UKMT

Senior Mathematical Challenge

Thursday 7 November 2013
Organised by the United Kingdom Mathematics Trust
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Solutions and investigations

These solutions augment the printed solutions that we send to schools. For convenience, the solutions sent to schools are conned to two sides of A4 paper and therefore in many cases are rather short. The solutions given here have been extended. In some cases we give alternative solutions, and we have included some exercises for further investigation. The Senior Mathematical Challenge (SMC) is a multiple-choice paper. For each question, you are presented with ve options, of which just one is correct. It follows that often you can nd the correct answers by working backwards from the given alternatives, or by showing that four of them are not correct. This can be a sensible thing to do in the context of the SMC. However, this does not provide a full mathematical explanation that would be acceptable if you were just given the question without any alternative answers. So for each question we have included a complete solution which does not use the fact that one of the given alternatives is correct.Thus we have aimed to give full solutions with all steps explained. We therefore hope that these solutions can be used as a model for the type of written solution that is expected when presenting a complete solution to a mathematical problem (for example, in the British Mathematical Olympiad and similar competitions).
These solutions may be used freely within your school or college. You may, without futher permission, post these solutions on a website that is accessible only to sta and students of the school or college, print out and distribute copies within the school or college, and use them in the classroom. If you wish to use them in any other way, please consult us. UKMT November 2013

Enquiries about the Senior Mathematical Challenge should be sent to: UKMT, School of Mathematics Satellite, University of Leeds, Leeds LS2 9JT 0113 343 2339
1 2 3 4 5 6 7 8

enquiry@ukmt.org.uk

UK

A C D C E E B B A B A B D D B D A C C E B C A E D

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

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www.ukmt.org.uk

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Senior Mathematical Challenge 2013

Solutions and investigations

1. Which of these is the largest number? A 2+0+1+3 E 2013 Solution A We calculate the value of each of the given options in turn. (a) 2 + 0 + 1 + 3 = 6 (b) 2 0 + 1 + 3 = 0 + 1 + 3 = 4 (c) 2 + 0 1 + 3 = 2 + 0 + 3 = 5 (d) 2 + 0 + 1 3 = 2 + 0 + 3 = 5 (e) 2 0 1 3 = 0 So option A gives the largest number. Remarks You may have obtained the wrong answer if you interpreted 2 + 0 + 1 3 to mean ((2 + 0) + 1) 3 rather than 2 + 0 + (1 3) . It is a standard convention (sometimes known as BIDMAS or BODMAS) that in evaluating an expression such as 2 + 0 + 1 3, the multiplications are carried out before the additions. We do not just carry out the operations from left to right. So in calculating 2 + 0 + 1 3, the multiplication 1 3 is done before the additions. A decent calculator will produce the correct answer 5 if you press the keys
2 + 0 + 1 3 =

B 2 0+1+3

C 2+0 1+3

D 2+0+1 3

in this order. If your calculator produces a dierent answer, you should replace it!

2. Little John claims he is 2 m 8 cm and 3 mm tall. What is this height in metres? A 2.83 m Solution C One metre is 100 centimetres. So 1 cm = 0.01 m and 8 cm = 0.08 m. Similarly, one metre is 1000 millimetres. So 1 mm = 0.001 m and 3 mm = 0.003 m. Therefore Little Johns height is 2 m + 0.08 m + 0.003 m = 2.083 m. B 2.803 m C 2.083 m D 2.0803 m E 2.0083 m

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Solutions and investigations

3. What is the tens digit of 20132 2013? A 0 Solution D The tens digit of 20132 2013 is the same as that of 132 13. Since 132 13 = 169 13 = 156, the tens digit of 20132 2013 is 5. Remarks Our comment that the tens digit of 20132 2013 is the same as that of 132 13 uses the fact that 20132 2013 = (2000 + 13) 2 2013 = (20002 + 2 2000 13 + 132 ) (2000 + 13) = 4000000 + 52000 2000 + 132 13. It is now clear that only the last two terms, that is, 132 13, can have any eect on the tens and units digits of the answer. A more sophisticated way to say this is to use the language and notation of modular arithmetic, which you may already have met. Using this notation we write a b (mod n ) to mean that a and b have the same remainder when divided by n. For example, 2013 13 (mod 100) and 156 56 (mod 100) . Then we can say that 20132 2013 132 13 (mod 100) and 132 13 56 (mod 100) . It follows that 20132 2013 56 (mod 100) . Thus 20132 2013 has remainder 56 when divided by 100. So its last two digits are 5 and 6. In particular its tens digit is 5 and its units digit is 6. For investigation 3.1 Find the tens digits of (a) 20142 2014 and (b) 20133 20132 . 3.2 Find the tens and units digits of (a) 20112011 and (b) 20132013 . B 1 C 4 D 5 E 6

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Solutions and investigations

4. A route on the 3 3 board shown consists of a number of steps. Each step is from one square to an adjacent square of a dierent colour. How many dierent routes are there from square S to square T which pass through every other square exactly once? A 0 Solution B 1 C 2 D 3 E 4 S

C For convenience, label the other squares as in the left-hand gure. The rst move of a route from S to T must be either SM or SQ. It is easy to see that a route that visits all the squares must include both the sequence MKL and the sequence QRP. Hence, we see that there are just two routes that meet all the required conditions; these are shown in two gures on the the right. K L T M N P S Q R For investigation 4.1 Consider the analogous problem for a 4 4 board. How many dierent routes are there from square S to square T which pass through every other square exactly once? S 4.2 Now consider the analogous problem for a 5 5 board. 4.3 Now consider the general case of an n n board. note By considering the problem with a 4 4 board, you should see that the case where n is even is not dicult. However, the case where n is odd is seemingly much more dicult, and we dont know a general formula for this case. Please let us know if you manage to make any progress with this. T

5. The numbers x and y satisfy the equations x ( y + 2) = 100 and y ( x + 2) = 60. What is the value of x y ? A 60 Solution E The two equations expand to give xy + 2 x = 100 and xy + 2 y = 60. It follows that ( x y + 2 x ) ( xy + 2 y ) = 100 60. That is, 2 x 2 y = 40. Hence 2 ( x y ) = 40 and so x y = 20.
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B 50

C 40

D 30

E 20

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Solutions and investigations

For investigation 5.1 The wording of the question implies that there are numbers x and y which satisfy the equations x ( y + 2) = 100 and y ( x + 2) = 60. Check that this is correct by nding real numbers x and y which satisfy both the equations x ( y + 2) = 100 and y ( x + 2) = 60. 5.2 Show that there are no real numbers x and y such that x ( y +2) = 100 and y ( x +2) = 80. note There are complex number solutions of these equations, and, if x and y are complex numbers which satisfy these equations, then x y = 10. [If you dont know what complex numbers are, ask your teacher.]

6. Rebecca went swimming yesterday. After a while she had covered one fth of her intended distance. After swimming six more lengths of the pool, she had covered one quarter of her intended distance. How many lengths of the pool did she intend to complete? A 40 Solution
1 1 1 E We have 1 4 5 = 20 . So the six additional lengths make up 20 of Rebeccas intended distance. So the number of lengths she intended to complete was 20 6 = 120.

B 72

C 80

D 100

E 120

7. In a ninety nine shop all items cost a number of pounds and 99 pence. Susanna spent 65.76. How many items did she buy? A 23 Solution B Let k be the number of items that Susanna bought. The cost of these is a whole number of pounds and 99 k pence, that is, a whole number of pounds less k pence. Susanna spent 65.76, that is, a whole number of pounds less 24 pence. It follows that k pence is a whole number of pounds plus 24 pence. So k is 24 or 124 or 224 or . . . . . However, since each item costs at least 99 pence and Susanna spent 65.76 pence, she bought at most 66 items. So k is 24. For investigation 7.1 Is it possible to spend 20.76 in a ninety nine shop? 7.2 For which non-negative integers m and n with n < 100 is it possible to spend m pounds and n pence in a ninety-nine shop? B 24 C 65 D 66 E 76

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Solutions and investigations

8. The right-angled triangle shown has a base which is 4 times its height. Four such triangles are placed so that their hypotenuses form the boundary of a large square as shown. What is the side length of the shaded square in the diagram? C 3x D 2 3x A 2x B 2 2x E 15 x

x 4x

Solution B The side length of the large square is 4 x and hence the area of this square is 16 x 2 . Each 2 triangle has base 4 x and height x and hence has area 1 2 (4 x x ) = 2 x . So the total area of 2 2 2 these four triangles is 8 x 2 . Therefore the area of the shaded is 16 x 8 x = 8 x . square Therefore the side length of the shaded square is 8 x 2 = 8 x = 2 2 x .

9. According to a headline Glaciers in the French Alps have lost a quarter of their area in the past 40 years. What is the approximate percentage reduction in the length of the side of a square when it loses one quarter of its area, thereby becoming a smaller square? A 13% Solution A Suppose that a square of side length 1, and hence area 1, has side length x when it loses one 3 3 2 quarter of its area. Then x = 4 and so x = 2 . Now 1.72 = 2.89 and so 1.7 < 3. Hence

B 25%

C 38%

D 50%

E 65%

0.85 < 23 . So the length of the side of the smaller square is at least 85% of its original value. Therefore the reduction in its length is less than 15%. So, of the given options, it must be that 13% is correct.

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Solutions and investigations

For investigation 9.1 The above solution is good enough in the context of the SMC. However, if you couldnt assume that one of the options must be correct, then, to show that the approximate percentage reduction is 13%, you would need to show that the percentage reduction lies 3 between 12.5% and 13.5%. That is, you would need to show that 0.865 < 2 < 0.875, or, equivalently, that 1.73 < 3 < 1.75.
2 2 You could do this by doing a calculation to show that 1.73 < 3 < 1.75 . Alternatively, you could calculate 3 to the appropriate number of decimal places. This is easy with a calculator, but what could you do if you dont have a calculator? There is an old fashioned method for calculating square roots by hand which resembles long division. This was taught in schools until about fty years ago. See if you can nd out what this method is, perhaps using the internet. Then use this method to calculate 3 to three decimal places.

9.2 The sequence dened by x 1 = 1, x n +1 =

gives better and better approximations to 3. (We say that it converges to 3). The rst 1 +3 2+3 5 three terms are x 1 = 1, x 2 = 1 +1 = 2, and x 3 = 2+1 = 3 . (b) Show that if the sequence converges, then it converges to 3. (c) Find a similar rule for a sequence that converges to 5. 9.3 The Generalized Binomial Theorem, discovered by Isaac Newton, tells us that if | x | < 1, then ( 1) 2 ( 1)( 2) 3 (1 + x ) = 1 + x + x + x + 2! 3! Use the rst three terms of this series with x = 1 4 and = to
3 2 . 1 2

xn + 3 xn + 1

(a) Find the values of x 4 and x 5 .

to obtain an approximation

9.4 What method does your calculator use to evaluate square roots?

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Solutions and investigations

10. Franks teacher asks him to write down ve integers such that the median is one more than the mean, and the mode is one greater than the median. Frank is also told that the median is 10. What is the smallest possible integer that he could include in his list? A 3 Solution B The median of the ve numbers is 10, hence the mean is 9 and the mode is 11. For the mode to be 11, Franks list must include more 11s than any other integer. It cannot include three or more 11s, since then the median would be 11. So two of the integers are 11, and the other three integers are all dierent. For the median to be 10, 10 must be one of the integers, and the list must include two integers that are less than 10 and two that are greater than 10. Hence the ve integers are a, b, 10, 11, 11 where a, b are two dierent integers both less than 10. Say a < b. Since the mean is 9, we have a + b + 10 + 11 + 11 = 5 9 = 45 and therefore a + b = 13. The largest value that b can take is 9, and hence the smallest possible value for a is 4. B 4 C 5 D 6 E 7

11. The diagram shows a circle with centre O and a triangle OPQ. Side PQ is a tangent to the circle. The area of the circle is equal to the area of the triangle. P What is the ratio of the length of PQ to the circumference of the circle? A 1:1 Solution B 2:3 C 2: D 3:2

O Q E :2

A Suppose that the circle has radius r and that PQ has length x . The height of the triangle OPQ is the length of the perpendicular from O to PQ. Since PQ is a tangent to the circle, this perpendicular is a radius of the circle and so has length r . Therefore the area of triangle 1 2 2 OPQ is 1 2 xr . The area of the circle is r . Since these areas are equal 2 xr = r , and hence x = 2r . So the length of PQ is the same as the circumference of the circle. So the ratio of their lengths is 1 : 1.

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Solutions and investigations

12. As a special treat, Sammy is allowed to eat ve sweets from his very large jar which contains many sweets of each of three avours Lemon, Orange and Strawberry. He wants to eat his ve sweets in such a way that no two consecutive sweets have the same avour. In how many ways can he do this? A 32 Solution B Sammy has a choice of 3 avours for the rst sweet that he eats. Each of the other sweets he eats cannot be the same avour as the sweet he has just eaten. So he has a choice of 2 avours for each of these four sweets. So the total number of ways that he can make his choices is 3 2 2 2 2 = 48. For investigation 12.1 What would the answer be if Sammy had sweets with four dierent avours in his jar? 12.2 Find a formula for the number of ways if Sammy is allowed to eat k sweets and has sweets of n dierent avours in his jar. B 48 C 72 D 108 E 162

13. Two entrants in a schools sponsored run adopt dierent tactics. Angus walks for half the time and runs for the other half, whilst Bruce walks for half the distance and runs for the other half. Both competitors walk at 3 mph and run at 6 mph. Angus takes 40 minutes to complete the course. How many minutes does Bruce take? A 30 Solution D Angus walks for 20 minutes at 3 mph and runs for 20 minutes at 6 mph. 20 minutes is 1 one-third of an hour. So the number of miles that Angus covers is 3 1 3 + 6 3 = 6. Bruce covers the same distance. So Bruce walks 1 2 3 miles at 3 mph which takes him 30 minutes and runs the same distance at 6 mph which takes him 15 minutes. So altogether it takes Bruce 45 minutes to nish the course. B 35 C 40 D 45 E 50

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Solutions and investigations

14. The diagram shows a rectangle PQRS in which PQ : QR = 1 : 2. The point T on PR is such that ST is perpendicular to PR. What is the ratio of the area of the triangle RST to the area of the rectangle PQRS ? B 1:6 C 1:8 D 1 : 10 A 1:4 2 E 1 : 12

T R S

Solution D We can suppose that we have chosen units so that the length of RS is of PS 1. So the length is 2. It follows from Pythagoras theorem that the length of PR is 12 + 22 , that is, 5. The triangles RST and PRS are both right-angled and the angle at R is common to both triangles. Therefore the triangles are similar. Hence the ratio of their areas is the ratio of the squares of the lengths of corresponding sides. The lengths of their hypotenuses are in the ratio 1 : 5. Hence, the ratio of the area of triangle RST to the area of triangle PRS is 2 12 : 5 , that is, 1 : 5. The area of triangle PRS is half the area of the rectangle. Hence the ratio of the area of triangle RST to the area of the rectangle is 1 : 10. For investigation 14.1 The solution shows that the ratio of the area of triangle RST to the area of the rectangle PQRS is 1 : 10. Is it possible to prove this by dissecting the rectangle PQRS into 10 triangles each congruent to the triangle RST ?

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Solutions and investigations

15. For how many positive integers n is 4n 1 a prime number? A 0 E innitely many Solution B We give two methods for answering this question. Both methods rely on the following property of prime numbers: a positive integer p is a prime number if, and only if, the only way p can be expressed as a product of two positive integers, m and n, is where exactly one of m and n is 1.
Method 1

B 1

C 2

D 3

The rst method uses the fact that when n is a positive integer 4n 1 is the dierence of two squares and so may be factorized as 4n 1 = (2n ) 2 12 = (2n + 1)(2n 1) . So 4n 1 is not a prime, unless one of the factors 2n + 1 and 2n 1 is equal to 1. Now 2n + 1 cannot be equal to 1 and 2n 1 = 1 if, and only if, n = 1. When n = 1, we have 4n 1 = 3 and so 4n 1 is prime. So there is just one positive integer n, namely 1, for which 4n 1 is a prime number.
Method 2

The second method uses the fact when n is a positive integer x 1 is a factor of x n 1, since, for n 2, x n 1 = ( x 1)( x n1 + x n2 + + x + 1) . Hence, putting x = 4, we deduce that, for each positive integer n, 3 is a factor of 4n 1. Therefore 4n 1 is not prime except when n = 1 and 4n 1 = 3. So there is just one positive integer n for which 4n 1 is a prime number. For investigation 15.1 For how many positive integers n is 5n 1 a prime number? 15.2 For how many positive integers n is 6n 1 a prime number? 15.3 Prove the identity used in method 2, that is, that for all integers n 2, x n 1 = ( x 1)( x n1 + x n2 + + x + 1) .

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16. Andrew states that every composite number of the form 8n + 3, where n is an integer, has a prime factor of the same form. Which of these numbers is an example showing that Andrews statement is false? A 19 Solution D To show that Andrews statement is wrong we need to nd a composite number of the form 8n + 3 which does not have any prime factors of this form. This rules out 19 which is not composite, and both 33 and 85 which are not of the form 8n + 3. The number 99 is of the form 8n + 3, but it has two prime factors 3 and 11 which are also of this form. So 99 wont do either. This leaves 91. We see that 91 = 8 11 + 3, and so it is of the form 8n + 3, but 91 = 7 13 and neither of its prime factors, 7 and 13, is of this form. So the number 91 shows that Andrews statement is false. B 33 C 85 D 91 E 99

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17. The equilateral triangle PQR has side length 1. The lines PT and PU trisect the angle RPQ, the lines RS and RT trisect the angle QRP and the lines QS and QU trisect the angle PQR. What is the side length of the equilateral triangle STU ? cos 80 A B 1 C cos2 20 3 cos 20 cos 20 D Solution A Since the triangle PQR is equilateral, all its three angles are 60. So the trisectors divide these into three 20 angles. In particular QPU = 20 and U PT = 20. Because of the symmetry of the gure, QU = PU = PT . So the triangles PQU and PUT are isosceles. In these triangles, we let X be the foot of the perpendicular from U to PQ, and Y be the foot of perpendicular from P to UT . In the right-angled triangle PXU , X PU = 20, and so PX PU = cos 20 . Since the length of PX is 1 1 2 it follows that PU has length 2 cos 20. In the right-angled triangle PYU , U PY = 10 and therefore YU P = 80. Therefore, from triangle PYU , cos 80 =
UY PU 1 6

U S Q

E cos 20 cos 80

U Y S

and hence UY has length PU cos 80, that is,

1 cos 80 . 2 cos 20 Since the length of UY is half the length of UT , it follows that the length of UT is cos 80 . cos 20

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Solutions and investigations

Remarks In this problem, it is clear that, because the triangle PQR is equilateral, so also is the triangle STU . It is a remarkable fact, discovered by Frank Morley (18601937), that: In every triangle PQR, the triangle STU formed by the points where the trisectors of the angles of PQR meet is an equilateral triangle. P The triangle STU is called the Morley triangle of the triangle PQR. No very easy proofs of this theorem are known. A proof may be found on pages 186187 of The Geometry of the Triangle by Gerry Leversha, UKMT, 2013. This book may be ordered from the UKMT website. The virtue of the proof in this book is that it uses purely geometrical ideas. Other proofs use trigonometry and complex numbers. For investigation 17.1 In the solution above we have assumed that X and Y are the midpoints of PQ and UT , respectively, and that U PY = T PY . These facts both follow from the fact that if a triangle ABC is isosceles with AB = AC , and N is the foot of the perpendicular from A to BC , then the triangles ABN and AC N are congruent. Can you prove this? Note that it then follows that BN = C N and B AN = C AN . So we can deduce that X , Y are the midpoints of PQ and UT , C B N and that U PY = T PY , as required. 17.2 An alternative method is to apply the Sine Rule to the triangles PQU and PUT . This gives PU PQ UT PU = and = . sin 20 sin 140 sin 20 sin 80 Hence, since PQ has length 1, it follows that sin 20 sin 20 sin 20 sin2 20 UT = PU = = . sin 80 sin 140 sin 80 sin 140 sin 80 Show, without using a calculator, that this is the same as the previous answer cos 80 . cos 20 A R T U S Q

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Solutions and investigations

18. The numbers 2, 3, 12, 14, 15, 20, 21 may be divided into two sets so that the product of the numbers in each set is the same. What is this product? A 420 Solution C If all the numbers 2, 3, 12,14, 15, 20, 21 are multiplied the result will be the square of the common product of the two sets with the same product. Now 2 3 12 14 15 20 21 = 2 3 ( 22 3 ) ( 2 7 ) ( 3 5 ) ( 22 5 ) 3 7 = 26 34 52 72 = (23 32 5 7) 2 . Therefore the common product is 23 32 5 7 = 2520. For investigation 18.1 This solution only shows that if the common product exists then it equals 2520. A complete solution should also show that it is possible to split the numbers 2, 3, 12, 14, 15, 20, 21 into two sets, each of whose products in 2520. Show that this is possible, and in just one way. 18.2 Devise some other problems of the same type as this one. B 1260 C 2520 D 6720 E 6 350 400

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19. The 16 small squares shown in the diagram each have a side length of 1 unit. How many pairs of vertices are there in the diagram whose distance apart is an integer number of units? A 40 Solution C Each pair of distinct vertices taken from the same row or from the same column is an integer number of units (1, 2, 3 or 4 units) apart. There are 5 vertices in each row and each column, and so 2 vertices may be chosen from the same row or the same column in 5 2 = 10 ways. Since there are 5 rows and 5 columns this gives 50 + 50 = 100 pairs of vertices that are an integer number of units apart. In addition, since 32 + 42 = 52 , it follows from Pythagoras theorem that the opposite vertices of a 3 4 rectangle are 5 units apart. This gives a further 8 pairs of vertices (forming the end points of the 8 diagonals shown in the diagrams) which are an integer number of units apart. There are no other Pythagorean triples, that is, positive integers m , n , q such that m2 + n2 = q2 , where m and n are both less than 5. So we have found all the pairs of vertices that are an integer number of units apart. So there are altogether 100 + 8 = 108 pairs of such vertices. 20. The ratio of two positive numbers equals the ratio of their sum to their dierence. What is this ratio? A 1+ 3 :2 E 1+ 2 :1 Solution E Let the two numbers be x and y and suppose that x : y = k : 1, so that x = k y . Since the ratio x + y : x y is also k : 1, it follows that x + y = k ( x y ) and so k y + y = k ( k y y ) . Since y is positive, we may divide through by y to obtain k + 1 = k ( k 1) . It follows that k 2 2 k 1 = 0, and so, using the standard formula for the roots of a quadratic equation, we get 2 4+4 k= = 1 2. 2 Since x and y are both positive k > 0, and it follows that k = 1 + 2. B 2:1 C 1+ 5 :2 D 2+ 2 :1 B 64 C 108 D 132 E 16

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Solutions and investigations

21. The shaded design shown in the diagram is made by drawing eight circular arcs, all with the same radius. The centres of four arcs are the vertices of the square; the centres of the four touching arcs are the midpoints of the sides of the square. The diagonals of the square have length 1. What is the total length of the border of the shaded design? A 2 Solution B The arcs whose centres are the vertices of the square are semicircular arcs, and the arcs whose centres are the midpoints of the sides of the square form three-quarter circles. Let all these arcs have radius r . Then the length of the border is 4
1 2

5 2

C 3

7 2

E 4

2r + 4

3 4

2r = 10r . Q

Consider the line PQ joining the midpoints of two adjacent sides of the square as shown. Clearly the length of PQ is half of the length of the diagonal of the square. So PQ has length 1 2. PQ joins the midpoints of two touching arcs and so its length 1 is twice the radius of these arcs. Hence r = 4 . Therefore the 5 length of the border of the shaded design is 10 1 4 = 2 . For investigation 21.1 The above solution assumes that the arcs whose centres are the vertices of the square are semi-circles, and the arcs whose centres are the midpoints of the sides of the square form three-quarter circles. Give arguments to prove that these assumptions are correct.

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22. Consider numbers of the form 10n + 1, where n is a positive integer. We shall call such a number grime if it cannot be expressed as the product of two smaller numbers, possibly equal, both of which are of the form 10 k + 1, where k is a positive integer. How many grime numbers are there in the sequence 11, 21, 31, 41, . . . , 981, 991? A 0 Solution C Instead of checking each of the 99 numbers in the sequence 11, 21, . . . , 981, 991 in turn to see whether or not they are grime numbers, it is easier to count the numbers in this sequence that are not grime numbers. A number is not a grime number if, and only if, it is a product, say r s, where both r and s are numbers of the form 10 k + 1 , where k is a positive integer. We can assume that r s, and since we are only interested in numbers r s such that r s 991, we can also assume that r 31, because 322 > 991. We see that the only products of this form which are not greater than 991 are the twelve numbers 11 11, 11 21 , 11 31 , 11 41 , 11 51 , 11 61 , 11 71 , 11 81, 21 21 , 21 31 , 21 41 and 31 31. It can be seen, without calculating their values, that all these products are dierent as they have dierent prime factorisations. So there are 12 numbers in the sequence 11, 21, . . . , 981, 991 that are not grime numbers. Hence there are 99 12 = 87 grime numbers in this sequence. B 8 C 87 D 92 E 99

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23. PQRS is a square. The points T and U are the midpoints of QR and RS respectively. The line QS cuts PT and PU at W and V respectively. What fraction of the total area of the square PQRS is the area of the pentagon RTWVU ? 1 2 3 5 4 A B C D E 3 5 7 12 15 Solution

P W V S U

A The lines PQ and SR are parallel. Hence U SV = PQV , since they are alternate angles. Similarly SUV = PQV . It follows that the triangles U SV and QPV are similar. Now SU : PQ = 1 : 2 and so the heights of these triangles are in the same ratio. So the height of 1 triangle U SV is 1 3 of the side-length of the square. The base of this triangle is 2 of the side 1 1 1 1 of the square. Hence the area of this triangle is 2 2 3 = 12 of the area of the square.
1 Similarly the area of triangle QWT is 12 of the area of the square. The area of triangle PQS 1 is 2 of the area of the square. The area of the pentagon RTWVU is the area of the square minus the total areas of the triangles U SV , QWT and PQS , so its area, as a fraction of the 1 1 1 area of the square PQRS , is 1 12 12 2 =1 3.

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24. The diagram shows two straight lines PR and QS crossing at O . What is the value of x ? A 7 2 B 2 29 D 7 1 + 13 E 9 2 Solution E Let SOR = . Applying the Cosine Rule to the triangle ROS , we obtain 82 = 42 + 52 2 4 5 cos , from which it follows that cos = 42 + 52 82 23 = . 245 40 C 14 2 4 4

x 10 5 8

We also have that QOP = SOR = , since they are vertically opposite angles. Hence, applying the Cosine Rule to triangle POQ, we get x 2 = 42 + 102 2 4 10 cos 23 = 16 + 100 80 40 = 116 + 46 = 162. Therefore x = 162 = 81 2 = 9 2.

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Senior Mathematical Challenge 2013

Solutions and investigations

25. Challengeboroughs underground train network consists of six lines p, q, r , s, t and u, as shown. Wherever two lines meet there is a station which enables passengers to change lines. On each line, each train stops at every station. Jessica wants to travel from station X to station Y . She does not want to use any line more than once, nor return to station X after leaving it, nor leave station Y after reaching it. How many dierent routes, satisfying these conditions, can she choose? A 9 B 36 C 41 D 81 Solution

r s

X Y

t u E 720

D A route is specied by giving the sequence of lines that Jessica travels on. We call the lines s, t and u the X -lines and the lines p, q and r the Y -lines. It follows from the layout of the network and Jessicas conditions, that she can change trains between any X -line and any Y -line and vice versa, but she cannot change between two X -lines, or between two Y -lines. So her route from X to Y is given by a sequence of lines starting with an X -line, alternating between X -lines and Y -lines, and ending with a Y-line. So it will consist of an even number of lines. Since Jessica does not wish to use any line more than once, a possible route for Jessica consists of 2, 4 or 6 lines. We count these routes according to the number of lines involved. 2 lines A route of 2 lines will be of the form g , h, where g is an X -line and h is a Y -line. There are 3 choices for g and 3 choices for h, and hence 3 3 = 9 routes of this form. 4 lines A route of 4 lines will be of the form g , h, i , j , where g and i are two dierent X -lines and h and j are two dierent Y -lines. Since there are 3 choices for g and then 2 choices for i and likewise for h and j , there are 3 3 2 2 = 36 routes of this form. 6 lines A route of 6 lines will have the form g , h, i , j , k , l , where g , i and k are X -lines and h, j and l are Y -lines. As before there are 3 choices for g and then 2 choices for i , leaving just one choice for k , and likewise for the choices of h, j and l . So there are 3 3 2 2 1 1 = 36 routes of this form. So, altogether, there are 9 + 36 + 36 = 81 routes that satisfy Jessicas conditions. For investigation 25.1 How many dierent routes would there be if there were four lines passing through X and four lines passing through Y , but otherwise the conditions are the same? 25.2 Can you nd an expression which gives the number of dierent routes if there are m lines passing through X and n lines passing through Y , but otherwise the conditions are the same?

14 October 2013

UKMT November 2013

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