Volume 10, Number 5 December 2005 – January 2006
Using Tangent Lines to Prove Inequalities
Kin-Yin Li
Olympiad Corner
Below is the Czech-Polish-Slovak Match held in Zwardon on June 20-21, 2005.
Problem 1. Let n be a given positive integer. Solve the system of equations
, 3 3 2 2 1 x x x n x n n = + + + + L 2 ) 1 ( 3 2 2 3 1 + = + + + + x x nx n n x L n
in the set of nonnegative real numbers x1, x2, …, xn.
Problem 2.
Let a convex quadrilateral ABCD be inscribed in a circle with center O and circumscribed to a circle with center I, and let its diagonals AC and BD meet at a point P. Prove that the points O, I and P are collinear.
Problem 3. Determine all integers n ≥ 3 such that the polynomial W(x) = xn −
3xn−1 + 2xn−2 + 6 can be expressed as a
product of two polynomials with positive degrees and integer coefficients.
Problem 4. We distribute n ≥ 1 labelled balls among nine persons A, B, C, D, E, F, G, H, I. Determine in how many ways
(continued on page 4)
Editors: Ի ஶ(CHEUNG Pak-Hong), Munsang College, HK
ଽ υ ࣻ (KO Tsz-Mei)
గ ႀ ᄸ (LEUNG Tat-Wing)
፱ (LI Kin-Yin), Dept. of Math., HKUST
֔ ᜢ ݰ (NG Keng-Po Roger), ITC, HKPU
Artist: ྆ ؾ ़ (YEUNG Sau-Ying Camille), MFA, CU
Acknowledgment: Thanks to Elina Chiu, Math. Dept.,
HKUST for general assistance.
On-line:
http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is February 12, 2006.
For individual subscription for the next five issues for the 03-04 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:
Dr. Kin-Yin LI Department of Mathematics
The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong
Fax: (852) 2358 1643 Email: [email protected]
For students who know calculus, sometimes they become frustrated in solving inequality problems when they do not see any way of using calculus. Below we will give some examples, where finding the equation of a tangent line is the critical step to solving the problems.
Example 1. Let a,b,c,d be positive real
numbers such that a + b + c + d = 1. Prove that
6(a3+b3+c3+d3) ≥ (a2+b2+c2+d2) + 1/8. Solution. We have 0 < a, b, c, d < 1. Let
f(x) = 6x3 – x2. (Note: Since there is
equality when a = b = c = d = 1/4, we consider the graph of f(x) and its tangent line at x = 1/4. By a simple sketch, it seems the tangent line is below the graph of f(x) on the interval (0,1). Now the equation of the tangent line at x = 1/4 is y = (5x – 1)/8.) So we claim that for 0 < x < 1, f(x) = 6x3 – x2 ≥ (5x – 1)/8. This
is equivalent to 48x3 − 8x2 − 5x + 1 ≥ 0.
(Note: Since the graphs intersect at x = 1/4, we expect 4x − 1 is a factor.) Indeed, 48x3 − 8x2 − 5x + 1 = (4x − 1)2
(3x + 1) ≥ 0 for 0 < x < 1. So the claim is true. Then f(a) + f(b) + f(c) + f(d) ≥ 5(a + b + c + d)/8 − 4/8 = 1/8, which is equivalent to the required inequality.
Example 2. (2003 USA Math Olympiad)
Let a,b,c be positive real numbers. Prove that . 8 ) ( 2 ) 2 ( ) ( 2 ) 2 ( ) ( 2 ) 2 ( 2 2 2 2 2 2 2 2 2 ≤ + + + + + + + + + + + + + + b a c b a c a c b a c b c b a c b a
Solution. Setting a' = a/(a + b + c), b' =
b/(a + b + c), c' = c/(a + b + c) if necessary, we may assume 0 < a, b, c < 1 and a + b + c = 1. Then the first term on the left side of the inequality is equal to
. 1 2 3 1 2 ) 1 ( 2 ) 1 ( ) ( 2 2 2 2 2 + − + + = − + + = a a a a a a a a f
(Note: When a = b = c = 1/3, there is equality. A simple sketch of f(x) on [0,1] shows the curve is below the tangent line
at x = 1/3, which has the equation y = (12x + 4)/3.) So we claim that 3 4 12 1 2 3 1 2 2 2 + ≤ + − + + a a a a a
for 0 < a < 1. Multiplying out, we see this is equivalent to 36a3 − 15a2 − 2a + 1
≥ 0 for 0 < a < 1. (Note: Since the curve and the line intersect at a = 1/3, we expect 3a−1 is a factor.) Indeed, 36a3 −
15a2 − 2a + 1 = (3a − 1)2(4a + 1) ≥ 0 for
0 < a < 1. Finally adding the similar inequality for b and c, we get the desired inequality.
The next example looks like the last example. However, it is much more sophisticated, especially without using tangent lines. The solution below is due to Titu Andreescu and Gabriel Dospinescu.
Example 3. (1997 Japanese Math
Olympiad) Let a,b,c be positive real numbers. Prove that
. 5 3 ) ( ) ( ) ( ) ( ) ( ) ( 2 2 2 2 2 2 2 2 2 ≥ + + − + + + + − + + + + − + c b a c b a b a c b a c a c b a c b
Solution. As in the last example, we
may assume 0 < a, b, c < 1 and a + b + c = 1. Then the first term on the left
become . ) 2 1 ( 1 2 2 ) 1 ( ) 2 1 ( 2 2 2 2 a a a a − + − = + − −
Next, let x1 = 1 − 2a, x2 = 1 − 2b, x3 = 1 −
2c, then x1 + x2 + x3 = 1, but −1 < x1, x2,
x3 < 1. In terms of x1, x2, x3, the desired
inequality is . 10 27 1 1 1 1 1 1 2 3 2 2 2 1 ≤ + + + + +x x x
(Note: As in the last example, we consider the equation of the tangent line to f(x) = 1/(1 + x2) at x = 1/3, which is y
= 27(−x + 2)/50.) So we claim that f(x) ≤ 27(−x + 2)/50 for −1 < x < 1. This is equivalent to (3x − 1)2(4 − 3x) ≥ 0.
Hence the claim is true for −1 < x < 1. Then f(x1) + f(x2) + f(x3) ≤ 27/10 and the
Mathematical Excalibur, Vol. 10, No. 5, Dec. 05- Jan. 06 Page 2
Schur’s Inequality
Kin Yin Li
Sometimes in proving an inequality, we do not see any easy way. It will be good to know some brute force methods in such situation. In this article, we introduce a simple inequality that turns out to be very critical in proving inequalities by brute force.
Schur’s Inequality. For any x, y, z ≥ 0
and r > 0,
xr(x–y)(x–z) + yr(y–x)(y–z)
+ zr(z–x)(z–y) ≥ 0.
Equality holds if and only if x = y = z or two of x, y, z are equal and the third is zero.
Proof. Observe that the inequality is
symmetric in x, y, z. So without loss of generality, we may assume x ≥ y ≥ z. Then xr(x – y)(x – z) ≥ yr(x – y)(y – z) so
that the sum of the first two terms is nonnegative. As the third term is also nonnegative, so the sum of all three terms is nonnegative. In case x ≥ y ≥ z, equality holds if and only if x = y first and z equals to them or zero.
In using the Schur’s inequality, we often expand out expressions. So to simplify writing, we introduce the symmetric sum notation
∑
f(x,y,z) tosym
denote the sum of the six terms f(x,y,z), f(x,z,y), f(y,z,x), f(y,x,z), f(z,x,y) and f(z,y,x). In particular,
∑
sym x3 = 2x3 +2y3+2z3,∑
symx2y= x2y+x2z+y2z+y2x+z2x+z2y and
∑
sym
xyz = 6xyz.
Similarly, for a function of n variables, the symmetric sum is the sum of all n! terms, where we take all possible permutations of the n variables.
The r = 1 case of Schur’s inequality is x(x–y)(x–z) + y(y–x)(y–z) + z(z–x)(z–y) = x3 + y3 + z3 – (x2y + x2z + y2x + y2z + z2x + z2y) + 3xyz ≥ 0. In symmetric sum notation, it is
∑
− + ≥ sym xyz y x x 2 ) 0 ( 3 2 .By expanding both sides and rearranging terms, each of the following inequalities is equivalent to the r = 1 case of Schur’s inequality. These are common disguises. a) x3+y3+z3+3xyz ≥ xy(x+y)+yz(y+z)
+zx(z+x), b) xyz ≥ (x+y–z)(y+z–x)(z+x–y), c) 4(x+y+z)(xy+yz+zx) ≤ (x+y+z)3+9xyz. Example 1. (2000 IMO) Let a, b, c be
positive real numbers such that abc = 1. Prove that . 1 ) 1 1 )( 1 1 )( 1 1 ( − + − + − + ≤ a c c b b a
Solution. Let x = a, y = 1, z = 1/b = ac.
Then a = x/y, b = y/z and c = z/x. Substituting these into the desired inequality, we get , 1 ) ( ) ( ) ( ≤ + − + − + − x y x z z x z y y z y x
which is disguise b) of the r = 1 case of Schur’s inequality.
Example 2. (1984 IMO) Prove that
0 ≤ yz + zx + xy – 2xyz ≤ 7/27, where x, y, z are nonnegative real numbers such that x + y + z = 1.
Solution. In Schur’s inequality, all terms
are of the same degree. So we first change the desired inequality to one where all terms are of the same degree. Since x + y + z = 1, the desired inequality is the same as
. 27 ) ( 7 2 ) )( ( 0 3 z y x xyz xy zx yz z y x+ + + + − ≤ + + ≤
Expanding the middle expression, we get xyz+
∑
xsym
2y, which is clearly nonnegative
and the left inequality is proved. Expanding the rightmost expression and subtracting the middle expression, we get
). 7 5 7 12 ( 54 7 x3 x2y xyz sym + −
∑
(1)By Schur’s inequality, we have
∑
− + ≥ (2) sym xyz y x x 2 ) 0 ( 3 2 .By the AM-GM inequality, we have
, ) ( 6 6 6 6 1/6 2
∑
∑
≥ = sym sym xyz z y x y xwhich is the same as
∑
− ≥ (3) sym xyz y x ) 0. ( 2Multiplying (3) by 2/7 and adding it to (2), we see the symmetric sum in (1) is nonnegative. So the right inequality is proved.
Example 3. (2004 APMO) Prove that
) ( 9 ) 2 )( 2 )( 2 (a2+ b2+ c2+ ≥ ab+bc+ca
for any positive real numbers a,b,c.
Solution. Expanding and expressing in
symmetric sum notation, the desired inequality is (abc)2+
∑
(a sym 2b2+2a2)+8 ≥ 2 9∑
sym ab. As a2+b2≥2ab, we get∑
sym a2 ≥∑
sym ab. As a2b2 + 1 ≥ 2ab, we get∑
sym a2b2 + 6 ≥ 2∑
sym ab. Using these, the problem is reduced to showing(abc)2 + 2 ≥
∑
(ab –sym 2
1 a2).
To prove this, we apply the AM-GM inequality twice and disguise c) of the r = 1 case of Schur’s inequality as follow:
(abc)2 +2 ≥ 3(abc)2/3 ≥ 9abc/(a+b+c) ≥ 4(ab+bc+ca) – (a+b+c)2 = 2(ab+bc+ca) – (a2+b2+c2) =
∑
(ab – sym 2 1 a2).Example 4. (2000 USA Team Selection
Test) Prove that for any positive real numbers a, b, c, the following inequality holds 3 3 abc c b a+ + − }. ) ( , ) ( , ) max{( a− b 2 b− c 2 c− a 2 ≤ (continued on page 4)
)
Problem Corner
We welcome readers to submit their solutions to the problems posed below for publication consideration. The solutions should be preceded by the solver’s name, home (or email) address and school affiliation. Please send submissions to Dr. Kin Y. Li, Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon, Hong Kong. The deadline for submitting solutions is
February 12, 2006.
Problem 241. Determine the smallest ossible value of
p
S = a1·a2·a3 + b1·b2·b3 + c1·c2·c3,
if a1, a2, a3, b1, b2, b3, c1, c2, c3 is a
permutation of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9. (Source: 2002 Belarussian Math. Olympiad)
Problem 242. Prove that for every positive integer n, 7 is a divisor of 3n +
n3 if and only if 7 is a divisor of 3nn3 + 1.
(Source: 1995 Bulgarian Winter Math Competition)
Problem 243. Let R+ be the set of all positive real numbers. Prove that there
s no function f : R
i + →R+ such that
(
f(x))
2 ≥ f(x+ y)(
f(x)+ y for arbitrary positive real numbers x and y. (Source: 1998 Bulgarian Math Olympiad)Problem 244. An infinite set S of coplanar points is given, such that every three of them are not collinear and every two of them are not nearer than 1cm from each other. Does there exist any division of S into two disjoint infinite subsets R and B such that inside every triangle with vertices in R is at least one point of B and inside every triangle with vertices in B is at least one point of R? Give a proof to your answer. (Source: 2002 Albanian Math Olympiad)
Problem 245. ABCD is a concave quadrilateral such that ∠BAD =∠ABC =∠CDA = 45˚. Prove that AC = BD.
*****************
Solutions
****************
Problem 236. Alice and Barbara order a pizza. They choose an arbitrary point
P, different from the center of the pizza and they do three straight cuts through P, which pairwise intersect at 60˚ and divide the pizza into 6 pieces. The center of the pizza is not on the cuts. Alice chooses one piece and then the pieces are taken clockwise by Barbara, Alice, Barbara, Alice and Barbara. Which piece should Alice choose first in order to get more pizza than Barbara? (Source: 2002 Slovenian National Math Olympiad)
Solution. (Official Solution)
Let Alice choose the piece that contains the center of the pizza first. We claim that the total area of the shaded regions below is greater than half of the area of the pizza.
O B P C B' C' A D A' D' P' P"
Without loss of generality, we can assume the center of the pizza is at the origin O and one of the cuts is parallel to the x-axis (that is, BC is parallel to AD in the picture). Let P’ be the intersection of the x-axis and the 60˚-cut. Let A’D’ be parallel to the 120˚-cut B’C’. Let P’’ be the intersection of BC and A’D’. Then ∆PP’P” is equilateral. This implies the belts ABCD and A’B’C’D’ have equal width. Since AD > A’D’, the area of the belt ABCD is greater than the area of the belt A’B’C’D’. Now when the area of the belt ABCD is subtracted from the total area of the shaded regions and the area of A’B’C’D’ is then added, O B P C B' C' A D A' D' P' P"
we get exactly half the area of the pizza. Therefore, the claim follows.
Problem 237. Determine (with proof) all polynomials p with real coefficients such that p(x) p(x + 1) = p(x2) holds for
every real number x. (Source: 2000 Bulgarian Math Olympiad)
Solution. YEUNG Wai Kit (STFA
Leung Kau Kui College, Form 5). Let p(x) be such a polynomial. In case p(x) is a constant polynomial, p(x) must be 0 or 1. For the case p(x) is nonconstant, let r be a root of p(x). Then setting x = r and x + 1 = r in the equation, we see r2 and (r − 1)2 are also roots of
p(x). Also, r2 is a root implies (r2 − 1)2 is
also a root. If 0 < |r| < 1 or |r| > 1, then p(x) will have infinitely many roots r, r2,
r4, …, a contradiction. So |r| = 0 or 1 for
every root r.
The case |r| = 1 and |r − 1| = 1 lead to
2 / ) 3 1 ( i r= ± , but then |r2 − 1| ≠ 0 or 1, a
contradiction. Hence, either |r| = 0 or |r − 1| = 0, that is, r = 0 or 1.
So p(x) = xm(x−1)n for some nonnegative
integers m, n. Putting this into the equation, we find m = n. Conversely, p(x) = xm(x − 1)m is easily checked to be a
solution for every nonnegative integer m. Problem 238. For which positive integers n, does there exist a permutation (x1, x2, …, xn) of the
numbers 1, 2, …, n such that the number x1 + x2+ ⋯ + xk is divisible by k
for every k∈{1,2, …, n}? (Source: 1998 Nordic Mathematics Contest)
Solution. G.R.A. 20 Math Problem
Group (Roma, Italy), LEE Kai Seng (HKUST), LO Ka Wai (Carmel Divine Grace Foundation Secondary School, Form 7), Anna Ying PUN (STFA Leung Kau Kui College, Form 7) and YEUNG Wai Kit (STFA Leung Kau Kui College, Form 5). For a solution n, since x1 + x2 + ⋯ + xn
= n(n + 1)/2 is divisible by n, n must be odd. The cases n = 1 and n = 3 (with permutation (1,3,2)) are solutions. Assume n ≥ 5. Then x1 + x2 + ⋯ + xn−1 = n(n + 1)/2 − xn ≡ 0 (mod n − 1) implies xn ≡ (n + 1)/2 (mod n − 1). Since 1 ≤ xn ≤ n and 3 ≤ (n + 1)/2 ≤ n − 2, we get xn = (n + 1)/2. Similarly, x1 + x2 + ⋯ + xn−2 = n(n + 1)/2 − xn − xn−1 ≡ 0 (mod n − 2) implies xn−1 ≡ (n + 1)/2 (mod n − 2).
Then also xn−1 = (n + 1)/2, which leads
to xn = xn−1, a contradiction. Therefore,
Mathematical Excalibur, Vol. 10, No. 5, Dec. 05- Jan. 06 Page 4 Problem 239. (Due to José Luis
Díaz-Barrero, Universitat Politècnica de Catalunya, Barcelona, Spain) In any acute triangle ABC, prove that
⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 2 cos 2 cos 2 cos A B B C C A . 2 2 2 2 2 2 2 2 ⎟⎟⎠ ⎞ ⎜⎜ ⎝ ⎛ + + + + + + + + ≤ a c a c c b c b b a b a
Solution. (Proposer’s Solution)
By cosine law and the AM-GM inequality, bc a c b A A 2 cos 2 sin 2 1 2 2 2 2 = = + − − . 1 2 2 2 2 2 2 2 2 c b a c b a c b + − = + − + ≥ So . ) ( 2 2 sin 2 2 c b a A + ≤
By sine law and cos(A/2) = sin((B+C)/2), we get = + = + B C A c b a sin sin sin . ) 2 cos( ) 2 / sin( ) 2 cos( ) 2 sin( 2 ) 2 / cos( ) 2 / sin( 2 C B A C B C B A A − = − + Then . 2 2 2 sin ) 2 cos( 2 2 c b c b A a c b C B + + ≤ + = −
Adding two similar inequalities, we get the desired inequality.
Commended solvers: Anna Ying PUN (STFA Leung Kau Kui College, Form 7) and YEUNG Wai Kit (STFA Leung Kau Kui College, Form 5).
Problem 240. Nine judges independently award the ranks of 1 to 20 to twenty figure-skaters, with no ties. No two of the rankings awarded to any figure-skater differ by more than 3. The nine rankings of each are added. What is the maximum of the lowest of the sums? Prove your answer is correct. (Source: 1968 All Soviet Union Math Competitions)
Solution. WONG Kwok Kit (Carmel
Divine Grace Foundation Secondary School, Form 7) and YEUNG Wai Kit (STFA Leung Kau Kui College, Form 5). Suppose the 9 first places go to the same figure skater. Then 9 is the lowest sum.
Suppose the 9 first places are shared by two figure skaters. Then one of them gets at least 5 first places and that skater’s other rankings are no worse than fourth places. So the lowest sum is at most 5 × 1 + 4 × 4 = 21. Suppose the 9 first places are shared by three figure skaters. Then the other 18 rankings of these figure skaters are no worse than 9 third and 9 fourth places. Then the lowest sum is at most 9(1 + 3 + 4)/3 = 24.
Suppose the 9 first places are shared by four figure skaters. Then their rankings must be all the first, second, third and fourth places. So the lowest sum is at most 9(1 + 2 + 3 + 4)/4 < 24.
Suppose the 9 first places are shared by k > 4 figure skaters. On one hand, these k skaters have a total of 9k > 36 rankings. On the other hand, these k skaters can only be awarded first to fourth places, so they can have at most 4 × 9 = 36 rankings all together, a contradiction.
Now 24 is possible if skaters A, B, C all received 3 first, 3 third and 3 fourth places; skater D received 5 second and 4 fifth places; skater E received 4 second and 5 fifth places; and skater F received 9 sixth places, …, skater T received 9 twentieth places. Therefore, 24 is the answer.
Olympiad Corner
(continued from page 1)
Problem 4. (Cont.) it is possible to distribute the balls under the condition that A gets the same number of balls as the persons B, C, D and E together.
Problem 5. Let ABCD be a given convex quadrilateral. Determine the locus of the point P lying inside the quadrilateral ABCD and satisfying
[PAB]·[PCD] = [PBC]·[PDA], where [XYZ] denotes the area of triangle XYZ.
Problem 6. Determine all pairs of integers (x,y) satisfying the equation
y(x + y) = x3 − 7x2 + 11x − 3.
Schur’s Inequality
(continued from page 2)
Solution. From the last part of the
solution of example 3, we get
3(xyz)2/3 ≥ 2(xy + yz + zx) – (x2 + y2 + z2)
for any x, y, z > 0. (Note: this used Schur’s inequality.) Setting
, a
x = y = b and z = c
and arranging terms, we get a + b + c − 3 abc3 ) ( 2 a+b+c− ab− bc− ca ≤ 2 2 2 ( ) ( ) ) ( a− b + b− c + c− a = }. ) ( , ) ( , ) max{( 3 a− b 2 b− c 2 c− a2 ≤
Dividing by 3, we get the desired inequality.
Example 5. (2003 USA Team Selection
Test) Let a,b,c be real numbers in the interval (0, π/2). Prove that
) sin( ) sin( ) sin( sin ) sin( ) sin( ) sin( sin a c a b c b b c b c a b a a + − − + + − − . 0 ) sin( ) sin( ) sin( sin ≥ + − − + b a b c a c c
Solution. Observe that
sin(u – v) sin(u + v) = (cos 2v – cos 2u)/2
= sin2 u – sin2v.
Setting x = sin2a, y = sin2b, z = sin2c, in
adding up the terms, the left side of the inequality becomes . ) sin( ) sin( ) sin( ) )( ( ) )( ( ) )( ( b a a c c b y z x z z x y z y y z x y x x + + + − − + − − + − −
This is nonnegative by the r = 1/2 case of Schur’s inequality.
For many more examples on Schur’s and other inequalities, we highly recommend the following book.
Titu Andreescu, Vasile Cîrtoaje, Gabriel Dospinescu and Mircea Lascu, Old and New Inequalities, GIL Publishing House, 2004.
Anyone interested may contact the publisher by post to GIL Publishing House, P. O. Box 44, Post Office 3, 450200, Zalau, Romania or by email to [email protected].
Czech-Polish-Slotiak Match Zwardon, June 20-21,2005
M@
1. Let n be a given positive integer. Solve the system of equations !
/ x1+x;+x;+...4-s::=n,
n(n + 1)
Xl+222+323+...+nx,=- 2
1
in the set of nonnegative real numbers xl, x2,. . . ,x,,.Solution. Suppose x1, x2, . . , , x,, satisfy the equations above. Then we have O=s~+x~+xf...+x~- n-(xl+222+323f...+nx,-&n(n+l))=
=(2;-2x~+2-l)+(x~-3xs+3-l)+...+(x”,-nx,+n-l).
However, the expressions in the brackets are nonnegative. Indeed, for k > 2 and x > 0 we have, by the AM-GM inequality,
and the equality holds if and only if x = 1. Therefore we have 2s = x3 = , . . = z, = 1 and, by the first equation, x1 = 1.
2. Let a convex quadrilateml ABCD be inscribed in a circle tith center 0 and circumscribed to a circle with center I, and let ita diagonals AC and BD meet at a point P. Prove that the points 0, I and P hre collinear.
I
I .
Solution. Assume that the lines AI, Si, CI, DI meet the circumcircle of the quadri- lateral ABCD at E, F, G, H7 respectively. Since the lines AI, BI, CI, DI are the bisectors of the respective angles of the quadrilateral ABCD, the lines EG and FH
are the diameters of the circumcircle of ABCD. Thus EG and FH meet at 0. Denote by X the point of intersection of EB and CH. ’
Using Pascal’s theorem for the hexagon ACHDBE We see that P, X and I are collinear. Once again Pascal’s theorem applied to the hexagon GCHFBE yields that 0, X and I are collinear. Thus 0, I and - P are collinear, as desired.
3. Determine all integers n > 3 such that the polynomial W(x) = xn - 3x*-’ + 2x”-’ + 6
x3-3x2+2x+6=(x+1)(x2-4x+6). Suppose that for n = 4 we have ,.‘
x4-3x3+2x2+6=(xa+ux+b)(x2+cs+d).
Then, comparing the coeEicients we obtain
a-I-c=-3, ac+b+d=2, bd=6.
The first equation implies that a and c are of different parity and therefore, by the second equality, b and d are of the same parity. This contradicts the third equation.
Hence we may a&me that n 2 5. Suppose that we have
W(x) = WQb), (1)
where
P(x) = akxk f ak-lxk-’ f . . . + ulx + uo, Q(x) = b,,_hxn-k + bn-k-~Xk-l + . . . -I- bls + bo,
and%=&,-k=fl. Withnolossofgeneralitywemayassumethatk<(fnj<n~2
(because n Z 5). Comparing the coefficients of both sides of (1) we obtain the following system of equations:
_ aobo = 6, aobl + aibo = 0,
aobk + albk-1 + . . . + Uk_lbl + ukbl = 0.
Now we can easily prove by induction that a0 divides al, us, :. . , ak. Indeed, having established this fact for al, as,. . . , al we write
0 = uo(u&+~ + ulbz + . . . + azbl + uz+lbo) =
= a:bz+l f uoulbz + . . . + aouzb~ -I- 6uz+1,
and hence
6az+1= -(u;bz+l I- uoulbz + . . . + uouzbl).
We note that aJ1 the summands on the right hand side, are divisible by a$, therefore the left hand side also has this property. Hence a0 ] al+l.
But we have ak = fl; hence ac = 4~1 and with no loss of generality we may take uc = 1 and then we have bo = 6.
Now if we repeat the above arguments for the coefficients of Q, then we see that
bo = 6 divides bl, b2, . . . , br;_3 (we set, if necessary, bz = 0 if’1 > 7t.k k). Then we obtain a contradiction (b,.,-l; = &l), u&as n - k > n - 3. We consider two cases.
The case of k = 2. Then we arrive-at d
aoh,+ + ‘Jlb,+k-1 +. . . + an--k--lb1 + Gdb = 2,
can be expressed us a product of two polynomials tith posit&e degrees and integer coe&ients.
-_. . .-
a contradiction, because on the left hand side ‘all the summands, except for the 6rst one, are divisibie by 6 and hence even, and the first summapd is equal to fl.
The case of k = 1. Then the problem reduces to finding an integer root of the ‘polynomial W; it is easy to check that if n is even there are no such roots; if n is odd, W(-1) = 0.
Therefore the answer to the problem is: n is odd.
4. We distribute n 2 1 labelled balls among nine persons A, B, C, R, E, F, G, H, I. Determine in how many ways it is possible to distribute the balls under the
condition that A gets the same number of balls as the persons B, C, D and E
together.
Solution. Consider the polynomial (x+2)‘“=(x2+4x+4)“.= ..
=(xs+x+x+x+x+1+1+1’+1)(xs+x+x+x~+x+1+1+1f1)... (x2+x+x+x+x+1+1+1+1)
and suppose that we multiply out the brackets, obtaining 9” summa&. We show the one_tcFone correspondence between the number of x’% and the number of the distributions we deal with in the problem.
Suppose we have such a distribution. If the k-th ball goes to A, we pick x2 from the k-th bracket. If it goes to B, C, D, E, we pick the first, second, third or fourth 1, respectively, from the k-th bracket. If it goes to F, G, H, I, then we take the first, second, third or fourth x from the k-th bracket. Now if we multiply the factors we have chosen, we see, that the result is equal to x” if and only if A gets the same number of bails as B, C, D, E jointly.
Therefore, the number of the distributions we are interested in is equal to the co&cient at xn in the polynomial (x + 2)2n, that is,
0
2n
.2”n -
5. Let ABCD be a given convex quadrilateral. Determine the l&s of the points P lying inside the quadrilateral ABCD and satishing
[PAB] . [PCD] = [PBC] . [PDA], where [XYZ] denotes the area of a ttingJe XYZ.
Solution. If P lies on one of the diagonals AC or BD, let’s say on AC, then
[PAB] AP [PDA] [pBc1=pc=[pcDl’
..-___- .--.
IPABI
AQ andipDA1
_ AR,
[PBCl=&C
[PCD] RCwhich, since Q # R, ,implies that the given equality cannot hold.
Therefore the desired locus of the points P consists of the diagonals AC and BD. 6. Determine all pairs of integers (0,~) sattifing the equation
y(x+y)=x3-7x2+112-3. Solution. The considered equation is equivalent to
(2yjx)2=4x3-27x2+44x-12=(x-2)(4x2-19x+6)= = (x - 2)((x -2)(4x - 11) - 16).
The expression above must be a perfect square. Therefore we have either x = 2 (and
Y = -l), or (x - 2) = ks”, where k E (-2, -l,l, 2 and s E N; indeed, if for some prime p and nonnegative integer m the number p2m J r divides (x - 2) but pa”‘+ does not, then we have p 1 (2 - 2)(4x - 11) - 16, so p 1 16 and p = 2.
We will consider three cases separately.
The case of k = f2. Then we have 4x2 - 19x + 6 = f2u2 for some integer u, or, equivalently,
(8x - 19)’ - 265 = f32u2, a contradiction module 5.
The case of k = 1. Then 4x2 - 19x + 6 = u2 for some integer u, which leads to 265 = (8x - 19)2 - 16u2 = (8x -‘19 - 421)(8x - 19 + 4~).
-__ ___--
We easily check that x = 6 is the only solution of this equation (we simply consider
all possible decompositions: 265 = 1.265 = 5 - 53 = . . . , etc, and take into account the fact that x - 2 = 5’). Therefore, we obtain two solutions of the original equation: (x, Y) E C(6,3), (6, -9)).
to
The caseofk = -1. As before, we have 4x2 -19x+6 = -u2, WE& is qtivdat
265 = (8x - 19)2 + (4~)~
and we check all possibiities with u < 4: for u = 0, 1,2 there are no solutions. If u = 3, then we obtain (8x - 19)2 = 121= 112, which leads to x = 1, which gives two solutio~:~(x,Y) E ((1, l), (1, -2)). Finahy, for u = 4 we arrive at (8x- 19)2 = 9 = 3s and x = 2, which gives (x,Y) = (2, -1).
Therefore, the set of solutions is ss fol.Iows:
1(6,3), (6, -9), (1, l), (1, -2)s (2, -1)).
which is the desired equality. We prove that no other point lying inside ABCD satisfies the conditions of the problem.
Denote by 0 the point of the intersection of the diagonals AC and BD and suppose that P lies inside the triangle ABO. Let moreover BP and AC meet at Q and DP and AC meet at R. Then
