HCMOP Open Incidence Matrix Solutions

Solutions to Problems on Incidence MatrixProblem 1. 8 singers participate in an art festival where m songs are performed. Eachsong is performed by 4 singers, and each pair of singers performs together in the samenumber of songs. Determine the minimum value of m.

Solution. Notice that for every song performed, there are(4

2

)= 6 ways to pair up the

four singers. Thus there were 6 pairs who performed together in each song. Supposeevery pair of singer sang together in k songs. There are a total of

(82

)= 28 pairs of

singers, so by counting the total number of times each pair of singers perform, we havethe equality

28k = 6m

which means 14 | m thus m ≥ 14. Finally, we give a construction for m = 14. Bynumbering the singers from 1 to 8, divide them into 14 sets of 4 people each, to representthe singers for each of the 14 songs as follows:

{1,2,3,4}, {3,4,5,6}, {5,6,7,8}, {1,2,7,8}, {1,2,5,6}, {2,3,6,7}, {3,4,7,8},{1,4,5,8}, {1,3,5,7}, {2,4,6,8}, {1,3,6,8}, {2,3,5,8}, {2,4,5,7}, {1,4,6,7}.

Problem 2. (Baltic Way 2001) A set of 8 problems was prepared for an examination. Eachstudent was given 3 of them. No two students received more than one common problem.What is the largest possible number of students?

Solution. Denote the problems by A,B,C,D,E,F,G,H, then 8 possible problem setsare ABC,ADE,AFG,BDG,BFH,CDH,CEF,EGH. Hence, there could be 8 students.Suppose that some problem (e.g., A) was given to 4 students. Then each of these 4students should receive 2 different “supplementary” problems, and there should be atleast 9 problems, a contradiction. Therefore each problem was given to at most 3 stu-dents, and there were at most 8×3 = 24 “awardings” of problems. As each student was“awarded” 3 problems, there were at most 8 students.

Problem 3. 17 students took part in a mathematical contest with 9 problems. It is giventhat every problem was answered correctly by exactly 11 students. Prove that there weretwo students who, between them, solved all 9 problems.

Solution. We set up an incidence matrix with columns P1,P2, . . . ,P9 to represent theproblems and rows a1,a2, . . . ,a17 to represent the students. The cell in row i and col-umn j is filled with ‘1’ if student ai solved problem Pj, and ‘0’ otherwise. Since everyproblem was answered correctly by exactly 11 students, there must be exactly six ‘0’sin each column. The number of ‘column pairs’ of ‘0’s is therefore

(62

)×9 = 135. Now

suppose that no two students solved all 9 problems between them. This means that forevery pair of students ai and a j, 1 ≤ i < j ≤ 17, there is at least one problem they bothdidn’t solve, contributing to at least one ‘column pair’ of ‘0’s. Thus total number ofcolumn pairs ≥

(172

)= 136, a contradiction.

Problem 4. (SMO(S)2001 Round 2) The numbers {1,2, . . . ,n2} are randomly arranged ona n2× n2 board, such that each number appears exactly n2 times. Prove that there is arow or a column that contains at least n distinct numbers.

Solution. Set up an incidence matrix with columns corresponding to the numbers 1 ton2 and rows r1,r2, . . . ,rn2 ,c1,c2, . . . ,cn2 to represent the rows and columns of the board

Prepared by Ho Jun WeiHwa Chong Math Olympiad Programme (Open)

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respectively. Fill cell in column k with a ‘1’ if the number k appears on that particularrow ri or column c j of the board, and ‘0’ if it does not appear in that row/column of theboard. Every number appears exactly n2 times on the board. If a certain number appearsin exactly a columns and b rows of the board, then we must have ab≥ n2⇒ a+b≥ 2nby AM-GM inequality. Then in every column of our incidence matrix, there must beat least 2n ‘1’s, thus the total number of ‘1’s in the matrix ≥ 2n3. Since there are 2n2

rows in our incidence matrix, by pigeonhole principle, there must be at least one rowthat contains at least n ‘1’s. This means that at least one row or column on the boardcontains n distinct numbers.

Problem 5. In a chess club, n people gathered to play chess against each other, as theypleased. No two people played against each other more than once. At the end of theday, it was observed that a total of 3n games had been played. Moreover, if we choseany two players, say A and B, there would be at most one other player who had playedwith both A and B. Prove that n > 30.

Solution. Set up an incidence matrix with n rows and n columns, and fill the cell inrow i and column j with a ‘1’ if person i played with person j, and ‘0’ otherwise (aperson cannot play against himself). Let ci represent the number of ‘1’s in column i.Each match contributes to two ‘1’s, thus ∑ci = 6n. By Jensen’s inequality, the function(x

2

)= x2−x

2 is concave upwards, thus the total number of column pairs is

n

∑i=1

(ci

2

)≥ n

(∑ci

n2

)= n

(62

)= 15n.

Also, for every pair of people A and B, at most one person played with both A and B,thus between every two rows, there can be at most one column pair, so total numberof column pairs ≤

(n2

)= n(n−1)

2 . Combining both inequalities, we obtain 15n ≤ n(n−1)2

which implies n≥ 31 as desired.

Problem 6. (SMO(O)2004 Round 2) Let m,n be integers so that m≥ n > 1. Let F1, . . . ,Fk bea collection of n-element subsets of {1, ...,m} so that Fi∩Fj contains at most 1 element,1≤ i < j ≤ k. Show that

k ≤ m(m−1)n(n−1)

Solution. Let S be the set of all possible two-element subsets {X ,Y} ⊂ {1,2,3, . . . ,m}.Clearly |S| =

(m2

). Each Fi contains n elements, there are exactly

(n2

)elements in S that

are subsets of Fi. Also, note that |Fi ∩Fj| ≤ 1, thus no element in S can be a subset oftwo sets Fi and Fj. This means k

(n2

)≤ |S| ⇒ k ≤ m(m−1)

n(n−1) .

Problem 7. (SMO(O)2006 Round 2) Let n be a positive integer. Let S1,S2, . . . ,Sk be acollection of 2n-element subsets of {1,2,3,4, . . . ,4n− 1,4n} so that Si∩ S j contains atmost n elements for all 1≤ i < j ≤ k. Show that

k ≤ 6(n+1)/2

Solution. Let A be the set of all possible (n+1)-element subsets of {1,2,3,4, . . . ,4n−1,4n}. Clearly |A| =

( 4nn+1

). Since each Si contains 2n elements, there will be

( 2nn+1

)elements in A that are subsets of Si. Furthermore, each element in A can be a subset of


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at most one Si, since Si∩S j < n+1. Thus we obtain

k(

2nn+1

)≤ |A|=

(4n

n+1

)⇒ k ≤

( 4nn+1

)( 2nn+1

) =4n× (4n−1)× . . .×3n2n× (2n−1)× . . .×n

.

Finally, we prove by induction that

4n× (4n−1)× . . .×3n2n× (2n−1)× . . .×n

≤ 6(n+1)/2.

For n = 1 the result is trivially true. Now suppose it holds for n = r.

Let µ =4r× (4r−1)× . . .×3r2r× (2r−1)× . . .× r

≤ 6(r+1)/2,

⇒ (4r +4)× (4r +3)× . . .× (3r +3)(2r +2)× (2r +1)× . . .× (r +1)

=(4r +4)(4r +3)(4r +2)(4r +1)(r)(3r +2)(3r +1)(3r)(2r +2)(2r +1)

·µ

≤ (4r +4)(4r +3)(4r +2)(4r +1)(r)(3r +2)(3r +1)(3r)(2r +2)(2r +1)

·6(r+1)/2

=(4r +4)(4r +2)(2r +2)(2r +1)

· (4r +3)(4r +1)(3r +2)(3r +1)

· r3r·6(r+1)/2

≤ 4 · 169· 1

3·6(r+1)/2

=6427·6(r+1)/2

≤√

6 ·6(r+1)/2

= 6(r+2)/2

The induction is complete. Thus

k ≤ 4n× (4n−1)× . . .×3n2n× (2n−1)× . . .×n

≤ 6(n+1)/2 for all integers n.

Problem 8. Consider the set A = {1,2,3, . . . ,100}. Let S1, S2, S3, . . . , S5n be subsets ofA, such that every element of A appears in exactly 4n of these subsets. Prove that thereexists i, j, k, such that Si∪S j ∪Sk = A.

Solution. We set up an incidence matrix with 100 columns and rows S1,S2, . . . ,S5n.The cell in row Si and column j is filled with ‘1’ if ‘j’ appears in Si and ‘0’ otherwise.Since each number appears in exactly 4n subsets, thus every column has exactly 4n ‘1’sand n ‘0’s. We define a ‘column triplet’ as a triplet of ‘0’s occuring in a column. Thusif there are ci ‘0’s in column i, the number of column triplets in that column is

(ci3

). The

total number of ‘column triplets’ is therefore 100(n

3

). Suppose to the contrary that for

all i < j < k, we have Si∪S j∪Sk 6= A, then each selection of (i, j,k) must admit at leastone column triplet. Thus there are at least

(5n3

)column triplets. We therefore obtain the

inequality (5n3

)≤ 100

(n3

)⇒

(5n3

)(n3

) ≤ 100⇒ 5n(5n−1)(5n−2)n(n−1)(n−2)

≤ 100,

but this is a contradiction, since5n(5n−1)(5n−2)

n(n−1)(n−2)≥ 5n(5n−5)(5n−10)

n(n−1)(n−2)= 125.


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Problem 9. (USAMO 1984) A math exam has two papers, where each paper consists of atleast one question and both papers have 28 questions altogether. Each pupil attempted 7questions. Each pair of questions was attempted by just two pupils. Show that one pupilattempted either zero or at least four questions in the first paper.

Solution. Each pupil attempts 7 questions and hence 21 ‘pairs’ of questions. Thereare

(282

)= 378 pairs of questions in total and each is attempted by 2 pupils. So there

must be 378×221 = 36 pupils. Suppose n pupils solved question 1. Each solved 6 ‘pairs’

involving question 1, so there must be 3n pairs involving question 1. But there are 27pairs involving question 1, so n = 9. The same applies to any other question. So eachquestion was solved by 9 pupils.

Suppose the result is false. Suppose there are m questions in the first paper, and that thenumber of pupils solving 1, 2, 3 questions in the first paper is a, b, c respectively. Soa + b + c = 36, a + 2b + 3c = 9m. Now consider pairs of problems in the first paper.There are m(m−1)

2 such pairs. Pupils solving just 1 solve no pairs, those solving 2 solve 1pair and those solving 3 solve 3 pairs, so we have b+3c = m(m−1). Solving for b weget

b =−2m2 +29m−108 =−2(m− 294

)2− 238

< 0.

This is a contradiction, so the result must be true.

Problem 10. (China 1994) Given that S = {1,2,3, . . . ,10} and A1, A2, . . . , Ak are subsetsof S satisfying |Ai|= 5 for 1≤ i≤ k and |Ai ∩ A j| ≤ 2 for 1≤ i < j ≤ k, find the maxi-mum value of k.

Solution. We construct an incidence matrix of 10 columns and rows A1 to Ak. Wefill each cell in column i and row A j with ‘1’ if the number i appears in A j, and ‘0’otherwise. Let ci be the number of ‘1’s in column i. Since |Ai| = 5, we have a totalof 5k ‘1’s in the matrix, thus ∑ci = 5k. Next we count column pairs of ‘1’s. For eachpair (Ai,A j), we have at most 2 column pairs since they contain at most two commonelement. Therefore,

10

∑i=1

(ci

2

)≤ 2

(k2

)= k(k−1).

However, by Jensen’s inequality, the function(x

2

)= x2−x

2 is concave upwards, thus thetotal number of column pairs is

10

∑i=1

(ci

2

)≥ 10

(∑ci102

)= 10

( k22

)= 5(

k2)(

k2−1).

Combining the inequalities, we obtain

5(k2)(

k2−1)≤ k(k−1)⇒ k ≤ 6,

Finally, we can easily verify that the following 6 sets satisfy the condition.

{1,2,3,4,5},{1,2,6,7,8},{1,3,6,9,10},{2,4,7,9,10},{3,5,7,8,10},{4,5,6,8,9}

Thus the maximum value of k is 6.


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Problem 11. (IMO 2001 Problem 3) Twenty-one girls and twenty-one boys took part ina mathematical competition. It turned out that each contestant solved at most six prob-lems, and for each pair of a girl and a boy, there was at least one problem that was solvedby both the girl and the boy. Show that there is a problem that was solved by at leastthree girls and at least three boys.

Solution. We set up a 21×21 matrix with rows corresponding to the boys b1,b2, . . . ,b21and columns corresponding to the girls g1,g2, . . . ,g21. For each pair (bi,g j), we fill thecorresponding cell with question number of a problem that both bi and g j solved (forinstance, if b1 and g1 both solved question 9, we fill this cell with a ‘9’). Next, considerthe row b1. Since each boy solved at most six problems, there are at most six differentquestion numbers that appear in the row b1. For every question number that appearsmore than three times in that row, we shade the cell blue. By pigeonhole principle,there will be a problem that was solved by at least 4 girls. Thus there are at most5 question numbers in the row whose cells are unshaded, and each of these questionnumbers occupies at most 2 cells (otherwise it will be shaded), thus there will at leastbe 11 cells shaded blue in the row b1. This argument extends to all rows, thus there willbe at least 21× 11 = 232 blue cells. Next, considering the column g1, we shade a cellpink if the question number appears at least three times in that column. By a similarargument, there are at least 11 pink cells in column g1, and at least 21×11 = 232 pinkcells in the entire matrix. Since there are only 21× 21 = 441 < 232 + 232 cells in thematrix, there must be a cell that is shaded both pink and blue. This problem was solvedby at least 3 girls (since it is shaded blue) and at least 3 boys (since it is shaded pink).


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HCMOP Open Incidence Matrix Solutions