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Intermediate Mathematical Challenge Organised by the United Kingdom Mathematics Trustsupported by Solutions and investigations 1 February 2018These 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 (or, occasionally, left as an exercise). We therefore hope that these solutions can be used as a model for the type of written solution that is expected when a complete solution to a mathematical problem is required (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 January 2018 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 B 2 B 3 C 4 D 5 B 6 A 7 B 8 A 9 E 10 E 11 B 12 E 13 C 14 C 15 E 16 C 17 D 18 A 19 D 20 E 21 D 22 E 23 A 24 B 25 DUKMT UKMTUKMT

Intermediate Mathematical Challenge 2018 Solutions and investigations1. Which of these is the sum of the cubes of two consecutive integers? A 4B 9C 16D 25E 36 Solution B We see that 9= 13 + 23 , and hence 9 is the sum of the cubes of two consecutive integers. NoteIn this context, it is sufficient to find one option that is correct, as we are entitled to assume that in the IMC there is just one correct option. It is, in fact, straightforward to check by listing the first few cubes that none of the other options is correct. We leave this as an exercise. For investigation 1.1 Show that none of the other numbers given as options, that is, 4, 16, 25 and 36, is the sum of two consecutive cubes. 1.2 Calculate the following sums. (a) 13 + 23 + 33 . (b) 13 + 23 + 33 + 43 . (c) 13 + 23 + 33 + 43 + 53 . 1.3 What do you notice about your answers to Problem 1.2? 1.4 Find a general formula for the sum of the cubes of the first npositive integers. 1.5 Prove that the formula given by your answer to Problem 1.4 is correct. Deduce that the sum of the first npositive cubes is always a square. 1.6 Other than 13 + 23+ 33 = 62, there is just one example where the sum of the cubes of three consecutive positive integers is a square. Can you find this example? Note It is known that 13 + 23 = 32is the only case where the sum of the cubes of two consecutive positive integers is a square. However, we do not know a short elementary proof of this fact. If you know of one, please send it to us (at enquiry@ukmt.org.uk). In order to establish this fact we would need to show that the only solution of the equation x3 + (x + 1)3 = y2 in which xand yare positive integers is x= 1, y = 3. The equation x3 + (x + 1)3 = y2 is an example of an equation of an elliptic curve . The theory of the integer solutions of such equations is complicated. The theory of elliptic curves was one of the ingredients of the proof by Andrew Wiles of “Fermat’s Last Theorem”. By the way, elliptic curves are not ellipses. Search the web if you wish to know how they come to have this name. 1.7 Sketch the graph of the curve given by the equation x3 + (x + 1)3 = y2 . © UKMT January 2018 www.ukmt.org.uk 2

Intermediate Mathematical Challenge 2018 Solutions and investigations2. How many of these four integers are prime? 1 11 111 1111 A 0B 1C 2D 3E 4 Solution B We begin by recalling the definition of a prime: A positive integer nis said to be primeifn, 1and the only divisors of nare 1 and n.It follows from the definition that 1 is not prime. It is easy to see that 11 is prime. Since 111 = 3× 37 , it follows that 111 is not prime. Since 1111 = 11 × 101 , it follows that 1111 is not prime. Therefore just one of the given integers is prime. Note It is just a convention not to regard 1 as a prime number, but it is a standard convention that it is important to remember. “In case you’re wondering, I’ll mention that 1 is not prime. That’s not because of some interesting philosophical point, or because it’s only got one number that divides into it rather than two, or anything like that. Definitions (such as that of a prime number) don’t get handed to mathematicians on stone tablets. Rather, part of the job of mathematicians is to make good definitions, ones that lead to interesting mathematics, and it turns out to be better to define 1 not to be prime.” Vicky Neale, Closing the Gap: the quest to understand prime numbers , Oxford University Press, 2017. For investigation 2.1 If ncan be factorized as r × s , where rand sare integers, at least one of rand swill be at most √ n . [Why?] Therefore, to test whether an integer n, with n > 1, is prime it is sufficient to see whether nis divisible by a prime pin the range 2≤ p≤ √ n . The integer n is prime if, and only if, it has no prime divisor in this range. Use this criterion to show that 11 is prime and to test whether the following are primes. (a) 91, (b) 107, (c) 899, (d) 901, (e) 907. 2.2 Find the prime factorizations of the following integers. (a) 11 111 , (b) 111 111 , (c)1 111 111 . 2.3 A positive integer which may be written as a string of 1s is called a repunit (short for rep eated unit s). Show that if nis not a prime number, then the repunit consisting of a string of n1s is not a prime number. 2.4 We have already noted that the repunit 111 consisting of a string of three 1s is not a prime. This shows that the converse of Problem 2.3 is not true in general; if nis a prime, the repunit consisting of a string of n1s need not be a prime. Find the least prime number n, with n > 2, such that the repunit consisting of a string of n 1s is a prime. (Note: as the repunits you will need to check are quite large, you will need an electronic aid to test whether they are primes.) © UKMT January 2018 www.ukmt.org.uk 3

Intermediate Mathematical Challenge 2018 Solutions and investigations3.In September 2016 a polymer £5 note was introduced. The Bank of England issued 440 million of them. What is the total face value of all these notes? A £ 220 000 000 B £440 000 000 C £2 200 000 000 D £ 4 400 000 000 E £22 000 000 000 Solution C One million is 1 000 000 . Therefore 440 million is 440 000 000 . It follows that the total face value of 440 million £5 notes is £ (5 × 440 000 000 )= £2 200 000 000 . 4. A kite is made by joining two congruent isosceles triangles, as shown. What is the value of x? A 36B 54C 60D 72E 80 Solution D Let the vertices of the kite be labelled as shown. The question tells us that the triangles PQ R and P RS are isosceles. It follows that ∠ Q RP =∠RQ P =x° and that ∠ S RP =∠RS P =x°. Hence ∠S RQ =2x°. The sum of the angles of a quadrilateral is 360 °. [You are asked to prove this in Problem 4.1 below.] Therefore, from the quadrilateral PQ RSwe have x ° + x° + 2x° + x° = 360 °. Therefore 5x° = 360 °and hence x= 72 . For investigation 4.1 Prove that the sum of the angles of a quadrilateral is 360°. 4.2 The solution also uses the following theorem The angles at the base of an isosceles triangle are equal to one another. Find a proof of this theorem. 4.3 The theorem stated in Problem 4.2 is Proposition 5 of Euclid’s Elements , Book 1. In the days when Euclid’s Elements was used as a textbook in schools this proposition was known as the pons asinorum . Find out what this means, and why the theorem was given this name. © UKMT January 2018 www.ukmt.org.uk 4x ° x° x °

Intermediate Mathematical Challenge 2018 Solutions and investigations5. The adult human body has 206 bones. Each foot has 26 bones.Approximately what fraction of the number of bones in the human body is found in one foot? A 1 6 B 1 8 C 1 10 D 1 12 E 1 20 Solution B The required fraction is 26 206 = 13 103 . Since 103=7× 13 +12 , we have 13 103 = 13 7 × 13 +12 = 1 7 + 12 13 . Because the integer closest to 7+ 12 13 is 8, we deduce that 26 206 is approximately equal to 1 8 . For investigation 5.1 For which integer ndoes the fraction 1 n give the best approximation to 23 2018 ? 6. In 2014, in Boston, Massachusetts, Eli Bishop set a world record for the greatest number of claps per minute. He achieved 1020 claps in one minute. How many claps is that per second? A 17B 16.5C 16D 15.5E 15 Solution A There are 60 seconds in one minute. Therefore 1020 claps per minute is the same as 1020 ÷ 60 claps per second. Now 1020 60 = 102 6 = 17 . Therefore Eli Bishop clapped at 17 claps per second. For investigation 6.1 Suppose that each time Eli clapped he moved each of his hands through 10 cm , that is, 5cm in moving the hands apart and then another 5cm bringing them together again. On this assumption, what was the average speed of movement of his hands in kilometres per hour? © UKMT January 2018 www.ukmt.org.uk 5

Intermediate Mathematical Challenge 2018 Solutions and investigations7.How many two-digit squares have the property that the product of their digits is also a square? A 0B 1C 2D 3E 4 Solution B In the following table we have listed the two-digit squares together with the products of their digits. product of the digits n n 2 ofn2 4 16 6 5 25 10 6 36 18 7 49 36 8 64 24 9 81 8 From this table we see that the only two-digit square which has the property that the product of its digits is a square is 49. Therefore the number of two-digit squares with this property is 1. For investigation 7.1 How many three-digit squares have the property that the product of their digits is the square of a positive integer? 7.2 There are just two four-digit squares which have the property that the product of their digits is the square of a positive integer. Can you find them? © UKMT January 2018 www.ukmt.org.uk 6

Intermediate Mathematical Challenge 2018 Solutions and investigations8.The diagram shows a square of perimeter 20 cm inscribed inside a square of perimeter 28 cm. What is the area of the shaded triangle? A 6 cm2 B 7cm2 C 8cm2 D 9cm2 E 10 cm2 Solution A A quick solution in the context of the IMC goes as follows: Clearly the four triangles in the corners of the larger square are congruent, and so have the same area. The squares have perimeters of lengths 20 cm and 28 cm . Hence their side lengths are 5cm and 7cm , so their areas are 52 cm 2 = 25 cm 2 and 72 cm 2 = 49 cm 2 . The area of the shaded region is one quarter of the difference between the areas of the squares. So the area of the shaded region is 1 4 ( 49 −25 )cm 2 = 6 cm 2 . However, for a full solution we need to spell out in detail how we know that the four triangles are congruent. We give two methods for doing this. Method 1 We let P,Q,R and Sbe the vertices of the larger square, and T,U,V and W be the vertices of the smaller square, arranged as shown in the figure. Because PQ RS is a square, ∠ T QU = 90 °. Therefore, applying the fact that the sum of the angles in a triangle is 180 °to triangle T QU , we have ∠QT U +∠T UQ +90 °= 180 °. Hence, ∠ T UQ =90 °− ∠QT U .(1 ) Because T UV W is a square, ∠ UT W = 90 °. Therefore, applying the fact that the sum of the angles on a line is 180°to the angles at T, we have ∠QT U +90 °+ ∠PT W =180 °. Hence ∠ PT W =90 °− ∠QT U .(2 ) By (1) and (2), ∠ T UQ =∠PT W . Similarly, ∠QT U =∠T W P . Because T UV W is a square, T U =W T . It follows that the triangles T QU and W PT are congruent. Similarly, these triangles are also congruent to the triangles V SW and U RV . Therefore all these four triangles have the same area. It follows that the area of the shaded region is one quarter of the area of the square PQ RS less the area of the square T UV W. The square PQ RS has perimeter 28 cm and hence it has side length 7cm . The square T UV W has perimeter 20 cm and hence side length 5cm . Therefore these squares have areas 49 cm2 and 25 cm 2 , respectively. Hence the area of the shaded region is given by 1 4 ( 49 cm 2 − 25 cm 2 ) = 1 4 ( 24 cm 2 ) = 6 cm 2 . © UKMT January 2018 www.ukmt.org.uk 7

Intermediate Mathematical Challenge 2018 Solutions and investigationsMethod 2Let O be the centre of the square T UV W . It is left as an exercise (see Problem 8.1) to show that Ois also the centre of the square PQ RS. Consider the rotation of both squares through a quarter turn anticlockwise about the point O. The effect of this rotation is to move P to Q,Q to R,R to Sand Sto P. It also moves Tto U,U toV,V to W and WtoT. It follows that this rotation moves the triangle PT W to the position of the triangle QUT ,QUT to RV U ,RV U to SW V and SW V to PT W . It follows that these four triangles all have the same area. It then follows, as in Method 1, that the area of each of these triangles is a quarter of the difference of the areas of the two squares. Hence the area of the shaded triangle is 6 cm2 . For investigation 8.1 Show that the two squares PQ RSandT UV W have the same centre. Note One way to do this is to suppose that O is the centre of the square T UV W . We then let K be the point where the perpendicular from O to S R meets S R , and let Lbe the point where the perpendicular from Oto PS meets PS. Now show that the triangles OV K and O LW are congruent. It will follow that OK=O L . Deduce from this that Olies on the diagonal SQ. In a similar way it may be shown that O lies on the diagonal P R , and hence that O is the centre of the square PQ RS. 9. Which integer nsatisfies 3 10 < n 20 < 2 5 ? A 3B 4C 5D 6E 7 Solution E If we multiply both sides of an inequality by a positive number, we obtain an equivalent inequality. Therefore, by multiplying by the positive number 20, we see that the given inequalities are equivalent to 6< n< 8. The only integer that satisfies these inequalities is 7. © UKMT January 2018 www.ukmt.org.uk 8

Intermediate Mathematical Challenge 2018 Solutions and investigations10. Which of these integers cannot be expressed as the difference of two squares? A 5B 7C 8D 9E 10 Solution E Method 1 We see that 5= 32 − 22 , 7= 42 − 32 , 8= 32 − 12 and 9= 52 − 42 .Therefore the first four integers given as options can be expressed as the difference of two squares. In the context of the IMC it is safe to deduce that the remaining option, 10, cannot be expressed as the difference between two squares. Method 2 The first six squares are 1, 4, 9, 16, 25, 36. Once we get beyond 25 the difference between consecutive squares is greater than 10. Hence if 10 could be expressed as the difference of squares, the two squares in question would have both to be at most 25. However, it is straightforward to check that there does not exist a pair of numbers taken from the list 1, 4, 9, 16, 25 whose difference is 10. Therefore we deduce that 10 cannot be expressed as the difference of two squares. A third approach to Question 10 is to find exactly which positive integers are the difference of two squares. Problems 10.1 and 10.2 lead you in this direction. For investigation 10.1 Prove that every odd integer can be expressed as the difference of two squares. 10.2 Determine which even integers can be expressed as the difference of two squares. 10.3 Deduce the answer to Question 10 from your answers to the above two problems. © UKMT January 2018 www.ukmt.org.uk 9

Intermediate Mathematical Challenge 2018 Solutions and investigations11.The diagram shows a regular hexagon which has been divided into six regions by three of its diagonals. Two of these regions have been shaded. The total shaded area is 20 cm2 . What is the area of the hexagon? A 40 cm2 B 48cm2 C 52cm2 D 54cm2 E 60 cm2 Solution B Let the area of the hexagon be acm 2 . In the figure on the right we have added two more of the diagonals of the hexagon, and a line joining the midpoints of two opposite edges. The lines in the figure divide the hexagon into 12 congruent triangles. The shaded region is covered by 5 of these triangles. Hence the shaded area makes up five-twelfths of the total area of the hexagon. The area of the shaded region is 20 cm2 . Therefore, 5 12 a = 20 . It follows that a= 12 5 × 20 =48 . Hence, the area of the hexagon is 48 cm2 . For investigation 11.1 Prove the assertion made in the above solution that the 12 triangles in the above figure are all congruent. 11.2 What is the length of each of the edges of the hexagon? © UKMT January 2018 www.ukmt.org.uk 10

Intermediate Mathematical Challenge 2018 Solutions and investigations12. Someone has switched the numbers around on Harry’s calculator!The numbers should be in the positions shown in the left-hand diagram, but have been switched to the positions in the right-hand diagram. Which of the following calculations will not give the correct answer when Harry uses his calculator? A 79 ×97 B78 ×98 C147 ×369 D123 ×321 E 159 ×951 Solution E Six of the numbers have been switched round in pairs, as follows: 1 ←→ 3 4 ←→6 7 ←→9. The numbers 2, 5 and 8 are unchanged. We consider the effect of these interchanges on each of the given calculations in turn. 79×97 becomes 97×79 . We have 97×79 =79 ×97 . 78 ×98 becomes 98×78 . We have 98×78 =78 ×98 . 147 ×369 becomes 369×147 . We have 369×147 =147 ×369 . 123 ×321 becomes 321×123 We have 321×123 =123 ×321 . 159 ×951 becomes 357×753 . However 753×357 ,159 ×951 . Therefore it is calculation E that does not give the correct answer. 13. The diagram shows a rhombus and two sizes of regular hexagon. What is the ratio of the area of the smaller hexagon to the area of the larger hexagon? A 1 : 2 B1 : 3 C1 : 4 D1 : 8 E1 : 9 Solution C From the figure in the question we see that the rhombus and the smaller hexagon have the same side lengths. We also see that the side length of the larger hexagon is equal to the sum of the side lengths of the rhombus and the smaller hexagon. Therefore the ratio of the side length of the smaller hexagon to that of the larger hexagon is 1 : 2. Hence the ratio of the area of the smaller hexagon to that of the larger hexagon is 12: 2 2, that is, 1 : 4 . For investigation 13.1 Explain why it is true that if the lengths of corresponding edges of two similar polygons are in the ratio a:b, then their areas are in the ratio a2 :b2 . © UKMT January 2018 www.ukmt.org.uk 111122 33 44 5566 77 88 99

Intermediate Mathematical Challenge 2018 Solutions and investigations14. Which of these is equal to 10 9 + 9 10 ? A 1B 2C 2.0 ? 1 D2.? 1 E2.? 2 Solution C Putting the two fractions over a common denominator, we have 10 9 + 9 10 = 10 ×10 +9× 9 90 = 100 +81 90 = 181 90 = 2+ 1 90 . Now, as a decimal, 1 9 is the recurring decimal 0.? 1 . Hence, as a decimal, 1 90 is 0.0 ? 1 . Therefore, 2 + 1 90 = 2+ 0.0 ? 1 = 2.0 ? 1 . For investigation 14.1 Work out each of the fractions 10 9 and 9 10 as decimals. Then add the two decimals. 15.How many of these four shapes could be the shape of the region where two triangles overlap? equilateral triangle square regular pentagon regular hexagon A 0B 1C 2D 3E 4 Solution E We see from the figures below that each of the four given shapes can be obtained as the intersection of two triangles. equilateral triangle square regular pentagon regular hexagon For investigation 15.1 Find examples of shapes that cannot be the shape of the region where two triangles overlap. © UKMT January 2018 www.ukmt.org.uk 12

Intermediate Mathematical Challenge 2018 Solutions and investigations16.The diagram shows a triangle with edges of length 3, 4 and 6. A circle of radius 1 is drawn at each vertex of the triangle. What is the total shaded area? A 2π B9 π 4 C 5 π 2 D 11 π 4 E 3π Solution C We let the vertices of the triangle be X,Y,Z and the angles be x°, y°, z°, as shown in the figure. Because the circle with centre X has radius 1, it has area π12, that is, π. Of this area, the fraction that is not shaded is x 360 . Hence the fraction that is shaded is 1− x 360 . Hence the area that is shaded is  1− x 360  π . Similarly the areas of the other two circles that are shaded are  1− y 360  π and  1 − z 360  π . It follows that the total area that is shaded is given by 1 − x 360  π +  1 − y 360  π +  1 − z 360  π =  3 − x + y+ z 360  π . Now, because x°, y° and z° are the three angles of a triangle, x+ y+ z= 180 . Hence  3 − x + y+ z 360  π =  3 − 180 360  π =  3 − 1 2  π = 5 π 2 . Hence the total shaded area is 5 π 2 . [Note that this answer is the same whatever the lengths of the edges of the triangle provided that they are sufficiently large that the circles do not overlap.] For investigation 16.1 The figure shows four circles each with radius 1 whose centres are the vertices of a quadrilateral. What is the total shaded area? 16.2 What is the total shaded area in the general case where the quadrilateral is replaced by a polygon with nvertices? © UKMT January 2018 www.ukmt.org.uk 13

Intermediate Mathematical Challenge 2018 Solutions and investigations17. How many three-digit numbers are increased by 99 when their digits are reversed? A 4B 8C 10D 80E 90 Solution DWe use the notation ‘ cba ’ to denote the number with hundreds digit c, tens digit band units digit a. Thus ‘ cba’ denotes the number 100c+ 10 b+ a. The number obtained by reversing the digits of ‘cba ’ is ‘abc ’ . The increase that results when its digits are reversed is therefore ‘abc ’− ‘cba ’= (100 a+ 10 b+ c) − ( 100c+ 10 b+ a) = 99 (a − c). Hence, the condition that the three-digit number ‘cba ’ is increased by 99 when its digits are reversed is that a− c= 1, that is, a= c+ 1. Now let ‘cba ’ be one such three-digit number. The hundreds digit ccannot be 0. Because a≤ 9, and a= c+ 1, we see that c ≤ 8. Therefore there are 8 choices for the hundreds digit c, namely, 1, 2, 3, 4, 5, 6, 7 and 8. The tens digit bcan be any of the 10 digits. However, once chas been chosen, there is just l choice for a, since a= c+ 1. Therefore the total number of choices for a three-digit number ‘cba ’ which is increased by 99 when its digits are reversed is 8× 10 ×1= 80 . For investigation 17.1 How many three-digit numbers are increased by 100 when their digits are reversed? 17.2 How many three-digit numbers are increased by 198 when their digits are reversed? 17.3 How many four-digit numbers are increased by 1089 when their digits are reversed? 17.4 Let nbe a three-digit number, and let n ∗ be the number obtained from nby reversing the order of its digits. What values can n− n∗ take? 17.5 Let nbe a four-digit number, and let n ∗ be the number obtained from nby reversing the order of its digits. What values can n− n∗ take? © UKMT January 2018 www.ukmt.org.uk 14

Intermediate Mathematical Challenge 2018 Solutions and investigations18.The diagram shows a regular pentagon and an equilateral triangle placed inside a square. What is the value of x? A 24B 26C 28D 30E 32 Solution A We let P,Q,R,S,T and U be the vertices as labelled in the figure on the right. Our strategy is to calculate all the angles in the hexagon PQ RST U other than the angle x ° . We can then use the fact that the sum of the angles in a hexagon is 720 °to work out the value of x. We let ˆ P,ˆ Q,ˆ R,ˆ S,ˆ Tand ˆ Ube the interior angles of the hexagon PQ RST U at the corresponding vertices. First, ˆ P = ˆ Q =90 °because they are angles of a square. Because it is an exterior angle of a regular pentagon, ˆ R = 72 °. Because the interior angle of a regular pentagon is 108°, we have ˆ S = 360 °− 108 °= 252 °. Because the interior angle of a regular pentagon is 108 °and the interior angle of an equilateral triangle is 60°, we have ˆ T =360 °− ( 108 °+ 60 °) = 192 °. Finally, ˆ U =x°. Since the sum of the angles in a hexagon is 720°, it follows that ˆ P + ˆ Q + ˆ R + ˆ S + ˆ T + ˆ U =720 °. Therefore, by the above facts about the angles in this equation, 90°+ 90 °+ 72 °+ 252 °+ 192 °+ x° = 720 °, and hence 696+x= 720 . It follows that x= 24 . For investigation 18.1 Prove the fact used in the above solution that the exterior angle of a regular pentagon is 72 °. 18.2 Prove the fact used in the above solution that the sum of the angles in a hexagon is 720°. © UKMT January 2018 www.ukmt.org.uk 15x °

Intermediate Mathematical Challenge 2018 Solutions and investigations19. The three rectangles shown below all have the same area. What is the value of x+ y? A 4B 6C 8D 10E 12 Solution D Method 1Because the rectangle with side lengths x + 4and y− 3has the same area as the rectangle with side lengths xand y, it follows that (x + 4)( y− 3) = xy. (1 ) Similarly, because the rectangle with side lengths x + 8 and y− 4 has the same area as the rectangle with side lengths xand y, ( x + 8)( y− 4) = xy. (2 ) By expanding the left hand sides of equations (1) and (2), we obtain xy − 3x + 4y − 12 =xy (3 ) and xy − 4x + 8y − 32 =xy. (4 ) By subtracting xy from both sides of equations (3) and (4), it follows that −3x + 4y = 12 (5 ) and −4x + 8y = 32 . (6 ) We now subtract twice equation (5) from equation (6). This gives (−4x + 8y) − (− 6x + 8y) = 32 −24 . That is, 2x = 8, and hence x= 4. Substituting this value for xin equation (5) we deduce that −12 +4y = 12 . Hence 4y = 24 and it follows that y= 6. Therefore x+ y= 4+ 6= 10 . © UKMT January 2018 www.ukmt.org.uk 16x y x + 4 y 3 x + 8 y 4

Intermediate Mathematical Challenge 2018 Solutions and investigationsMethod 2In the figure on the right we have drawn the rectangle with side lengths x + 4and y− 3overlapping the rectangle with side lengths xand y. This produces three rectangles. We let their areas be R,S and T, as indicated in the figure. As the two original rectangles have the same area, R + S= R+T , and therefore S= T. We see from the figure that Sis the area of a rectangle with side lengths xand 3, and T is the area of a rectangle with side lengths 4 and y− 3. Therefore 3 x = 4(y − 3). This equation may be rearranged as −3x + 4y = 12 . (1 ) Similarly, by drawing the rectangle with side lengths x + 8 and y− 4overlapping the rectangle with side lengths xand y, as shown in the figure on the right, we obtain a rectangle with side lengths 4 and x, and a rectangle with side lengths 8 and y− 4which have equal areas. Therefore 4x = 8(y − 4), and hence −4x + 8y = 32 . (2 ) Equations (1) and (2) are the same as equations (5) and (6) of Method 1. Therefore, as in Method 1, we may deduce from these equations that x+ y= 4+ 6= 10 . For investigation 19.1 The three rectangles shown below all have the same area. Find the values of xand y. © UKMT January 2018 www.ukmt.org.uk 17

Intermediate Mathematical Challenge 2018 Solutions and investigations20. A particular integer is the smallest multiple of 72, each of whose digits is either 0 or 1. How many digits does this integer have? A 4B 6C 8D 10E 12 Solution EIn this question we use the notation ‘ . . . cba ’ for the positive integer which has units digit a, tens digit b, hundreds digit c, and so on. We first note that 72 = 8× 9. Hence, as 8 and 9 have no common factors (other than 1), the test for whether a number is divisible by 72 is that it should be divisible both by 8 and by 9. Next we consider the criteria for divisibility by 8 and by 9. These are as follows. • For the number ‘ . . . cba ’ to be divisible by 8, the number ‘cba ’ made up of its last three digits has to be divisible by 8. • For the number ‘ . . .cba ’ to be divisible by 9, the sum of its digits has to be divisible by 9. It is easy to check that if each of the digits a,band cis either 0 or 1, and ‘cba ’ is divisible by 8, then ‘ cba’= ‘000 ’. The sum of the digits of an integer each of whose digits is 0 or 1 is equal to the number of its digits that are 1s. So if the integer is to be divisible by 9, the number of its digits that are 1s has to be a multiple of 9. The fewer the number of digits, the smaller the number will be. The smallest positive multiple of 9 is, of course, 9 itself. So the integer we seek is the integer with the smallest number of digits which ends with three 0s and includes nine 1s. This the integer which consists of nine 1s followed by three 0s. Therefore the required integer is 111 111 111 000. This has 12 digits.For investigation 20.1 Check the truth of the statement in the above argument that if each of the digits a,band c is either 0 or 1, and ‘ cba’ is divisible by 8, then ‘ cba’= ‘000 ’. 20.2 Explain why the criterion given in the above argument for a number to be divisible by 8 is correct. 20.3 Explain why the criterion given in the above argument for a number to be divisible by 9 is correct. 20.4 Which is the smallest integer, each of whose digits is either 0 or 1, which is greater than 111 111 111 000 and is divisible by 72? 20.5 Factorize 111 111 111 000 into prime factors. © UKMT January 2018 www.ukmt.org.uk 18

Intermediate Mathematical Challenge 2018 Solutions and investigations21. For certain values of x, the list x, x + 6and x2 contains just two different numbers. How many such values of xare there? A 1B 2C 3D 4E 5 Solution DWhatever value xtakes, xis not equal to x + 6. Therefore, the values of xfor which there are just two different numbers in the list are those for which either x2 = xor x2 = x+ 6. The equation x2 = x may be rearranged as x2 − x= 0. The left-hand side of this last equation may be factorized to give x(x − 1) = 0. From this we see that there are two values of xfor which x2 = x, namely 0 and 1. The equation x2 = x+ 6 may be rearranged as x2 − x− 6= 0. The left-hand side of this last equation may be factorized to give (x + 2)( x− 3) = 0. From this we see that there are two values of xfor which x2 = x+ 6, namely −2 and 3. Therefore there are altogether 4 different values for xfor which there are just two different numbers in the list, namely, −2, 0, 1 and 3. © UKMT January 2018 www.ukmt.org.uk 19

Intermediate Mathematical Challenge 2018 Solutions and investigations22.Three squares, with side-lengths 2, are placed together edge-to-edge to make an L-shape. The L-shape is placed inside a rectangle so that all five vertices of the L-shape lie on the rectangle, one of them at the midpoint of an edge, as shown. What is the area of the rectangle? A 16B 18C 20D 22E 24 Solution E Method 1 Let d be the length of the diagonals of the squares. Each diagonal is the hypotenuse of a right-angled triangle whose other sides are each of length 2. Therefore, by Pythagoras’ Theorem d 2 = 22 + 22 = 8. Hence d= √ 8 = 2√ 2 . We therefore see from the figure that the width of the rectangle is 2 √ 2 + √ 2 , that is, 3 √ 2 . Also, the height of the rectangle is 2 √ 2 + 2√ 2 , that is, 4√ 2 . Hence the area of the rectangle is given by 3√ 2 × 4√ 2 = 3× 4× √ 2 2 = 3× 4× 2= 24 . Method 2 We divide the rectangle into 12 congruent smaller squares as shown in the figure. From the figure we see that the three 2×2squares cover 12 triangles each of which is half of one of 12 smaller squares. It follows that the three larger squares cover half the area of the rectangle. Therefore the area of the rectangle is twice the area of the three 2× 2squares. Therefore the area of the rectangle is 2× 3 × ( 2× 2)
= 24 . For investigation 22.1 Both the above methods tacitly assume that the angles between the sides of the squares and the sides of the rectangle are all 45°. Prove that this assumption is correct. © UKMT January 2018 www.ukmt.org.uk 20

Intermediate Mathematical Challenge 2018 Solutions and investigations23.The diagram shows a hexagon. All the interior angles of the hexagon are 120 °. The lengths of some of the sides are indicated. What is the area of the hexagon? A 20√ 3 B21√ 3 C22√ 3 D23√ 3 E 24√ 3 Solution A We give two methods. Both use the fact that the area of an equilateral triangle with side length s is √ 3 4 s2 . The proof of this is left as an exercise (See Problem 23.1). Method 1 The solution by Method 1 may seem rather long. But this is only because we have given the geometrical details. In the context of IMC, where you are not asked to explain your solutions, the answer could be obtained using this method quite quickly. We have labelled the vertices of the hexagon as shown in the figure. The points where the sides of the hexagon Q P ,RS and T U meet when extended have been labelled X,Y and Z, as shown. Because all the interior angles of the hexagon are 120 °and the angles at a point have sum 180 °, we have ∠ RQY =∠Q RY = 60 °. Because the sum of the angles in a triangle is 180 °, it follows that ∠RY Q = 60 °. Therefore the triangle QY R is equilateral, and hence Y R=Y Q =Q R =1. Similarly, the triangle T S Z is equilateral and S Z =ZT =ST =2. Similarly, the triangle X PUis equilateral. We work out the length of its sides as follows. We see that Y Z has length 1+8+2, that is, 11. Also, because ∠RY Q =∠T Z S = 60 °, the triangle XY Z is equilateral. Therefore XY=X Z =Y Z = 11 . It follows that P X = 6. Hence X U and PUalso have length 6. The area of the hexagon PQ RST U is equal to the area of the equilateral triangle XY Z , less the areas of the equilateral triangles QY R ,T S Z and X PU . Therefore, using the formula √ 3 4 s2 for the area of an equilateral triangle with side length s, we deduce that the area of the hexagon is given by √ 3 4 ( 11 2 ) − √ 3 4 ( 1 2 ) − √ 3 4 ( 2 2 ) − √ 3 4 ( 6 2 ) = √ 3 4 ( 11 2 − 12 − 22 − 62 ) = √ 3 4 ( 121 −1− 4− 36 )= √ 3 4 ( 80 )= 20 √ 3 . © UKMT January 2018 www.ukmt.org.uk 21

Intermediate Mathematical Challenge 2018 Solutions and investigationsMethod 2We divide the hexagon into equilateral triangles all of which have side length 1, as shown in the figure. By counting we see that there are 80 of these triangles. It follows from the formula √ 3 4 s2 for the area of an equilateral triangle with side length sthat each of 80 equilateral triangles with side length 1 has has area √ 3 4 . Therefore the total area of these triangles is 80×√ 3 4 = 20 √ 3 . For investigation 23.1 Prove that the area of an equilateral triangle with side length sis √ 3 4 s2 . [A proof using the formula 1 2 ab sin θ for the area of a triangle with two sides with lengths aand band included angle θis complete only if you show that this formula is correct, and that sin 60 °= √ 3 2 .] 23.2 The hexagon of this question is a hexagon all of whose interior angles are 120 °but all of whose sides have different lengths. Give another example of a hexagon with these properties. 23.3 Find a condition on a sequence a,b,c,d,e,fof positive integers for there to be a hexagon all of whose angles are 120 °and whose side lengths are a,b,c,d,eand fas you go round the hexagon anticlockwise. [Note that the existence of regular hexagons shows that, for each positive integer k, the sequence k,k,k,k,k,khas the required property. Also, the hexagon of this question shows that the sequence 1, 8, 2, 3, 6, 4 has this property.] © UKMT January 2018 www.ukmt.org.uk 22

Intermediate Mathematical Challenge 2018 Solutions and investigations24. A list of 5 positive integers has mean 5, mode 5, median 5 and range 5. How many such lists of 5 positive integers are there? A 1B 2C 3D 4E 5 Solution B We suppose that the positive integers in the list are a, b, c, d and e, where a≤ b≤ c≤ d≤ e. Because the median is 5, we have c= 5.Because the mode is 5, there must be at least two 5s in the list. Therefore, either b = 5or d = 5. Because the range is 5, we have e= a+ 5. Hence the list is either a , 5, 5, d, a + 5 or a, b, 5, 5,a + 5. Because the 5 numbers in the list have mean 5, their sum is 5× 5, that is, 25. Therefore in the first case a + 5+ 5+ d+ (a + 5)= 25 and hence d = 10 − 2a, and in the second case, a+ b+ 5+ 5+ (a + 5) = 25 and hence b= 10 −2a . Hence the list is either a, 5, 5, 10 −2a, a + 5 or a, 10 −2a, 5, 5,a + 5. In the first case we have 5≤ 10 − 2a ≤ a+ 5, and hence 2a ≤ 5and 5≤ 3a. It follows that, as a is an integer, 2 is the only possible value for a. In the second case we have a ≤ 10 −2a≤ 5, and hence 3a ≤ 10 and 5≤ 2a. It follows that, as a is an integer, 3 is the only possible value for a. It follows that the only lists that satisfy the given conditions are 2, 5, 5, 6, 7 and 3, 4, 5, 5, 8. Therefore there are just 2 lists that meet the given conditions. For investigation 24.1 How many lists are there of 6 positive integers which have mean 6, mode 6, median 6 and range 6? 24.2 How many lists are there of 7 positive integers which have mean 7, mode 7, median 7 and range 7? © UKMT January 2018 www.ukmt.org.uk 23

Intermediate Mathematical Challenge 2018 Solutions and investigations25.The diagram shows two equilateral triangles. The distance from each point of the smaller triangle to the nearest point of the larger triangle is √ 3 , as shown. What is the difference between the lengths of the edges of the two triangles? A 2√ 3 B41 2 C 3√ 3 D 6 E 4√ 3 Solution D We label the vertices of the two triangles as shown in the figure. We let V and W , respectively, be the points where the perpendiculars from Tand UtoQ R meet Q R. Because each point of the triangle ST U is at the distance √ 3 from the nearest point of the triangle PQ R , we have T V =UW =√ 3 . Similarly, the distance from Tto PQ is√ 3 . Therefore Tis equidistant from the sides Q R and Q P of the triangle PQ R . Hence the line QT bisects the angle of the equilateral triangle at Q. [You are asked to prove this in Problem 25.1.] Therefore ∠V QT =30 °. It follows that the triangle T QV has one right angle and its other angles are 30 °and 60 °. Therefore QT =2T V =2√ 3 . By Pythagoras’ Theorem applied to this triangle, QT 2 = QV 2 + T V 2 . It follows that QV 2 = QT 2 − T V 2 = (2 √ 3 )2 − √ 3 2 = 12 −3= 9. Hence QV=3. Similarly W R=3. Because T V W Uis a rectangle, T U=V W . Hence, we have Q R =QV +V W +W R =3+ T U +3= T U +6. It follows that the difference between the lengths of the edges of the two triangles is 6. For investigation 25.1 Prove that, because T is equidistant from the sides Q R and Q P of the triangle PQ R , it follows that QTbisects the angle of the triangle at Q. 25.2 Explain why it follows from the facts that ∠ QV T = 90 °and ∠T QV = 30 °that QT= 2T V . 25.3 Prove that T V W Uis a rectangle. © UKMT January 2018 www.ukmt.org.uk 24p 3