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Intermediate Mathematical Challenge Organised by the United Kingdom Mathematics Trustsupported by Solutions and investigationsThese solutions augment the printed solutions that we send to schools. For convenience, the solutions sent to schools are confined 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. We welcome comments on these solutions. Please send them to enquiry@ukmt.org.uk. The Intermediate Mathematical Challenge (IMC) is a multiple-choice paper. For each question, you are presented with five options, of which just one is correct. It follows that often you can find 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 IMC. 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 Intermediate Mathematical Olympiad and similar competitions). These solutions may be used freely within your school or college. You may, without further permission, post these solutions on a website that is accessible only to staff 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 February 2017 Enquiries about the Intermediate Mathematical Challenge should be sent to: IMC, UKMT, School of Mathematics Satellite, University of Leeds, Leeds LS2 9JT T 0113 343 2339 enquiry@ukmt.org.uk www.ukmt.org.uk1 D 2 B 3 A 4 D 5 B 6 D 7 C 8 D 9 B 10 A 11 E 12 D 13 A 14 B 15 C 16 D 17 B 18 E 19 E 20 D 21 A 22 C 23 D 24 C 25 AUKMT UKMTUKMT

Intermediate Mathematical Challenge 2017 Solutions and investigations1. What is the value of 2 5 + 2 50 + 2 500 ? A 0.111B 0.222C 0.333D 0.444E 0.555 Solution DWe have 2 5 = 0.4. Multiplying by 1 10 is the same as dividing by 10, and therefore has the effect of moving the decimal point one place to the left. Therefore 2 50 = 1 10 × 2 5 = 0.04 and 2 500 = 1 10 × 2 50 = 0.004 .Hence 2 5 + 2 50 + 2 500 = 0.4 + 0.04 +0.004 =0.444 . For investigation 1.1 Write as a decimal the answer to the sum 2 5 + 2 50 + 2 500 + 2 5000 + 2 50 000 + 2 500 000 + 2 5 000 000 . 1.2 What is the decimal representation of the fraction 4 9? How does this compare with your answer to 1.1? What conclusion do you draw? 2. Each of the diagrams below shows a circle and four small squares. In each case, the centre of the circle is the point where all four squares meet. In one of the diagrams, exactly one third of the circle is shaded. Which one? A B C D E Solution B The straight edges of the shaded sector which occupies one third of the area of the circle will make an angle equal to one third of a complete revolution, that is, an angle of 1 3 × 360 °= 120 °. In diagram A the straight edges of the shaded sector make an angle of 90 °. In diagram C these edges make an angle equal to 90 ° + 45 ° = 135 °. The angles in diagrams D and E are even larger. So in none of these cases is exactly one third of the circle shaded. This leaves diagram B where the angle is between 90 °and 135 °. Therefore, this is the diagram where the angle is 120 °. Hence it is the diagram where the shaded area occupies one third of the circle. © UKMT February 2017 www.ukmt.org.uk 2

Intermediate Mathematical Challenge 2017 Solutions and investigations3. How many squares have 7 as their units digit? A 0B 1C 2D 3E 4 Solution AThe units digit of the square of an integer nis the same as that of the square of the units digit of n. [You are asked to prove this in Problem 3.1.] For example, the units digit of 237 2is the same as that of 72 , namely 9. Therefore, to find the possible units digits of squares, we need only consider the squares of the one-digit numbers. We have 02 = 0,12= 1,22 = 4,32= 9,42 = 16 ,52= 25 ,62= 36 ,72 = 49 ,82= 64 and 92 = 81 . This shows that the units digit of a square can only be 0, 1, 4, 5, 6 or 9. In particular, there are are no squares which have 7 as their units digit. For investigation 3.1 Prove that the units digit of the square of an integer nis the same as the units digit of the square of the units digit of n. [Hint: Show that if n,m and uare positive integers with n = 10 m+u, then n2 and u2 have the same remainder when they are divided by 10.] 3.2 What are the possible remainders when x 2 + 3xy + y2 , is divided by 10, where xand y are positive integers? Deduce that the equation x 2 + 3xy + y2 = 222 222 has no positive integer solutions. 4. Which of the following is notthe sum of two primes? A 5B 7C 9D 11E 13 Solution D The sum of two odd numbers is even. Therefore, no odd number is the sum of two odd primes. The only even prime is 2. It follows that if any of the odd numbers 5, 7, 9, 11 and 13 is the sum of two primes, it is the sum of an odd prime and 2. We see that 5= 3+ 2, 7= 5+ 2, 9= 7+ 2, 11=9+ 2 and 13=11 +2. The numbers 3, 5, 7 and 11 are all primes, so each of 5, 7, 9 and 13 is the sum of two primes. However, 9 is not a prime and therefore 11 is not the sum of two primes. Note As we show above, the only odd prime numbers that are the sum of two primes are those of the form p + 2, where pis an odd prime. In this case the numbers p,p+ 2form what is called aprime pair , that is, they are consecutive odd numbers that are both prime. It is an unsolved problem as to whether there are infinitely many prime pairs. According to the Goldbach Conjecture every even number greater than 2 is the sum of two primes. This has not been proved but, at the time of writing, it has been verified for all even numbers up to 4× 10 18. Some partial results are known. For example, Chen’s Theorem (proved by Chen Jingrun in 1973) says that all but a finite number of even numbers can be expressed as the sum of a prime and a number which is either prime or the product of two primes. © UKMT February 2017 www.ukmt.org.uk 3

Intermediate Mathematical Challenge 2017 Solutions and investigations5.The diagram shows two circles with the same centre. The radius of the outer circle is twice the radius of the inner circle. The region between the inner circle and the outer circle is divided into six equal segments as shown. What fraction of the area of the outer circle is shaded? A 3 7 B 3 8 C 3 9 D 3 10 E 3 11 Solution B We suppose that the radius of the inner circle is r. It follows that the radius of the outer circle is 2 r. Therefore the inner circle has area πr2 and the outer circle has area π(2 r)2 , that is, 4π r2 . Hence, the area of the region between the inner and outer circles is 4π r2 − πr2 = 3πr2 . Half of this region is shaded. Therefore the shaded area is 3 2 π r2 . It follows that the fraction of the area of the outer circle that is shaded is given by 3 2 π r2 4 π r2 = 3 8 . 6. The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. What is the difference between the largest angle and the smallest angle? A 30 ° B 40° C 50° D 60° E 70° Solution D Because the angles are in the ratio 3 : 4 : 5 : 6 , and 3+ 4+ 5+ 6= 18 , the largest angle makes up 6 18, that is, 1 3of the sum of the angles of the quadrilateral. Similarly, the smallest angle makes up 3 18 , that is, 1 6 of the sum of the angles of the quadrilateral. It follows that the difference between the largest and the smallest angle is 1 3 − 1 6 , that is, 1 6of the sum of the angles of the quadrilateral. The sum of the angles of a quadrilateral is 360 °. It follows that the difference between the largest angle and the smallest angle of the quadrilateral is 1 6 × 360 °= 60 °. For investigation 6.1 What are the four angles of a quadrilateral whose angles are in the ratio 3 : 4 : 5 : 6? 6.2 The angles of a pentagon are in the ratio 3 : 4 : 5 : 6 : 7 . What is the difference between the largest angle and the smallest angle? 6.3 Prove that the sum of the angles of a quadrilateral is 360°. © UKMT February 2017 www.ukmt.org.uk 4

Intermediate Mathematical Challenge 2017 Solutions and investigations7. Four different positive integers are to be chosen so that they have a mean of 2017. What is the smallest possible range of the chosen integers? A 2B 3C 4D 5E 6 Solution CWe recall that the range of a set of numbers is the difference between the largest and smallest numbers in the set. The smallest range of a set of four different positive integers is 3. This occurs when the integers are consecutive, and only in this case. We first show that we cannot have four consecutive integers with a mean of 2017. Assume, to the contrary, that nis a positive integer such that the consecutive integers n,n + 1, n+ 2and n+ 3have 2017 as their mean. Then 1 4 ( n + (n + 1) + (n + 2) + (n + 3)) =2017 , and hence n+ (n + 1) + (n + 2) + (n + 3) = 4× 2017 . It follows that 4n + 6= 4× 2017 , and hence 4n = 4× 2017 −6. Therefore n= 2017 −3 2 , contradicting our assumption that nis an integer. This contradiction shows that there do not exist four consecutive integers whose mean is 2017. So the smallest possible range is not 3. On the other hand, we see that the four integers 2015, 2016, 2018 and 2019 have mean 2017, because 2015+2016 +2018 +2019 =(2015 +2019 )+ (2016 +2018 ) = 2× 2017 +2× 2017 = 4× 2017 . The range of these numbers is 2019 −2015 , that is, 4. We deduce that the smallest possible range is 4. For investigation 7.1 Five different positive integers are to be chosen so that their mean is 2017. What is the smallest possible range of the chosen integers? 7.2 Note that there is nothing special in this context about the number 2017. Show how the argument above may be adapted to show that the mean of four consecutive integers is never an integer. 7.3 Investigate for which positive integers kit is possible to find kconsecutive integers whose mean is an integer. © UKMT February 2017 www.ukmt.org.uk 5

Intermediate Mathematical Challenge 2017 Solutions and investigations8. Which of the following numbers is the largest? A 1.3542B 1.354 ? 2 C1.35 ? 4 ? 2 D1.3 ? 54 ? 2 E1.? 354 ? 2 Solution D To compare the numbers, we give them to 8 decimal places, as follows. 1.3542 ≈1.3542 0000 1 .354 ? 2 ≈ 1.3542 2222 1 .35 ? 4 ? 2 ≈ 1.3542 4242 1 .3 ? 54 ? 2 ≈ 1.3542 5425 1 .? 354 ? 2 ≈ 1.3542 3542These numbers agree in their first four decimal places. The number 1.3? 54? 2has a larger digit in the fifth decimal place than the others. Therefore 1.3 ? 54 ? 2 is the largest of the listed numbers. 9. The number ‘ t u ’ is the two-digit number with units digit uand tens digit t. The digits a and bare distinct, and non-zero. What is the largest possible value of ‘ ab’− ‘ba ’? A 81B 72C 63D 54E 45 Solution B The place-value notation means that ‘ ab’= 10 a+ band ‘ ba’= 10 b+ a. Therefore ‘ ab ’− ‘ba ’= (10 a+ b) − ( 10b+ a) = (10 a− a) + (b − 10 b) = 9a − 9b = 9(a − b). The largest possible value of 9(a − b) occurs when a − b has its largest possible value. Because a and bare different non-zero digits, the largest value of a− b is 8 when a= 9and b= 1. In this case ‘ ab’− ‘ba ’= 91 −19 =72 . Therefore the largest possible value of ‘ ab’− ‘ba ’= 72 . Note In Problems 9.1 and 9.2 we use the notation ‘ ht u ’ for the three-digit number with units digit u, tens digit tand hundreds digit h. For investigation 9.1 The digits a,band care distinct and non-zero. What is the largest possible value of ‘abc ’− ‘cba ’? 9.2 Let a,band cbe digits with a > c . Suppose that ‘abc ’− ‘cba ’= ‘de f ’ , where d,eand fare digits. Investigate the possible values of ‘ de f’+ ‘f ed ’. © UKMT February 2017 www.ukmt.org.uk 6

Intermediate Mathematical Challenge 2017 Solutions and investigations10. The diagram shows three rectangles. What is the value of x? A 108B 104C 100 D 96E 92 Solution AWe let the angles in the triangle that is formed by the rectangles be p°, q ° and r°, as shown in the figure. The angles of a rectangle are 90 °, as shown. The angles on the line at the points where the bottom rectangle meets the other two are each 180 °, as shown. The total of the angles at a point is 360 °. Therefore, from the angles at the points where the rectangles meet, we have the following three equations. p + 90 +90 +x= 360 , q + 180 +90 +43 =360 , and r+ 180 +90 +29 =360 . It follows that p= 180 −x, q = 47 , and r= 61 . Therefore, because the angles in a triangle total 180°, ( 180 −x) + 47 +61 =180 , from which it follows that 180−x+ 108 =180 , and hence x= 108 . For investigation 10.1 What is the value of xin the case where the angles between the lower rectangle and the other two rectangles are 37°and 31°? 10.2 Let the angles between the lower rectangle and the other rectangles be a ° and b° . Find a formula for xin terms of aand b. © UKMT February 2017 www.ukmt.org.uk 7x ° 29° 43 °

Intermediate Mathematical Challenge 2017 Solutions and investigations11.The diagram shows four equilateral triangles with sides of length 1, 2, 3 and 4. The area of the shaded region is equal to n times the area of the unshaded triangle of side-length 1. What is the value of n? A 8B 11C 18D 23E 26 Solution E We suppose that the equilateral triangle with sides of length 1 has area a. Because the areas of similar figures are proportional to the squares of their side lengths, it follows that the equilateral triangles with side lengths 2, 3 and 4, have areas 4a ,9 a and 16a, respectively. It follows that the areas of the parts of these triangles that are shaded are 4a − a ,9a− a and 16 a− a, that is, 3a ,8 a and 15a, respectively. Therefore, the total shaded area is 3a + 8a + 15 a= 26 a. Hence n= 26 . For investigation 11.1 Note that we solved this problem without the need to work out the area of an equilateral triangle with side length 1. What is the area of this triangle? 11.2 Show that if two similar triangles have corresponding side lengths in the ratio k:m, then their areas are in the ratio k2 :m 2 . 12. The combined age of Alice and Bob is 39. The combined age of Bob and Clare is 40. The combined age of Clare and Dan is 38. The combined age of Dan and Eve is 44. The total of all five ages is 105. Which of the five is the youngest? A AliceB BobC ClareD DanE Eve Solution D Eve’s age is the total of all their ages, less the ages of Alice, Bob, Clare and Dan. Therefore we can find Eve’s age by subtracting from the total of all five ages both the combined total of the ages of Alice and Bob, and the combined total of the ages of Clare and Dan. Therefore Eve’s age is 105−39 −38 =28 . Since the combined age of Dan and Eve is 44, we deduce that Dan is 16. Since the combined age of Clare and Dan is 38, it follows that Clare is 22. Since the combined age of Bob and Clare is 40, Bob is 18. Finally, since the combined age of Alice and Bob is 39, Alice is 21. Hence the youngest is Dan. © UKMT February 2017 www.ukmt.org.uk 81 2 3 4

Intermediate Mathematical Challenge 2017 Solutions and investigations13.The diagram shows a quadrilateral PQ RS made from two similar right-angled triangles, PQ R and P RS . The length PQ is 3, the length Q Ris 4 and ∠P RQ =∠PS R . What is the perimeter of PQ RS? A 22B 225 6 C 27D 32E 451 3 Solution A Applying Pythagoras’ Theorem to the right-angled triangle PQ R gives RP 2 = PQ 2 + Q R 2 = 3 2 + 42 = 9+ 16 =25 . Therefore RPhas length 5. The corresponding sides PQ and P R of the similar triangles PQ R and P RS have lengths 3 and 5. It follows that the ratio of the perimeter of PQ Rto that of P RSis3 : 5 . The perimeter of the triangle PQ R is PQ+Q R +RP = 3+ 4+ 5= 12 . Let the perimeter of P RS bep. Then 12 :p= 3 : 5 . Hence 12 p = 3 5 . Hence p= 5 3 × 12 =60 3 = 20 . The perimeter of PQ RS is PQ +Q R +RS +S P . This is equal to ((PQ +Q R +RP ) −RP )+ ((RS + S P +RP ) − RP . Thus the perimeter equals the sum of the perimeters of PQ R and P RS less twice the length of RP. We deduce that the perimeter of PQ RSis12 +20 −2× 5= 32 −10 =22 . 14. For what value of xis 64 x equal to 5125 ? A 6B 7.5C 8D 16E 40 Solution B We note first that 64 = 26and 512 = 29. Therefore 64 x = ( 26 )x = 26x, and 512 5 = ( 29 )5 = 2 9 × 5 = 245 . It follows that the equation 64 x= 512 5is equivalent to the equation 26x = 245. We deduce from this that 6x = 45 . Hence x= 45 6 = 15 2 = 7.5 . For investigation 14.1 Find the value of xin each of the following equations. (a) 32x = 1024 3 . (b) 16×32 x = 128 x . (c) 64x = 2. (d) 9x = 243 . © UKMT February 2017 www.ukmt.org.uk 9P Q R S 3 4

Intermediate Mathematical Challenge 2017 Solutions and investigations15.In the diagram shown, PQ =SQ =Q R and ∠S PQ = 2 × ∠RSQ . What is the size of angle Q RS? A 20 ° B 25° C 30° D 35° E 40 ° Solution C Method 1 We let ∠ Q RS =x° . Because SQ =Q R , the triangle SQ R is isosceles. Therefore ∠RSQ = ∠ Q RS =x°. Because ∠ S PQ = 2×∠RSQ , we deduce that ∠S PQ = 2x° . Since PQ =SQ , the triangle S PQ is isosceles . Therefore ∠PSQ =∠S PQ = 2x° . It follows that ∠RS P =2x° + x° = 3x°. We now apply the fact that the angles of a triangle have sum 180 °to the triangle P RS . This gives x + 2x + 3x = 180 . Hence 6x = 180 . Therefore x= 30 . It follows that ∠Q RS is30 °. Method 2 We let ∠ Q RS =x° . Because SQ =Q R , the triangle SQ R is isosceles. Therefore ∠RSQ = ∠ Q RS =x° . Because ∠ S PQ = 2×∠RSQ , we deduce that ∠S PQ =2x°. Because PQ =SQ =Q R , the point Slies on the circle with centre Q which goes through P and R. It follows that ∠RS P is the angle in a semicircle and hence is 90°. We now apply the fact that the angles of a triangle have sum 180 °to the triangle P RS . This gives x + 2x + 90 =180 . Hence 3x = 90 . Therefore x= 30 . It follows that ∠Q RS is30 °. For investigation 15.1 Suppose that, as in the question, PQ =SQ =Q R , but ∠S PQ = 3×∠RSQ . Find the size of angle Q RS. 15.2 Suppose that, as in the question, PQ =SQ =Q R , but ∠S PQ =k× ∠RSQ , where kis a positive number. Find the size of angle Q RSin terms of k. 15.3 Our second method used the theorem that the angle in a semicircle is a right angle . Give a proof of this theorem. © UKMT February 2017 www.ukmt.org.uk 10P Q R S

Intermediate Mathematical Challenge 2017 Solutions and investigations16.The product of two positive integers is equal to twice their sum. This product is also equal to six times the difference between the two integers. What is the sum of these two integers? A 6B 7C 8D 9E 10 Solution D We let the two positive integers be mand n, with m≥n. From the information in the question we deduce that mn=2(m +n) and mn=6(m −n). It follows that 6(m −n) = 2(m +n). Hence 6m −6n = 2m +2n . It follows that 4m =8n, and therefore m=2n . Substituting 2n for min the equation mn=2(m +n), gives 2 n2 = 6n . Because n, 0, we can divide both sides of this last equation by 2n to deduce that n = 3. Therefore m=6. Hence m+n= 6+ 3 = 9. Therefore the sum of the two integers is 9. For investigation 16.1 The product of two positive numbers m and nis equal to three times their sum and nine times their difference. Find the values of mand n. 16.2 Let kand lbe positive numbers with l > k . The product of the positive numbers m and n is ktimes their sum and ltimes their difference. (a)Find formulas for mand nin terms of kand l. (b) Check that if you substitute k = 3and l= 9in the formulas that you found in (a) above, you obtain the same values for mand nthat you found in 16.1. © UKMT February 2017 www.ukmt.org.uk 11

Intermediate Mathematical Challenge 2017 Solutions and investigations17.The diagram shows two rectangles and a regular pentagon. One side of each rectangle has been extended to meet at X. What is the value of x? A 52B 54C 56D 58E 60 Solution B Note The value of xmay be found in several ways. We describe three of these here. There are others. Method 1 We use the fact that the sum of the angles of a pentagon is 540 °. Hence, each interior angle of a regular pentagon is 108 °. We also know that the interior angles of the rectangles are each 90°. We now consider the pentagon T UV W X , as shown in the figure. The interior angles of this polygon at T and W are each 90 °. The interior angle at U is the interior angle of the regular polygon, namely 108 °. The interior angle at V is the sum of an interior angle of the regular pentagon and a right angle, that is, 108°+ 90 °. Therefore 90 + 108 + ( 108 + 90 )+ 90 + x = 540 . Hence, x+ 486 = 540 . Therefore x= 540 −486 =54 . Method 2 We label the vertices of the pentagon and one of the rectangles as shown in the fig- ure. Let Y Z be a line through S that is parallel to UV . It follows that the alternate angles, ∠ S X U and ∠ X S Z are equal. Hence ∠X S Z =x°. By the symmetry of the pentagon about the line through Sperpendicular to Y Z , we have ∠ Y ST =∠RS Z . Let ∠Y ST =∠RS Z =y°. The angles on the line at the point Shave sum 180 °. Because it is the angle of a regular pentagon. ∠ T S R =108 °. Because it is the angle of a rectangle ∠T S X =90 °. Therefore 2 y + 108 =180 © UKMT February 2017 www.ukmt.org.uk 12x ° X

Intermediate Mathematical Challenge 2017 Solutions and investigationsand x+ y+ 90 =180 . From the first of these equations it follows that y= 1 2 ( 180 −108 ) = 1 2 ( 72 )= 36 . Therefore from the second equation we deduce that x= 180 −36 −90 =54 . Method 3Let the rectangles be PQ RS and T UV W , labelled as shown in the figure. Let the edge PQ when extended meet W V at Y and T U at Z . The lines S X and P Z are extensions of opposite edges of the rectangle PQ RS and hence they are parallel. Therefore the corresponding angles ∠S X U and ∠ P ZT are equal. Therefore ∠P ZT =x°. Because the edges W V and T U of the rectangle T UV W are parallel, the corresponding angles ∠ PY W and ∠P ZT are equal. Therefore ∠PY W =x°. We now consider the other two angles of the triangle PY W . First, ∠ Y W P = 108 °since it is an angle of the regular pentagon. Next, ∠W PY is the angle ∠W PS of the regular pentagon minus the angle ∠Q PS of the rectangle PQ RS. Therefore ∠W PY =108 °− 90 °= 18 °. Because the angles of the triangle PT Whave sum 180°, x + 108 +18 =180 and therefore x= 180 −108 −18 = 54 . For investigation 17.1 We have used the fact that the sum of the interior angles of a pentagon is 540 °and hence that each angle of a regular pentagon is 108°. Show that this is correct. 17.2 In Method 2, we took it for granted that by symmetry ∠ Y ST =∠RS Z . Show that this is correct. © UKMT February 2017 www.ukmt.org.uk 13

Intermediate Mathematical Challenge 2017 Solutions and investigations18.A water tank is 5 6 full. When 30 litres of water are removed from the tank, the tank is 4 5 full. How much water does the tank hold when full? A 180 litresB 360 litresC 540 litresD 720 litresE 900 litres Solution E The 30 litres of water removed from the tank is the difference between 5 6ths and 4 5ths of the capacity of the tank. Now 5 6 − 4 5 = 5 × 5− 4× 6 5 × 6 = 25 −24 30 = 1 30 . We therefore see that 30 litres amounts to 1 30th of the capacity of the tank. It follows that when the tank is full the number of litres that it holds is 30×30 =900 . 19. PQ RS is a square. Point T lies on PQ so that PT :T Q= 1 : 2 . Point U lies on S R so that SU:U R =1 : 2 . The perimeter of PT U Sis 40cm. What is the area of PT U S? A 40 cm2 B 45cm2 C 48cm2 D 60cm2 E 75cm2 Solution E Because PQ RS is a square, S R =PQ and PS =PQ . Because the ratio PT :T Q is 1 : 2 , we deduce that PT =1 3 PQ . Similarly, we have SU =1 3 S R =1 3 PQ =PT . The lines PT and and SU are parts of opposite sides of the square PQ RS . Therefore they are parallel. Because they are also of equal length, it follows that PT U Sis a parallelogram. Therefore T U=PS =PQ . It follows that the perimeter of PT U S is PT +T U +U S +S P =1 3 PQ +PQ +1 3 PQ +PQ =8 3 PQ . Therefore 8 3 PQ =40 cm . Hence PQ=3 8 × 40 cm =15 cm . Therefore PT=1 3 × 15 cm =5 cm . We also note that, because ∠S PT =∠PSU =90 °, we can deduce that PT U Sis a rectangle. It follows that the area of PT U Sis given by PT ×T U =5 cm ×15 cm =75 cm 2 . For investigation 19.1 In the above solution we said that because the lines PT and SU are parallel and of equal length, it follows that PT U S is a parallelogram. Prove that this is the case by showing that S P is parallel to UT. © UKMT February 2017 www.ukmt.org.uk 14

Intermediate Mathematical Challenge 2017 Solutions and investigations20.The diagram shows seven circular arcs and a heptagon with equal sides but unequal angles. The sides of the heptagon have length 4. The centre of each arc is a vertex of the heptagon, and the ends of the arc are the midpoints of the two adjacent sides. What is the total shaded area? A 12π B14π C16π D18π E20π Solution D The shaded area is made up of seven sectors of circles each of radius 2. The area of a circle with radius 2 is π× 22 = 4π . The area of a sector of a circle is directly proportional to the angle subtended by the sector at the centre of the circle. A full circle subtends an angle of 360 °at its centre. Therefore, if the total of the angles subtended by the shaded sectors shown in the figure is x × 360 °, the total shaded area is x× 4π . The internal angles of a polygon with nsides add up to (n − 2) × 180 °. Therefore, the internal angles of a heptagon add up to 5× 180 °. The total of the angles at a point is 360 °. Therefore the total of the angles at the seven vertices of the heptagon is 7× 360 °. Since the internal angles have total 5× 180 °, the total of the external angles is 7× 360 °− 5× 180 °= 7× 360 °− 5 2 × 360 °= 9 2 × 360 °. Therefore the total shaded area is 9 2 × 4π , that is, 18π. For investigation 20.1 Explain why the internal angles of a polygon with nsides add up to (n − 2) × 360 °. 20.2 The above solution uses the fact that, if the total of the angles subtended by sectors of circles with the same radius is x × 360 °, then the total area of the sectors is x × the area of one circle with the same radius. Explain why this is true. 20.3 What would the total shaded area be in the similar problem in which the heptagon is replaced by an nonagon with sides all of length 2? 20.4 Let nbe a positive integer with n ≥ 3. Find the formula, in terms of n, for the total shaded area in the similar problem in which the heptagon is replaced by a polygon with nsides each with length 2. © UKMT February 2017 www.ukmt.org.uk 15

Intermediate Mathematical Challenge 2017 Solutions and investigations21.Brachycephalus frogs are tiny – less than 1 cm long – and have three toes on each foot and two fingers on each ‘hand’, whereas the common frog has five toes on each foot and four fingers on each ‘hand’. Some Brachycephalus and common frogs are in a bucket. Each frog has all its fingers and toes. Between them they have 122 toes and 92 fingers. How many frogs are in the bucket? A 15B 17C 19D 21E 23 Solution A Let bbe the number of Brachycephalusfrogs in the bucket and let cbe the number of common frogs in the bucket. A Brachycephalus frog has three toes on each foot and two fingers on each ‘hand’. Therefore, in total, it has 6 toes and 4 fingers. A common frog has, in total, 10 toes and 8 fingers. Therefore, between them bBrachycephalus frogs and ccommon frogs have 6b + 10 ctoes and 4b + 8c fingers. Hence, from the information given in the question, we have 6b + 10 c= 122 , and 4b + 8c = 92 . Subtracting the second equation from the first, we obtain 2b + 2c = 30 . Dividing the last equation by 2, we deduce that b+ c= 15 . Therefore, the total number of frogs in the bucket is 15. For investigation 21.1 Note that in the above solution we calculated the total number of frogs without working out the number of frogs of each of the two species. How many Brachycephalus frogs are in the bucket, and how many common frogs? 21.2 How many frogs would there have been in the bucket if between them they had 62 toes and 44 fingers? 21.3 How many frogs would there have been in the bucket if between them they had ttoes and f fingers? 21.4 Assume that a bucket contains an assortment of Brachycephalus and common frogs, and that each frog has all its fingers and all its toes. What are the possible values for the number tof toes and the number fof fingers that they have between them? © UKMT February 2017 www.ukmt.org.uk 16

Intermediate Mathematical Challenge 2017 Solutions and investigations22.The diagram shows an arc PQ of a circle with centre O and radius 8. Angle QO P is a right angle, the point M is the midpoint of O P and N lies on the arc PQ so that M N is perpendicular to O P. Which of the following is closest to the length of the perimeter of triangle P N M? A 17B 18C 19D 20E 21 Solution C Because N M is perpendicular to O P , the triangle O M N has a right angle at M. Therefore, by Pythagoras’ Theorem, O M2 + N M 2 = O N 2 . Since M is the midpoint of O P and the circle has radius 8, it follows that O M = 4. Also O N= 8. Therefore, by the above equation N M2 = 82 − 42 = 64 −16 =48 . It follows that N M=√ 48 . In the right-angled triangle N M P , we have N M 2 = 48 and M P 2 = 42= 16 . Therefore, by Pythagoras’ Theorem applied to this triangle, N P 2 = M N 2 + M P 2 = 48 +16 = 64 , and therefore N P =8. It follows that the perimeter of the triangle P N M is M P +P N +N M = 4+8+√ 48 = 12 +√ 48 . Because √ 48 is close to 7, the perimeter is close to 12+7= 19 . For investigation 22.1 In the argument above we used Pythagoras’ Theorem to show that P N = 8. It is also possible to prove that P N=O N = 8, by showing that the triangles O M N and P M N are congruent. Explain how this may be proved. 22.2 The argument above relied on the rather vague statement that because √ 48 is close to 7, it follows that 12 + √ 48 is close to 19. To give a more precise argument that, of the given options, 12 + √ 48 is closest to 19, we need to show that 18 .5< 12 + √ 48 < 19 .5. These inequalities are equivalent to 6.5 < √ 48

Intermediate Mathematical Challenge 2017 Solutions and investigations23.Two brothers and three sisters form a single line for a photograph. The two boys refuse to stand next to each other. How many different line-ups are possible? A 24B 36C 60D 72E 120 Solution D We first count the number of ways in which two brothers and three sisters can stand in line without two brothers being next to each other, without regard to which particular brother and which particular sister is in any given position. Counting from the left, the first brother can be in the first, second or third position. (He cannot be in fourth position as then the second brother from the left would be next to him in fifth position.) If the first brother is in first position, the second brother can be in third, fourth or fifth position. If the first brother is in second position, the second brother can be in fourth or fifth position. If the first brother is in third position, the second brother can only be in fifth position. It follows that there are six ways to arrange where the brothers and the sisters are positioned. These are listed below using Bto represent a brother and Sto represent a sister. B , S, B, S, S B , S, S, B, S B , S, S, S, B S , B, S, B, S S , B, S, S, B S , S, B, S, B For any given two positions which the two brothers occupy, they can be put in these positions in 2 ways. If the brothers are B1 and B2, these are B1, B 2 and B2, B 1. For any given three positions which the three sisters occupy, they can be put in these positions in 6 ways. If the sisters are S1,S2and S3, these are S1,S 2,S 3;S1,S 3,S 2; S2,S 1,S 3;S2,S 3,S 1; S 3, S 1, S 2 and S3, S 2, S 1. Since the 6 choices of which positions the brothers occupy, the 2 choices of how the brothers are put in these positions, and the 6 choices of how the sisters are put in the remaining positions may be made independently, the total number of different line-ups is given by 6× 2× 6= 72 . For investigation 23.1 How many different line-ups are possible in the similar problem with two brothers and four sisters? 23.2 Let nbe a positive integer. Find the formula, in terms of n, for the number of line-ups that are possible in the similar problem with two brothers and nsisters. © UKMT February 2017 www.ukmt.org.uk 18

Intermediate Mathematical Challenge 2017 Solutions and investigations24.The nth term of a certain sequence is calculated by multiplying together all the numbers r 1 + 1 k , where ktakes all the integer values from 2 to n + 1inclusive. For example, the third term in the sequence is r 1 + 1 2 × r 1 + 1 3 × r 1 + 1 4 . What is the smallest value of nfor which for the nth term of the sequence is an integer? A 3B 5C 6D 7 E more than 7 Solution C Let nbe a positive integer. The nth term of the sequence is r 1 + 1 2 × r 1 + 1 3 × r 1 + 1 4 × · · · × r 1 + 1 n + 1. This expression may be rewritten as r 3 2 × r 4 3 × r 5 4 × · · · × r n + 2 n + 1, which is equivalent to r 3 2 × 4 3 × 5 4 × · · · × n + 2 n + 1. In the product above we may cancel all the terms other than the denominator of 3 2 and the numerator of n + 2 n + 1. In this way the above expression may be simplified to r n + 2 2 . For this to be an integer, we require that n + 2 2 be a square and hence that n + 2is twice a square. Because n+ 2> 2, the least possible value of n+ 2is 2× 22, that is, 8. Now n+ 2= 8for n = 6. Therefore 6 is the smallest value of nfor which the nth term of the sequence is an integer. For investigation 24.1 Which is the least value of n, such that n > 6and the nth term of the sequence is an integer? 24.2 Show that there are infinitely many different positive integers nfor which r n + 2 2 is an integer. © UKMT February 2017 www.ukmt.org.uk 19

Intermediate Mathematical Challenge 2017 Solutions and investigations25.The diagram shows a circle with radius 2 and a square with sides of length 2. The centre of the circle lies on the perpendicular bisector of a side of the square, at a distance xfrom the side, as shown. The shaded region—inside the square but outside the circle—has area 2. What is the value of x? A π 3 + √ 3 2 − 1 Bπ 3 + √ 3 4 − 1 C π 3 + 1 2 D π 3 + 1 E π 3 Solution A We let O be the centre of the circle, and the points J, K ,L ,M ,N ,P ,Q and Tbe as shown in the figure. The area that is shaded is the area of the square less the area of the part of the square that is inside the circle. Therefore we begin by working out this latter area. The part of the square that is inside the circle is made up of the segment of the circle cut off by the chord J M and the rectangle J K L M . We let the areas of these regions be A S and A R, respectively. From the figure above we see that A S equals the area of the sector of the circle subtended by the arc J M at the centre Oof the circle, less the area of the triangle J O M. In the triangle J O M ,J O and M O have length 2, because they are radii of the circle, and J M has length 2, because it has the same length as the sides of the square. It follows that the triangle J O M is equilateral. Hence ∠J O M = 60 °. Because 60 °is one-sixth of a complete revolution, the area of the sector of the circle subtended by the arc J M is one-sixth of the area of the circle of radius 2. Hence the area of this sector is 1 6 × π(2 2 ) = 2 π 3 . We now work out the area of the equilateral triangle J O M which has side length 2. We note that T is the point on J M such that OT is perpendicular to J M . Hence the triangles J OT and M OT are congruent. Therefore JT =T M , and hence JT has length 1. By Pythagoras’ Theorem applied to the triangle JT O , We have OT 2 = J O 2 − JT 2 = 22− 12 = 3. Therefore OT has length √ 3 . Therefore the triangle J O M has a base of length 2 and height √ 3 . Hence the area of this triangle is 1 2 × 2× √ 3 = √ 3 . © UKMT February 2017 www.ukmt.org.uk 20x

Intermediate Mathematical Challenge 2017 Solutions and investigationsIt follows that A S = 2 π 3 −√ 3 . We now work out the value of A R which is the area of the rectangle J K L M.We have already noted that J M has length 2. The length of K J is the same as the length of NT . This length is OT−O N =√ 3 − x. So K J=√ 3 − x. Therefore A R = J M ×J K = 2(√ 3 − x). Therefore the area of the square that is inside the circle is AS + A R =  2 π 3 −√ 3  + 2(√ 3 − x) = 2 π 3 +√ 3 − 2x. The area of the square is 4. Therefore for the area outside the circle to be 2, we require that the area inside the circle is also 2. That is, we require that 2= 2 π 3 +√ 3 − 2x. This last equation may be rearranged as 2x = 2 π 3 +√ 3 − 2, from which it follows that x= π 3 + √ 3 2 − 1. For investigation 25.1 Prove the following statements that are used in the above solution. (a) J K L M is a rectangle. (b) J Mhas length 2. (c)The triangles J OTandM OT are congruent. (d) ∠N PO =∠J PK =60 °. (e) Nlies on the line OT. (f ) NT =K J . © UKMT February 2017 www.ukmt.org.uk 21