Senior Challenge ‘20 The Solutions

University of Southampton Entries

485 students

31 schools

41 prizes Top Schools

School Number of prizes

Wilson’s School 9

Brookfield Community School 5

King Edward VI Camp Hill School for Girls 3

King’s School Winchester 2 (including 1st and a 2nd prize)

Well done to all schools! 1. Flicks for Four Families

Four families go to the cinema. The Reid family take 2 adults and 2 children for £18; the McGhee family take 1 senior, 2 adults and 1 child for £18.50; and the Griffiths family take 1 senior and 3 adults for £19.50. How much will it cost for the Linton family to take 1 senior, 2 adults and 3 children? 1. Flicks for Four Families

① Reid 2a + 2c = 18 ② McGhee s + 2a + c = 18.50 ③ Griffiths s + 3a = 19.50 ④= ③ – ② (s+3a) – (s+2a+c) = a – c 19.50 – 18.50 = 1 So ④ is a – c = 1 1. Flicks for Four Families

① Reid 2a + 2c = 18 ② McGhee s + 2a + c = 18.50 ③ Griffiths s + 3a = 19.50 ④ a – c = 1 ① +2 × ④ (2a + 2c) + 2 × (a – c) = (2a + 2c) + (2a – 2c) = 4a 18 + 2 × 1 = 20 so 4a = 20, so a = 5 1. Flicks for Four Families

① Reid 2a + 2c = 18 a = £5.00 ② McGhee s + 2a + c = 18.50 ③ Griffiths s + 3a = 19.50 Substitute a=5 into ① gives 2 × 5 + 2c = 18, so 10 + 2c = 18, so 2c = 8, so c = 4 1. Flicks for Four Families

① Reid 2a + 2c = 18 a = £5.00 ② McGhee s + 2a + c = 18.50 c = £4.00 ③ Griffiths s + 3a = 19.50 Substitute a=5 into ③ gives s + 3 × 5 = 19.50, so s + 15 = 19.50, so s = 4.50

So a=£5, c=£4, s=£4.50 1. Flicks for Four Families a=£5, c=£4, s=£4.50 Lintons are s + 2a + 3c = 4.50 + 2 × 5 + 3 × 4 = 4.50 + 10 + 12 =£26.50 2. I Can See Here From My House Sarah claimed she could see Russia from her house in Wasilla (Alaska) although the nearest point in Alaska to Russia (called , oddly enough) is 635 miles away from Wasilla. If you square 635, and subtract 1 from the result, you get a number divisible by four (try it). Show that the same is true of 7. Can you prove that the same is true of ANY odd number? 2. I Can See Here From My House 635² – 1 = 403,225 – 1 = 403,224 403,224=4×100,806

7² – 1 = 49 – 1 = 48 48=4×12

A number of entries lost marks here because they omitted the first part of the question or tried to divide 403,224 by 7.

We also had far too many people who said ‘0 is not divisible by 4’! 2. I Can See Here From My House

An even number is 2n for some integer n. An odd number is one more than an even number, so can be written as 2n+1 The square of an odd number is then (2n+1)×(2n+1)= 4n²+4n+1 Subtract 1 gives 4n²+4n This can be factorised to give 4(n²+n) which shows that 4 is a factor, so 4n²+4n is divisible by 4 3. Spectacular! Six students, Oleg, Bonnie, Pete, Georgia, Rachel and Simon, all need new glasses. The opticians have 6 pairs of glasses, each with a different coloured frame: red, blue, green, silver, pink and orange. Each of the boys has 2 colours that he’d prefer and each girl has 1 colour she’d prefer. Oleg prefers silver and pink, Simon prefers green and red, Pete prefers orange and blue, Bonnie prefers orange, Rachel prefers pink, and Georgia prefers blue. One of the girls (who has an r in her name) realises that not everyone can have their preference. How do they know? She then offers to choose a different colour so that everyone else can have their preference. Who was it and what colour did they choose? Which colour does each student ultimately select? 3. Spectacular!

Colours Students Simon is the only person who likes Red and Green Blue Bonnie

Green Georgia If Oleg has Silver, then

Orange Oleg Rachel has Pink, so her changing wouldn’t help. Pink Pete Bonnie doesn’t have an r Red Rachel So Georgia changes her mind. Silver Simon 3. Spectacular!

Colours Students This gives us two possible answers, with these 4 Blue Bonnie matches fixed Green Georgia If Simon has red, Georgia gets green and vice versa. Orange Oleg

Pink Pete

Red Rachel

Silver Simon 3. Spectacular!

Without the ‘r’ condition, Colours Students Bonnie could change, this gives these 4 as the fixed Blue Bonnie matches

Green Georgia If Simon has red, Bonnie gets Orange Oleg green and vice versa.

Pink Pete We marked this question so that students got the more Red Rachel generous mark from the two versions. Silver Simon 4. Whatever the Weather In a network of weather stations, each station is required to know how many hours of daylight there were at each of the others. Unfortunately, their email system has been infected by a virus and they can only communicate via phone calls. For three stations, show that three separate calls are required so that all three sets of results are known at all three stations. For five stations, show that at least six separate calls are required. What is the minimum number of calls if there are seven stations? Extend your reasoning for n stations. 4. Whatever the Weather

Calls Stations 0 1 2 3

A A AB AB ABC

B B AB ABC ABC

C C C ABC ABC A

C B 4. Whatever the Weather Calls Stations 0 1 2 3 4 5 6

A A A AB AB AB AB AB ABCDE

B B AB ABC ABC ABCDE ABCDE ABCDE B C C C C ABC ABC ABC ABCDE ABCDE

D D D D DE ABCDE ABCDE ABCDE D E

E E E E DE DE ABCDE ABCDE 4. Whatever the Weather Calls Statio 0 1 2 3 4 5 6 7 8 9 10 ns F ABCD ABCD ABCD A A A AFG ABFG ABFG ABFG ABFG ABFG EFG EFG EFG ABCF ABCF ABCD ABCD ABCD ABCD ABCD G B B B B ABFG G G EFG EFG EFG EFG EFG ABCF ABCF ABCF ABCD ABCD ABCD ABCD C C C C C G G G EFG EFG EFG EFG A ABCD ABCD ABCD ABCD ABCD D D D D D D DE EFG EFG EFG EFG EFG ABCD ABCD ABCD ABCD E E E E E E DE DE B C EFG EFG EFG EFG ABCD F F FG FG FG FG FG FG FG FG FG EFG D E ABCD ABCD G G FG AFG AFG AFG AFG AFG AFG AFG EFG EFG 4. Whatever the Weather

For 4 stations, the information can be B C shared with 4 calls as shown here.

Each additional station needs 2 calls: one D E to put their information into the loop, and one to take everyone else’s information back out. So for n ≥ 4, the number of calls is 2n - 4 5. Staying Out of the Spotlight Andrew is standing in the centre of a circle. Above the edge of the circle there are 3 spotlights, pointed straight down. These are equally spaced around the circle and can be moved around it in either direction by Nicole whilst remaining the same distance from each other. The spotlights can move twice as fast as Andrew. Andrew gets caught if a spotlight hits him. Determine whether Andrew can get out of the circle without being caught. 5. Staying Out of the Spotlight

Andrew can’t make it out by simply running for the edge. He moves r, the light 2휋푟 has to move 6 The light is moving at twice his speed, so catches him 5. Staying Out of the Spotlight

TheThis dottedmeans circlehe can is halfwayreach the to halfway the edge. Withinpoint, whilst this circle, keeping Andrewone of the can lights move behind aroundhim. faster than the lightsHe then as hemakes has aless dash thanfor the half edge the anddistance can toescape. cover for a given angle. 6. Say What You See

Lilly, Shaun and Ellen are stood in a line, one behind the other, looking in the same direction along the line. Lilly can see both Shaun and Ellen; Shaun can only see Ellen, who cannot see either of the other two. They are blindfolded and a hat is placed on each of their heads. They are told that the hats are selected at random from a box containing two red hats and three green hats. The blindfolds are removed, and they are asked to state, without turning around or conferring, what colour hat they themselves are wearing. After a pause, during which no-one has spoken, Ellen announces that she knows the colour of her hat. What is the colour of her hat and how does she know this? 6. Say What You See

Lilly Shaun Ellen

Looking this way

Ellen does not turn round She doesn’t take her hat off There is no mirror There is no box with the other hats in! 6. Say What You See

Looking this way Lilly Shaun Ellen Lilly can see both red hats, so knows hers is green

If Lilly is silent, she must be able to see at least one green hat. If Shaun can see a red hat, he knows his must be green

If Shaun is silent, he cannot see a red hat, so Ellen’s hat must be green. 7. Clwydian Conundrum

Lesley is standing at the hill fort on the top of Foel Fenlli, looking out along the of hills. The information sign next to her tells her that the two hilltops she can see are in fact both 2613m away from her (in a straight line). She notes that is on a bearing of 352° and Moel Gyw is on a bearing of 166°. What is the straight-line distance between the two hill tops? It is a bright sunny day. What is the furthest away visible object?

Welcome to Foel Fenlli 7. Clwydian Conundrum

Moel Famau Moel Famau is 2613m away on a bearing of 352° Moel Gyw is 2613m away on Foel Fenlli a bearing of 166° So this is the distance we need to find. Moel Gyw 3° 7. Clwydian Conundrum

This angle is 352-166=174° This makes these two angles each 3°, since the triangle is isosceles. 174° We can find the distance between the two Moels in at least 3 ways.

3° 3° 7. Clwydian Conundrum

푎 푏 Using the sine rule: = 푠푖푛 퐴 푠푖푛 푏 푑푖푠푡푎푛푐푒 2613 = 푠푖푛 174 푠푖푛 3 174° 푑푖푠푡푎푛푐푒 2613 = 0.105 0.052 푑푖푠푡푎푛푐푒 = 49927 0.105 푑푖푠푡푎푛푐푒 = 5218.83푚

3° 3° 7. Clwydian Conundrum Using the cosine rule: 푐2 = 푎2 + 푏2 − 2푎푏푐표푠 퐶 푑푖푠푡푎푛푐푒2 = 26132 + 26132 − 2 × 2613 × 2613 × 푐표푠 174 푑푖푠푡푎푛푐푒2 = 6827769 + 6827769 − 13655538 × −0.995 174° 푑푖푠푡푎푛푐푒2 = 13655538 + 13580732 푑푖푠푡푎푛푐푒2 = 27236270 푑푖푠푡푎푛푐푒 = 5218.83푚

3° 7. Clwydian Conundrum Using basic trigonometry and Pythagoras’ theorem.

2613m Foel Fenlli to Moel Famau 8° 2613 sin 8 = 363.66 2613 cos 8 = 2587.57

2613mFoel Fenlli to Moel Gyw 14° 2613 sin 14 = 632.14 2613 cos 14 = 2535.38 7. Clwydian Conundrum Using basic trigonometry and Pythagoras’ theorem.

2587.57 Moel Famau to Moel Gyw + 2535.38 2587.57 + 2535.38 = 5122.95 363.66 + 632.14 = 995.80

5122.952 + 995.802 = 푐2 26244616 + 991618 = 푐2 27236234 = 푐2 363.66 푐+ = 5218.83푚 632.14 7. Clwydian Conundrum

In real life, the point at which this works is about 36m along the footpath from the cairn on the top of Foel Fenlli. I had planned to go and take pictures to illustrate this, but Covid-19 got in the way! The furthest away object which is visible on a bright sunny day is the sun. 8. Catching Crabs Michael (an amateur physicist) has decided to go spearfishing. His friend Rob loves crab so he decides to try to spear one. Michael sees a crab on the seabed, which he knows is 3m below the water’s surface; however, his knowledge of physics tells him that things may not be as they appear. Michael knows that when light leaves the water, it ‘refracts’ (changes direction) at the surface. Thinking back to his physics lessons, Michael remembers that 푛1 sin 휃1 = 푛2 sin 휃2 (for water n=4/3 and for air n=1) Michael is looking at the water at an angle of 45° from the vertical. At what angle from the vertical is the light travelling through the water? Michael knows he will need to aim his spear closer to himself than where the crab appears to be. How much closer? 8. Catching Crabs

푛1 sin 휃1 = 푛2 sin 휃2 (for water n=4/3 and for air n=1) 4 sin 휃 = 1 × sin 45 3 1 3 1 sin 휃1 = × = 0.5303 4 2 휃1 = 32.07° 8. Catching Crabs

45 Water 휃 휃1 = 32.07° 1

xm Sea bed 3m 32.07° 8. Catching Crabs 푥 푡푎푛 32.07 = 3 푥 = 3푡푎푛 32.07 = 1.877푚 3m

So the crab is 3 – 1.877 = 1.123m closer to Michael than it appears to be.

xm Thank you for watching!