an inside dividing line, either horizontally or vertically. There are only six of these, so it can cut only seven squares in all.

A3 The total value of the 5 parrots was 5 x €6000 = €30 OOO. After one has flown away, the total value is 4 x €5000 = €20 OOO. So the value of the parrot that escaped was €10000. A4 If the original number is 19, then crossing out 9 leaves 1, and 1 x 19 = 19. If 'ab' (meaning 1Oa + b) is the original two-digit number, then crossing out b leaves a. If the original number was more than 19 times larger than a, then 1Oa + b > 19a leading to b > 9a. This is impossible because b is a single-digit number and a is not zero. The greatest possible value of x is 19. AS The factors of 42 are 2, 3 and 7. If the perimeter was 18 then the base was 2 by 7. Thus the height was 3 cm.

Section B

Bl The number 'sss' equals s

x 111 = s x 3 x 37. Now 37 is prime, so one of the two numbers 'pq' and 'rq' is 37 or 74. The case 74 is not possible, since then q = 4, giving s = 6 and so 'sss' equals 9 x 74. The case 37 gives q = 7, s = 9 and 'sss' = 27 x 37, so that p, r = 2, 3, in either order.

B2 Let LAQP = x, so that LAQR = x since A AQ bisects LPQ R. Then LKRP = 2x + 80°, since LKRP is an exterior angle of triangle PQR and so equals the sum of the opposite interior angles. But AR bisects LKRP, so that R LARK = x + 40°. K Q However, LARK = x + LRAQ, since LARK is an exterior angle of triangle RAQ. It follows thatLRAQ = 40°.

61 B3 It is possible to complete a table showing the numbers of pupils in each category, as shown. hockey

swimming

total

126

54

180

badminton

99

21

total

225

tennis

120 300

There are 300 children, 60% of them play tennis and 40% play badminton, so 180 play tennis and 120 play badminton. Also, 30% of the tennis players swim, so there are 54 tennis players who swim. Hence there are 180 - 54 = 126 tennis players who play hockey. Now 56% of the hockey players also play tennis, so 56% of the number of hockey players= 126, therefore the number of hockey players=~ x 100 = 225. Hence the number of hockey players who play badminton is 225 - 126 = 99 and so the number of badminton players who swim is 120 - 99 = 21.

B4 Let the circles with diameters PQ and RS have radius x and the circle with diameter QR have radius y. Then the radius of the circle with diameter PS is y + 2x, so that the shaded area is

t.n (y + 2x)2 - .nx2 + ~.ny2

i.n (y2 + 4.xy + 4x 2) - .nx2 + i.nl i.nl + 2.nxy + 2.nx2 - .nx2 +

i.nJ

.ny2 + 2.nxy + .nx2 .n(x + y)2 which is the area of a circle with radius x + y. But MN= (y + 2x) + y = 2y + 2x, so that the circle with diameter MN has radius x + y, as required. [The shaded figure is known as a salinon and this result about its area appears in Archimedes, Liber Assumptorum, Proposition 14.]

BS (a) The total of the numbers 1 to 9 is 45. Let the numbers placed at the vertices of the triangle be a, b and c. The numbers along each side of the triangle add up to T, so that adding the three sides together gives 3T. This total includes all nine numbers, but with a, b and c included twice. So, 3T = 45 + a + b + c. The smallest value for a + b + c is 1 + 2 + 3 = 6 and the largest is 7 + 8 + 9 = 24. Hence 45 + 6 .;;;; 3T .;;;; 45 + 24 and so 17 .;;;; T .;;;; 23. (b) We know from above that 7, 8 and 9 are placed at the vertices. The diagram shows one possible solution. (c) Now, a + b + c = 15 and so the only possible choices for the values of a, b and c are: 9 5 1; 9 4 2; 8 6 1; 8 5 2; 8 4 3; 7 6 2; 7 5 3; 6 5 4. Some of these may lead to a solution and some may not, but we can conclude that there are at most these 8 possible choices.

® CD ® ®

25

Solutions to the Olympiad Cayley Paper 1. How many four-digit multiples of 9 consist of four different odd digits? First solution There are five odd digits: 1, 3, 5, 7 and 9. The sum of the four smallest odd digits is 16 and the sum of the four largest is 24. Hence the digit sum of any four-digit number with different odd digits lies between 16 and 24, inclusive. However, the sum of the digits of a multiple of 9 is also a multiple of 9, and the only multiple of 9 between 16 and 24 is 18. Hence the sum of the four digits is 18. Now 1 + 3 + 5 + 9 = 18, so that the four digits can be 1, 3, 5 and 9. If 7 is one of the four digits then the sum of the other three is 11, which is impossible. So 7 cannot be one of the digits and therefore the four digits can only be 1, 3, 5 and 9. The number of arrangements of these four digits is 4 × 3 × 2 × 1 = 24. Hence there are 24 four-digit multiples of 9 that consist of four different odd digits. Second solution The sum of all five odd digits is 1 + 3 + 5 + 7 + 9 = 25. Subtracting 1, 3, 5, 7 and 9 in turn we get 24, 22, 20, 18 and 16, only one of which is a multiple of 9, namely 18 = 25 − 7. Since the sum of the digits of a multiple of 9 is also a multiple of 9, it follows that the four digits can only be 1, 3, 5 and 9. The number of arrangements of these four digits is 4 × 3 × 2 × 1 = 24. Hence there are 24 four-digit multiples of 9 that consist of four different odd digits.

2. A hexagon is made by cutting a small equilateral triangle from each corner of a larger equilateral triangle. The sides of the smaller triangles have lengths 1, 2 and 3 units. The lengths of the perimeters of the hexagon and the original triangle are in the ratio 5 : 7. What fraction of the area of the original triangle remains?

First solution Let the side length of the large equilateral triangle be x units; this triangle therefore has a perimeter of length 3x units. Now consider the hexagon, which has sides of lengths 1, x − 3, 2, x − 5, 3 and x − 4 units. Hence the hexagon has perimeter length 3x − 6 units. Since the ratio of the perimeter lengths of the hexagon and the large triangle is 5 : 7, we have 3x − 6 5 = . 3x 7 Rearranging and solving for x we obtain x = 7.

(∗)

26 Now, in order to find the area of the large equilateral triangle, we determine the height h units using Pythagoras' theorem: h2 = 72 − ( 27 )

2

= 49 (1 −

1 4

)

= 49 × 34 .

3 3 = 7 × . 4 2 Therefore the area of the large equilateral triangle is Hence

h =

49 ×

1 7 3 49 3 × 7 × = . 2 2 4 We may find the areas of the three small equilateral triangles in a similar way. These areas are 3 4 3 9 3 , and . 4 4 4 The area of the hexagon is the area of the large equilateral triangle minus the areas of the three small equilateral triangles, that is,

(

)

49 3 3 4 3 9 3 35 3 = − + + . 4 4 4 4 4 Finally, the fraction of the original equilateral triangle remaining is 35 3 49 3 5 ÷ = . 4 4 7 Second solution Having established that the large triangle has sides of length 7 (equation (*) in the solution above), we may proceed as follows: The four equilateral triangles in the problem are similar. Now the ratio of the areas of similar figures is equal to the ratio of the squares of their sides. Hence the four triangles have areas in the ratio 12 : 22 : 32 : 72 = 1 : 4 : 9 : 49. Hence the ratio of the areas of the hexagon and the large triangle is 49 − (1 + 4 + 9) : 49 = 35 : 49 = 5 : 7. This may be illustrated by dividing the large triangle into 49 small triangles, as shown.

Note: The observant reader will have noticed that the answer to this problem is surprising: the ratio of the areas is the same as the ratio of the perimeters. There is no reason to expect this to happen.

27 3. In the rectangle ABCD the midpoint of AB is M and AB : AD = 2 : 1. The point X is such that triangle MDX is equilateral, with X and A lying on opposite sides of the line MD. Find the value of ∠XCD.

Solution The key to this solution is to draw MC and consider triangle MCX. We are given that ABCD is a rectangle, so that BC = AD and ∠DAM = 90° = ∠MBC. We are also given that AB = 2AD and that M is the midpoint of AB. Therefore DA = AM = MB = BC. X

C

D

A

M

B

It follows that triangles DAM and MBC are congruent (SAS) and we deduce that DM = MC. But triangle MDX is equilateral, so MX = DM and hence MX = MC . In other words, triangle MCX is isosceles. Now consider the angles at M. 1. Triangle DAM is right-angled with ∠DAM = 90°. It is also isosceles, so ∠AMD = ∠ADM = 45°, since the angle sum is 180°. 2. Similarly, from triangle MBC , ∠CMB = 45°. 3. Finally, because triangle MDX is equilateral, ∠DMX = 60°. Hence

∠CMX = 180° − 45° − 45° − 60° = 30°

since angles on a straight line add up to 180°. Lastly, we consider the angles at C. We know that triangle XCM is isosceles and that ∠CMX = 30°. Hence each base angle is 12 (180° − 30°) = 75°; in particular, ∠XCM = 75°. Also, ABCD is a rectangle, so ∠DCB = 90°, and triangle MBC is right-angled and isosceles, so ∠MCB = 45°. Therefore ∠DCM = ∠DCB − ∠MCB = 90° − 45° = 45°. We can now calculate the value of ∠XCD. We have

∠XCD = ∠XCM − ∠DCM = 75° − 45° = 30°.

28 4. The number N is the product of the first 99 positive integers. The number M is the product of the first 99 positive integers after each has been reversed. That is, for example, the reverse of 8 is 8; of 17 is 71; and of 20 is 02. Find the exact value of N ÷ M.

First solution From the given definition we have N = (1 × 2 ×… × 9) × 10 × (11 × 12 ×… × 19) × 20 ×… × 90 × (91 ×… × 99) , which rearranges to N = (1 × 2 ×… × 9) × (11 × 12 ×… × 19) ×… × (91 ×… × 99) × (10 × 20 ×… × 90) . Also M = (1 × 2 ×… × 9) × 01 × (11 × 21 ×… × 91) × 02 ×… × 09 × (19 ×… × 99) which rearranges to M = (1 × 2 ×… × 9) × (11 × 12 ×… × 19) ×… × (91 ×… × 99) × (01 × 02 ×… × 09) = (1 × 2 ×… × 9) × (11 × 12 ×… × 19) ×… × (91 ×… × 99) × (1 × 2 ×… × 9) . Comparing these arrangements for M and N , we see that M has the same terms as N except that the product 10 × 20 ×… × 90 is replaced by the product 1 × 2 ×… × 9. Thus when we divide N by M all the common terms cancel and we are left with N 10 × 20 ×… × 90 = M 1 × 2 ×… × 9 = 109. Second solution We may place the numbers from 1 to 99 into three categories, determined by how they are transformed when they are reversed: 1. single digit numbers ‘a’ are unchanged; 2. a two-digit number ‘ab’, where neither a nor b is zero, is transformed to the twodigit number ‘ba’; and 3. a multiple of 10 such as ‘a0’ is transformed to ‘0a’ = a, a single-digit number. Thus there is a correspondence between the factors in N and M, as shown in the table: N

M

‘a’ ‘aa’ ‘ab’ and ‘ba’

‘a’ ‘aa’ ‘ba’ and ‘ab’

‘a0’

‘a’

Single-digit numbers are unchanged; two-digit numbers with a repeated digit are unchanged; pairs of two-digit numbers, with different digits and neither digit zero, are unchanged as a pair; the multiples of 10 in N are replaced by single-digit numbers in M. Thus when we divide N by M all the identical factors cancel and we are left with N 10 × 20 ×… × 90 = M 1 × 2 ×… × 9 = 109.

29 5. A kite has sides AB and AD of length 25 cm and m 39 c

m

cm

C

25 c

24

sides CB and CD of length 39 cm. The perpendicular distance from B to AD is 24 cm. The perpendicular distance from B to CD is h cm. Find the value of h.

h cm

B

39 c m

25

A

cm

D

First solution As shown in the figure below, let the perpendicular from B to the line AD meet the line AD at the point X; let the perpendicular from B to the line CD meet the line CD at the point Y and let the distance DY be y cm. B 39

25 h

C

24

A X

39 − y

25

Y y D

Considering triangle ABX and using Pythagoras' Theorem we obtain AX =

252 − 242 cm

= 7 cm. Similarly, from triangle BDX we have BD =

242 + (25 − 7)2 cm

= 30 cm . Now from triangles BDY and BCY , again by Pythagoras' theorem, we deduce that

(∗)

h2 + (39 − y)2 = 392 and Subtract to get

h2 + y2 = 302.

(1)

(39 − y)2 − y2 = 392 − 302,

which simplifies to

78y = 900,

so that

y=

150 . 13

Finally, by substituting in equation (1), we find h =

360 . 13

Second solution Another solution uses the length of BD obtained in (*) above to find the area of isosceles triangle BCD. Once the area is known the value of the height h may be found from area = 12 × 39 × h. Can you see how to find the area of triangle BCD and so complete the solution? Note: Triangle DYB is a ‘5, 12, 13’ triangle.

30 6. A regular tetrahedron ABCD has edges of length 2 units. The midpoint of the edge AB is M and the midpoint of the edge CD is N . Find the exact length of the segment MN.

A M B C N D

First solution We make use of the following result. Theorem (Median of isosceles triangle): The line joining the apex to the midpoint of the base of an isosceles triangle is perpendicular to the base. That is, in the following figure, ∠PSR = 90°.

R

P

S

Q

Applying the theorem to triangle ABC, we find that ∠AMC = 90°. Similarly, in triangle ABD, ∠AMD = 90°. Now applying Pythagoras' theorem to the triangles AMC and AMD we get CM2 = AC2 − AM2 = 22 − 12 2

2

and DM2 = AD2 − AM2 = 2 − 1 . Hence CM = 3 and DM = 3, so triangle CMD is isosceles. Now apply the theorem to triangle CMD to obtain ∠CNM = 90°. Then by Pythagoras' theorem in triangle CNM MN 2 = CM2 − CN 2 = 3 − 1. Therefore MN =

2.

Second solution A tetrahedron may be formed by joining face diagonals of a cube, as shown below. Since the faces of the cube are congruent squares the face diagonals have equal length and so the tetrahedron is regular. Now M and N are midpoints of opposite edges of the tetrahedron. Therefore they are midpoints of opposite face diagonals of the cube, that is, centres of opposite faces of the cube. Hence MN = AR. Letting the sides of the cube have length a, from Pythagoras' theorem in triangle ARC we get AC2 = AR2 + RC2 2

2 = a2 + a2

so that

= 2a2. Hence a =

2 and therefore MN =

2.

31 P A

M B

Q

D

R N S

C