Solutions to the Shortlisted Problems of IMO 1983

4.24 Solutions to the Shortlisted Problems of IMO 1983

1. Suppose that there are n airlines A 1 ,...,A n and N >2 n cities. We shall prove that there is a round trip by at least one A i containing an odd number of stops. For n = 1 the statement is trivial, since one airline serves at least 3 cities and

hence P 1 P 2 P 3 P 1 is a round trip with 3 landings. We use induction on n, and assume that n > 1. Suppose the contrary, that all round trips by A n consist of an even number of stops. Then we can separate the cities into two nonempty classes

Q = {Q 1 ,...,Q r } and R = {R 1 ,...,R s } (where r + s = N), so that each flight by

A n runs between a Q-city and an R-city. (Indeed, take any city Q 1 served by A n ; include each city linked to Q 1 by A n in R, then include in Q each city linked by

A n to any R-city, etc. Since all round trips are even, no contradiction can arise.) At least one of r , s is larger than 2 n −1 , say r >2 n −1 . But, only A 1 ,...,A n −1 run between cities in {Q 1 ,...,Q r }; hence by the induction hypothesis at least one of them flies a round trip with an odd number of landings, a contradiction. It only remains to notice that for n = 10, 2 n = 1024 < 1983.

Remark. If there are N =2 n cities, there is a schedule with n airlines that contain no odd round trip by any of the airlines. Let the cities be P k ,k = 0, . . . , 2 n −1, and

write k in the binary system as an n-digit number a 1 ...a n (e.g., 1 = (0 . . . 001) 2 ). Link P k and P l by A i if the ith digits k and l are distinct but the first i − 1 digits are the same. All round trips under A i are even, since the ith digit alternates.

2. By definition, σ (n) = ∑ d |n d =∑ d |n n /d = n ∑ d |n 1 /d, hence σ (n)/n = ∑ d |n 1 /d. In particular, σ (n!)/n! = ∑ d |n! 1 /d ≥ ∑ n k =1 1 /k. It follows that the sequence σ (n)/n is unbounded, and consequently there exist an infinite number of in- tegers n such that σ (n)/n is strictly greater than σ (k)/k for k < n.

3. (a) A circle is not Pythagorean. Indeed, consider the partition into two semi- circles each closed at one and open at the other end. (b) An equilateral triangle, call it PQR, is Pythagorean. Let P ′ ,Q ′ , and R ′ be the points on QR, RP, and PQ such that PR ′ :R ′ Q = QP ′ :P ′ R = RQ ′ : Q ′ P = 1 : 2. Then Q ′ R ′ ⊥ PQ, etc. Suppose that PQR is not Pythagorean, and consider a partition into A , B, neither of which contains the vertices of

a right-angled triangle. At least two of P ′ ,Q ′ , and R ′ belong to the same class, say P ′ ,Q ′ ∈ A. Then [PR] \ {Q ′ } ⊂ B and hence R ′ ∈ A (otherwise, if R ′′ is the foot of the perpendicular from R ′ to PR, △RR ′ R ′′ is right-angled with all vertices in B). But this implies again that [PQ] \ {R ′ } ⊂ B, and thus

B contains vertices of a rectangular triangle. This is a contradiction.

4. The rotational homothety centered at C that sends B to R also sends A to Q; hence the triangles ABC and QRC are similar. For the same reason, △ABC and △PBR are similar. Moreover, BR = CR; hence △CRQ ∼ = △RBP. Thus PR = QC = AQ and QR = PB = PA, so APQR is a parallelogram.

5. Each natural number p can be written uniquely in the form p =2 q (2r − 1). We call 2r − 1 the odd part of p. Let A n = (a 1 ,a 2 ,...,a n ) be the first sequence. Clearly the terms of A n must have different odd parts, so those parts must be at

468 4 Solutions least 1 , 3, . . . , 2n − 1. Being the first sequence, A n must have the numbers 2n −

1 , 2n − 3,...,2k + 1 as terms, where k = [n + 1/3] (then 3(2k − 1) < 2n − 1 <

3 (2k + 1)). Smaller odd numbers 2s + 1 (with s < k) obviously cannot be terms of A n . In this way we have obtained the n − k odd numbers of A n . The other k terms must be even, and by the same reasoning as above they must be precisely the terms of 2A k (twice the terms of A k ). Therefore A n is defined recursively as

A n = {2n − 1,2n − 3,...,2k + 1} ∪ 2A k .

6. The existence of r: Let S = {x 1 +x 2 + ··· + x i − 2i | i = 1,2,...,n}. Let max S be attained for the first time at r ′ . If r ′ = n, then x 1 +x 2 + ··· + x i − 2i < 2 for 1 ≤ i ≤ n − 1, so one can take r = r ′ . Suppose that r ′ < n. Then for l < n − r ′ we have x r ′ +1 +x r ′ +2 + ··· + x r ′ +l = (x 1 + ··· + x r ′ +l − 2(r ′ + l)) − (x 1 + ··· + x r ′ − 2r ′ ) + 2l ≤ 2l; also, for i < r ′ we have (x r ′ +1 + ···+ x n ) + (x 1 + ···+ x i − 2i) < (x r ′ +1 + ···+ x n ) + (x 1 + ···+ x r ′ − 2r ′ ) = (x 1 + ··· + x n ) − 2r ′ = 2(n − r ′ )+2⇒x r ′ +1 + ··· + x n +x 1 + ··· + x i ≤

2 (n + i − r ′ ) + 1, so we can again take r = r ′ . For the second part of the problem, we relabel the sequence so that r = 0 works. Suppose that the inequalities are strict. We have x 1 +x 2 + ··· + x k ≤ 2k, k =

1 , . . . , n − 1. Now, 2n + 2 = (x 1 + ···+ x k ) + (x k +1 + ···+ x n ) ≤ 2k + x k +1 + ···+ x n ⇒x k +1 + ···+ x n ≥ 2(n− k)+2 > 2(n− k)+1. So we cannot begin with x k +1 for any k > 0. Now assume that there is an equality for some k. There are two cases:

(i) Suppose x 1 +x 2 + ··· + x i ≤ 2i (i = 1,...,k) and x 1 + ··· + x k = 2k + 1, x 1 + ··· + x k +l ≤ 2(k + l) + 1 (1 ≤ l ≤ n − 1 − k). For i ≤ k − 1 we have x i +1 + ··· + x n = 2(n + 1) − (x 1 + ··· + x i ) > 2(n − i) + 1, so we cannot take r = i. If there is a j ≥ 1 such that x 1 +x 2 + ··· + x k +j ≤ 2(k + j), then also x k + j+1 + ··· + x n > 2(n − k − j) + 1. If (∀ j ≥ 1) x 1 + ··· + x k +j =

2 (k + j) + 1, then x n = 3 and x k +1 = ··· = x n −1 = 2. In this case we directly verify that we cannot take r = k + j. However, we can also take r = k: for k + l ≤ n − 1, x k +1 + ··· + x k +l ≤ 2(k + l) + 1 − (2k + 1) = 2l, also

x k +1 + ··· + x n = 2(n − k) + 1, and moreover x 1 ≤ 2, x 1 +x 2 ≤ 4,... . (ii) Suppose x 1 + ···+x i ≤ 2i (1 ≤ i ≤ n −2) and x 1 + ···+x n −1 = 2n − 1. Then we can obviously take r = n − 1. On the other hand, for any 1 ≤ i ≤ n − 2, x i +1 + ···+ x n −1 +x n = (x 1 + ···+ x n −1 ) − (x 1 + ···+ x i ) + 3 > 2(n − i)+1, so we cannot take another r 6= 0.

7. Clearly, each a n is positive and √a n +1 = √a n a +1+ a n +1 a . Notice that √

a n √ +1 +1= a +1 a n +1+ a a n . Therefore

( a +1− a )( a n √ +1− a n )

p √ =( a +1 a n +1+ a a n )−( a n a +1+ a n +1 a ) p

= a n √ +1 +1− a n +1 .

4.24 Shortlisted Problems 1983 469 √

a +1− a . Similarly, √a n +1 + √a n = √

By induction, √a n +1 − a n =

a +1+ a . Hence, √

a +1+ a −

a +1− a ,

2 from which the result follows.

8. Situations in which the condition of the statement is fulfilled are the following: S 1 :N 1 (t) = N 2 (t) = N 3 (t) S 2 :N i (t) = N j (t) = h, N k (t) = h + 1, where (i, j, k) is a permutation of the set {1,2,3}. In this case the first student to leave must be from row k. This

leads to the situation S 1 .

S 3 :N i (t) = h, N j (t) = N k (t) = h + 1, ((i, j, k) is a permutation of the set {1,2,3}). In this situation the first student leaving the room belongs to row

j (or k) and the second to row k (or j). After this we arrive at the situation S 1 . Hence, the initial situation is S 1 and after each triple of students leaving the room the situation S 1 must recur. We shall compute the probability P h that from

a situation S 1 with 3h students in the room (h ≤ n) one arrives at a situation S 1 with 3 (h − 1) students in the room:

(3h) · (2h) · h 3!h 3 P h =

. (3h) · (3h − 1) · (3h − 2)

3h (3h − 1)(3h − 2) Since the room becomes empty after the repetition of n such processes, which

are independent, we obtain for the probability sought

(3!) n (n!) 3

h =1

(3n)!

9. For any triangle of sides a , b, c there exist 3 nonnegative numbers x, y, z such that

a = y + z, b = z + x, c = x + y (these numbers correspond to the division of the sides of a triangle by the point of contact of the incircle). The inequality becomes

(y + z) 2 2 (z + x)(y − x) + (z + x) 2 (x + y)(z − y) + (x + y) (y + z)(x − z) ≥ 0. Expanding, we get xy 3 + yz 3 + zx 3 ≥ xyz(x + y + z). This follows from Cauchy’s

)(z + x + y) ≥ 2 xyz (x + y + z) with equality if and only if xy 3 /z = yz 3 /x = zx 3 /y, or equivalently x = y = z, i.e., a = b = c.

inequality (xy 3 + yz 3 + zx 3 √

10. Choose P (x) = p

q (qx − 1) +1 ,I = [1/2q, 3/2q]. Then all the coefficients of P are integers, and

2n +1

q 2 2n +1 , for x ∈ I. The desired inequality follows if n is chosen large enough.

q (qx − 1)

2n +1

470 4 Solutions

11. First suppose that the binary representation of x is finite: x = 0, a 1 a 2 ...a n = ∑ n j =1 a j 2 −j ,a i ∈ {0,1}. We shall prove by induction on n that

−b if a ∑ k b

f (x) =

0 ...b j −1 a j , where b k =

1 − b if a k = 1. (Here a 0 = 0.) Indeed, by the recursion formula,

a 1 = 0 ⇒ f (x) = b f ∑ a j +1 2 −j =b ∑ b 1 ...b j a j +1

hence f (x) = ∑ b 0 ...b j a j +1 as b 0 =b 1 = b;

a 1 = 1 ⇒ f (x) = b + (1 − b) f ∑ a j +1 2 −j = ∑ b 0 ...b j a j +1 ,

Clearly, f

j =0 a j 2 −j , and for k ≥ 2, v = x + 2 −n−k+1 ,u =x+2 −n−k = (v + x)/2. Then f (v) = f (x) +

(0) = 0, f (1) = 1, f (1/2) = b > 1/2. Assume x = ∑ n

b 0 ...b n b k −2 and f (u) = f (x) + b 0 ...b n b k −1 > ( f (v) + f (x))/2. This means that the point (u, f (u)) lies above the line joining (x, f (x)) and (v, f (v)). By induc- tion, every (x, f (x)), where x has a finite binary expansion, lies above the line joining (0, 0) and (1/2, b) if 0 < x < 1/2, or above the line joining (1/2, b) and (1, 1) if 1/2 < x < 1. It follows immediately that f (x) > x. For the second in- equality, observe that

f (x) − x = ∑ (b 0 ...b j

∑ b (b j −2 −j )a j < ∑ (b −2 −j )=

− 1 = c.

1 −b By continuity, these inequalities also hold for x with infinite binary representa-

12. Putting y = x in (i) we see that there exist positive real numbers z such that

f (z) = z (this is true for every z = x f (x)). Let a be any of them. Then f (a 2 )=

f (a f (a)) = a f (a) = a 2 , and by induction, f (a n )=a n . If a > 1, then a n → +∞ as n → ∞, and we have a contradiction with (ii). Again, a = f (a) = f (1 · a) =

af (1), so f (1) = 1. Then, a f (a −1 ) = f (a −1 f (a)) = f (1) = 1, and by induction,

f (a −n )=a −n . This shows that a 6< 1. Hence, a = 1. It follows that x f (x) = 1, i.e., f (x) = 1/x for all x. This function satisfies (i) and (ii), so f (x) = 1/x is the unique solution.

13. Given any coloring of the 3 × 1983 − 2 points of the axes, we prove that there is a unique coloring of E having the given property and extending this coloring.

4.24 Shortlisted Problems 1983 471 The first thing to notice is that given any rectangle R 1 parallel to a coordinate

plane and whose edges are parallel to the axes, there is an even number r 1 of red vertices on R 1 . Indeed, let R 2 and R 3 be two other rectangles that are translated from R 1 orthogonally to R 1 and let r 2 ,r 3 be the numbers of red vertices on R 2 and R 3 respectively. Then r 1 +r 2 ,r 1 +r 3 , and r 2 +r 3 are multiples of 4, so

r 1 = (r 1 +r 2 +r 1 +r 3 −r 2 −r 3 )/2 is even.

Since any point of a coordinate plane is a vertex of a rectangle whose remaining three vertices lie on the corresponding axes, this determines uniquely the color- ing of the coordinate planes. Similarly, the coloring of the inner points of the par-

allelepiped is completely determined. The solution is hence 2 3 ×1983−2 =2 5947 .

14. Let T n

be the set of all nonnegative integers whose ternary representations consist of at most n digits and do not contain a digit 2. The cardinality of T n is 2 n , and the greatest integer in T is 11 ...1=3 0 n +3 1 +···+3 n −1 = (3 n −1)/2. We claim that there is no arithmetic triple in T n . To see this, suppose x , y, z ∈ T n and 2y = x + z. Then 2y has only 0’s and 2’s in its ternary representation, and a number of this form can be the sum of two integers x ,z∈T n in only one way, namely x = z = y. But

|T 10 10 |=2 10 = 1024 and max T 10 = (3 − 1)/2 = 29524 < 30000. Thus the answer is yes.

15. There is no such set. Suppose that M satisfies the conditions (i) and (ii) and let q n = |{a ∈ M : a ≤ n}|. Consider the differences b − a, where a,b ∈ M and

10 < a < b ≤ k. They are all positive and less than k, and (ii) implies that they are q k −q 10 2 different integers. Hence q k −q 10

2 < k, so q k ≤ 2k + 10. It follows from (i) that among the numbers of the form a + b, where a, b ∈ M, a ≤ b ≤

n , or a ≤ n < b ≤ 2n, there are all integers from the interval [2,2n + 1]. Thus

q n +1

2 +q n (q 2n −q n ) ≥ 2n for every n ∈ N. Set Q k = 2k + 10. We have

which is less than n for n large enough, a contradiction.

16. Set h n ,i (x) = x i + ···+ x n −i , 2i ≤ n. The set F(n) is the set of linear combinations with nonnegative coefficients of the h n ,i ’s. This is a convex cone. Hence, it suf-

fices to prove that h n ,i h m ,j ∈ F(m + n). Indeed, setting p = n − 2i and q = m − 2 j and assuming p ≤ q we obtain

n −i+ j

i n +p ,i (x)h m ,j (x) = (x + ··· + x )(x j

j + ··· + x +q )= ∑ h

m +n,k ,

k =i+ j

which proves the claim.

472 4 Solutions

17. Set a = minP i P j ,b = max P i P j . We use the following lemma. √ Lemma. There exists a disk of radius less than or equal to b /

3 containing all the P i ’s. Assuming that this is proved, the disks with center P √ i and radius a /2 are disjoint and included in a disk of radius b / 3 + a/2; hence comparing areas,

√ n π · < π · √ + a/2

4 /2 · ( n 3 − 1)a. Proof of the lemma. If a nonobtuse triangle with sides a ≥ b ≥ c has a circum- √

and

scribed circle of radius R, we have R = a/(2 sin α ) ≤ a/

3. Now we show that there exists a disk D of radius R containing A = {P 1 ,...,P n } whose border C is such that C ∩ A is not included in an open semicircle, and hence contains either two diametrically opposite points and R √ ≤ b/2, or an acute- angled triangle and R ≤ b/ 3. Among all disks whose borders pass through three points of A and that con- tain all of A, let D be the one of least radius. Suppose that C ∩A is contained in an arc of central angle less than 180 ◦ , and that P i ,P j are its endpoints. Then there exists a circle through P i ,P j of smaller radius that contains A, a contradiction. Thus D has the required property, and the assertion follows.

18. Let (x 0 ,y 0 ,z 0 ) be one solution of bcx + cay + abz = n (not necessarily nonnega-

tive). By subtracting bcx 0 + cay 0 + abz 0 = n we get

bc (x − x 0 ) + ca(y − y 0 ) + ab(z − z 0 ) = 0. Since (a, b) = (a, c) = 1, we must have a|x − x 0 or x −x 0 = as. Substituting this

in the last equation gives bcs + c(y − y 0 ) + b(z − z 0 ) = 0. Since (b, c) = 1, we have b|y − y 0 or y −y 0 = bt. If we substitute this in the

last equation we get bcs + bct + b(z − z 0 ) = 0, or cs + ct + z − z 0 = 0, or z − z 0 = −c(s+t). In x = x 0 + as and y = y 0 + bt, we can choose s and t such that 0 ≤ x ≤

a − 1 and 0 ≤ y ≤ b − 1. If n > 2abc − bc − ca − ab, then abz = n − bcx − acy > 2abc − ab − bc − ca− bc(a− 1) −ca(b −1) = −ab or z > −1, i.e., z ≥ 0. Hence, it is representable as bcx + cay + abz with x, y, z ≥ 0. Now we prove that 2abc − bc − ca − ab is not representable as bcx + cay + abz with x , y, z ≥ 0. Suppose that bcx + cay + abz = 2abc − ab − bc− ca with x,y,z ≥

0. Then

bc (x + 1) + ca(y + 1) + ab(z + 1) = 2abc with x + 1, y + 1, z + 1 ≥ 1. Since (a,b) = (a,c) = 1, we have a|x + 1 and thus

a ≤ x + 1. Similarly b ≤ y + 1 and c ≤ z + 1. Thus bca + cab + abc ≤ 2abc, a contradiction.

19. For all n, there exists a unique polynomial P n of degree n such that P n (k) = F k for n + 2 ≤ k ≤ 2n + 2 and P n (2n + 3) = F 2n +3 − 1. For n = 0, we have F 1 =

4.24 Shortlisted Problems 1983 473

F 2 = 1, F 3 = 2, P 0 = 1. Now suppose that P n −1 has been constructed and let P n

be the polynomial of degree n satisfying P n (X + 2) − P n (X + 1) = P n −1 (X ) and P n (n + 2) = F n +2 . (The mapping R n [X ] → R n −1 [X ] × R, P 7→ (Q,P(n + 2)), where Q (X) = P(X + 2) − P(X + 1), is bijective, since it is injective and those two spaces have the same dimension; clearly deg Q = degP−1.) Thus for n+2 ≤ k ≤ 2n + 2 we have P n (k + 1) = P n (k) + F k −1 and P n (n + 2) = F n +2 ; hence by induction on k, P n (k) = F k for n + 2 ≤ k ≤ 2n + 2 and

P n (2n + 3) = F 2n +2 +P n −1 (2n + 1) = F 2n +3 − 1.

Finally, P 990 is exactly the polynomial P of the terms of the problem, for P 990 − P has degree less than or equal to 990 and vanishes at the 991 points k = 992, . . ., 1982.

20. If (x 1 ,x 2 ,...,x n ) satisfies the system with parameter a, then (−x 1 , −x 2 , . . . , −x n ) satisfies the system with parameter −a. Hence it is sufficient to consider only

a ≥ 0. Let (x 1 ,...,x n ) be a solution. Suppose x 1 ≤ a, x 2 ≤ a,...,x n ≤ a. Summing the equations we get

− a) 2 =0 and see that (a, a, . . . , a) is the only such solution. Now suppose that x k ≥ a for

(x 1 − a) 2 + ··· + (x n

some k. According to the kth equation, x

k +1 |x k +1 |=x k − (x k − a) = a(2x k − a) ≥ a ,

which implies that x k +1 ≥ a as well (here x n +1 =x 1 ). Consequently, all x 1 ,x 2 , ..., x n are greater than or equal to a, and as above (a, a, . . . , a) is the only solution.

21. Using the identity

n −1

a n −b n = (a − b) ∑ a n −m−1 b m

m =0

with a =k 1 /n and b

1 = (k − 1) /n one obtains

1 < k 1 /n

1 − (k − 1) /n nk 1 −1/n for all integers n > 1 and k ≥ 1. This gives us the inequality k 1 /n−1 <nk 1 /n − (k − 1) 1 /n if n > 1 and k ≥ 1. In

1 −k /n <k 1 /n−1 if n > 1 and k ≥ 1. Hence for n > 1 and m > 1 it holds that

a similar way one proves that n (k + 1) 1 /n

n ∑ (k + 1) 1 /n 1 −k /n < ∑ k 1 /n−1

1 ∑ /n − (k − 1) + 1,

k 1 /n

k =2

or equivalently,

474 4 Solutions

n (m + 1) 1 /n

< k 1 /n−1 <n m 1 −1 /n ∑ −1 + 1.

k =1

The choice n = 1983 and m = 2 1983 then gives

1983 < ∑ k 1 /1983−1 < 1984.

k =1

Therefore the greatest integer less than or equal to the given sum is 1983.

22. Decompose n into n = st, where the greatest common divisor of s and t is 1 and where s > 1 and t > 1. For 1 ≤ k ≤ n put k = vs + u, where 0 ≤ v ≤ t − 1 and 1 ≤ u ≤ s, and let a k =a vs +u

be the unique integer in the set {1,2,3,...,n} such that vs + ut − a vs +u is a multiple of n. To prove that this construction gives

a permutation, assume that a k 1 =a k 2 , where k i =v i s +u i ,i = 1, 2. Then (v 1 − v 2 )s + (u 1 −u 2 )t is a multiple of n = st. It follows that t divides (v 1 −v 2 ), while |v 1 −v 2 | ≤ t − 1, and that s divides (u 1 −u 2 ), while |u 1 −u 2 | ≤ s − 1. Hence, v 1 =v 2 ,u 1 =u 2 , and k 1 =k 2 . We have proved that a 1 ,...,a n is a permutation of {1,2,...,n} and hence

2 π a k t −1

k cos

= ∑ ∑ (vs + u) cos

s Using ∑ s u =1 cos (2 π u

k =1

v =0 u =1

/s) = ∑ s u =1 sin (2 π u /s) = 0 and the additive formulas for cosine, one finds that

2 π a t −1

∑ k cos k = ∑ cos ∑ u cos − sin

∑ sin

= ∑ u cos

− ∑ u sin

∑ sin

23. We note that ∠O 1 KO 2 = ∠M 1 KM 2 is equivalent to ∠O 1 KM 1 = ∠O 2 KM 2 . Let S

be the intersection point of the common tangents, and let L be the second

point of intersection of SK and W 1 .

Since △SO 1 P 1 ∼ △SP 1 M

1 , we have

SK · SL = SP 2 1 = SO 1 · SM 1 which im-

plies that points O 1 , L, K, M 1 lie on a

circle. Hence ∠O 1 KM 1 = ∠O 1 LM 1 =

Q ∠O 2

2 KM 2 . Q 1

24. See the solution of (SL91-15).

4.24 Shortlisted Problems 1983 475

25. Suppose the contrary, that R 3 =P 1 ∪P 2 ∪P 3 is a partition such that a 1 ∈R + is not realized by P 1 ,a 2 ∈R + is not realized by P 2 and a 3 ∈R + not realized by P 3 , where w.l.o.g. a 1 ≥a 2 ≥a 3 . If P 1 = /0 = P 2 , then P 3 =R 3 , which is impossible. If P 1 = /0, and X ∈ P 2 , the sphere centered at X with radius a 2 is included in P 3 and a 3 ≤a 2 is realized, which is impossible. If P 1 6= /0, let X 1 ∈P 1 . The sphere S centered in X 1 , of radius a 1 is included in P 2 ∩ P 3 . Since a 1 ≥a 3 ,S 6⊂ P 3 . Let X 2 ∈P 2 ∩ S. The circle {Y ∈ S | d(X 2 q ,Y ) = a 2 } is

√ included in P 3 , but a 2 ≤a 1 ; hence it has radius r =a 2 1 −a 2 2 /(4a 2 1 )≥a 2 3 /2

and a 3 ≤a 2 ≤a 2 3 < 2r; hence a 3 is realized by P 3 .

476 4 Solutions