bmo1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/bmo_1114.pdf ·...

52
British Mathematical Olympiad Round 1 United Kingdom Mathematics Trust 20 to 201

Upload: duongthu

Post on 24-Jul-2018

279 views

Category:

Documents


3 download

TRANSCRIPT

Page 1: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

United Kingdom Mathematics Trust

British MathematicalOlympiad

2003 to 2008

Round 2Round 1

2011 to 2011

United Kingdom Mathematics Trust

2011 to 2014

amtejc
Typewritten Text
© UKMT 2015
Page 2: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

United Kingdom Mathematics Trust

British Mathematical OlympiadRound 1 : Friday, 2 December 2011

Time allowed 3 12 hours.

Instructions • Full written solutions – not just answers – arerequired, with complete proofs of any assertionsyou may make. Marks awarded will depend on theclarity of your mathematical presentation. Workin rough first, and then write up your best attempt.Do not hand in rough work.

• One complete solution will gain more credit thanseveral unfinished attempts. It is more importantto complete a small number of questions than totry all the problems.

• Each question carries 10 marks. However, earlierquestions tend to be easier. In general you areadvised to concentrate on these problems first.

• The use of rulers and compasses is allowed, butcalculators and protractors are forbidden.

• Start each question on a fresh sheet of paper. Writeon one side of the paper only. On each sheet ofworking write the number of the question in thetop left hand corner and your name, initials andschool in the top right hand corner.

• Complete the cover sheet provided and attach it tothe front of your script, followed by your solutionsin question number order.

• Staple all the pages neatly together in the top lefthand corner.

• To accommodate candidates sitting in other time-zones, please do not discuss the paper on theinternet until 8am GMT on Saturday 3 December.

Do not turn over until told to do so.

5

Page 3: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

United Kingdom Mathematics Trust

2011/12 British Mathematical Olympiad

Round 1: Friday, 2 December 2011

1. Find all (positive or negative) integers n for which n2 + 20n+ 11 is aperfect square. Remember that you must justify that you have foundthem all.

2. Consider the numbers 1, 2, . . . , n. Find, in terms of n, the largestinteger t such that these numbers can be arranged in a row so that allconsecutive terms differ by at least t.

3. Consider a circle S. The point P lies outside S and a line is drawnthrough P , cutting S at distinct points X and Y . Circles S1 and S2

are drawn through P which are tangent to S at X and Y respectively.Prove that the difference of the radii of S1 and S2 is independent ofthe positions of P , X and Y .

4. Initially there are m balls in one bag, and n in the other, where m,n >0. Two different operations are allowed:

a) Remove an equal number of balls from each bag;b) Double the number of balls in one bag.

Is it always possible to empty both bags after a finite sequence ofoperations?

Operation b) is now replaced with

b′) Triple the number of balls in one bag.

Is it now always possible to empty both bags after a finite sequence ofoperations?

5. Prove that the product of four consecutive positive integers cannot beequal to the product of two consecutive positive integers.

6. Let ABC be an acute-angled triangle. The feet of the altitudes fromA,B and C are D, E and F respectively. Prove that DE+DF ≤ BCand determine the triangles for which equality holds.

The altitude from A is the line through A which is perpendicular toBC. The foot of this altitude is the point D where it meets BC. Theother altitudes are similarly defined.

6

Page 4: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

United Kingdom Mathematics Trust

British Mathematical OlympiadRound 1 : Friday, 30 November 2012

Time allowed 3 12 hours.

Instructions • Full written solutions – not just answers – arerequired, with complete proofs of any assertionsyou may make. Marks awarded will depend on theclarity of your mathematical presentation. Workin rough first, and then write up your best attempt.Do not hand in rough work.

• One complete solution will gain more credit thanseveral unfinished attempts. It is more importantto complete a small number of questions than totry all the problems.

• Each question carries 10 marks. However, earlierquestions tend to be easier. In general you areadvised to concentrate on these problems first.

• The use of rulers, set squares and compassesis allowed, but calculators and protractors areforbidden.

• Start each question on a fresh sheet of paper. Writeon one side of the paper only. On each sheet ofworking write the number of the question in thetop left hand corner and your name, initials andschool in the top right hand corner.

• Complete the cover sheet provided and attach it tothe front of your script, followed by your solutionsin question number order.

• Staple all the pages neatly together in the top lefthand corner.

• To accommodate candidates sitting in other time-zones, please do not discuss the paper on theinternet until 8am GMT on Saturday 1 December.

Do not turn over until told to do so.

5

Page 5: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

United Kingdom Mathematics Trust

2012/13 British Mathematical Olympiad

Round 1: Friday, 30 November 2012

1. Isaac places some counters onto the squares of an 8 by 8 chessboard sothat there is at most one counter in each of the 64 squares. Determine,with justification, the maximum number that he can place withouthaving five or more counters in the same row, or in the same column,or on either of the two long diagonals.

2. Two circles S and T touch at X. They have a common tangent whichmeets S at A and T at B. The points A and B are different. Let APbe a diameter of S. Prove that B, X and P lie on a straight line.

3. Find all real numbers x, y and z which satisfy the simultaneousequations x2 − 4y + 7 = 0, y2 − 6z + 14 = 0 and z2 − 2x− 7 = 0.

4. Find all positive integers n such that 12n−119 and 75n−539 are bothperfect squares.

5. A triangle has sides of length at most 2, 3 and 4 respectively.Determine, with proof, the maximum possible area of the triangle.

6. Let ABC be a triangle. Let S be the circle through B tangent to CAat A and let T be the circle through C tangent to AB at A. Thecircles S and T intersect at A and D. Let E be the point where theline AD meets the circle ABC. Prove that D is the midpoint of AE.

6

Page 6: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

United Kingdom Mathematics Trust

British Mathematical OlympiadRound 1 : Friday, 29 November 2013

Time allowed 3 1

2hours.

Instructions • Full written solutions – not just answers – arerequired, with complete proofs of any assertionsyou may make. Marks awarded will depend on theclarity of your mathematical presentation. Workin rough first, and then write up your best attempt.Do not hand in rough work.

• One complete solution will gain more credit thanseveral unfinished attempts. It is more importantto complete a small number of questions than totry all the problems.

• Each question carries 10 marks. However, earlierquestions tend to be easier. In general you areadvised to concentrate on these problems first.

• The use of rulers, set squares and compassesis allowed, but calculators and protractors areforbidden.

• Start each question on a fresh sheet of paper. Writeon one side of the paper only. On each sheet ofworking write the number of the question in thetop left hand corner and your name, initials andschool in the top right hand corner.

• Complete the cover sheet provided and attach it tothe front of your script, followed by your solutionsin question number order.

• Staple all the pages neatly together in the top lefthand corner.

• To accommodate candidates sitting in other timezones, please do not discuss the paper onthe internet until 8am GMT on Saturday 30November.

Do not turn over until told to do so.

Page 7: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

United Kingdom Mathematics Trust

2013/14 British Mathematical Olympiad

Round 1: Friday, 29 November 2013

1. Calculate the value of

20144 + 4× 20134

20132 + 40272−

20124 + 4× 20134

20132 + 40252.

2. In the acute-angled triangle ABC, the foot of the perpendicular fromB to CA is E. Let l be the tangent to the circle ABC at B. The footof the perpendicular from C to l is F . Prove that EF is parallel toAB.

3. A number written in base 10 is a string of 32013 digit 3s. No otherdigit appears. Find the highest power of 3 which divides this number.

4. Isaac is planning a nine-day holiday. Every day he will go surfing,or water skiing, or he will rest. On any given day he does just one ofthese three things. He never does different water-sports on consecutivedays. How many schedules are possible for the holiday?

5. Let ABC be an equilateral triangle, and P be a point inside thistriangle. Let D,E and F be the feet of the perpendiculars from P tothe sides BC, CA and AB respectively. Prove that

a) AF +BD + CE = AE +BF + CD andb) [APF ] + [BPD] + [CPE] = [APE] + [BPF ] + [CPD].

The area of triangle XY Z is denoted [XY Z].

6. The angles A,B and C of a triangle are measured in degrees, and thelengths of the opposite sides are a, b and c respectively. Prove that

60 ≤aA+ bB + cC

a+ b+ c< 90.

Page 8: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

United Kingdom Mathematics Trust

British Mathematical OlympiadRound 1 : Friday, 28 November 2014

Time allowed 3 1

2hours.

Instructions • Full written solutions – not just answers – arerequired, with complete proofs of any assertionsyou may make. Marks awarded will depend on theclarity of your mathematical presentation. Workin rough first, and then write up your best attempt.Do not hand in rough work.

• One complete solution will gain more credit thanseveral unfinished attempts. It is more importantto complete a small number of questions than totry all the problems.

• Each question carries 10 marks. However, earlierquestions tend to be easier. In general you areadvised to concentrate on these problems first.

• The use of rulers, set squares and compassesis allowed, but calculators and protractors areforbidden.

• Start each question on a fresh sheet of paper. Writeon one side of the paper only. On each sheet ofworking write the number of the question in thetop left hand corner and your name, initials andschool in the top right hand corner.

• Complete the cover sheet provided and attach it tothe front of your script, followed by your solutionsin question number order.

• Staple all the pages neatly together in the top lefthand corner.

• To accommodate candidates sitting in other timezones, please do not discuss the paper onthe internet until 8am GMT on Saturday 29November.

Do not turn over until told to do so.

Page 9: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

United Kingdom Mathematics Trust

2014/15 British Mathematical Olympiad

Round 1: Friday, 28 November 2014

1. Place the following numbers in increasing order of size, and justifyyour reasoning:

334

, 343

, 344

, 433

and 434

.

Note that abc

means a(bc

).

2. Positive integers p, a and b satisfy the equation p2 + a2 = b2. Provethat if p is a prime greater than 3, then a is a multiple of 12 and2(p+ a+ 1) is a perfect square.

3. A hotel has ten rooms along each side of a corridor. An olympiadteam leader wishes to book seven rooms on the corridor so that notwo reserved rooms on the same side of the corridor are adjacent. Inhow many ways can this be done?

4. Let x be a real number such that t = x + x−1 is an integer greaterthan 2. Prove that tn = xn + x−n is an integer for all positiveintegers n. Determine the values of n for which t divides tn.

5. Let ABCD be a cyclic quadrilateral. Let F be the midpoint of the arcAB of its circumcircle which does not contain C or D. Let the linesDF and AC meet at P and the lines CF and BD meet at Q. Provethat the lines PQ and AB are parallel.

6. Determine all functions f(n) from the positive integers to the positiveintegers which satisfy the following condition: whenever a, b and c arepositive integers such that 1/a+ 1/b = 1/c, then

1/f(a) + 1/f(b) = 1/f(c).

Page 10: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

Introduction to the problems and full solutions

The ‘official’ solutions are the result of many hours’ work by a large numberof people, and have been subjected to many drafts and revisions. The con-testants’ solutions included here will also have been redrafted several timesby the contestants themselves, and also shortened and cleaned up somewhatby the editors. As such, they do not resemble the first jottings, failed ideasand discarded pages of rough work with which any solution is started.

Before looking at the solutions, pupils (and teachers) are encouraged tomake a concerted effort to attack the problems themselves. Only by doing sois it possible to develop a feel for the question, to understand where the diffi-culties lie and why one method of attack is successful while others may fail.Problem solving is a skill that can only be learnt by practice; going straightto the solutions is unlikely to be of any benefit.

It is also important to bear in mind that solutions to Olympiad problemsare not marked for elegance. A solution that is completely valid will receive afull score, no matter how long and tortuous it may be. However, elegance hasbeen an important factor influencing our selection of contestants’ answers.

13

Page 11: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

BMO Round 1

Problem 1 (Proposed by Dr Gerry Leversha.) Find all (positive or negative)integers n for which n2 + 20n+ 11 is a perfect square. Remember that you mustjustify that you have found them all.

Solution (Rafi Dover, King David High School): Let n2 + 20n + 11 = a2

where a is a positive integer. Then

n2 + 20n+ 11 = a2

(n+ 10)2 − 100 + 11 = a2

(n+ 10)2 = a2 + 89

(n+ 10)2 − a2 = 89

So solutions correspond to pairs of squares that differ by 89.We claim that

∑nr=1(2r−1) = n2 – the proof is by induction. When n = 1,

this is clearly true, and if we suppose that∑k

r=1 = k2 then

k+1∑r=1

(2r − 1) =

k∑r=1

(2r − 1) + 2k + 1 = (k + 1)2

and so our claim is true for all n.So for positive integers x and y,

x2 − y2 =

x∑r=y+1

(2r − 1)

So whenever we have a solution to the original equation, we have a set ofconsecutive positive odd integers that sum to 89. The sum of an even numberof odd integers is even, and as 89 is odd this means that there must be an oddnumber of integers in the sum. But then there is a ‘middle’ integer in the list,which divides the sum. But as 89 is prime, the only way we can make 89 asa sum of consecutive odd integers is 89 itself.

So we must have a = 44 and hence (n + 10)2 = 452. This gives us twopossibilities (corresponding to positive and negative square roots) for n: 35and −55. We can check that these do both give us n2 + 20n+ 11 = 442.

14

Solutions2011

Page 12: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

Problem 2 (Proposed by Dr Geoff Smith.) Consider the numbers 1, 2, . . . , n.Find, in terms of n, the largest t such that these numbers can be arranged in a rowso that all consecutive terms differ by at least t.

Solution (Natalie Behague, Dartford Grammar School for Girls): Firstconsider the case when n is even, n = 2m. Then the sequence

m, 2m,m− 1, 2m− 1, . . . 1,m+ 1

has every consecutive term differing by m or m+ 1. So we can take t = m =n2 . We can’t have t > m because the sequence would have to contain m,which is distance at most m from all the other terms.

Now consider the case where n is odd; n = 2m + 1. We can take thesequence

2m+ 1,m+ 1, 1,m+ 2, 2,m+ 3, . . . , 2m,m

which has every term differeing by m or m+1. So we can take t = m = n−12 .

We can’t have t > m because m+1 differs by at most m from each other termin the sequence.

The best value of t is therefore

t =

{n2 if n is evenn−12 if n is odd.

15

Page 13: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

Problem 3 (Proposed by Dr Gerry Leversha.) Consider a circle S. The pointP lies outside of S and a line is drawn through P , cutting S at distinct points Xand Y . Circles S1 and S2 are drawn through P which are tangent to S at X and Yrespectively. Prove that the difference of the radii of S1 and S2 is independent of thepositions of P , X and Y .

The first solution given below makes elegant use of the fact that if twocircles are tangent then there is an enlargement centered on the point of tan-gency taking one to the other. The second solution is similar at heart butinstead uses several similar right-angled triangles.

Solution 1 (Adam Goucher, Netherthorpe School): Assume, without lossof generality, that PX < PY . Denote by R, R1 and R2 the radii of S, S1 andS2 respectively. S2 and S are then internally tangent at Y , so that there is anenlargement about Y with positive scale factor R2

R sending S to S2. Similarly,there is an enlargement about X with negative scale factor −R1

R sending S

to S1. The first enlargement maps XY to PY , so that PYXY = R2

R . The secondenlargement maps XY to XP , so that PX

XY = R1

R . Subtracting these twoequations gives

R2 −R1

R=

PY − PX

XY= 1.

Therefore |R2 −R1| = R is independent of the positions of P , X and Y .

Solution 2 (B.T.G. de Jager, Eton College): Without loss of generality, as-sume that PX < PY . Let O be the center of S, and let R, R1 and R2 be

16

Page 14: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

as in the previous solution. Let Y O meet S again at D and S2 again at A,and let B be diametrically opposite X on S1. Then B, X and O are collinearbecause S1 and S are tangent, and AY is a diameter of S2 because S2 and Sare tangent. Therefore, because the angle in a semicircle is 90◦, we have that∠DXY , ∠APY , and ∠XPB are all 90◦.

Let ∠AY P = θ. Then, because OX = OY , triangle OXY is isosceles andso ∠PXB = ∠Y XO = ∠OYX . From right-angled triangle APY , we findthat 2R2 = PY

cos θ . Similarly, from right-angled triangle BPX , we find that2R1 = PX

cos θ . Subtracting these we get

2(R2 −R1) =PY − PX

cos θ=

XY

cos θ.

But from right-angled triangle DXY this is equal to DY = 2R. So |R2−R1| =R, which is independent of the positions of P , X and Y .

17

Page 15: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

Problem 4 (Proposed by Dr Gerry Leversha.) Initially there are m balls in onebag, and n in the other, where m,n > 0. Two different operations are allowed:

a) Remove an equal number of balls from each bag;

b) Double the number of balls in one bag.

Is it always possible to empty both bags after a finite sequence of operations?

Operation b) is now replaced by

b’) Triple the number of balls in one bag

Is it now always possible to empty both bags after a finite sequence of operations?

Solution (Andrew Carlotti, Sir Roger Manwood’s School): We claim inthe first case that it is always possible. If m = n then we can just empty bothbags. Otherwise, we may as well assume m < n. We can double the numberof balls in the bag with m balls repeatedly (if necessary) until there are atleast n

2 balls in that bag. There are now m′ and n (where n2 ≤ m′ < n) balls

in the two bags. We can then remove 2m − n from each bag, at which pointthere are n−m and 2(n−m) balls in each bag. Then we double the numberof balls in the bag with fewer balls in it, and finally empty both bags.

With the modified operation, it is not always possible. Operations a) andb’) both alter the total number of balls by an even number, and so if thereare an odd number of balls in total (say 2 balls in the first bag, and 1 in thesecond), we cannot empty them both.

18

Page 16: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

Problem 5 (Proposed by Robin Bhattacharyya) Prove that the product of fourconsecutive positive integers cannot be equal to the product of two consecutive posi-tive integers.

The main difficulty with this problem is finding a way to work with thefact that we are looking for positive integer solutions, as any approach whichattempts to solve the below equations without this cannot work. Most so-lutions given were variants on one of the two similar solutions shown here,where we ‘sandwich’ possible solutions between two consecutive integers.

Solution 1 (Katya Richards, The School of St Helen and St Katharine):Let the four consecutive integers by b, b + 1, b + 2, b + 3. Then their prod-uct is

b(b+ 3)(b+ 1)(b+ 2) = (b2 + 3b)(b2 + 3b+ 2)

Let c = b2 + 3b. Then this product is equal to c(c + 2). If this product wereequal to the product of two consecutive integers, a and (a+1), then we wouldhave a(a+ 1) = c(c+ 2).

We clearly can’t have a = c or a = c + 1, since c(c + 1) < c(c + 2) <(c+1)(c+2). We also can’t have c < a because then also (a+1) < (c+2) andso a(a+1) < c(c+2). But we also can’t have a > (c+1) as then (a+1) > (c+2)and so a(a + 1) > c(c + 2). So there are no solutions to a(a + 1) = c(c + 2),and hence we can’t write the product of four consecutive positive integers asthe product of two consecutive positive integers.

Solution 2 (Oliver Feng, Eton College): Suppose that

n(n+ 1) = m(m+ 1)(m+ 2)(m+ 3)

= (m+ 1)(m+ 2) ·m(m+ 3)

= (m2 + 3m+ 2)(m2 + 3m)

= (m2 + 3m+ 1)2 − 1

So setting M = m2 + 3m+ 1 we have n(n+ 1) = M2 − 1. Then

M2 = n(n+ 1) + 1 = n2 + n+ 1

However,

n2 < n2 + n+ 1 < n2 + 2n+ 1 = (n+ 1)2

for n a positive integer. Then M2 is a square number strictly between n2 and(n+ 1)2, which is clearly impossible. So no such m and n exist.

19

Page 17: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

Problem 6 (Proposed by Dr David Monk.) Let ABC be an acute-angled trian-gle. The feet of the altitudes from A, B and C are D, E and F respectively. Provethat DE + EF ≤ BC and determine the triangles for which equality holds.

The altitude through A is the line through A which is perpendicular toBC. The foot of this altitude is the point D where it meets BC. The otheraltitudes are similarly defined.

Most people who solved this problem used trigonometry to reduce theproblem to an algebraic inequality, which could then be solved with a varietyof methods – solution 1 is a nice example of this approach. It is also possibleto give a more ‘geometric’ proof of the inequality, as in solution 3. Solution 2is an interesting hybrid of the two approaches.

An angle-chase common to all three solutions is this: because ∠AEB and∠ADB are right angled, AEDB is cyclic with diameter AB. So ∠EDC =180 − ∠BDE = ∠BAE = ∠BAC by the fact that opposite angles in a cyclicquadrilateral add to 180◦. Similarly, ∠FDC = ∠CAB.

In all solutions, let A, B and C be the angles of triangle ABC and let a, band c be the lengths of BC, CA and AB respectively.

Solution 1 (Yu Wan): First we will compute DE and DF . From right-angled triangles BEC and ADC we have that CE = a cosC. By the sinerule in triangle CED, DE = sinC CE

sinEDC = sinC a cosCsinA by the previous sen-

tence and the angle chase in the introduction. By the sine rule in triangleABC, a

sinA = csinC so that DE = c cosC. Similarly, DF = b cosB.

So we have DE + DF = c cosC + b cosB, while BC = BD + DC =c cosB + b cosC. Therefore

DE +DF −BC = c cosC + b cosB − c cosB − b cosC

= (c− b)(cosC − cosB).

20

Page 18: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

If c > b, then C > B, so cosC < cosB and therefore DE+DF < BC. If c = bthen DE +DF = BC. If c < b, then C < B, so cosC > cosB and thereforeDE +DF < BC.

We have shown that DE +DF ≤ BC with equality if, and only if, b = c.

Solution 2 (Joshua Lam, The Leys School): Let X and Y be the feet of theperpendiculars from F and E to BC, respectively. As in the introduction, wehave ∠FDX = A and ∠EDY = A. So DF = DX

cosA and DE = DYcosA , so that

DF +DE = DX+DYcosA = XY

cosA . So it suffices to show that XY ≤ a cosA.

But XY is the orthogonal projection of FE onto the side BC, so thatFE ≥ XY (essentially, because the side opposite the right angle is the longestside of a right-angled triangle). Equality holds whenever FE is parallel toXY . We will show that FE = a cosA, which gives XY ≤ a cosA.

This calculation is exactly the same as the calculation of DE in the previ-ous solution, and the result is that FE = a cosA as required.

Equality holds if and only if FE is parallel to XY . This happens if andonly if ∠AFE = ∠ABC. But the cyclic quadrilateral BFEC gives that∠AFE = ∠ACB. So equality occurs if, and only if, ∠ABC = ∠ACB i.e.that AB = AC.

Solution 3 (by the markers): As before, we have that ∠EDC = ∠FDB =A. This means that if we reflect F in the line BC to get F ′, then ∠F ′DB =∠EDC, so that F ′DE is a straight line. But because ∠BF ′C = ∠BEC = 90◦,F ′BEC is cyclic with diameter BC. The diameter is the greatest distancebetween two points on the circumference of a circle, so that BC ≥ F ′E =F ′D +DE = DF +DE, as required.

21

Page 19: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

Equality holds if, and only if, EF ′ is a diameter of F ′BEC, which in turnhappens if, and only if, ∠F ′CE = 90◦. But ∠F ′CE = ∠F ′CB + ∠ECB =∠FCB + ∠ECB = 90−B + C. So equality occurs if, and only if, B = C i.e.AB = AC.

22

Page 20: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

BMO Round 1

Problem 1 (Proposed by Dr Jeremy King.) Isaac places some counters into thesquares of an 8 × 8 chessboard so that there is at most one counter in each of the64 squares. Determine, with justification, the maximum number that he can placewithout having five or more counters in the same row, or in the same column, or ineither of the two long diagonals.

The key to this problem is to realise that, if we focus only on the constraintthat there can be only four counters in each row, the maximum number ofcounters we can put on the board is 32. (Of course, we could also do thesame by considering only the columns). This can be attained in a variety ofways.

Solution by Abigail Hayes, Stretford Grammar School: We know thatthere must be fewer than 5 counters in each row, column and long diago-nal, so the maximum in each is 4 counters.

If we had more than 32 counters, then there would be at least 5 in onerow, which cannot happen. Hence, there can be at most 32 counters.

Here is an example which attains this bound:

14

SSolutions2012

Page 21: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

Problem 2 (Proposed by Richard Freeland.) Two circles S and T touch at X .They have a common tangent which meets S at A and T at B. The points A and Bare different. Let AP be a diameter of S. Prove that B,X and P lie on a straightline.

There are a few different ways to approach this problem, mostly using thealternate segment theorem and other standard circle facts. Whilst a good di-agram is always helpful, it is important not to accidentally assume that PXBis a straight line as part of the proof, just because it looks that way in thediagram!

Solution 1 by Harry Metrebian, Winchester College: Let the centres ofcircles S and T be O1 and O2 respectively; let the other end of the diameterof T through B be Q and let the common tangent to both circles at X passthrough a point L on the same side of X as the line AB and a point M on theother side of X .

We know that O1XO2 is a straight line as the lines O1X and O2X are bothperpendicular to the tangent at X .

Since tangents and radii meet at right angles,

∠PAB = ∠ABQ = 90◦.

Therefore the lines AP and BQ are parallel. So, by alternate angles,

∠PO1X = ∠BO2X.

Since the angle subtended by a chord at the centre of a circle is twice thatsubtended by the same chord at the circumference,

∠PAX =1

2∠PO1X =

1

2∠BO2X = ∠BQX.

15

Page 22: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

By the alternate segment theorem,

∠PXM = ∠PAX = ∠BQX = ∠BXL.

Since LXM is a straight line, BXP is also a straight line by the converse ofthe vertically opposite angles theorem.

Solution 2 by Aatreyee Mimi Das, Heckmondwike Grammar School: Letthe other end of the diameter of T through B be Q. Let the tangent to circlesS and T at X meet AB at V , and let M be a point on XV the other side of Xfrom V .

Note that ∠PXA = 90◦ because AP is a diameter. Similarly ∠BXQ =90◦. Let ∠APX = y. Then by the alternate segment theorem, ∠BAX = y.Also let ∠XQB = z; then ∠XBA = z also by the alternate segment theorem.

Using the alternate segment theorem again, it must be the case that ∠AXV =y and ∠V XB = z. Then ∠AXB = y + z. But then the sum of the angles intriangle BAX is

∠XBA+ ∠BAX + ∠AXB

= z + y + (y + z)

= 2(y + z).

Since this must be 180◦, it follows that ∠AXB = y + z = 90◦. Therefore

∠PXB = ∠PXA+ ∠AXB = 90◦ + 90◦ = 180◦,

and so PXB is a straight line.

16

Page 23: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

Problem 3 (Proposed by Dr David Monk.) Find all real numbers x, y and zwhich satisfy the simultaneous equations x2 − 4y + 7 = 0, y2 − 6z + 14 = 0 andz2 − 2x− 7 = 0.

It might not seem obvious where to start with this curious looking set ofsimultaneous equations, but there turns out to be a useful trick that leads tothe solution. Most successful solutions used the method below.

Solution by the markers: Sum all three equations, to yield

x2 − 4y + 7 + y2 − 6z + 14 + z2 − 2x− 7 = 0.

This can be rearranged to form

(x− 1)2 + (y − 2)2 + (z − 3)2 = 0.

Since all squares are nonnegative, this can only happen if all three terms areequal to 0. Therefore, if there is any solution, it must be that x = 1, y = 2 andz = 3.

We must check against the original equations to see if this is indeed asolution, and it turns out that x = 1, y = 2, z = 3 does satisfy them. Thereforethis is the unique solution.

17

Page 24: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

Problem 4 (Proposed by Richard Freeland.) Find all positive integers n suchthat 12n− 119 and 75n− 539 are both perfect squares.

The key in this question is to somehow reduce the problem to a finite setof cases which we can then check each of. A typical way of doing this is tofind some factorisation of an integer, as the solution below does.

Solution by Alex Harris, Perse School: Firstly, let a2 = 12n − 119 andb2 = 75n− 539.

We have n =a2 + 119

12, from the definition of a. So we can use the defini-

tion of b to derive

75(a2 + 119)

12− 539 = b2,

25(a2 + 119)− 4× 539 = 4b2,

and hence

25a2 − 4b2 = −819.

This can be factorised to give

(2b− 5a)(2b+ 5a) = 819.

So we know that the integers 2b−5a and 2b+5a multiply together to give 819,so, since we can assume that a and b are nonnegative, we have a limited setof possibilities for (2b− 5a, 2b+ 5a): (1, 819), (3, 273), (7, 117), (9, 91), (13, 63)and (21, 39).

Only the second, third and fifth give integer values for a, so (a, b) can be(27, 69), (11, 31) or (5, 19).

We can substitute into the formula for n above to determine its possiblevalues. The first does not give an integer, so we are left with n = 20 or n = 12.We can easily check that both these values of n work.

18

Page 25: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

Problem 5 (Proposed by Dr Jeremy King.) A triangle has sides of length atmost 2, 3 and 4 respectively. Determine, with proof, the maximum possible area ofthe triangle.

Whilst there are some very neat, short solutions to this problem, it is im-portant to carefully justify why the given triangle is maximal.

Solution 1 by Gavin O’Connell, Bristol Grammar School: Let θ be thelargest angle of the triangle. Thus, θ is the angle between the two shortersides, of lengths a and b.

The area of the triangle is given by A = 12ab sin θ, where a ≤ b are the

shorter sides. But we have

a ≤ 2,

b ≤ 3,

and

sin θ ≤ 1.

Hence A ≤ 3, with equality when the triangle is right angled with legs 2and 3. This gives a hypotenuse of length

√13 < 4, which is valid.

Thus, the maximum area is 3.

Solution 2 by Warren Li, Fulford School: If the sides of the triangle are oflengths a, b and c, and the circumradius is R, then it is well known that thearea of the triangle is A = abc

4R .Let a ≤ b ≤ c, without loss of generality. We have c ≤ 2R, a ≤ 2 and

b ≤ 3, so A ≤ ab2 ≤ 3.

The same example from before works.

19

Page 26: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

Problem 6 (Proposed by Dr Gerry Leversha.) Let ABC be a triangle. Let S bethe circle through B tangent to CA at A and let T be the circle through C tangentto AB at A. The circles S and T intersect at A and D. Let E be the point where theline AD meets the circle ABC. Prove that D is the midpoint of AE.

There are a wide variety of ways to approach this problem. Some stu-dents used a similar triangles argument along the lines of the first solutionbelow; some other methods involved considering the centres of circles S andT , as in the second solution. Another method is to observe that the circum-centre of ABC lies on the circle through B,D,C, and extend the line ADE tomeet this circle again.

Solution 1 by Oliver Feng, Eton College: Let ∠ABD = α and ∠DCA = β.

Then ∠CAD = ∠ABD = α and ∠DAB = ∠DCA = β, by the alternatesegment theorem. So �ABD is similar to �CAD.

Also, ∠CBE = ∠CAE = α and ∠ECB = ∠EAB = β by the theorem ofangles in the same segment. Similarly ∠BED = ∠BEA = ∠BCA. Further,∠DBE = ∠CBE + ∠DBC = ∠ABD + ∠DBC = ∠ABC. So �DBE and�ABC are similar.

20

Page 27: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

Therefore, ABBD = AC

AD and so AD = AC×BDAB . Also, DB

DE = ABAC , and so

DE = AC×BDAB = AD.

So D is the midpoint of AE.

Solution 2 by Ian Fan, Dr Challoner’s Grammar School Denote the cen-tres of circles S and T respectively by OS and OT , and let ∠DAB = θ,∠CAE = φ.

We can see that

∠OTOSO = 90◦ − ∠BXOS = ∠DAB = θ

and

∠OOTOS = 90◦ − ∠OTY C = ∠CAE = φ.

Then

∠OTOSD = 90◦ − ∠OSDA = 90◦ − ∠DAOS = φ

since the line AC is tangent to circle S. Similarly, ∠DOTOS = θ. Combiningthis with what we deduced above,

∠OTOSO = ∠DOTOS

21

Page 28: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

and

∠OOTOS = ∠OTOSD.

Therefore the two triangles �OTOSD and �OSOTO share the side OSOT

and the two angles at OS and OT (but the other way round). So one is areflection of the other. Consider this reflection. It maps OS to OT and viceversa, so it must be a reflection in a line perpendicular to OSOT . Since Dmaps to O (and vice versa), this line must also be perpendicular to the lineDO. So DO is parallel to OSOT .

Since OSOT is perpendicular to the line ADE, it must also be the casethat DO is perpendicular to ADE. As �AOE is isosceles with AO = EO,this implies that D is the midpoint of AE.

22

Page 29: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

����������� ��������������������������� !"#$%#"&'&(')"#%'*+,-./01/2,-.30,-.341/-,54 6,-.,01/2,-.30,-.341/-,74 89:�;��<�<:������=�>�?�<���?=���<:?<���@�A=����<�<:��:�=��B����������<��>��CDA�<���EF?��?�=�GHIJKL���H�MA�F?=��<=�HIJKI��IJKN OP�@�<:��B������������?�?=����?�@C��>�<��>A@:>����?<A�?=<��B�?���A<<:�<��>�?����>�=�C�Q��������<<���?�����:�@:<:����A=<����B��������@?����:���<�F?���:O

R� �ST���UV�WXY��Z[�\� ]RT ���Y �RZX�� _ �<ab,-.3O9:��<:���F���B��������@?������<<��?�ca1.d01/a0a41c,a1.d46ca6.d01/a0a41c,a6.d4ba01/ae1fa41/a1.1/a0a41/a41/a1. 6a46/ae1fa46/a1.1/a4a41/a46/a1.�:�@:@?�<:������>�=�g��<�7a01/ae1fa41/a1.7a41/a1. 67a06/ae1fa46/a1.7a46/a1.ba4c7a41/a1.d17a41/a1.7a41/a1. 6a4c7a46/a1.d17a46/a1.7a46/a1.bca41.d6ca41.d?����<:�?�������JOR� �ST��h�U���U��YSX]iX�j�Y�W�RZX�� ]kY�Sl���mn TWXS U��TS��o_�<pb,-.3Oq�@?�<:�����<�cp1.d01/p0p41c,p1.d46cp6.d01/p0p41c,p6.d4brcp1.d01/p0srp41c,p6.d4s6rcp6.d01/p0srp41c,p1.d4srp41c,p1.d4srp41c,p6.d4s 8q�<:��C�@A����B�?������A<<:��A>��?<��O9:�g��<�?�<�C<:�����MA?=<� p4cp1.d01/pt1/p0c,p6.d41cp1.d0c,p6.d48

KN

uvwxyzv{|}}}}}~���

Page 30: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

�B�?�����<:���?@;�<�?�����A����=�;�<��>�H<:�����MA?=<�,7pt1.�p067p41.8P�<:��<:��:?��H�B�?������A<<:���@����?���C��?@;�<���F��A�p4cp6.d01/pt1/p0c,p1.d41cp6.d0c,p1.d48P�@�?�?��H��@?��B�?���A<<:���?@;�<�CA�<:��?�����A�=�;�<��>�H<����=� ,7pt1.�p067p41.8��<:�?��F��B�����������MA?=<�r,7pt1.�p067p41.s6r,7pt1.�p067p41.srp41c,p1.d4srp41c,p6.d4s�:�@:���MA?=<�JO

K�

Page 31: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

��� ��h������������?F�����; ��"$%&'�"��#'�&��"��#'����&('+**&*+&('�'��'���$%#"�+�*��&*�������'&��'&('&"��'�&&*&('$��$#'���"&���('+**&*+&('�'��'���$%#"�+�*��&*�������*)'&("&�����"�"##'#&*������<�A@@���CA=��=A<����<�<:������=�>��F�=F��<:�?=<���?<����>��<<:����><���<:����<:��>����A=<�?��A<?��=����?@�@=�@MA?���=?<��?=O

R� �ST���U�Y Y�T��TnX�Y]RSY�\� RZX�� _ ���<��<�@�<:?<������?@�@=�@MA?���=?<��?=���@�¡���1¡���b.¢-£O9:���C���H¡���b¡�����<:�<:����>�C?��=����<:��?>����>��<O�=��H��<:�?=<���?<����>��<<:����>H¡���b¡���O9:A�¡���b¡���O����<:�@��F�����C<:�@������������?��=��<:����>H=������?����?���?�?==�=O

Page 32: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

��� ��¥���������������>�¦��� §�%��'� ��&&'����"�'.-��"�&����*+34©ªe����&3��«**&('�����&"��'"���¬���&('(��('�&�*'�*+3 (�$(��)��'�&(���%��'�����<��=A<����<�<:������=�>��F�=F��?>�<:����>�=?�<�<:�g��<��­=A<���������<��®��@A���F�=����?;���<:��A>���������<�?����A@<�C>?��<��>�H�?@:��F����=���3�A<��<���O9:���@�����=A<�����?�A�­A�A?=�A<��?<?=<���?<�F�OR� �ST���U �Y�°kY�][� \TZX±� �W�_ �<<:��A>�����MA��<������O²�<�@�<:?<ªe���?�<�����C34©ªe.�O³�������>��������?==�<:��A>���� �:�@:@�����<��C?�<�����C3 ����<.�O�<µ´��<:��A>���C��>���C?����<.Hc3 6.d@����@A<�F�����<�-H?��<:������<.H?��<:��c3 6.d@����@A<�F�����<�-?��<:��?��<:��.O²�<�@�<:?<� ¶µ´b� ·ªO���@�<:�����<?=�A>�Cµ´��3H�<����F����=���3O ���F��H���@�<:������<��F����=����Hµ´����<��F����=����O��� ·ª����F����=����B?@<=����:��:��������C3<:?�� O²���ªb...����F����=���3ª�A<��<��34H?����� ����F����=���3 �A<��<��3·ªO¹��?�<�@A=?�H<:��>�?��<:?<ªe�����F����=���34©ªe�A<��<��34©ª0O ��@������F����=���34©ª0H�A<����:��:��������C3OR� �ST��h�U�YnTYºY����S��]��TWYS�j�Y��Y�RZX�� _³?==<:��A>­�����MA��<���»O9:��3»1.b.-e¼½¾¿bc341.de¼½¾¿bc34de¼½¾¿1À34©ªe. Ác34de¼½¾¿Âª1¶¶¶1À 34©ªe34©ªe6,Ác34d41À 34©ªe34©ªe6.Á341.?����

3»bc34de¼½¾¿1À34©ªe. Ác34de¼½¾¿Âª1¶¶¶1À 34©ªe34©ªe6,Ác34d41À 34©ªe34©ªe6.Á348

Page 33: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

²��@�������<:�����F��A?=<��>�� bÀ 34©ªe34©ªe6pÁc34dbc34©ªedc34©ªe6.d¶¶¶c34©ªe6p1.dpcp6.dcp6,d¶¶¶. ¶34bc34©ªe6.d¶¶¶c34©ªe6p1.dcp6.dcp6,d¶¶¶. ¶3p¶3 ¶34©ªebc34©ªe6.d. ¶c34©ªe6,d, ¶¶¶c34©ªe6cp6.ddcp6.d ¶3p¶3 ¶34©ªe8q���==��F��<��?<��:�@:�������C3��F���� O ���?@:ÄÅpH�C34©ªe6Ä����F����=���?<=�?�<?�>?���������CL?�Ä��O9:���C��������C<:�<��>�e¼½¾¿ÂÆÆ :?F�?���������C3��<:�����>��?<��O��� ����F����=���?<=�?�<?����?<?������C3?�eǶ3¶34©ªe��O ���F��H3 ����F����=���?<=�?�<?����?<?������C3?�p��H?���������F����=���?<=�?�<?����?<?������C3?�34©ªe· Oq�@?�@:�@;<:?<�ªb 34©ªÈ?��<:?<34©ªtÉ�4OÊ�<:�?��F����A=<H34©ªtÉ� C��?==pË,H?����3»b�ª1�41¶¶¶1�e¼½¾¿����F����=���34©ªÈH�A<��<��34©ªtO��<:�:��:��<������C3<:?<��F����»��34©ª0O

Page 34: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

��� ��Í���������������>�¦��� ��""$���#"�����"���'��"Î(*#��"Î�Ï)�'�Î�"Î(' �##�*�%�Ð���*� "&'��Ñ������*�(' �##�'�&�Ò�"�Î��)'��"Î('�*'�Ó%�&*�'*+&('�'&(�''&(�����Ô'�')'��*'���++'�'�& "&'����*�&�*�$*��'$%&�)'�"Î��Ô* �"�Î�$('�%#'�"�'�*����#'+*�&(��(*#��"ÎÕ9:���?��?F?���<��C�?���C?����?@:���<:������=�>O�>����<�A@­@���CA=��=A<����H����C<:�>��<@�>>���?�<���<A�?��@A����@�H@?=­@A=?<���<:��A>�����C�?��<:?<¹�??@@�A=������<:�g��<p�?���C:��:�=��?�O���@���?����?@:��F�=F����=�<<���<:�:�=��?�A���<��=�@;��C<:����?��H?��@�A�<���<:��A>����C�?��<:?<<:����=�@;�@�A=������@��<���<:��OR� �ST���USX�n�SS��n_����<���Öcpd<:��A>����C������=�:�=��?��<:?<@��C��><�¹�??@G��A=��H?��=?�<p�?��Oq�?��?�;��<�g��Öc�dO¹C:����<���<:�p<:�?�H<:��:�@?��A�CH�;������<��<:�cp1.d<:�?�®<:���@:��@��O¹C:��?<���;�����<:�p<:�?�H<:��:�@?����<���?<���;����<:�cp1.d<:�?�®<��@:��@��O��>�=?�=��C:��A�C�H<:��:�@?���<:���A�C�����<O��<:��A>����C������=�:�=��?���C=���<:p1.���MA?=<�<:���<�>��<:��A>����C:�=��?���C=���<:p<:?<�����<:?���<�?�H�=A�<��<�>��<:��A>����C:�=��?���C=���<:p<:?<�����<:�A�g��H�=A�<��<�>��<:��A>����C:�=��?���C=���<:p<:?<�����<:�?<���;����O9:A�<:��A>����C:�=��?���F��cp1.d�?�����MA?=<�<��@�<:��A>­����C:�=��?���F��p�?��H�=A�<:��A>����C:�=��?���Cp�?��<:?<�����<:?���<�?�O ���F��H<:��A>����C:�=��?����<:p�?��<:?<�����<:?���<���MA?=<�<:��A>����C:�=��?���C=���<:cp6.dO9:?<��<��?�H<:?<Öcp1.db,Öcpd1Öcp6.d8q�@?�>?�A?==�g��<:?<Öc.db3?��Öc,db5O9:���?C<��H��@?�@?=@A­=?<�Öc3db.5HÖc/db/.HÖc7db��HÖcfdb,3�HÖc5db755HÖc¢db.3�3?��Öc�db33f3O��<:�?�������33f3OR� �ST��h�U× �ØY�����Y]kY���\RZX�� _��=�<<:�Ù�?����<�<:�����<��CLOq��<���EÚGC�����<���HE�GC���A�g��?��EqGC���?<���;����H<:���?��KÃ������=��@:��A=��C��?<:����?��<��<@:H�?>�=����H��ÚH�ÚÚH�Ú�H�ÚqHqqqHqqÚHqÚÚHqÚqHqÚ�HÚÚÚHÚÚ�HÚ�ÚHÚ��HÚÚqHÚqÚHÚqq��>?���?��@?������@�<:����C<:���<���<:��Û¹C�����@�C�<:�=?�<�?��C�?@:�C<:�g��<<����<��C<:���H<:����@?�@?=@A=?<�<:��A>����C

Page 35: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

������=��?��­<:����<��@<����?��<:?<�C<:�g��<��<������?��H<:��<:���@�����<>A�<��������ÚH?���C�<������qH<:��<:���@�����<>A�<�����q��ÚO ���<��<�������H��@������Ü72/2.,b,/-O ���<��<�������H��@�����ÚÜ7272.5b/,7O ���<��<�������H��@�����qÜ7232.,b.¢-O ���<��<������ÚH��@������Ü5272.,b/,-O ���<��<������ÚH��@�����ÚÜ5252.5b¢33O ���<��<������ÚH��@�����qÜ5272.,b/,-O ���<��<������qH��@������Ü7232.,b.¢-O ���<��<������qH��@�����ÚÜ7272.5b/,7O ���<��<������qH��@�����qÜ72/2.,b/,-O9:�<�<?=�A>����C������=��@:��A=����<:��<:��A>�C<:��������A>­����H ,/-1/,71.¢-1/,-1¢331/,-1.¢-1/,71/,-b33f38

IJ

Page 36: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

��� ��Ý�����������¦?�<:�;9?���?�? �'&����'"�'Þ%�#"&'�"#&��"��#'�"��ß�'"�*��&�����'&('&��"��#'��'&àá�"����'&('+''&*+&('�'��'���$%#"��+�*�ß&*&('���'���á��"�����'��'$&�)'#Î���*)'&("&"â��1�à1��b��1��1�à�"���âr�ß�s1r�ßàs1r�ß�sbr�ß�s1r�ß�s1r�ßàs�9:�>��<@�>>��?����?@:<�<:������=�>�?�<������F�<:?<<:���­MA�����MA?=�<���:�=��:��ß����>��?<A�?=����<�A@:?�<:�@��<���C<:�<��?��=�H?��<:���:��<:?<?�ß>�F��?��A��H<:�<�������?��A�?C­C�@<��Oq�?=��������<?F�����CC����<H@=�?�?����?@:<:?<?F����?������C��@?=@A=?<���OR� �ST���UiTY���TãT]±��Z���±� �W�_¹Cß��<:�@��<���C<��?��=����H<:��<:����A=<:�=������>>�<��Oq�@?���?@:?�?���<�?������<ß������������<?�<���?<<:�@��<��ß©?��>�F���?=���<:�=���ßà?��<:��?=���<:�=���ß��C����>����<?�@�����<:������@<��� Oq���==���F�<:?<<:��MA?=�<�? :�=�������F���<:?<�<��������F�����A@:>�F�>��<��CßO

³�����������<�ß?��ßä�A@:<:?<ßä=�����<:����>��<ß�H?��:�����<:�?��F���?��?>O�<<:�C��<�C<:���������@A=?��C��>ßä<���?������àä?���ä�����@<�F�=�O�=��=�<å��<:�C��<�C<:���������@A=?�C��>ß<�ßä�äOÊ�@����������MA?���=?<��?=ßä���äH¡åßßäb¡���äbf-£O��ßåbßßäæçèf-£OÊA<ßå�ä���?��@<?��=�?������äbßßäæçèf-£OÊA<����>>�<��Hààäbßßäæçèf-£b��äO9:A� ��ä1�àä1��b��6��ä1�à1ààä1��b��1�à1��IK

Page 37: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

?�� ����1����ä1�à�àäb����1����ä1����ä1�à�à6ààààäb����1����1�à�à?�����C����1�à�à1����b����1����1�à�à<:��?=������ä1�à�àä1����b����1����ä1�à�àäO9:���MA?=�<���?=��������F���:��>�F���?=���<:�=���ß�ß�""ÎC��>�H��@����������<:��?>���<A?<�����<:ß?��ßä��F�����O��>�=?�=��<��������F���:��>�F���?=���<:�=���ßàßàO���@��<:�=���:��ß��<:�@��<�����C���H�<:�=��C��?==����<�ß���������O9����F��?�<� H��G==A��<:��?>�>�<:��®<:����A=<��@=�?�=�<�A��:��ß��<:�@��<�����C<:�<��?��=�H?������==�:��<:?<<:��MA?=�<���������F����>�F�>��<�?=���<:�=����ßàßà?��ß�ß�O

 ���?���C��<?<���H��G==@?==<:�=���<:�ßàßàbéªáßáßäàäbéäªáß�bé4?������?�����@?<����<:���?��?>O�=��H=�<ßßßßäbêO ���<�C?==��<�<:?<H?�¡ßßßßäåbf-£Hßäåbßäëbì4?����éª6éäªbì4bée6éäeO�²�<�<:?<�<C�==���C��><:��<:?<éª1é41éebéäª1éä41éäeH�A<���:?==��<?@<A?==�A��<:��C?@<O Ê�����?<��=�A����<:�C��>A=?E?��?bª4�?��2:���:<GHr�ß�ßä�äs1r�ß�ßäàäs1r�ß�ßä�s6r�ß�s6r�ßàs6r�ß�sbcbcéäe����ä1éäª�à�àä1é4����6ée����6éª�à�à6é4����dbcbcée1ª4êdcdc����6ì4íîæ3-£d6ée����1c1céª1ª4êdcdc�à�à6ì4íîæ3-£d6éª�à�à1c1cé46êd����6é4����bì4c����1�à�à6,����1ï3c3céª6éedddd8 �K q�G==?�><�A���@;<:��=?�<<��>Héª6éeO ��>�A��?�=������;Héª6éebéäª6éäeO9:��:�=��C��?==ßä?=���<:�=���ß�ß�H?������@�A=���@;ßääääb�OIIII

Page 38: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

9:A�H�:��éääääª��<:���������@A=?����<?�@�C��>�<�<�����?��éääääe��<:���������@A=?����<?�@�C��>�<�<�����H��:?F�<:?<céª6éedbcéääääª6éääääedO

²�����@�¡àä����bf-£H��;���<:?<éääääªb����æçèf-£?����>�=?�=�éääääeb����æçèf-£O��céª6éedbdbï3,c����6c����6����ddbï3����6ï3,��8q�;���C��>�?�<? <:?<����1�à�àbe4����6����H?������:?F�<:?<�K ���MA?=<�

ì4ce4����63����1313����6e4����db-89:A� r�ß�ßä�äs1r�ß�ßäàäs1r�ß�ßä�sbr�ß�s1r�ßàs1r�ß�s?���� r�ß�ßä�s1r�ß�ßä�s1r�ß�ßäàsbr�ß�s1r�ß�s1r�ßàs8³�>������<:���<��H�����<:?<�C<:���������MA?=�<�:�=��C��ßH<:���<?=������C��ßäO����C���H�<C�==���<:��<:?<�<:�=��C��?��@:��@�ßä��<:�=���ß�ß�H?��<:��C��?��ßä�����������>�F���?=���<:�=�������?��<:��<:�=����à�àOR� �ST��h�USX�n�SS��n_�<?�<�CC����?����<:���=����<:��A�:ßH�?�­?==�=<��?@:�C<:�<:��������®ëðëð�?�?==�=<�����Hñòñò�?�?==�=<�����?��óôóô�?�?==�=<�����O�9:���B����<�ëáñáóáðáò?��ô?����<:�������C<:�<��?��=�?��:�����<:���?��?>O

ILIL

Page 39: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

���@�������MA�=?<��?=H�?@:�C<:���<:���=����>��<�<:�������C<:�<��?��=�?<?f-£?��=�O��<:�<��?��=��ßôñHßðó ?��ßòë?��?==�MA�­=?<��?=O��?���A=<H<:���������@A=?�ßà@A<�<:�=������>��<ôñ��:?=CH?����>�=?�=�C��ß�?��ß�Oq�<:A�:?F�<:?<ñàbàôá �I ó�b�ðá?�� �L ë�b�ò8 �N ���F����?=��;���<:?<�ëð���?����@�=��<�?��õ�A>O���ë?��ð�?��<:��?>�=���<:O���=����<:��?>�=���@<�<:�<�?��õ�?�ñò�?���óô���F��A�<:?< �ëbð�á �� �ñbò�á?�� �¤ �ó bô�8 �à q�@?���F���A��?@:�C<:���B=���<:���MA��<���?�C�==���Ü��b�ò1ò��àb�ô1ôà��b�ð1ð���b�ó 1ó���b�ë1ë��àb�ñ1ñà9:A�<:�@�>���?<����C�I H�L H�N H�� H�¤ ?���à <�==�A��B?@<=�<:?<��1�à1��b��1��1�àáIN

Page 40: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

?�<:�MA��<�����MA����O ���?�<� H��G==?��<:�=�����ßá�ß?���ß<�<:���?��?>O

9:�MA?���=?<��?=�òßó ��?�?�?==�=���?>O9:���C���H<:�<��<��?��=��ß�ó ?���ßò?��@����A��<O��>�=?�=�H<:�<��?��=��ß�ë?���ßô?��@����A��<H?��<:�<��?��=��ß�ñ?���ßð?��@����A��<O�B?@<=�?����?�<? H��;���<:?<<:�=���ßà��=�<�<:�<��?��=�ßñô��:?=CH?����<��?��=��ßàñ?��ßàô?��@����A��<O��>�=?�=�H<��?��=��ß�ó ?��ß�ð?��@����A��<H?��<��?��=��ß�ë?��ß�ò?��@����A��<Oq�:?F�=�?��<<:?<��B�?����C<��?��=��?��@����A��<H?�����?@:�?��:?��MA?=?��?Ü rß�ósbr�ßòsá �Ì rß�ësbr�ßôsá �Ù rß�ñsbr�ßðsá �KJ rßàñsbrßàôsá �KK rß�ósbrß�ðsá?�� �KI rß�ësbrß�òs8 �KL �A>>���A�<:�=�C<?�����:<������C�Ì H�Ù H�KJ H�KK H�KI ?���KL <�==�A��B?@<=�<:?<r�ß�s1r�ßàs1r�ß�sbr�ß�s1r�ß�s1r�ßàsá?���MA����O

I�

Page 41: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

��� ��ö�����������÷�?:?> �?�� �('"��#'��á�"���*+"&��"��#'"�'�'"�%�'����'��''��"��&('#'��&(�*+&('*��*��&'���'�"�'aáø"��ù�'��'$�&�)'#Î���*)'&("& f-úa�1ø�1ù�a1ø1ù ú�-9:�<���?�<��C<:�<:�����MA?=�<�@?���<?@;=�����?�?<�=�O���<�A@­@���CA=��=A<����A���<:���?��?���>��<���MA?=�<�C��<:�=�C<����H?��<:�<��?��=����MA?=�<�C��<:����:<����H?�<:���=A<���������<����=������O

R� �ST���U ����T�û T�W\��SX]�YW�Y ��±� �W�RZX�� _q�<:�A<=����C�����?=�<���>?�?��A>�<:?<aü øü ùO���@�<:�=?����<?��=���������<�<:�=�����<������?��<��?��=�H?��<:��>?==��<?��=���������<�<:��:��<��<����H��<:��?=��:?F�<:?<�ü�ü�O9:?<��<��?�H<:?<aáøáù?���á�á�?����<:���­��@��?������MA��@��O9:���<C�==���C��><:���?��?���>��<���MA?=�<�<:?<a�1ø�1ù�üa�1ø�1ù�a�1ø�1ù�üa�1ø�1ù�8�A>>���<:���<�����MA?=�<���H?��?�����a�1ø�1ù�<��?@:������F��3ca�1ø�1ù�düa�1a�1a�1ø�1ø�1ø�1ù�1ù�1ù�bca1ø1ùdc�1�1�db.¢-ca1ø1ùd8���@�<:�?��=����?<��?��=��A><�.¢-£O��F�����<:��A�:��3ca1ø1ùd��F��A� a�1ø�1ù�a1ø1ù üf-8²�������F�<:��<:�����MA?=�<�Oq�@��<��A�<�?��A>�<:?<aüøüù?��@����MA��<=�<:?<�ü�ü�O9:�<��?��=����MA?=�<��<?<��<:?<aÅø1ùH?���� a�Åcø1ùd�bø�1ù�8q�?=��;���<:?<ø�úa�?��ù�úa�O9:A�a�1ø�1ù�Åa�1a�1ø�1ù�Åa�1a�1ø�1ø�1ù�1ù�

Page 42: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

9:A� ,ca�1ø�1ù�dÅa�1a�1a�1ø�1ø�1ø�1ù�1ù�1ù�bca1ø1ùdc�1�1�db.¢-ca1ø1ùd8��F�����<:��A�:��,ca1ø1ùd��F��A�<:?<a�1ø�1ù�a1ø1ù Å�-?���MA����O

Page 43: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

����������� ������������������� �����!"#$%&'(&)*##*+,-.-/01&23,-,-%2&$34,-.*25&2*)3,6&7$-58/3',)99*/22&$3*-,-.:;<=>;?@>;?=>A<@$-5A<=B*'&'($'CDE0&$-3CFDEGH� �IJ���KLM��K��I���JM�NOJ�PQ�RI��S� �T�UVW�XY���Z�������;[\>;]?>;_]>A`���A[\ab������;]?c;[\c;_]de�fA`ghA<ijgkAj���;]?gh;?i\]glm\]an�o�����pW�pkAjclm\]����� A`c;]?c;[\c;_]dq��o�A[\r;[\�p��Z����p�o�Z����A[\���;_]as�f�Y��A[\ghA<i gkA`���;_]gh;?i]?glm]?����p��o����pW�pkA`clm]?aq�A`c;]?c;[\cA[\c;_]>pW�p��t A<@c;?@c;<=cA<=c;?=dH� �IJ��u�K��KM�vTN���w���HPQ�� Ux�o��o���p���t;<= g;[\t;?@ g;]?t;?= g ;_]tA<@ g A`���A<= gA[\an�o��p�y�pW�p��pW���p�z���XY���Z����f�pW��p�zz�op��{pW��������{t������o�pW�����|��Y����pp���Y����{pW��}�����p���p��t�p��Z����p����ppW�z����f��{��Z����~;[�\>;]�?>;_�]>A� ���A[�\dVW�� k�kmg;[c;[�\c;jgm�kl;>���g;]c;]�?c;g�ml�>mkgAcA� cA<gkA>k��;kgA[cA[�\cAjg�k�mAA>��� �k�mAAck�km<gh;[i<g;?c;_�]d

��

�����������������

Page 44: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

���ZpW���f�o�������zzpW������~A� c;]�?c;[�\cA[�\c;_�]����� A<@c;?@c;<=cA<=c;?=d

��

Page 45: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

��� ��u����������� ������Y���W�!"*3,',�&,-'&.&23�>C$-5�3$',3)9'(&&�/$',*-��C g��"2*�&'($',)�,3$�2,0&.2&$'&2'($-;7'(&-C,3$0/#',�#&*)m�$-5�h��C�mi,3$�&2)&%'3�/$2&�H� �IJ���KLM��K��I���JM�NOJ�PQ�RI��S� �T�U�������{��{���z�o�p������{tf�W�Y�pW�p�g� C gh��Cih� CidVW���������p�Y�z�op����z����mt�����an�o����pW�Y�pW�p��Cg� Cg����o�C������p�Y�t���pZ��p��pW�o���pW�p��Cg����� Cgmaq� �g��m� ���Cg� m����pW���z��� �Cg� mgh��mih� midq��o�������t��pW��m���� m����Y��t���Z����Y������zpW�Z����Y��������Aas��o�l��Y�����Caq��o������p��Y��������;t����z��m���� m����Y��������;�����;��Y�����Ca¡�l���;���o����Z�t�pz����f�pW�p�A��Y�����C���m���Y����Ct����|�����ae�f �h��C�mig�h��¢£¤\^ �mig���� m��g�����mgh��mifW�oW��p����pW�p�h��C�mi������z�op�|����t����|�����aH� �IJ��u�K¥M��R��¦��N§¦I��S���I �M��M�HPQ�� ©R JTQI K��JI��ªUb���������{pW��|��p���Z�����;t�������{pW�z�oppW�p���o��{����pp���pW��«��mZ�����;tf����pW�p�p����p����z����C ��o��{����pp�«Z�����;a¬po��­p���t���pW���z���C ®« °±;���� ®�®m°±;a¬����p�o����t;��Y����Cab���������{pW��|��p���Z�����At�������{pW�z�oppW�p���o��{�����pp���pW��«��mZ�����Atf����pW�p�p����p����z����C��o��{����pp�«Z�����Aa¬po��­p���t���pW���z���C ®« °±At�����C���Y����p����������pW���a��op������{pW����{�����|��p���tC gh� �ih���iaq��o���pW�����������t����zh� �i���h���iZ��p����Y��������At������o�pW��pW����

�²

Page 46: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

�����Y��t�pz����f�pW�pl��Y����CaVW���z���A��Y����Can�W�Y��W�f�pW�pC����Y����������pW;���At������o�;���A���o����Z�t�pz����f�pW�pC����Y��������m�a��op������{pW����{�����|��p����{���t� gh� Cih��CiaVW���������p�Y�z�op����z����mt�����an�o����pW�Y�pW�p��Cg� Cg����o�C������p�Y�t���pZ��p��pW�o���pW�p��Cg����� Cgmaq���Cg����� CgmaVW����C�mg���a��p­�o�������������Z��³pW�p��Y������a¬z³����p����t³o����p��Y���� ����pW���pf������Y���pW���zz����o�h���i h� �ig��!�����³��Y��������}�op����Z���p�Z�����p��Y����CtfW�oW�����Y����Z����zp�Z��a¬po����p��pW�o���pW�p���Y�������t��pW���pf������Y�����pW����Ct��p� Cgm������f������Y���man�y��fpW�p�}�op������zh���i���h� �i��o��{����pp��Z�����At������o��Z��p��Y���C���Y����Z����zp�Z��t���Y������oW�z������� ��������Z����zp�Z��aq����Y�����h���i���Y����Z����zp�Z��aVW���z����Y������Z�pW�p��Y�����h���ig�h��C�mi��Y�����p���Y����Z����zp�Z��aVW���z����h��C�mi������z�op�|����a

Page 47: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

��� ��µ�����������¶����� ������!·(*'&#($3'&-2**03$#*-.&$%(3,5&*)$%*22,5*2�·-*#90�,$5'&$0#&$5&2+,3(&3'*1**3&�&-2**03*-'(&%*22,5*23*'($'-*'+*2&3&2�&52**03*-'(&3$0&3,5&*)'(&%*22,5*2$2&$58$%&-'�¹-(*+0$-9+$93%$-'(,31&5*-&ºH� �IJ���K»T�J¼��M���¼��N ��½�»PM���KUe�Z���pW����Z�����oW�����zpW�o��������p��¾an�­����Z���p��p��f����fW�oW���o����oy¿������À�o��p���Z�����������zpW�o�������a��oy��p�{���C\>C>dddCÁ��oWpW�pmÂC\cC cÃÃÃcCÁÂmm ¿aVW����oy���Z�C\>C�m>C<��>dddCÁ�¿ man�o�����pW�ppW������Z��Z���������o������{t������o�CÄÅ\rCÄthCÄÅ\�Æi hCÄ�Æ miÇ�aq���pf���oW���Z������À�o��pab��Y������t�������pW�pf�W�Y��������{�Z��p�z¿������À�o��p���Z�����������zpW�o�������ax��������{pW�Z����o������{���Z���!�����t������p��op��{hÆ miz��ZpW���Z����zpW�ÆpW���Ztf�{�p���|���o��\>�>ddd�Áf�pWmÂ�\c� cÃÃÃc�ÁÂmm ¿t������������������{��Z��p��z¿������À�o��p���Z��������pW�f�����o��������Y�aVW������È\\¤ÁÁ Éf���p���oyC\>dddCÁ�����È\\¤ÁÁ Éf���p���oy¿������À�o��p���Z�����������zpW�o�������a�e�p�pW�p�p����p��������p���oy��}��Z���������À�o��p���Z�����������zpW�o�������!a¬zf������pW�pf�������zpW�o�������Ê���ËtpW��f�W�Y�z���oW��o��~��oy��{����Z�z��ZÊ���Ìz��ZËt��oy��{����Z�z��ZÊ���Íz��ZËt��oy��{Í���Z�z��ZÊ�������Z�z��ZËt����oy��{Ì���Z�z��ZÊ����z��ZËaq��o�fW�pf�������������zpW�o�������������p�zz�oppW��pW������tpW������Îk�ÏÃÎ��Ï�Î�AÏÃÎl;Ï�Îl;ÏÃÎ�AÏ�Î��ÏÃÎk�Ïg�mk�m�k«�m�k«��mkgA;��f����z����{���{pW����Z�a

�Ð

Page 48: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

��� ��Ñ���������������Z�Ò��{!Ó&'Ô1&$2&$#-/01&23/%('($'ÕgÔ�Ô¤\,3$-,-'&.&2.2&$'&2'($-Ö�"2*�&'($'ÕÁgÔÁ�Ô¤Á,3$-,-'&.&2)*2$##�*3,',�&,-'&.&23¿�×&'&20,-&'(&�$#/&3*)¿)*2+(,%(Õ5,�,5&3ÕÁ�H� �IJ���KLM�½�KØM�NØRT� ÙKww�K�ÚMwU����p��tÕ\gÕfW�oW������p�{��aVW��ÕgÔ ���Ô¤^ �gÕ �fW�oW����������p�{��aÛ���{��������t ÕÜgÔÜ�Ô¤ÜgÔÜ�Ôܤ^�Ô¤Ü�ԤܠÔܤ^ Ô¤ÜghÔ�Ô¤\iÃhÔܤ\�Ô\¤Üi hÔܤ^�Ô¤ÜigÕÃÕܤ\ Õܤ^VW���z����zÕܤ\���Õܤ^�����pW��p�{���tÕÜ����������p�{��an�W�Y��W�f�pW�pÕ\���Õ�����p�{���t������pz����f�������op���pW�pÕÁ������p�{��z�����¿an�o���ZpW�pÕ��Y����ÕÁ�z��������z¿�����aq��o�Õr�tf�o�����pW�pÕ������p��Y���Õaq������pW�pÝ�����t���Õ��Y����Õܤ^��p��pÕܤ\aVW��t��ÕÜgÕÃÕܤ\ Õܤ^�pz����f�pW�pÕ��Y����ÕÜaÞ�pW��pW��W���t�������pW�pÝ���Y��t���Õ��Y����Õܤ\��p��pÕܤ a¡{���tÕÜgÕÃÕܤ\ Õܤ tz��ZfW�oW�pz����f�pW�pÕ������p��Y���ÕÜaq����o���Z��p���������op���a

�ß

Page 49: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

��� ��à�����������¶�Y��Û��y!Ó&'ÊËáâ1&$%9%#,%�/$52,#$'&2$#�Ó&'ã1&'(&0,5�*,-'*)'(&$2%ÊË*),'3%,2%/0%,2%#&'($'5*&3-*'%*-'$,-á*2â�Ó&''(&#,-&3âã$-5Êá0&&'$'ä$-5'(&#,-&3áã$-5Ëâ0&&'$'å�"2*�&'($''(&#,-&3äå$-5ÊË$2&�$2$##&#�H� �IJ���KH���J�T¥M�TN�M��J�I��HPQ�� Ux�pW�pW����Z�z��{�����pW���Z���{Z��ptæËáãgæËâãax�p���o�pW�oW���ãÊW��pW���Z����{pW��pW�oW���ËãtpW���{�����p�������pW�oW���ãÊ���|���p�pW���{�����p�������pW�oW���Ëãaq�æãáÊgæËáãa

¡�æåâägæËâãgæËáãgæãáÊgæåáät�pz����f���pW�o��Y�����zpW�pW����Z�z��{�����pW���Z���{Z��ppW�p|�������p����äåáâ��o�o��oaVW��æáäågæáâå��pW�pW����Z�z��{�����pW���Z���{Z��pt���æáâågæáâËgæáÊË��pW���Z�pW����Zaq�æáäågæáÊËa���ZpW���pz����f�t��pW�o��Y�����zpW�pW����Z�zo�����������{��{���tpW�p�����äå���ÊË�����������aH� �IJ��u�KLM��K��I���JM�NOJ�PQ�RI��S� �T�U��pç��pW�����p�z��p����op����zÊá���Ëâaq��o�oW����ãÊ���Ëã����z�|������{pWtæãâÊgæËâãgæãáÊgæËáã��pW�pW����Z�z��{�����pW���Z���{Z��pan�W�Y�pW��pW�pæçâÊgæËáç���æÊçâgæËçá�Y��p��o����������p���{���!aVW���z���p����{��Êçâ����Z����p�p����{��Ëçáa

̾

Page 50: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

q�èéêé g èëìëax�p����t��pW���{������op��pW����Z��p����{���Ëçá���Êçâ�����op�Y���tèíêí g èéêé ���èîîì g èëìëab�Z�����{pW���pf�tèíêí g èîìîaVW��êíèí g îìèî ����������{mgèíèí g èîèî p���pW�����f�W�Y�pW�pèêèí gèìèîaq��o�æËçÊgæåçät�pz����f�pW�pp����{���ËçÊ���åçä�����Z�����aq�æçäågæçÊË�����pW������ÊË���äå�������������pW�o��Y�����zpW�pW����Z�zo�����������{��{���a

�

Page 51: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

��� ��ï����������������� �����!×&'&20,-&$##)/-%',*-3ðh¿i)2*0'(&�*3,',�&,-'&.&23'*'(&�*3,',�&,-'&.&23+(,%(3$',3)9'(&)*##*+,-.%*-5,',*-:+(&-4&�&2C7�$-5ñ$2&�*3,',�&,-'&.&233/%('($'\ò�\Dg\ó7'(&-mðhCi� mðh�ig mðhñidH� �IJ���KSQJôMRQJ�JR��NÚQ�õ�KRHPQ�� Un�o���ZpW�pðhö¿igöðh¿iz���������p�Y���p�{���ö���¿an���Zp����Y�pW��������op�����öa¬p��o������p����z�����¿!fW��ögmaq������pW�ppW����p����z�����¿!fW��ögÝa�����Z�����p����¿t��pCghÝ�mi¿t�gÝhÝ�mi¿���ñgÝ¿ae�p�pW�pt��z�op������{tf�X��pW�Y�pW�p mC�m�g mhÝ�mi¿� mÝhÝ�mi¿gm¿Î mÝ�m� mÝhÝ�miÏ���pW�����o�\Üg \ÜÅ\� \ÜFÜÅ\Gt�pz����f�pW�pmC�m�gm¿ÃmÝgmñ�����f�o�����pW�o����p���aVW��mðhCi� mðh�ig mðhñimðhhÝ�mi¿i� mðhÝhÝ�mi¿ig mðhÝ¿imðhhÝ�mi¿i� mÝðhhÝ�mi¿ig mÝðh¿ifW���f����pW�����op���W���pW����p�{�z��ZpW���o�������p�pW�pW���ax�ptz�op������{tÝ�mÝ Ã mðhhÝ�mi¿igmÝà mðh¿imðhhÝ�mi¿ig mhÝ�miðh¿i�����ðhhÝ�mi¿ighÝ�miðh¿ito�Z���p��{pW�����op����p��a

ÌÌ

Page 52: BMO1 past papers - learninglighthouses.webs.comlearninglighthouses.webs.com/BMO_1114.pdf · UnitedKingdomMathematicsTrust. 2011/12 British Mathematical Olympiad Round 1: Friday, 2

q�f�W�Y�pW�pðhö¿igöðh¿iz���������p�Y���p�{���ö���¿a¬����p�o����tðhöigöðhmi��������z��op����ðpW�p��p��z�pW��o����p�������zpW�z��Zðh¿igñ¿z����Z�����p�Y���p�{��o���p��pña¬p��o����pW�p�����oWz��op��������������p��z�pW�o����p���a

ÌÍ