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I ntermediate Mathematical Challenge Thursday 5th February 2015 Organised by the United Kingdom Mathematics Trust supported by S olutions 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 sta ffand 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 2015] 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 A 2 E 3 E 4 C 5 B 6 D 7 E 8 B 9 B 10 C 11 D 12 D 13 A 14 C 15 C 16 D 17 D 18 A 19 A 20 C 21 D 22 B 23 B 24 E 25 DUKMT UKMTUKMT

I ntermediate Mathematical Challenge 2015 Solutions and investigations1. What is the value of 1 −0.2 + 0.03 −0.004? A 0.826B 0.834C 0.926D 1.226E 1.234 S olution AWe have 1 − 0.2= 0.8 and so 1 − 0.2+ 0.03 = 0.8+ 0.03 = 0.83, and therefore 1 − 0.2+ 0.03 − 0 .004 =0.83 −0.004 =0.826. Note that in the context of the IMC where we just need to decide which of the options is correct, it isn’t really necessary to do this exact arithmetic. Looking at the terms making up the sum it is evident that the answer is just greater than 0.8, and the presence of the term −0.004 shows that the digit in the thousandths column is 6. So A is the correct option. F or investigation 1.1 What is the value of 1 −0.2+0.03 −0.004 +0.000 5 −0.000 06 +0.000 007 −0.000 000 8 + 0 .000 000 09? 2. Last year, Australian Suzy Walsham won the annual women’s race up the 1576 steps of the Empire State Building in New York for a record fifth time. Her winning time was 11 minutes 57 seconds. Approximately how many steps did she climb per minute? A 13B 20C 80D 100E 130 S olution E We see that 11 minutes 53 seconds is just under 12 minutes, and 1576 is just under 1600. So the number of steps per minute is approximately 1600 12 = 400 3 ∼ 133 . We conclude that, of the given options, 130 is the best approximation. [There are other methods.] 3. What is a half of a third, plus a third of a quarter, plus a quarter of a fifth? A 1 1440 B 3 38 C 1 30 D 1 3 E 3 10 S olution E A half of a third is 1 2 × 1 3 = 1 6 , a third of a quarter is 1 3 × 1 4 = 1 12 , and a quarter of a fifth is 1 4 × 1 5 = 1 20 . It follows that the required answer is 1 6 + 1 12 + 1 20 = 10 +5+ 3 60 = 18 60 = 3 10 . © UKMT February 2015 www.ukmt.org.uk 2

I ntermediate Mathematical Challenge 2015 Solutions and investigations4. The diagram shows a regular pentagon inside a square. What is the value of x? A 48B 51C 54D 60E 72 S olution C M ethod 1Let the vertices of the square be labelled K,L,M ,N and those of the regular pentagon P,Q ,R ,S,T , as shown in the diagram. We use the fact that the interior angles of a regular pentagon are each 108 °. You are asked to show this in Problem 4.1. It follows that ∠RST =∠ST P = 108 °. Because it is the angle of a square, ∠S N T = 90 °. By the Exterior Angle Theorem, ∠RST =∠S N T + ∠ N T S . Therefore, 108 ° = 90 °+ ∠N T S . It follows that ∠N T S = 108° −90° =18°. Finally, because the angles at Ton the line K Nhave sum 180°, ∠ N T S +∠ST P +∠PT K =180° , that is, 18°+108° +x° = 180° . We conclude that x= 180 −18 −108 =54 . M ethod 2 We extend STto the point Uas shown. The external angle of a regular pentagon is 72 °. Hence ∠T S N = ∠ PT U = 72 °. Therefore, from the right-angled triangle ST N , we can deduce that ∠ST N = 90 °− 72 ° = 18 °. So the vertically opposite angle ∠K T U is also 18 °. It follows that ∠ PT K =∠PT U −∠K T U = 72 °− 18 ° = 54 °. Therefore x = 54. F or investigation 4.1 Show that the interior angles of a regular pentagon are each 108 °and that the exterior angles are each 72°. [Note that it might be helpful to look at Problem 9.1, below.] 4.2 Find a formula for the interior angles of a regular polygon with nsides. [Again, Problem 9.1, below, is relevant.] 4.3 What would xbe in the case where the regular pentagon is replaced by (a) a regular hexagon? (b) a regular heptagon? (c) a regular octagon? 4.4 What happens in the case where the regular pentagon is replaced by a regular nonagon? [Note that a nonagonis a polygon with 9 sides.] © UKMT February 2015 www.ukmt.org.uk 3x ° x ° KL MN RS T P Q x ° KL MN RS T U P Q

I ntermediate Mathematical Challenge 2015 Solutions and investigations5. Which of the following numbers is not a square? A 1 6 B 25 C 34 D 43 E 52 S olution B We have 2 5 = 32 which is not a square.In the context of the IMC we can assume that only one option is correct and that it is su fficient just to find it. However, for a complete solution we need to check that none of the other options is a square. Note that we can show this without evaluating any of them, as follows. We have 16 = ( 13 )2 , 3 4 = ( 32 )2 , 4 3 = ( 22 )3 = 26= ( 23 )2 , and so these are all squares, and it is immediate that 5 2 is a square. F or investigation 5.1 Which of the following are squares:(a)2015 2015 ,(b)49 2015 ,(c)2015 2016 ? 5.2 Suppose that the positive integer m is a square. For which positive integers nis m n a square? 5.3 Suppose that the positive integer m is not a square. For which positive integers nis m n a square? 6. The equilateral triangle and regular hexagon shown have perimeters of the same length. What is the ratio of the area of the triangle to the area of the hexagon? A 5 : 6B 4 : 5C 3 : 4D 2 : 3E 1 : 1 S olution D We can assume that we have chosen units so that each side of the regular hexagon has length 1. It follows that the length of the perimeter of this hexagon is 6. Therefore the length of the perimeter of the equilateral triangle is 6. Hence each of its sides has length 2. It follows that we can divide the equilateral triangle into 4 smaller equilateral triangles each with side length 1 as shown. The hexagon may be divided into 6 equilateral triangles with side length 1 as shown. It follows that the ratio of their areas is 4 : 6, that is, 2 : 3. F or investigation 6.1 What is the area of an equilateral triangle with side length a? 6.2 What is the area of an equilateral triangle with a perimeter of length 6 a? 6.3 What is the area of a regular hexagon with a perimeter of length 6 a? 6.4 Check that your answers to 6.2 and 6.3 are in the ratio 2 : 3. © UKMT February 2015 www.ukmt.org.uk 4

I ntermediate Mathematical Challenge 2015 Solutions and investigations7.A tetrahedron is a solid figure which has four faces, all of which are triangles. What is the product of the number of edges and the number of vertices of the tetrahedron? A 8B 10C 12D 18E 24 S olution E We can think of a tetrahedron as a pyramid with a triangular base. The base has 3 vertices and there is 1 more vertex, called the apex , at the peak of the pyramid. So a tetrahedron has 4 vertices. The base has 3 edges, and there are 3 more edges joining the vertices of the base to the apex. Therefore a tetrahedron has 3 +3= 6 edges. Finally, we observe that 4 ×6= 24. 8. How many two-digit squares di ffer by 1 from a multiple of 10? A 1B 2C 3D 4E 5 S olution B The two-digit squares are 16, 25, 36, 49, 64 and 81. Of these, just 49 =50 −1 and 81 =80 +1 di ffer by 1 from a multiple of 10. So there are 2 such two-digit squares. F or investigation 8.1 How many three-digit squares di ffer by 1 from a multiple of 10? 8.2 How many four-digit squares di ffer by 1 from a multiple of 10? 8.3 How many ten-digit squares di ffer by 1 from a multiple of 10? 8.4 Let nbe a positive integer. How many squares with 2 n − 1 or 2 ndigits di ffer by 1 from a multiple of 10? © UKMT February 2015 www.ukmt.org.uk 5

I ntermediate Mathematical Challenge 2015 Solutions and investigations9.What is the value of p + q+ r+ s+ t+ u+ v+ w +x+ y in the diagram? A 540B 720C 900D 1080E 1440 S olution B We use the fact that for every polygon the sum of its exterior angles is 360 °. See the note below for how this statement should be interpreted. The shaded angles in the diagram on the right form one set of external angles. Therefore q+ s+ u+ w +y= 360 . The other marked angles form another set of external angles and therefore r+ t+ v+ x+ p= 360 . [Alternatively, we could argue that the angles marked q° and r° are vertically opposite and therefore q= r, and, similarly, s= t, u = v, w =xand p= y, and therefore r+ t+ v+ x+ p= q + s+ u+ w +y= 360.] It follows that p+ q+ r+ s+ t+ u+ v+ w +x+ y+ z= (q + s+ u+ w +y) + (r + t+ v+ x+ p) = 360 +360 = 720 . N ote The figure on the right shows part of the pentagon. One of the interior angles of this pentagon has been marked a° . There are two exterior angles associated with this interior angle, those marked q° and r° . When we talk about the sum of the exterior angles we mean the sum obtained by choosing, for each vertex, just one of the two exterior angles associated with it. Because these are vertically opposite angles, they are equal, and so it does not matter which one we choose. For example, with the pentagon of this question both q + s+ u+ w +y and r+ t+ v+ x+ pgive the sum of the exterior angles. F or investigation 9.1 Show that for every polygon the sum of its exterior angles is 360°. © UKMT February 2015 www.ukmt.org.uk 6q ° p ° s ° r ° u° t ° w ° v ° y ° x° q ° p ° s ° r ° u° t ° w ° v ° y ° x° q ° r° a °

I ntermediate Mathematical Challenge 2015 Solutions and investigations10. What is the remainder when 2 2 × 33 × 55 × 77 is divided by 8? A 2B 3C 4D 5E 7 S olution CFirst, we note that 3 3 × 55× 77is a positive odd integer greater than 1 and so is equal to 2 n + 1, for some positive integer n. It follows that 22 × 33 × 55 × 77 = 4× (3 3 × 55 × 77 ) = 4(2 n + 1) = 8n + 4. Therefore when 2 2 × 33 × 55 × 77 is divided by 8, the quotient is nand the remainder is 4. F or investigation 10.1 What is the remainder when 2 2 × 33 × 55 × 77 × 11 11 is divided by 8? 10.2 What is the remainder when 2 2 × 33 × 55 × 77 is divided by 16? 11. Three di fferent positive integers have a mean of 7. What is the largest positive integer that could be one of them? A 15B 16C 17D 18E 19 S olution D Because the three integers have mean 7, their sum is 3 × 7 = 21. For the largest of the three positive integers to be as large as possible the other two need to be as small as possible. Since they are di fferent positive integers, the smallest they can be is 1 and 2. The largest of the three integers is then 21 −(1 + 2) = 18. So 18 is the largest any of them could be. F or investigation 11.1 Five di fferent positive integers have a mean of 10. What is the largest positive integer that could be one of them? 11.2 A set of di fferent positive integers has mean 10. The largest number in the set is 50. What is the maximum number of positive integers that there could be in the set? © UKMT February 2015 www.ukmt.org.uk 7

I ntermediate Mathematical Challenge 2015 Solutions and investigations12.An ant is on the square marked with a black dot. The ant moves across an edge from one square to an adjacent square four times and then stops. How many of the possible finishing squares are black? A 0B 2C 4D 6E 8 S olution D The figure shows the part of the layout that is relevant. Some of the squares have been labelled so that we may refer to them. A movement of the ant will be indicated by the sequence of labels of the four squares, in order, that the ant travels to after leaving the square labelled P. For example, QTUV is one such sequence. In this example the ant ends up on the black square V. The ant starts on a white square. If at each move the colour of the square it is on changes, after four moves it will end up on a white square. So the only way the ant can end up on a black square is if on one of its moves the colour of the square does not change. That is, during its sequence of moves it either moves from T to U, or from U to T, or from W to X, or from X to W. We therefore see that the only movements which result in the ant ending up on a black square are QTUR, QTUV, QTUX, QTWX, RUTQ, RUTS, RUTW and RUXW. From the final letters of these sequences we see that there are exactly 6 black squares that the ant can end up on after four moves, namely, Q, R, S, V, W and X. 13. What is the area of the shaded region in the rectangle? A 21 cm 2 B 22 cm2 C 23 cm2 D 24 cm 2 E more information needed S olution A We divide the rectangle into four smaller rectangles by the vertical dashed lines as shown in the figure. It is then evident that the shaded area is the sum of exactly half the area of each of the smaller rectangles. Therefore the total shaded area is half the total area of the original rectangle. So the shaded area is 1 2 ( 3 × 14 )cm 2 = 21 cm 2 . We could also solve this question algebraically. We let the top edges of the two shaded triangles have lengths a cm and bcm , respectively, so that a + b= 14. The two shaded triangles have areas 1 2 ( a × 3)cm 2 = 3 2 a cm 2 and 1 2 ( b × 3)cm 2 = 3 2 b cm 2 , respectively. Hence their total area is  3 2 a + 3 2 b  cm 2 = 3 2 ( a + b) cm 2 = 3 2 × 14 cm 2 = 21 cm 2 . © UKMT February 2015 www.ukmt.org.uk 8 PQ RS T U VW X 14 cm3 cm

I ntermediate Mathematical Challenge 2015 Solutions and investigations14.In a sequence, each term after the first two terms is the mean of all the terms which come before that term. The first term is 8 and the tenth term is 26. What is the second term? A 17B 18C 44D 52E 68 S olution C Here it is actually a little easier to see what is going on if we think about the general case of a sequence in which every term after the second is the mean of all the terms that come before it. Consider a sequence generated by this rule whose first two terms are aand b. The mean of these terms is a + b 2 , and so the first three terms are a, b, a + b 2 . The sum of these three terms is the same as the sum of the three terms a+ b 2 , a + b 2 , a + b 2 and it follows that the mean of the first three terms is also a + b 2 . So the first four terms of the sequence are a, b, a + b 2 , a + b 2 . These have the same sum as the four terms a+ b 2 , a + b 2 , a + b 2 , a + b 2 . It follows that the mean of the first four terms is also a + b 2 and that therefore this is the fifth term of the sequence, and so on. We thus see that each term of the sequence after the first two is equal to the mean of the first two terms. In the sequence we are given in the question the first term is 8 and the tenth term is 26. It follows that 26 is the mean of the first two terms. Hence their sum is 2 × 26 = 52. Since the first term is 8, we deduce that the second term is 52 −8= 44. © UKMT February 2015 www.ukmt.org.uk 9

I ntermediate Mathematical Challenge 2015 Solutions and investigations15.A flag is in the shape of a right-angled triangle, as shown, with the horizontal and vertical sides being of length 72 cm and 24 cm respectively. The flag is divided into 6 vertical stripes of equal width. What, in cm 2 , is the di fference between the areas of any two adjacent stripes? A 96B 72C 48D 32E 24 S olution C The lines we have added in the figure on the right divide the flag into 15 congruent rectangles and 6 congruent triangles. Let the area of each of the 15 rectangles be x cm 2 . The area of each of the 6 congruent triangles is half that of the rectangles, that is, 1 2 x cm 2 . The di fference between the area of the adjacent vertical stripes of the flag is the area of one of the rectangles. From its dimensions, we see that the area of the flag is 1 2 ( 72 × 24 ) cm 2 = ( 36 × 24 )cm 2 . It is made up of 15 rectangles, each with area xcm 2 , and 6 triangles, each with area 1 2 x cm 2 . Therefore 15x+ 6(1 2 x ) = 36 ×24 , that is, 18x= 36 ×24 , and it follows that x= 36 ×24 18 = 2× 24 =48 . Hence the di fference between the area of adjacent stripes of the flag is 48 cm 2 . F or investigation 15.1 If the flag were divided into 8 vertical stripes of equal width, what would be the di fference between the areas of any two adjacent stripes? © UKMT February 2015 www.ukmt.org.uk 1072 cm 24 cm 72 cm 24 cm

I ntermediate Mathematical Challenge 2015 Solutions and investigations16.You are asked to choose two positive integers, m and nwith m >n , so that as many as possible of the expressions m+n, m −n, m ×nand m÷nhave values that are prime. When you do this correctly, how many of these four expressions have values that are prime? A 0B 1C 2D 3E 4 S olution D With m = 2 and n= 1, we have m +n= 3, m −n= 1, m ×n= 2 and m ÷n= 2. So it is possible for three of the expressions to have values that are prime. [Note that 1 is not a prime, so in this case m−nis not prime. See the Notebelow for a discussion of this point.] We now show that this is the best that can be achieved by showing that it is not possible for all four of the expressions to have values that are primes. Since m and nare positive integers with m>n , we have m > 1. Therefore m×n is prime if, and only if, m is prime and n= 1. Then m +n= m + 1 and m−n= m − 1. Therefore, if all four expressions are primes, m− 1, m and m + 1 would be three consecutive primes. However a trio of consecutive positive integers that are all primes does not exist (see Problem 16.3). We conclude that not all four of the expressions can have values that are primes. F or investigation 16.1 For which values of m and n, other than m = 2 and n= 1, do three of the expressions have values that are primes? 16.2 In our example with m = 2 and n= 1 three of the expressions have values that are primes, but these values include just two di fferent primes, namely, 2 and 3. Is it possible for three of the expressions m+n, m −n, m ×n, m ÷nto take three di fferent prime values? 16.3 Prove there is no positive integer msuch that m−1, mand m+1 are all primes. N ote The standard definition of a prime number is that it is a positive integer which has exactly two factors. You can see that, according to this definition, 1 is not a prime number. It is only a convention that makes us now adopt a definition according to which 1 is not a prime. (Earlier writers were not consistent about this.) However, it is a sensible convention. One reason, among many, for this is that it enables us to state The Fundamental Theorem of Arithmetic in a simple way. This theorem says that “each positive integer greater than 1 can be expressed as a product of prime numbers in just one way”. Of course, we can often write the prime factors of a given integer in more than one order. For example, 12 may be expressed as 2 × 2× 3, 2 × 3× 2 and 3 × 2× 2, but in each case the prime factor 2 occurs twice, and the prime factor 3 occurs once. This is what the Fundamental Theorem means by “in just one way”. We would need to state this theorem in a more complicated way if we regarded 1 as a prime number. (If 1 counted as a prime then, for example, 12 would have infinitely many di fferent factorizations into primes: 2 × 2× 3, 1 × 2× 2× 3, 1 × 1× 2× 2× 3, and so on, with any number of 1s.) © UKMT February 2015 www.ukmt.org.uk 11

I ntermediate Mathematical Challenge 2015 Solutions and investigations17.The football shown is made by sewing together 12 black pentagonal panels and 20 white hexagonal panels. There is a join wherever two panels meet along an edge. How many joins are there? A 20B 32C 60D 90E 180 S olution D Each of the 12 pentagons has 5 edges, and each of the 20 hexagons has 6 edges. So, altogether, the panels have 12 ×5+20 ×6= 180 edges. Each join connects 2 edges, So there are 1 2 ( 180 ) = 90 joins. N ote There is a good discussion of the shapes of footballs in the book Beating the odds: the hidden mathematics of sport by Rob Eastaway and John Haigh. The following problems are based on this book. F or investigation 17.1 When a football is inflated, the panels become curved, so a football is not a polyhedron with flat faces. However a football can be thought of as an inflated polyhedron. The football illustrated in this question is obtained by inflating a polyhedron made up of 12 regular pentagons and 20 regular hexagons. There are other possibilities. (a)An inflated dodecahedronmade up of 12 panels, all regular pentagons. (b)An inflated icosahedronmade up of 20 panels, all equilateral triangles. (c) An inflated rhombicosidodecahedron made up of 20 equilateral triangles, 30 squares and 12 regular pentagons. For each of these shapes, work out how many joins would be needed to make it. 17.2 The football in the question is obtained by inflating the polyhedron which is called a truncated icosahedron . Why does it have this name? 17.3 Why is the design in the question preferred by some manufacturers of footballs? © UKMT February 2015 www.ukmt.org.uk 12

I ntermediate Mathematical Challenge 2015 Solutions and investigations18.The total weight of a box, 20 plates and 30 cups is 4.8 kg. The total weight of the box, 40 plates and 50 cups is 8.4 kg. What is the total weight of the box, 10 plates and 20 cups? A 3 kgB 3.2 kgC 3.6 kgD 4 kgE 4.2 kg S olution A The di fference between the box, 20 plates, 30 cups and the box, 40 plates, 50 cups, is 20 plates and 20 cups. From the information in the question we see that the extra weight of 20 plates and 20 cups is 8 .4kg − 4.8kg = 3.6kg . Therefore the weight of 10 plates and 10 cups is 1 2 ( 3 .6 ) kg =1.8 kg. If we remove 10 plates and 10 cups from the box, 20 plates and 30 cups we end up with the box, 10 plates and 20 cups. Removing the 10 plates and 10 cups reduces the original weight of 4 .8kg by 1 .8 kg. Hence the weight of the box, 10 plates and 20 cups is 4 .8 kg −1.8 kg =3 kg. F or investigation 18.1 This problem could also have been solved using the language of algebra. We let x, y and zbe the weights, in kg, of the box, 1 plate and 1 cup, respectively. (a) Write down the two equations that correspond to the information given in the ques- tion. (b) Use these equations to find the value of x + 10 y+ 20 zwhich gives the weight, in kg, of the box, 10 plates and 20 cups. (c) Note that, as we have only two equations relating the three unknowns, x,yand z, it is not possible to solve them to obtain definite values for each of the unknowns. However, it is possible to use these equations to find the value of certain combinations of x,yand z. For example, we have already found the value of x + 10 y+ 20 z. Investigate the values of α,βand γfor which it is possible to calculate the value of αx + βy + γz. © UKMT February 2015 www.ukmt.org.uk 13

I ntermediate Mathematical Challenge 2015 Solutions and investigations19.The figure shows four smaller squares in the corners of a large square. The smaller squares have sides of length 1 cm , 2 cm , 3 cm and 4 cm (in anticlockwise order) and the sides of the large square have length 11 cm. What is the area of the shaded quadrilateral? A 35 cm 2 B 36 cm2 C 37 cm2 D 38 cm 2 E 39 cm2 S olution A We give two methods for answering this question. The first is a routine calculation. The second method is more elegant and involves fewer calculations. M ethod 1 We work out the area that is not shaded and subtract this area from that of the large square. The area that is not shaded is made up of the four corner squares and four trapeziums. The total area of the squares is, in cm 2 , 1 2 + 22 + 32 + 42 = 1+ 4+ 9+ 16 =30 . We now use the fact that the area of a trapezium is given by the formula 1 2 ( a + b)h , where aand bare the lengths of the parallel sides, and his their distance apart. You are asked to prove this in Problem 19.1. The values of a,band hfor the trapeziums P,Q,R and S, as labelled in the diagram, can be calculated from the side lengths of the squares, and are shown in the diagram. We therefore have that area ofP= 1 2 ( 1 + 2) × 8= 12 , area of Q=1 2 ( 2 + 3) × 6= 15 , area of R= 1 2 ( 3 + 4) × 4= 14 , and area of S= 1 2 ( 4 + 1) × 6= 15 . It follows that the total area of the trapeziums is 12+15 +14 +15 =56 . The large square has area 11 2 cm 2 = 121 cm 2 . Therefore the area of the shaded quadrilateral is, in cm 2 , 121−30 −56 =35 . © UKMT February 2015 www.ukmt.org.uk 14 11 1 1 8 P 2 2 2 26 Q 3 3 33 4 R 4 44 4 6 S

I ntermediate Mathematical Challenge 2015 Solutions and investigationsM ethod 2 We label the vertices of the shaded quadrilateral K,L ,M and N, as shown.Because K,L,M and N are vertices of the small squares it can be seen that they lie on the diagonals of the large square. The diagonals of a square meet at right angles. Therefore K Mis at right angles to L N. As the diagonals of the quadrilateral K L M N are at right angles, its area is half the product of the lengths of its diagonals. We ask you to prove this in Problem 19.2. Thus, to complete the answer all we need do is find the lengths of K MandL N. In Problem 19.3 we ask you to show that a square with side length shas diagonals of length s√ 2. It follows from this that the diagonals of the large square have length 11 √ 2 cm , and that the diagonals of the small squares have lengths √ 2 cm , 2 √ 2 cm , 3 √ 2 cm and 4 √ 2 cm . Therefore K M has length (11 −1−3) √ 2 cm = 7 √ 2 cm and L N has length (11 −2−4) √ 2 cm = 5 √ 2 cm . We can now deduce that the area of the shaded quadilateral K L M Nis 1 2 ( 7 √ 2 × 5√ 2 ) cm 2 = 35 cm 2 . F or investigation 19.1 Show that the area of a trapezium is 1 2 ( a + b)h , where a and bare the lengths of the parallel sides and his their distance apart. 19.2 Show that if the diagonals of a quadralateral meet at right angles, then the area of the quadrilateral is 1 2 de , where dand eare the lengths of the diagonals. 19.3 Show that if the side length of a square is s, then its diagonals have length s√ 2. 20. A voucher code is made up of four characters. The first is a letter: V, X or P. The second and third are di fferent digits. The fourth is the units digit of the sum of the second and third digits. How many di fferent voucher codes like this are there? A 180B 243C 270D 300E 2700 S olution C There are 3 choices for the first character, namely V, X or P. There are 10 choices for the second character, namely one of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9. Because the second and third characters are di fferent digits, the third character can be any one of the 10 digits, other than the second character. Therefore there are 9 choices for the third character. There is just 1 choice for the fourth character as this is the units digit of the sum of the second and third digits. It follows that the total number of voucher codes is 3 ×10 ×9× 1= 270. © UKMT February 2015 www.ukmt.org.uk 151 12 2 3 3 4 4 K L M N

I ntermediate Mathematical Challenge 2015 Solutions and investigations21.A rectangle is placed obliquely on top of an identical rectangle, as shown. The area X of the overlapping region (shaded more darkly) is one eighth of the total shaded area. What fraction of the area of one rectangle is X? A 1 3 B 2 7 C 1 4 D 2 9 E 1 5 S olution D The rectangles are identical. Therefore they have the same area. Let this area be Y. It follows that the area of each rectangle that is not part of the overlap is Y − X . Hence the total shaded area, made up of the two areas of the rectangles which are not part of the overlap, together with the overlap, is 2(Y − X)+ X =2Y − X. We are given that the area of the overlap is one eighth of the total shaded area. Therefore X 2 Y − X = 1 8 . This equation may be rearranged to give 8X =2Y − X, from which it follows that 9X =2Y , and therefore that X Y = 2 9 . F or investigation 21.1 If the area X is one tenth of the total shaded area, what fraction of the area of one rectangle is X? 21.2 If the area X is one quarter of the area of one rectangle, what fraction is X of the total shaded area? © UKMT February 2015 www.ukmt.org.uk 16X

I ntermediate Mathematical Challenge 2015 Solutions and investigations22.The diagram shows a shaded region inside a large square. The shaded region is divided into small squares. What fraction of the area of the large square is shaded? A 3 10 B 1 3 C 3 8 D 2 5 E 3 7 S olution B M ethod 1 We form a complete grid inside the large square by extending the sides of the small squares, as shown in the figure. In this way the large square is divided up into 61 small squares, 20 triangles along the edges whose areas are equal to half that of the small squares, and 4 triangles in the corners whose areas are equal to one quarter that of the small squares. So the area of the large square is equal to that of 61×1+ 20 ×1 2 + 4× 1 4 = 72 of the small squares. The shaded area is made up of 24 small squares. Therefore the fraction of the large square that is shaded is 24 72 = 1 3 . [Note that we could save some work by taking advantage of the symmetry of the figure, and just work out which fraction of, say, the top left hand corner of the square is shaded.] M ethod 2 We let sbe the side length of the large square, and tbe the side length of the small squares. It follows that the small squares have diagonals of length t√ 2. [See Problem 19.3, above.] It can be seen from the diagram that the side length of the large square is equal to the total length of the diagonals of 6 of the small squares. That is, s= 6× t√ 2 , and therefore t= s 6 √ 2 . It follows that t2 = s 2 72 . Hence the total area of the 24 shaded squares is 24×t2 = 24 ×s 2 72 = 1 3 s 2 . Therefore the fraction of the area of the large square that is shaded is 1 3 . © UKMT February 2015 www.ukmt.org.uk 17

I ntermediate Mathematical Challenge 2015 Solutions and investigations23.There are 120 di fferent ways of arranging the letters, U, K, M, I and C. All of these arrangements are listed in dictionary order, starting with CIKMU. Which position in the list does UKIMC occupy? A 110thB 112thC 114thD 116thE 118th S olution B M ethod 1 Because the sequence UKIMC comes towards the bottom of the list in dictionary order of all 120 arrangements of the letters U, K, M, I, C, we can most easily find its position by listing the arrangements in reverse dictionary order, until we reach UKIMC, as follows 120 UMKIC 119 UMKCI 118 UMIKC 117 UMICK 116 UMCKI 115 UMCIK 114 UKMIC 113 UKMCI 112 UKIMC We see from this that UKIMC is the 112th arrangement in the dictionary order list. M ethod 2 It would take rather too long to list all the arrangements in dictionary order from the start, but we can find the position of UKMIC by counting. Of the 120 arrangements of the letters U, K, M, I, C, there are 24 beginning with each of the 5 letters. It follows that, in dictionary order, the first 24 × 4= 96 of these arrangements begin with C, I, K or M. After these comes the first arrangement, UCIKM, beginning with U. This arrangement therefore occurs in position 97. Of the 24 arrangements beginning with U there are 6 beginning with each of the remaining 4 letters. Hence, starting from position 97, there are, first, 6 arrangements beginning with UC, and next, 6 beginning UI, before the first arrangement, UKCIM, which begins UK. This arrangement is therefore in position 97 +12 =109. Of the 6 arrangements beginning UK, there are 2 beginning with each of the remaining 3 letters. The first 2 of these, in positions 109 and 110, are UKCMI and UKCIM. Next there are the two arrangements beginning UKI. These are UKICM and UKIMC. There are in positions 111 and 112, respectively. Therefore UKIMC occurs in position 112. F or investigation 23.1 If the 120 arrangements of the letters U, S, M, I, C are put in dictionary order, in which position is the arrangement MUSIC? © UKMT February 2015 www.ukmt.org.uk 18

I ntermediate Mathematical Challenge 2015 Solutions and investigations24.In square RST U a quarter-circle arc with centre Sis drawn from T to R. A point P on this arc is 1 unit from T U and 8 units from RU. What is the length of the side of square RST U? A 9B 10C 11D 12E 13 S olution E Suppose that xis the length of the sides of the square RST U . We need to find the value of x. We let K be the point where the perpendicular from P to RS meets RS , and L be the point where the perpen- dicular from P to ST meets ST , as shown in the figure. Then P K has length x − 1, and so also does L S . Also, P L has length x− 8. Because P S is a radius of the quarter circle, its length is the same as that of the sides of the square, namely, x. Therefore S L P is a right-angled triangle in which the hypotenuse has length x, and the other two sides have lengths x− 1 and x− 8. Hence, by Pythagoras’ Theorem, (x − 1)2 + (x − 8)2 = x2 . Multiplying out the squares, we get (x 2 − 2x + 1) + (x 2 − 16 x+ 64 )= x2 , which we can rearrange as x2 − 18 x+ 65 =0. We can factorize the quadratic on the left-hand side of this equation to give the equivalent equation (x − 5) ( x− 13 )= 0, which has the solutions x = 5 and x= 13. Since x− 8 is the length of P L and therefore cannot be negative, x, 5. Therefore x= 13. F or investigation 24.1 How can we interpret the solution x = 5 of the quadratic equation which occurs in the solution above? © UKMT February 2015 www.ukmt.org.uk 19RS T U P 8 1 RS T U P K L 8 1 x x 8 x 1 x 1

I ntermediate Mathematical Challenge 2015 Solutions and investigations25.A point is marked one quarter of the way along each side of a triangle, as shown. What fraction of the area of the triangle is shaded? A 7 16 B 1 2 C 9 16 D 5 8 E 11 16 S olution D We give two methods for solving this problem. We use the notation △ X Y Z for the area of a triangle X Y Z. M ethod 1 We let P,Q and R be the vertices of the triangle, and let S,T and U be the points one quarter of the way along each side, as shown in the figure. We let K and L be the points where the perpendiculars from Pand T, respectively, meet Q R. The area of a triangle is half the base multiplied by the height of the triangle. Therefore △PQ R =1 2 ( Q R ×P K ). We now obtain an expression for △T S R . Because Sis one quarter of the way along Q R,S R =3 4 Q R . In the triangles P K R and T L R , the angles at K and L are both right angles, and the angle at R is common to both triangles. It follows that these triangles are similar. Therefore T L P K = T R P R = 1 4 , and hence T L=1 4 P K . It follows that △ST R =1 2 ( S R ×T L )= 1 2 (3 4 Q R ×1 4 P K )= 3 16  1 2 ( Q R ×P K ) = 3 16 △ PQ R . In Problem 25.1 you are asked to show that, similarly, △T PU =3 16 △ PQ R . It follows that the fraction of the area of triangle PQ Rthat is shaded is 1 − 3 16 − 3 16 = 5 8 . © UKMT February 2015 www.ukmt.org.uk 20 P Q R KLS T U

I ntermediate Mathematical Challenge 2015 Solutions and investigationsM ethod 2This, more elegant, approach uses the result that the ratio of the areas of two triangles with the same height is equal to the ratio of their bases. If you haven’t met this before, you are encouraged to find a proof (see Problem 25.2). We join Qand Tas shown in the figure. The triangles QT S and QT R have the same height. Therefore, using the above result, △QT S :△ QT R =Q S :Q R =1 : 4 . In other words △QT S =1 4 △ QT R . The triangles QT RandPQ R have the same height, and therefore △ QT R :△ PQ R =T R :P R =1 : 4 . In other words △QT R =1 4 △ PQ R . Therefore △QT S =1 4 △ QT R =1 4 (1 4 △ PQ R )= 1 16 △ PQ R . Similarly (see Problem 25.3), △QT U =9 16 △ PQ R . The shaded region is made up of the triangles QT SandQT U . Therefore its area is △ QT S +△QT U =1 16 △ PQ R +9 16 △ PQ R =  1 16 + 9 16  △ PQ R = 10 16 △ PQ R = 5 8 △ PQ R . F or investigation 25.1 Show that, as used in Method 1, △T PU =3 16 △ PQ R . 25.2 Prove that the ratio of the areas of two triangles with the same height is equal to the ratio of their bases. 25.3 Verify that, as used in Method 2, △QT U =9 16 △ PQ R . © UKMT February 2015 www.ukmt.org.uk 21P Q R S T U