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42nd International Mathematical Olympiad
Washington, DC, United States of America
July 8–9, 2001
Problems
Each problem is worth seven points.
Problem 1
Let ABC be an acute-angled triangle with circumcentre O. Let P on BC be the foot of the altitude from A.
Suppose that BCA ABC 30 ^.
Prove that CAB COP 90 ^.
Problem 2
Prove that
a
a^2 8 bc
b
b^2 8 ca
c
c^2 8 ab
for all positive real numbers a, b and c.
Problem 3
Twenty-one girls and twenty-one boys took part in a mathematical contest.
- Each contestant solved at most six problems.
- For each girl and each boy, at least one problem was solved by both of them.
Prove that there was a problem that was solved by at least three girls and at least three boys.
Problem 4
Let n be an odd integer greater than 1, and let k 1 , k 2 , …, k (^) n be given integers. For each of the n permutations a a 1 , a 2 , …, a (^) n of1 , 2, …, n , let
Sa
i 1
n
k i a i.
Prove that there are two permutations b and c, b c, such that n is a divisor of Sb Sc.
Problem 5
In a triangle ABC , let AP bisect BAC , with P on BC , and let BQ bisect ABC , with Q on CA.
It is known that BAC 60 ^ and that AB BP AQ QB.
What are the possible angles of triangle ABC?
Problem 6
Let a, b, c, d be integers with a b c d 0. Suppose that
ac bd b d a cb d a c.
Prove that ab cd is not prime.
Solution 3
We first show that R^2 CP CB. To this end, since CB 2 RsinΑ and CP ACcosΓ 2 RsinΒcosΓ , it suffices to show that ^14 sinΑsinΒcosΓ. We note that 1 sinΑ sinΓ Β sinΓcos Β sinΒcosΓ and ^1 2 ^ sinΓ^ Β^ ^ sinΓcos^ Β^ sin^ ΒcosΓ^ since 30
(^) Γ Β 90 (^). It follows that 1 1 4 sinΒcosΓ^ and that 4 sinΑsinΒcosΓ.
Now we choose a point J on BC so that CJ CP R^2. It follows from this and from R^2 CP CB that CJ CB , so that OBC OJC. Since OC CJ PC CO and JCO OCP, we have JCO OCP and OJC POC ∆. It follows that ∆ OBC 90 ^ Α or Α ∆ 90 ^.
Solution 4
On the one hand, as in the third solution, we have R^2 CP CB. On the other hand, the power of P with respect to the circumcircle of ABC is BP PC R^2 OP^2. From these two equations we find that
OP^2 R^2 BP PC PC CB BP PC PC^2 ,
from which OP PC. Therefore, as in the first solution, we conclude that Α ∆ 90 ^.
Problem 2
Prove that
a
a^2 8 bc
b
b^2 8 ca
c
c^2 8 ab
for all positive real numbers a, b and c.
Solution
First we shall prove that
a
a^2 8 bc
a
^4 3
a
^4
3 b
^4
3 c
^4 3
or equivalently, that
a
^4
3 b
^4
3 c
^4
2
a
^2
3 a^2 8 bc.
The AM-GM inequality yields
a
^4
3 b
^4
3 c
^4
2
a
^4
2
b
^4
3 c
^4
3 a
^4
3 a
^4
3 b
^4
3 c
^4
2 b
^2
3 c
^2
3 4 a
^2
3 b
(^1)
3 c
^1 3
8 a
^2
3 bc.
Thus
a
^4
3 b
^4
3 c
^4
2
a
^4
2
8 a
^2
3 bc
a
^2
3 a^2 8 bc,
so
a
a^2 8 bc
a
^4 3
a
^4
3 b
^4
3 c
^4
Similarly, we have
^ b
b^2 8 ca
b^
^4 3
a ^4 (^3) b ^4 (^3) c ^4 3
and
^ c
c^2 8 ab
c^
(^4) 3
a ^4 (^3) b ^4 (^3) c ^4 3
Adding these three inequalities yields
a
a^2 8 bc
b
b^2 8 ca
c
c^2 8 ab
Comment. It can be shown that for any a, b, c 0 and Λ 8 , the following inequality holds:
a
a^2 Λbc
b
b^2 Λca
c
c^2 Λab
Problem 3
Twenty-one girls and twenty-one boys took part in a mathematical contest.
- Each contestant solved at most six problems.
- For each girl and each boy, at least one problem was solved by both of them.
Prove that there was a problem that was solved by at least three girls and at least three boys.
Solution
Solution 1
We introduce the following symbols: G is the set of girls at the competition and B is the set of boys, P is the set of problems, Pg is the set of problems solved by g G, and Pb is the set of problems solved by b B. Finally, Gp is the set of girls that solve p P and Bp is the set of boys that solve p. In terms of this notation, we have that for all g G and b B,
a Pg 6 , Pb 6 , b Pg Pb .
We wish to prove that some p P satisfies Gp 3 and Bp 3. To do this, we shall assume the contrary and reach a contradiction by counting (two ways) all ordered triples p, q, r such that p Pg Pb. With T p, g, b : p Pg Pb , condition (b) yields
Solution 2
Let us use some of the notation given in the first solution. Suppose that for every p P either Gp 2 or Bp 2. For each p P , color p red if Gp 2 and otherwise color it black. In this way, if p is red then Gp 2 and if p is black then Bp 2. Consider a chessboard with 21 rows, each representing one of the girls, and 21 columns, each representing one of the boys. For each g G and b B, color the square corresponding to g, b as follows: pick p Pg Pb and assign p's color to that square. (By condition (b), there is always an available choice.) By the Pigeonhole Principle, one of the two colors is assigned to at least 441 2 221 squares, and thus some row has at least 221 21 11 black squares or some column has at least 11 red squares.
Suppose the row corresponding to g G has at least 11 black squares. Then for each of 11 squares, the black prob- lem that was chosen in assigning the color was solved by at most 2 boys. Thus we account for at least 11 2 6 distinct problems solved by g. In view of condition (a), g solves only these problems. But then at most 12 boys solve a problem also solved by g , in violation of condition (b).
In exactly the same way, a contradiction is reached if we suppose that some column has at least 11 red squares. Hence some p P satisfies Gp 3 and Bp 3.
Problem 4
Let n be an odd integer greater than 1, and let k 1 , k 2 , …, k (^) n be given integers. For each of the n permutations a a 1 , a 2 , …, a (^) n of1 , 2, …, n , let
Sa
i 1
n
k i a i.
Prove that there are two permutations b and c, b c, such that n is a divisor of Sb Sc.
Solution
Let Sa be the sum of Sa over all n permutations a a 1 , a 2 , …, a (^) n . We compute Sa mod n two ways, one of which depends on the desired conclusion being false, and reach a contradiction when n is odd.
First way. In Sa, k 1 is multiplied by each i 1 , …,n a total of n 1 times, once for each permutation of 1 , …,n in which a 1 i. Thus the coefficient of k 1 in Sa is
n 1 1 2 n n 1 2.
The same is true for all ki , so
Sa^ ^ (1)
n 1 2
i 1
n ki.
Second way. If n is not a divisor of Sb Sc for any b c, then each Sa must have a different remainder mod n . Since there are n permutations, these remainders must be precisely the numbers 0, 1, 2, …, n 1. Thus
Sa^ ^ (2)
n 1 n 2
mod n .
Combining (1) and (2), we get
n 1 2
i 1
n k (^) i
n 1 n 2
mod n .
Now, for n odd, the left side of (3) is congruent to 0 modulo n , while for n 1 the right side is not congruent to 0 (n 1 is odd). For n 1 and odd, we have a contradiction.
Problem 5
In a triangle ABC , let AP bisect BAC , with P on BC , and let BQ bisect ABC , with Q on CA.
It is known that BAC 60 ^ and that AB BP AQ QB.
What are the possible angles of triangle ABC?
Solution
Denote the angles of ABC by Α 60 ^ , Β, and Γ. Extend AB to P^ so that BP^ BP, and construct P^ on AQ so that AP^ AP^. Then BP^ P is an isosceles triangle with base angle Β 2. Since AQ QP^ AB BP^ AB BP AQ QB, it follows that QP^ QB. Since AP^ P^ is equilateral and AP bisects the angle at A, we have PP^ PP^.
Claim. Points B, P, P^ are collinear, so P^ coincides with C.
Proof. Suppose to the contrary that BPP^ is a nondegenerate triangle. We have that PBQ PP^ B PP^ Q Β 2. Thus the diagram appears as below, or else with P is on the other side of BP^. In either case, the assumption that BPP^ is nondegenerate leads to BP PP^ PP^ , thus to the conclusion that BPP is equilateral, and finally to the absurdity Β 2 60 ^ so Α Β 60 ^ 120 ^ 180 ^.
In view of this, there exists a positive integer k such that
a c d kb c,
b c d ka d.
Adding these equations, we obtain a b ka b c d and thus kc d k 1 a b. Recall that a b c d. If k 1 then c d , a contradiction. If k 2 then
k
k 1
a b
c d
a contradiction.
Since a contradiction is reached in both (i) and (ii), ab cd is not prime.
Solution 2
The equality ac bd b d a cb d a c is equivalent to
a^2 ac c^2 b^2 bd d^2. (1)
Let ABCD be the quadrilateral with AB a , BC d , CD b , AD c , BAD 60 ^ , and BCD 120 ^. Such a quadrilateral exists in view of (1) and the Law of Cosines; the common value in (1) is BD^2. Let ABC Α , so that CDA 180 ^ Α. Applying the Law of Cosines to triangles ABC and ACD gives
a^2 d^2 2 adcosΑ AC^2 b^2 c^2 2 bccosΑ.
Hence 2cosΑ a^2 d^2 b^2 c^2 ad bc , and
AC^2 a^2 d^2 ad
a^2 d^2 b^2 c^2
ad bc
ab cdac bd
ad bc.
Because ABCD is cyclic, Ptolemy's Theorem gives
AC BD^2 ab cd^2
It follows that
ac bda^2 ac c^2 ab cdad bc. (2)
(Note. Straightforward algebra can also be used obtain (2) from (1).) Next observe that
ab cd ac bd ad bc. (3)
The first inequality follows from a db c 0 , and the second from a bc d 0.
Now assume that ab cd is prime. It then follows from (3) that ab cd and ac bd are relatively prime. Hence, from (2), it must be true that ac bd divides ad bc. However, this is impossible by (3). Thus ab cd must not be prime.
Note. Examples of 4-tuples a, b, c, d that satisfy the given conditions are 21, 18, 14, 1 and 65, 50, 34, 11.
