Volume 18, Number 5 February 2014 – March 2014
Olympiad Corner
Below are the problems of the Fourth Round of the 53rd Ukrainian National Math Olympiad for 10-th Graders.
Problem 1. Suppose that for real x,y,z,t
the following equalities hold: {x+y+z} = {y+z+t} = {z+t+x} = {t+x+y} = 1/4. Find all possible values of {x+y+z+t}. (Here {x}=x−[x].)
Problem 2. Let M be the midpoint of
the side BC of ΔABC. On the side AB and AC the points F and E are chosen. Let K be the point of the intersection of BF and CE and L be chosen in a way that CL||AB and BL||CE. Let N be the point of intersection of AM and CL. Show that KN is parallel to FL.
Problem 3. It is known that for natural
numbers a,b,c,d and n the following inequalities hold: a+c < n and a/b+c/d < 1. Prove that a/b+c/d < 1−1/n3.
Problem 4. There are 100 cards with
numbers from 1 to 100 on the table. Andriy and Nick took the same number of cards in a way that the following condition holds: if Andriy has a card with a number n then Nick has a card with a number 2n+2. What is the maximal number of cards could be taken by the two guys?
(continued on page 4)
Using Tangent Lines to Prove Inequalities (Part II)
Ibragim Ibatulin and Adilsultan Lepes
The Republican Specialized Physics Mathematics Secondary Boarding School Named after O. Zhautykov, Almaty, Kazakhstan
We offer a continuation of the paper by Kin-Yin Li (cf. Math Excalibur, vol. 10, no. 5) where he considers using tangent lines to prove inequalities.
Example 1. Suppose that a, b, and c
are positive real numbers satisfying a+b+c=3. Find the minimum of the expression a4+2b4+3c4.
Solution. Let fk(x)=kx4, where x∈(0,3), k = 1, 2, 3. As fk"(x) = 12kx2 > 0, where x>0, so functions fk are convex, which means that their graphs do not fall below their tangents drawn at any point xk∈(0,3) (k=1,2,3). Points x1, x2 and x3 are chosen such that f1′(x1) = f2′(x2) = f3′(x3) and x1+x2+x3=3. That is,
4x13=8x23=12x33 and x1+x2+x3=3. Hence, 3 1 3 3 1 2 3 3 3 3 1 3 , 2 , 6 3 2 6 3 x x x x x
and for any x∈(0,3), we have the inequalities (k = 1,2,3, see Fig. 1) kx4 ≥ f
k(xk) + fk′(xk)(x–xk). (1)
Fig. 1
Adding inequalities (1) for x equals a, b and c, we obtain , ) 6 3 2 ( ) 2 2 3 3 3 6 ( 81 ) 3 )( ( ' 3 3 1 2 2 1 1 3 2 4 3 3 3 3 3 3 3 1 1 1 3 3 4 1 4 4 4
k k x x f x c b awhich is the minimum (with equality holding at a=x1, b=x2 and c=x3).
Example 2. Let a, b, c > 0 be real
numbers such that ab+bc+ca = 1. Prove the inequality . 2 3 2 2 2 a b c a c b c b a
Solution. Let S=a+b+c. Based on the
inequality (a+b+c)2 ≥ 3(ab+bc+ca), which is equivalent to (a–b)2 + (b–c)2 + (c–a)2 ≥ 0, we find that S 3. Let f(x) = x2/(S–x) for x∈(0,S). Let us construct the tangent equation at the point x0=S/3 (see Fig. 2a,b):
. 4 5 3 4 5 6 3 3 ' 3 S x S x S S x S f S f y Fig. 2a Fig. 2b
Since the inequality x2/(S–x) ≥ (5x–S)/4 is equivalent to (S–3x)2 ≥ 0 on the interval (0,S), applying it thrice, based on the previously proved inequality
, 3 S we find that . 2 3 2 4 3 ) ( 5 2 2 2 2 2 2 S S c b a c S c b S b a S a b a c a c b c b a (continued on page 2) Editors: 張 百 康(CHEUNG Pak-Hong), Munsang College, HK
高 子 眉 (KO Tsz-Mei)
梁 達 榮 (LEUNG Tat-Wing)
李 健 賢 (LI Kin-Yin), Dept. of Math., HKUST
吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU
Artist: 楊 秀 英 (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 April 12, 2014.
For individual subscription for the next five issues for the 13-14 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:
Dr. Kin-Yin LI, Math Dept., Hong Kong Univ. of Science and Technology, Clear Water Bay, Kowloon, Hong Kong
Fax: (852) 2358 1643 Email: [email protected]
© Department of Mathematics, The Hong Kong University of Science and Technology
Mathematical Excalibur, Vol. 18, No. 5, Feb. 14 – Mar. 14 Page 2
Example 3. Let a, b, c ≥ 0 be real
numbers. Prove the inequality
. ) ( 6 1 1 1 2 2 2 b c a b c a
Solution. Assume that S=a+b+c and 1
) (x x2
f for x∈(0,S). We form the tangent equation at the point x0=S/3 :
. 9 3 3 9 3 9 3 3 ' 3 2 2 2 S Sx S x S S S S x S f S f y
Since on the interval (0,S), the inequality 9 3 1 2 2 S Sx x (2)
is equivalent to the inequality (x – S/3)2 ≥ 0, we find that . ) ( 6 6 9 ) ( 9 9 1 1 1 2 2 2 2 2 2 c b a S S S c b a S S S c b a
Example 4. Let a, b and c be positive
real numbers such that a+2b+3c ≥ 20. Prove the inequality
. 13 4 2 9 3 c b a c b a
Solution. Note that if a=2, b=3, c=4,
the inequality becomes equality. Let f(x)=1/x for x >0. Then f is convex in the interval (0,+∞). Hence the graph of the function f does not go below the tangent line drawn at any point x0 > 0. Thus, the following inequalities are valid (see Fig. 3):
. 16 2 1 ) 4 ( 16 1 4 1 1 , 9 3 2 ) 3 ( 9 1 3 1 1 , 4 1 ) 2 ( 4 1 2 1 1 c c c b b b a a a Fig. 3
As given in the statement of the problem, we find that . 13 4 20 8 4 3 2 8 4 2 2 3 4 3 3 4 2 9 3 c b a c b a c b a c b a c b a
Example 5. (Pham Kim Hung) Let a, b
and c be positive real numbers such that a2+b2+c2=3. Prove the inequality
. 3 2 1 2 1 2 1 a b c
Solution. Note that when a=b=c=1, the
inequality becomes an equality. Consider f(x) = 1/(2–x) and g(x) = kx2+m, where x∈(0, 3 ). The numbers k and m are to be chosen so that f (1) = g (1) and f ′(1) = g′(1). That is, 1=k+m and 1=2k. Hence, k=m=1/2 and g(x)=(x2+1)/2. Since the inequality 1/(2 –x) ≥ (x2+1)/2 is equivalent to x(x–1)2 ≥ 0, it is true for any x∈(0, 3 ) (see Fig. 4). Hence,
. 3 2 3 2 1 2 1 2 1 2 2 2 c b a c b a Fig. 4
Example 6. Let a, b and c be positive real
numbers. Prove the inequality
. ) ( ) 3 )( 3 )( 3 (a5a2 b5b2 c5c2 abc3
Solution. Note that when a=b=c=1, the
inequality becomes an equality. Consider f(x)=x5–x2+3 and g(x) = kx3+m, where x>0. The numbers k and m are to be chosen so that f (1) = g (1) and f ′(1) = g′(1). That is, 3=k+m and 3=3k. Hence, k=1, m=1/2 and g(x)=x3+2. The inequality (see Fig. 5)
x
5
x
2
3
x
3
2
(3) is true for any x >0 as it can be represented in the form (x–1)2(x3+2x2+2x+1) ≥ 0.Fig. 5
Example 7. Let a, b, c, d and e be
positive real numbers such that
. 1 4 1 4 1 4 1 4 1 4 1 a b c d e
Prove the inequality
. 1 4 4 4 4 4 2 2 2 2 e2 e d d c c b b a a
Solution. Consider f(x) = x/(4+x2) and g(x) = m + k/(4+x), where x ≥ 0. The numbers k and m are to be chosen so that f (1) = g (1) and f ′(1) = g′(1). Hence k = –3 and m = 4/5. Since the inequality x x x 4 3 5 4 4 2 is equivalent to (x–1)2(x+1)≥0, it is true for any x ≥ 0 (see Fig. 6).
Fig. 6
Applying this inequality, we have
. 1 4 1 4 1 4 1 4 1 4 1 3 4 4 4 4 4 4 2 2 2 2 2 e d c b a e e d d c c b b a a
Finally, we have some exercises for the readers.
Exercise 1. (Gabriel Dospinescu) Let
a1, a2, …, an be positive real numbers such that a1a2⋯an = 1. Prove that
. 2 1 1 1 2 1 2 2 2 2 1 n n a a a a a a (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 sending solutions is April 12, 2014.
Problem 441. There are six circles on
a plane such that the center of each circle lies outside of the five other circles. Prove there is no point on the plane lying inside all six circles.
Problem 442. Prove that if n > 1 is an
integer, then n5+n+1 has at least two distinct prime divisors.
Problem 443. Each pair of n (n≥6)
people play a game resulting in either a win or a loss, but no draw. If among every five people, there is one person beating the other four and one losing to the other four, then prove that there exists one of the n people beating all the other n–1 people.
Problem 444. Let D be on side BC of
equilateral triangle ABC. Let P and Q be the incenters of ∆ABD and ∆ACD respectively. Let E be the point so that ∆EPQ is equilateral and D, E are on opposite sides of line PQ. Prove that lines BC and DE are perpendicular.
Problem 445. For each positive integer
n, prove there exists a polynomial p(x) of degree n with integer coefficients such that p(0), p(1), …, p(n) are distinct and each is of the form 22014k+3 for some positive integer k.
*****************
Solutions
****************Problem 436. Prove that for every
positive integer n, there exists a positive integer p(n) such that the interval [1, p(n)] can be divided into n pairwise disjoint intervals with each contains at least one integer and the sum of the integers in each of these intervals is the square of some integer.
Solution. Jerry AUMAN, Math Activity Center (Carmel Alison Lam
Foundation Secondary School), Jon
GLIMMS (Vancouver, Canada) and ZOLBAYAR Shagdar (Orchlon
International School, Ulaanbaatar, Mongolia).
We look for a pattern. Since 1=12, let p(1)=1. Since 2+3+4=32, let p(2)=1+3=4 and divide [1,4] into [1,1] and (1,4]. Since
5+6+7+8+9+10+11+12+13 = 92, let p(3) = 1+3+9=13 and divide [1,13] into [1,1], (1,4], (4,13].
This suggests we let p(n) = 1 + 3 + 32 + ⋯ + 3n–1 = (3n–1)/2 and divide [1, p(n)] into [1, p(1)], (p(1), p(2)], …, (p(n–1), p(n)]. The integers in (p(k), p(k+1)] are from (3k+1)/2 to (3k+1–1)/2, which sums to 32k. So we are done.
Other commended solvers: Kaustav
CHATTERJEE (MCKV Institute of
Engineering College, India) and SP47 (Hanoi, Vietnam).
Problem 437. Determine all real numbers
x satisfying the condition that cos x, cos 2x, cos 4x, …, cos 2nx, … are all negative.
Solution 1. Jerry AUMAN, T. W. LEE
(Alumni of New Method College) and
Math Activity Center (Carmel Alison
Lam Foundation Secondary School). For such x, we have 2nx=2π(k
n+θn), where kn∈ℤ and 1/4 < θn < 3/4. In base 2 this is .012< θ0=.d1d2d3…2<.10111…2. No dndn+1 can be 00 or 11, otherwise θn-1=.00…2 or .11…2 would not be in (1/4,3/4). So θ0 = .010101…2 =1/3 or .101010…2=2/3. Then x=2π(k0+1/3) or 2π(k0+2/3) and for all n = 0,1,2,⋯, cos 2nx = −1/2.
Solution 2. Ioan Viorel CODREANU
(Secondary School Satulung, Maramures, Romania) and GLIMMS (Vancouver, Canada).
Let t=cos 2θ. Suppose cos θ, cos 2θ and cos 4θ are negative. Then t<0 and 2t2–1<0 imply 2/2t2cos210. We get
. 4 1 2 2 2 cos
Suppose sn = cos 2nx < 0 for n = 0,1,2,3,⋯. Then sn ∈[–1, –1/4). So |sn–1/2| > 3/4. Using this and sn+1=2sn2–1, we have . 2 1 2 3 2 1 2 1 2 4 1 2 2 1 2 1 n n n n n s s s s s
Repeating this, since –1≤ sn+1<0, we get
. 2 1 2 3 2 1 2 1 0 1 s s n n
Then |s0+1/2| < (2/3)n. Taking limit, we see cos x = s0= –1/2, i.e. x = ±2π/3+2kπ, where k is integer. Conversely, s0= –1/2 implies sn= –1/2 for n = 1,2,3,⋯. Other commended solvers: Henry
LEUNG Kai Chung (Graduate of
HKUST Maths).
Problem 438. Suppose P(x) is a
polynomial with integer coefficients such that for every integer n, P(n) is divisible by at least one of the positive integers a1, a2,…, am. Prove that there exists one of the ai such that for all integer n, P(n) is divisible by that ai.
Solution. Jerry AUMAN, Jon GLIMMS (Vancouver, Canada) and Math Activity Center (Carmel Alison
Lam Foundation Secondary School). Assume the contrary that for each ai, there exists integer ni such that P(ni) is not divisible by ai. Consider the prime factorizations of ai and |P(ni)|. Then there exists a prime divisor pi of ai such that ei
i
i p
d is the greatest power of pi dividing ai, however di does not divide | P(ni)|. If two of the di’s are powers of the same prime, then eliminate the one with the larger exponent. (In this way, each of a1, a2,…, am is still divisible by one of the remaining di’s.)
By the Chinese remainder theorem, there exist integers n such that n ≡ ni (mod di) for the remaining di’s. Now P(n)–P(ni) is divisible by n–ni, which is divisible by di. Since P(ni) is is not divisible by di. So P(n) is not divisible by any di’s, contradicting P(n) is divisible by at least one of the positive integers a1, a2,…, am, hence also divisible by at least one di.
Problem 439. In acute triangle ABC, T
is a point on the altitude AD (with D on side BC). Lines BT and AC intersect at E, lines CT and AB intersect at F, lines EF and AD intersect at G. A line ℓ passing through G intersects side AB, side AC, line BT, line CT at M, N, P, Q respectively.
Prove that ∠MDQ =∠NDP.
Solution. William FUNG and Math Activity Center (Carmel Alison Lam
Mathematical Excalibur, Vol. 18, No. 5, Feb. 14 – Mar. 14 Page 4 Set the origin at D and A, B, C at (0,a),
(b,0), (c,0) respectively. A B D C T E F G M N P Q
Let T be at (0,1). The equations of the lines BT, CT, AB, AC are
y = – (x/b) + 1, y = – (x/c) + 1, y = – (ax/b) + a, y = – (ax/c) + a respectively. Since E = BT ∩ AC and F = CT ∩ AB, we can solve the equations of the lines to get
c ab c b a c ab bc a E ( 1) , ( ) and ( 1) , ( ). b ac b c a b ac cb a F
From the y-intercept of line EF, we get G=(0, 2a/(a+1)). Let the equation of ℓ be y=mx+2a/(a+1). Then M = ℓ ∩AB is at . ) 1 )( ( ) ) 1 ( 2 ( , ) 1 )( ( ) 1 ( a mb a mb a a a a mb a b a a
Using role symmetry of B and C, we can replace b by c in the coordinates of M to get coordinates of N. Similarly, P = ℓ ∩BT is at . ) 1 )( ( ) 1 ( 2 , ) 1 )( ( ) 1 ( a mb a mb a a a mb a b a
The coordinates of Q can be found by replacing b by c in the coordinates of P. Since D is the origin, the slopes of lines DM and DP can be found by taking the y-coordinates of M and P dividing by their respective x-coordinates, which turn out to be the negative of each other! So lines DM and DP are symmetric with respect to the y-axis! Similarly, lines DN and DQ are symmetric with respect to the y-axis. Therefore, ∠MDQ =∠NDP.
Comments: There is a pure geometry solution using a number of equations from applying Menelaus’ theorem to different triangles. There is also a solution using harmonic division and cross-ratios from projective geometry.
Other commended solvers: Georgios
BATZOLIS (Mandoulides High School,
Thessaloniki, Greece), Andrea
FANCHINI (Cantu, Italy), T. W. LEE
(Alumni of New Method College), SP47 (Hanoi, Vietnam), Titu ZVONARU (Comăneşti, Romania) and Neculai
STANCIU (“George Emil Palade’’
Secondary School, Buzău, Romania).
Problem 440. There are n schools in a city.
The i-th school will send Ci students to watch a performance at a field. It is known that 0 ≤ Ci ≤ 39 for i=1, 2, …, n and C1+C2+⋯+Cn=1990. The seats will be put in a rectangle arrangement with each row having 199 seats. Determine the least number of rows needed to satisfy the condition that all students from the same school must sit in the same row for all possibilities of the known conditions above.
Solution. Adnan ALI (9th Grade, Atomic Energy Central School 4 (AECS4), Mumbai, India), Jerry AUMAN and Jon
GLIMMS (Vancouver, Canada).
Let k be the minimal number of rows needed. For m =1, 2,…, k, let there be am students in row m. If there are no more than 160 students in some row, then since each school sends at most 39 students, we can put in students from one more school in that row. So we may assume am ≥ 161. Now
1990 = a1 + a2 + ⋯ + ak ≥ 161k, which implies k ≤ 12.
Next, we show 11 rows may not be enough. Suppose there are n = 80 schools with Ci = 25 for i = 1, 2, …, 79 and C80 = 15. This totals to 1990 students. Then there can only be one row with 25×7+15 = 190 students and the other 10 rows with 25×7=175 students. This only totals to 1940 students.
So the least number of rows needed to satisfy the condition that all students from the same school must sit in the same row for all possibilities of the known conditions is 12.
Other commended solvers: T. W. LEE (Alumni of New Method College) and
Math Activity Center (Carmel Alison
Lam Foundation Secondary School).
Olympiad Corner
(Continued from page 1)
Problem 5. Find the values of x such
that the following inequality holds
min{sin x,cos x} < min{1−sin x,1−cos x}. Problem 6. Find all pairs of prime
numbers p and q that satisfy the following equation
. 19 2
3pq qp1
Problem 7. Is it possible to choose 24
points in the space, such that no three of them lie on the same line and choose 2013 planes in a way that each plane passes through at least 3 of the chosen points and each triple of points belongs to at least one of the chosen planes?
Problem 8. Let M be the midpoint of
the internal bisector AD of Δ ABC. Circle ω1 with diameter AC intersects BM at E and circle ω2 with diameter AB intersects CM at F. Show that B, E, F, C belong to the same circle.
Using Tangent Lines …
(Continued from page 2)Exercises 2. Let a, b and c be
non-negative real numbers. Prove that
. 1 2 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 d c b a c b a d b a d c a d c b d c b a
Exercise 3. Let a, b and c be positive
real numbers. Determine the minimal value of . 5 4 3 b a c a c b c b a
Exercise 4. Let a, b and c be positive
real numbers such that ab+bc+ca=3. Prove that . 27 ) 3 )( 3 )( 3 (a7a4 b5b2 c4c References [1] Chetkovski, Z., Inequalities. Theorems, Techniques and Problems. Springer Verlag, Berlin Heidelberg, (2012).
[2] Pham Kim Hung, Secrets in Inequalities (volume 1). Editura Gil, Zalău, (2007).
1. Suppose that for real x, y, z, t the following equalities hold:
{x+ y+z} = {y+z+t} ={z+t+x} ={t+x+ y}
=t.
Find all possible values of {x+ y+z+t}. ·(Vyacheslav Yasinskyy) Answer: 0,
±,
f.
Solution. We have Thus, x+ y+z=[x+y+zJ+t, y+z+t=[y+z+tJ+t, z+t+x=[z+t-i-xJ+t, t+x+ y =[t+x+yJ+t. 3(x+ y-J;z+t)=[x+y+z]+[y+z+t]+[z+t+x]+[t+x+ y]+1.11lis implies that 3( x + y + z + t) is integer so the fraction part of X+ y + z + t is either 0 or
t,
f.
All these values can be achieved which follows from the following exan1ples:· x=y =z=t=.l x=y=z=t=.l., x=y=z=t=2.. 4' 12 12. 2. Let M be the midpoint of the side BC of
MBC. On the side AB and AC the points F and E are cl)osen. Let K be the point of the intersection of BF and CE, L is chosen in a way that C,L
II
AB and BLII
CE. Let N the point of the intersection of AM and CL. Showthat KN is parallel to FL.
' · · . (Vyacheslav Yasinskyy)
~· J:':lqte, that, BECL is a parallelogram. Since Alt
II
CN ~ rui'd M is a midpoint of B C we have that ABNC is .. ~ paiallelogram and thus AE = NL(F~g. 2.0). Let f be the point of the intersection of BF and Cl( sin~e F'CI(BN then MBN ~ !lPFC.
Th;m·. PF = PC so
PF
= PB-PC. Since PCII
BE we 1 PB· PN; . , . ·PNhave .. MCJ<: ~ MlEK. Therefore, ~~ = ~~ ,
A
have a proportion PP'K:F = PP'NL which leads to the similarity of MFL and /lPJ('T v1 • H ence, LPFL= .LPKN and so KN
II
FL. The rule now follows.3. It is known, tl1at for natural numbers a, b, c, d and n the following inequalities: hold: a +.c < n and f+~<l. Prove that f+~ <
1-1 :1 •
(Vyacheslav Yasinskyy) So_lution. Since n>a+c2::2 then n2::3. Also, a<b and c<d because a<b, c<d. We distinguish between the £allowing cases:
Case]. Let b 2.. nand d 2n then .!!.+.f_:O: £+£= a+c :0: n-l = 1-.l < 1-_j_.
- - ' b d 11 n n 11 11 ,n3
Case 2. Let b
s
n and d :0: n, then ineq1.1alityf
+ ~ < 1 implies that ad+ be < bd in other wordsad+bc+1sbd. Now we have f+7s1-b~ =1-~<1---l,-.
n n
Case 3. Let b<n<d. If d:O:n2 then bd<n3 and thus f+7:0:1-Jd<l-~.
,
If d>n2 thenfs "{ =1-~ because a<n-c:O:n-1, so a:O:n-2. Suppose that f+721-~. Then
n n 1J
1-E.<.f_ _ _j_ <1.-...L+_L b - d ,,J - w ,z : 113 < .l n . This implies that b > n(b -a)> n - which contradicts b < n < d .
· 2 3 I
Case 4. Let d < n < b . If .b
s
n then bd < n and thusf
+7
s
1-bd < 1- :3 • If b > n2 then ._dc:o;n-;2==1-~ because c<n-a:O:n-1, so c:O:n--,.2. Suppose that _ba+_dc21-_L. Then
• 11 n nJ
1-E. d S E.-b _j_ S n3 .l-.1_ + __]_ < .l. Thls implies that d > n(d -c) 2 n which contradicts d < n < b
c n n2 n3 n . · · Therefore, the result holds in all cases.
4. There are 100 cards with numbers from 1 to 100 on the table. Andriy and Nick took the same number of cards in a way that the following condition holds: if Andriy has a card
with a number n then Nick has a card with with a number 2n + 2 . What is the maximal
number of cards could be taken by the J:wo guys?
(Vyacheslav Yasinsky;1 Arzswer: 66 cards.
Solutiott. We are going to show that the total number of cards taken by the guys does not exceed 66, in
other words each of them could take not more than 33 cards.
Since 2n
+
2s
100 then 2n S 98 so n S 49. This implies that all Andry's numbers belong to the set {1, 2, ... , 49}. We break tills set into the folloWing pieces: {1, 4}, {3, 8}, { 5, 12}, ... , {23, 48} . (12 subsets): {2, 6}, {10, 22}, {14, 30} and {18, 38} (4 subsets); {25}, {27}. ... , { 49} (13 subsets); {26}, {34}, { 42}, { 46} (4 subsets). In total, we have 33 Subsets. Each two of these sets do not have common elements. If Andriy has at least 34 Cards then by the Pigeonhole principle at least two numbers will belong to the same subset. This leads to a contradiction.The following example shows that 33 cards could be achieved: Andriy takes the cards with numbers:
A= {1, 3, 5, ... , 23, 2, 10, 14, 18, 25, 27, 29, ... , 49, 26, 34, 42, 46}.
5. Find all values of x such that the following inequality holds
. min{sinx, cosx} < min{l-sinx, 1-cosx}.
lValentyn Lei[ur~ Answer:
~
3;
+27tn; 2Jrn), nEZ.·Solution. The condition of the problem is equivalent to the following[ sys{le~ of in~qualities: {
sinx<l-sinx, {cosx<l-sinx, smx<2,
· v :::;> cosx<!,
sinx<l-cosx, cosx < 1-cosx,
sinx+cosx<l.
6. Find all pairs of prime numbers p and q that satisfy the following equation
3pq -2qp-! =19.
(Vyacheslav Yasinskyy)
Aris1ver: p=3, q=2; p=7, q=2.
Solution. It is clear that p ~ 3. If p = q then we have (3 pP-2- 2 pp-3 )p2 = 19. which is impossible
whenever p ;::._ 3 is prime.
Suppose that p :;C q. By the Little Fermat's theorem qp-l -1
=
0 (mod p), pq-l -1=
0 (mod q).We write our equality~· the form 3 pq-2(qP-1-1) = 21 and since the left)llljid side i~ diVisible by p
we have that p=3 or p=7. Then, writing 3p(pq-1-1)-2qp-I =19-3p we get that 19-3p is
divisible by q.
7. Is it possible to choose 24 points in the space, such that no three of them lie on the same line and choose 2013 planes in a way that each plane passes through at least 3 of the
chosen points and each triple of points belongs to at least one of the chosen planes?
(V. Radchenko, V. Yasinskyy) Answer: impossibel.
Solution. Suppose that it is possible. Let 7r1, 7r2 , ... , 7r2013 be the planes and n1, ~' ... , n2013 be the
corresponding numbers of points that belong to the planes. By the gi:rens ni ;::.. 3 for each i
=
1, 2013 . Itis clear that c! + C,3, + ... + C,3, = 2024. In addition, n,. ~ 5 for all i = 1, 2013 becauseif among
I 2 2013 . nl'.~'."'' n2013 atleastoneisgreaterthan 5 thenwehave 2024=C~ +C! + ... +C! ;::..1+1+ ... +1+C:=2012+20=2032 I 2 2013 ~ 2012 whic;h.is false.
Suppose tllllt among the numbers np ~, ... , n2013 there is exactly a that are equal to 3, exactly b that
are. equal to 4 and exactly c that are equal to 5 . Then a+ b + c := 2013 and
aCj + bC! +
cc;
= 2024. Hence, 3b + 9c = 11 which is impossible for nonnegative b and c .8. Let M be the midpoint of the internal bisector AD of MBC. Circle m
1 with diameter
AC intersects BM atE and circle OJ2 with diameter AB intersect CM at F. Show that
B, E, F, c·belong to the same circle.
(Vyacheslav Yasinskyy) Soluti'on. If AB = AC then tl1e statement is obvious. Without loss of
generality we assume that AB < AC.
Let All be the common chord of the
given circles (Fig. 25). We draw the line
which passes through A and is
perpendicular 1o AD and denote by K
and
£
the points of the intersection ofthe given Jines with aJ 1 and OJ 2
respective! y.
We prbve that BL passes through M.
Let X be the point of the intersection of
BL and AD. Since KB II AD
II LC
we have the foliowing.proportions:
~=fi, ~i=~~·, ~~=f£.
L
Thus, AX= KBLKLA, 'DX =
cix.KA.
Also, LKAB = LLAC and triangles AKB and ALC are similar.Thus, 'f1 ='
?c ,
KA · LC = KB ·LA. Hence, AX= DX and X coincides with M. By analogy, wecan show that CK passes through M. We have
LDME
=
LLMA = LCLE = 180°-LDHE.This implies that E, M, D, H belong to the same circle. KBHF is inscribed in the circle m1 and thus
LDMF = LXMA = LMKB =180°-LBHF = LDHF,
whichimpliesthat M, H; D, F lie on the circle.
We haite shown that M, H, D, F, E lie on the same circle. In the right angled triangle HAD HM is
a med;an. Therefore, MD= MI. Thus,
LMDH = LMHD= LMED = LMFH.
Consider triangles MDE and MBD that have the common angle M and·
180°-LCFE = LMFE = LMDE
