Volume 19, Number 4 February 2015 – March 2015
Olympiad Corner
Below are the problems of the Team Selection Test 1 for the Dutch IMO team held in June, 2014.
Problem 1. Determine all pairs (a,b) of positive integers satisfying
a2+b | a2b+a and b2−a | ab2+b. Problem 2. Let ΔABC be a triangle. Let M be the midpoint of BC and let D be a point on the interior of side AB. The intersection of AM and CD is called E. Suppose that |AD|=|DE|. Prove that |AB|=|CE|.
Problem 3. Let a, b and c be rational numbers for which a+bc, b+ac and
a+b are all non-zero and for which we
have . 1 1 1 b a ac b bc a
Prove that (c3)(c1) is rational.
Problem 4. Let ΔABC be a triangle with |AC|=2|AB| and let O be its circumcenter. Let D be the intersection of the angle bisector of ∠A and BC. Let
E be the orthogonal projection of O on AD and let F≠D be a point on AD
satisfying |CD|=|CF|. Prove that ∠EBF=∠ECF.
(continued on page 4)
Polygonal Problems
Kin Yin Li
In geometry textbooks, we often come across problems about triangles and quadrilaterals. In this article we will present some problems about n-sided polygons with n > 4. This type of problem appears every few years in math olympiads of many countries.
Example 1. Prove that if ABCDE is a
convex pentagon with all sides equal and ∠A≥∠B≥∠C≥∠D≥∠E, then it is a regular pentagon. Solution. A B C D E Since , 2 sin 2 2 sin 2AB B CD D CE AC
we get ∠AEC≥∠EAC. Next,
. 2 180 90 2 90 2 2 180 AEC D E D E B A B A EAC
Hence, ∠EAC=∠AEC. Then equality holds everywhere above so that ∠A=∠E and we are done.
Example 2. (Bulgaria, 1979) In convex
pentagon ABCDE, ΔABC and ΔCDE are equilateral. Prove that if O is the center of ΔABC and M, N are midpoints of BD, AE respectively, then ΔOME∼
ΔOND. Solution. A C B D E P Q M N O
Let P, Q be the midpoints of BC, AC respectively. Observe that ∠COP=60°,
OC=2OP, PM||CD, ∠DCE=60° and EC
= DC = 2MP. Then rotating about O by 60° clockwise and follow by doubling distance from O, we see ΔOPM goes to
ΔOCE. Hence ∠EOM =∠COP =60°
and OE=2OM. Similarly we can rotate about O by 60° counterclockwise and double distance from O to bring ΔOQN to ΔOCD. Then ∠DON = 60°, OD = 2ON and so ΔOME∼ΔOND.
Example 3. (IMO 2005) Six points are
chosen on the sides of an equilateral triangle ABC: A1, A2 on BC, B1, B2 on
CA and C1, C2 on AB, so that they are the vertices of a convex hexagon
A1A2B1B2C1C2 with equal side lengths. Prove that A1B2, B1C2 and C1A2 are concurrent. Solution. A B C C1 B2 B1 A2 C2 A1 P
Let P be the point inside ΔABC such that ΔA1A2P is equilateral. Observe that
A1P||C1C2 and A1P=C1C2. So A1PC1C2 is a rhombus. Similarly, B1PB2B1 is a rhombus. So ΔC1B2P is equilateral. Let
α = ∠B2B1A2, β = ∠B1A2A1 andγ = ∠
C1C2A1. Then α and β are external angles of Δ CB1A2 with ∠C=60°. So
α+β=240°. Now∠B2PA2=α and ∠C1PA1 = γ. So α+γ=360°−(∠C1PB2 +∠A1PA2) =240°. So β=γ. Similarly, ∠C1B2B1=β. Hence, Δ A1A2B1, Δ B1B2C1 and Δ
C1C2A1 are congruent, which implies Δ
A1B1C1 is equilateral. Since sides of
A1A2B1B2C1C2 have equal lengths, lines
A1B2, B1C2 and C1A2 are the perpendicular bisectors of the sides of ΔA1B1C1 and the result follows. (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 10, 2015.
For individual subscription for the next five issues for the 14-15 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. 19, No. 4, Feb. 15 – Mar. 15 Page 2
Example 4. (Czechoslovakia, 1974)
Prove that if a circumscribed hexagon
ABCDEF satisfies
AB=BC, CD=DE and EF=FA,
then the area of ΔACE is less than or equal to the area of ΔBDF.
Solution. Let O be the circumcenter of
hexagon ABCDEF and R be the radius of the circumcircle. Let
. , , AEC ACE CAE
From the given conditions on the sides, we get . , , FOA EOF DOE COD BOC AOB
Let [XYZ] denote the area of ΔXYZ. We have . sin sin sin 2 4 sin 2 sin 2 sin 2 4 ] [ 2 R R R R R R AE CA EC ACE Similarly, . 2 sin 2 sin 2 sin 2 ] [BDF R2 Now for positive α, β, γ satisfying
α+β+γ = 180°, we have . 2 sin 2 sin 2 sin 2 ) cos( 1 2 ) cos( ) cos( ) sin )(sin sin )(sin sin (sin sin sin sin 2 2 2 2 2 2
cyc cyc Therefore, [ACE]≤[BDF].Example 5. (IMO 1996) Let ABCDEF
be a convex hexagon such that AB is parallel to DE, BC is parallel to EF and
CD is parallel to FA. Let RA, RC, RE be the circumradii of triangles FAB, BCD,
DEF respectively, and let P denote the
perimeter of the hexagon. Prove that . 2 P R R RA C E
Solution. Let a, b, c, d, e, f denote the
lengths of the sides AB, BC, CD, DE,
EF, FA respectively. By the parallel
conditions, we have ∠A=∠D, ∠B=∠E, ∠C=∠F.
Consider rectangle PQRS such that A is on
PQ; F,E are on QR; D is on RS and B,C are
on SP. b e f a c d Q P S R F A B C D E We have BF≥PQ=SR. So 2BF≥PQ+SR, which is the same as
). sin sin ( ) sin sin ( 2BF a Bf C c Cd B Similarly, ). sin sin ( ) sin sin ( 2 ), sin sin ( ) sin sin ( 2 C b A a A d C c DF A f B c B b A c BD
Next, by the extended sine law,
. sin 2 , sin 2 , sin 2 E DE R C BD R A BF RA C E
Then using the inequalities and equations above, we have . 2 2 sin sin sin sin 4 sin sin sin sin 4 P f e d c b a A F F A f B A A B a R R RA C E
Example 6. (Great Britain, 1988) Let four
consecutive vertices A, B, C, D of a regular polygon satisfy
. 1 1 1 AD AC AB
Determine the number of sides of the polygon.
Solution. Let the circumcircle of the
polygon have center O and radius R. Let α =∠AOB, then 0 < 3α =∠AOD < 360°. So 0 < α < 120°. Also, from , 2 3 sin 2 , sin 2 , 2 sin 2 R AD R AC R AB we get . 2 3 sin 1 sin 1 2 sin 1
Clearing denominators, we have
. 2 sin 4 sin 4 7 cos 2 4 3 cos 4 cos 4 7 cos 2 5 cos cos 2 cos 2 3 cos 2 1 2 cos cos 2 1 2 3 cos 2 cos 2 1 2 5 cos 2 cos 2 1 2 sin 2 3 sin sin 2 3 sin sin 0
Then 7α/4=90°, that is α=360°/7. So the polygon has 7 sides.
Example 7. (Austria, 1973) Prove that
if the angles of a convex octagon are all equal and the ratio of all pairs of adjacent sides is rational, then each pair of opposite sides has equal length.
Solution. Without loss of generality, we
may assume the sides of such a polygon A1A2…A8 are rational (since the conclusion is the same for octagons similar to such an octagon). Now the sum of all angles of the octagon is 6×180°. Hence each angle is 45°. Let vn be the vector from An to An+1 for
n=1,2,…,8 (with A9=A1). Then the angle between vn and vn+1 at the origin is 45°. Observe that the sum of these vectors is zero since we start at A1 and traverse the octagon once to return to
A1.
Let i and j be a pair of unit vectors perpendicular to each other at the origin. By rotation, we may assume v1 is a vector in the i direction and v3 is in the j direction. Then v1+v5=xi and v3+v7 = yj for some rational x and y. Also,
j r i r v v v v2 4 6 8 2 2
for some rational r. Then
. 0 ) 2 ( ) 2 ( 8 1
n n v j r y i r xSince, x and r are rational, we must have x = r = 0. That is, v5 = −v1. By rotating the i, j vectors by 45°, similarly we get v6= −v2. Then also v7= −v3 and
v8= −v4. The result follows.
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 10, 2015.
Problem 461. Inside rectangle ABCD, there is a circle. Points W, X, Y, Z are on the circle such that lines AW, BX, CY,
DZ are tangent to the circle. If AW=3, BX=4, CY=5, then find DZ with proof.
Problem 462. For all x1, x2, …, xn ≥ 0, let xn+1 = x1, then prove that
n k k k k n x x x 1 1 2 2 1 2 ( 1) 2. ) 1 ( 1Problem 463. Let S be a set with 20 elements. N 2-element subsets of S are chosen with no two of these subsets equal. Find the least number N such that among any 3 elements in S, there exist 2 of them belong to one of the N chosen subsets.
Problem 464. Determine all positive integers n such that for n, there exists an integer m with 2n−1 divides m2+289. Problem 465. Points A, E, D, C, F, B lie on a circle Γ in clockwise order. Rays AD, BC, the tangents to Γ at E and at F pass through P. Chord EF meets chords AD and BC at M and N respectively. Prove that lines AB, CD,
EF are concurrent.
*****************
Solutions
****************
Problem 456. Suppose x1, x2, …, xn are non-negative and their sum is 1. Prove that there exists a permutation σ of {1,2,⋯,n} such that
xσ(1)xσ(2)+xσ(2) xσ(3)+⋯+xσ(n)xσ(1) ≤ 1/n.
Solution. CHAN Long Tin (Cambridge
University, Year 3), Ioan Viorel CODREANU (Secondary School Satulung, Maramures, Romania), KWOK Man Yi (Baptist Lui Ming Choi Secondary School, S4), Samiron SADHUKHAN (Kendriya Vidyalaya,
India) and WONG Yat (G. T. (Ellen Yeung) College).
Assume the contrary is true. Let σ(n+1) =
σ(1) for all permutations σ. For 1≤i<j≤n,
the terms xixj and xjxi appear a total of 2n(n−2)! times in
.
1 ) 1 ( ) (
n S n n k k kx
x
So, we have . 1 )! 2 ( )! 2 ( )! 2 ( 2 ! 1 2 1 2 2 1 1 1 () ( 1)
n i i n i i n i i n j i i j S n k k k x n n x x n n x x n n x x n n n This simplifies to (*) 1. 1 2 n x n i i
However,by the Cauchy-Schwarz inequality,
, 1 1 2 1 1 2 1 2 1 2
n i i n i i n i n i i x x x n which contradicts (*).Problem 457. Prove that for each n = 1,2,3,…, there exist integers a, b such that if integers x, y are relatively prime, then
. ) ( )
(ax 2 by 2 n
Solution. Samiron SADHUKHAN
(Kendriya Vidyalaya, India) and WONG Yat (G. T. (Ellen Yeung) College).
There are (2n+1)2 ordered pairs (r,s) of integers satisfying |r|, |s| ≤ n. Assign a distinct prime number pr,s to each such (r,s). By the Chinese remainder theorem, there exist integers a,b such that for all integers r, s satisfying |r|, |s| ≤ n, we have
a≡r (mod pr,s) and b≡s (mod pr,s). Let integers x, y be relatively prime. Assume (x,y) has distance at most n from (a,b). Then |a−x|≤n and |b−y|≤n. Let
a−x=r and b−y=s. Then x=a−r and y=b−s are multiples of pr,s , contradicting gcd(x,y) = 1. Therefore, . ) ( ) (ax 2 by 2 n
Problem 458. Nonempty sets A1, A2, A3 form a partition of {1,2,…,n}. If x+y=z have no solution with x in Ai , y in Aj , z in
Ak and {i,j,k}={1,2,3}, then prove that A1,
A2, A3 cannot have the same number of elements.
Solution. Oliver GEUPEL (Brühl,
NRW, Germany) and John GLIMMS. Without loss of generality, say 1∈A1 and the smallest element in A2∪ A3 is b∈ A2. Let the elements in A3 be c1, c2, …, ck in increasing order.
Assume ci+1−ci=1 for some i. Then take i to be the smallest possible. Since b∈A2, the equations (ci−b)+b=ci and (ci−b+1)+b=ci+1 imply ci−b and ci−b+1 are both not in A1.
Since 1∈A1 and (ci−b)+1= ci−b+1, so either ci−b+1 and ci−b both are in A2 or both are in A3. Since i is smallest such that ci+1−ci=1, so ci−b+1 and ci−b cannot be in A3. However, ci−b+1 and ci−b in
A2, b−1 in A1 (by property of b) and (b−1)+(ci−b+1)=ci lead to contradiction. So ci+1−ci ≥ 2 for all i.
Finally, since 1+(ci−1)=ci , we get
ci−1∉B. Hence ci−1∈A. Then A1 contains 1, c1−1, c2−1, … , ck−1. Therefore, A1 has more elements than A3. Problem 459. H is the orthocenter of acute ΔABC. D,E,F are midpoints of sides BC, CA, AB respectively. Inside ΔABC, a circle with center H meets DE at P,Q, EF at R,S, FD at T,U. Prove
that CP=CQ=AR=AS=BT=BU.
Solution. John GLIMMS.
r r r r r r I=A,B,C D E F H P Q R S T U J
Let lines AH and FE meet at J. From
AH⊥BC and BC||FE, we get FE is
perpendicular to AJ and HJ. By folding along DE, EF and FD, we can make a tetrahedron having ΔDEF as the base and points A, B, C meet at a point
I.
Then FE is perpendicular to IJ and HJ. So FE is perpendicular to the plane through I,J,H. Then FE⊥IH. Similarly,
DE⊥IH. Then the plane through D,E,F
is perpendicular to IH. By Pythagoras’ theorem, IH2+r2 = CP2 = CQ2 = AR2 =
AS2 = BT2= BV2, where r is the radius of the circle.
Other commended solvers: Adnan
ALI (Atomic Energy Central School 4, Mumbai, India),Andrea FANCHINI
Mathematical Excalibur, Vol. 19, No. 4, Feb. 15 – Mar. 15 Page 4
(Cantú, Italy), William FUNG,Oliver GEUPEL (Brühl, NRW, Germany), MANOLOUDIS Apostolis (4 High School of Korydallos, Piraeus, Greece), Samiron SADHUKHAN (Kendriya Vidyalaya, India), Titu ZVONARU (Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania). Problem 460. If x,y,z > 0 and x+y+z+2 = xyz, then prove that
. 2 6 yz zx xy z y x Solution. Adnan ALI (Atomic Energy
Central School 4, Mumbai, India), CHAN Long Tin (Cambridge University, Year 3), Ioan Viorel CODREANU (Secondary School Satulung, Maramures, Romania), Oliver GEUPEL (Brühl, NRW, Germany), KWOK Man Yi (Baptist Lui Ming Choi Secondary School, S4), Vijaya Prasad NULLARI (Retired Principal, AP Educational Service, India), Nicuşor ZLOTA (“Traian Vuia” Technical College,Focşani, Romania) and Titu ZVONARU (Comăneşti,
Romania). Let . 1 1 , 1 1 , 1 1 z c y b x a
Using x+y+z+2 = xyz, we get a+b+c = 1. Then x = (1−a)/a = (b+c)/a and similarly y=(c+a)/b and z=(a+b)/c. By the AM-GM inequality, we have
. 2 ) )( ( 2 6 6 xy zx yz bc b a a c b b a c a c a c b z y x cyc cyc cyc
Other commended solvers: Paolo
PERFETTI (Dipartimento di Matematica, Università degli studi di Tor Vergata Roma, via della ricerca scientifica, Roma, Italy), WONG Yat (G. T. (Ellen Yeung) College).
Olympiad Corner
(Continued from page 1)
Problem 5. On each of the 20142 squares of a 2014×2014-board a light
bulb is put. Light bulbs can be either on or off. In the starting situation a number of light bulbs are on. A move consists of choosing a row or column in which at least 1007 light bulbs are on and changing the state of all 2014 light bulbs in this row or column (from on to off or from off to on). Find the smallest non-negative integer k such that from each starting situation there is a finite sequence of moves to a situation in which at most k light bulbs are on.
Polygonal Problems
(Continued from page 2)
Example 8. (IMO 1997) Equilateral
triangles ABK, BCL, CDM, DAN are constructed inside the square ABCD. Prove that the midpoints of the four segments KL, LM, MN, NK and the midpints of the eight segments AK, BK,
BL, CL, CM, DM, DN, AN are the twelve
vertices of a regular dodecagon.
Solution. D A B C M N L P2 O P1
Let us denote the midpoints of segments
LM, AN, BL, MN, BK, CM, NK, CL, DN, KL, DM, AK by P1, P2, P3, P4, P5, P6, P7, P8, P9, P10, P11, P12, respectively. To prove the dodecagon P1P2P3P4P5P6P7P8P9P10P11P12 is regular, we observe that BL=BA and ∠ABL=30°. Then ∠BAL=75°. Similarly ∠DAM =75°. So
∠LAM =∠BAL+∠DAM −∠BAD=60°. Along with AL=AM, we see triangle ALM is equilateral.
Looking at triangles OLM and ALN, we get OP1=½LM, OP2=½AL and OP2|| AL. Hence, OP1=OP2, ∠P1OP2=∠P1AL = 30°, ∠P2OM =∠DAL=15° and ∠P2OP3 = 2∠P2OM = 30°. By symmetry, we can conclude that the dodecagon is regular.
Example 9. (IMO 1992, Shortlisted Problem from India) Show that in the
plane there exists a convex polygon of 1992 sides satisfying the following conditions:
(i) its sides lengths are 1,2,3,…,1992 in some order;
(ii) the polygon is circumscribable about a circle.
Solution. For a positive number r, let us
draw a circle of radius r and let us draw a polygonal path A1A2…A1993 such that for i=1 to 1992, side AiAi+1 is tangent to the circle at a point Ti and T1992A1993 =
A1T1, T1A2 = A2T2, … , T1991A1992 = A1992T1992. O T1 A1 A2 A1992 A1993 T1992 T1991 T2
To achieve condition (i), we need A1A2,
A2A3, …, A1992A1993 to be a permutation of 1, 2, …, 1992. This can be done as follow:
If i≡1 (mod 4), then let AiTi=1/2. If i≡3 (mod 4), then let AiTi=3/2. If i≡0,2 (mod 4), then let AiTi=i−3/2. We can check that the lengths of AiAi+1 for i=1 to 1992 are 1, 2, 4, 3, 5, 6, 8, 7,…, 1989,1990,1992,1991.
To achieve condition (ii), we define a function . arctan 2 ) ( 1992 1 1992 1 1
i i i i i i r T A OA A r fObserve that f (r) is a continuous function on (0,∞). As r tends to 0, f (r) tends to infinity and as r tends to infinity, f (r) tends to 0. By the intermediate value theorem, there exists r such that f (r) = 2π. Then
A1993=A1 and A1A2…A1992 is a desired polygon.
We remark that if 1992 is replaced by other positive integers of the form 4k, then there are such 4k-sided polygon.
.). . ' .
1. From a2 + b
I
~2b +a it follows .th11t~
2~
bl
(a2b
+a,)-' b(a2 + 6)=
~-""'
b2. _.··. ' . '' . . \ . . . .
From b2 - a 1 ab2+ b it follows that
b~-
aI
(ab2+.6)-"
a(b2 :._a)=·b
+ a2. ·..:' ...
.Hence
we
hii.ve'a2+b I a.-b21 a2+b. This Di~~ri.s that ~2+bis
equ~(to ~:-p2,up to sign: We distinguish tWo case8: a2
+
b.= b2-.a_
~J.tid, a2 +b.~. a~ li~,,:rn
the latter ~a.s~·we have a2 + b2 =:/a~ b;. But a2 ~ a, and)J.~ ~-b
..
> .:::b, 4!=!nc~this is iiilpb'ssible. Thei'\3fore w~ tnust be iiithe former .. ca8e: a2t-b.=: b,?-
-'-a;
·This yields a?.-"'
l:P
::.t'-(1;.:.. b1 hence {<a b)(a-:- b) =_-(a:_fb).;·A~:a+,b}spositive, we
may
divide byit and·w~_geta;-
_b ~,:::l, he.pce.b_ ~ q. ~-hAllpairs that could ·possibly satisfy. the ~omlit~miS. ar~ of the f(l~~. (a;
a+
1) fot'a:positive i:hteger a. d , \ 'i•. · ·· We consider these pai.Is. We h~ve a2+b=
a2+~+1. and a2b+a=
a2(a+1)+a =a3
+
a2 +a=
a( a2 +a+. 1), hence. the first di'YislWity. conditiq,n,js satisfie<;L ·Furthetmdte, •we have b~ :"'a.= (a +1)~1~}i T a2
+.at
1;,~~:a_b'?-;+b =.a(a+1)2+(ci+l) =a3+2a2+2a+l= a(!k2
+a-i-1)+a2;l:-a+.l =,{a:+:J.,},(q2tafl),
hence also the second; divisibility: condition i§ ~a.tisfied.., 1H~D:9P
..
WE!.
P.!rlis
(a,
a+
1) satisfy the: conditions .~d theyiate .. ex.actly the i>aitS._sati,sfyihg the conditions. , · · ·. · : : ·. · · . • . · · · · . o. ;
2.
we'
applyM.~J;lelaos's
theorem to. the line through A, E andM
inside triangle.BeD'.
This yields ' , · . · . ·. r. · •. · · . . ·· · . .. . . .
IBMI IC'EI · ·. jDAl . ·
IMCI : IEDI ·,
r!ABI
= 1. . : ,',BecaUse
M
iS the ril.idpoint of EC;We
'h~~ j!~l
=
1.Flirtherm~re;.
it is given thatIADI
=
IDEI.
Altogether this yieldsICE!= IABI.
o·. ·i
26
i. II
f
i. ·jl
1
} 3. We have . 1 1 (b+ac)+(a+bc) a+b+ac+bc - - + - -==
-a+ be b + ac (a+ bc)(b + ac) ab + a2c+ b2c+ abc2 ·
Hence, the problem statement~s equality-yields
(a+ b)(a + b + ac+ be)= ab+ a2c+ b2c + abc2
,
or equivalently,
;;,2 + ab + a2c+ abc+ ab + b2 +abc+ b2c = ab + a2c+ b2c + abc2,
or equival(lntly,
We can consider this as a quadratic equation in a, of whi~h we know that it has a rational solution. Hi:nice, the discriminant of this equation must be the square of a rational riur.iJ.bet:. This discrin:iinant equals
D = (2bc + b ~ br:.,2)2 - 4b2 = (2bc+ b~ bc2 - 2b)(2bc+ b- bc2 + 2b) · ,;, b2(2c
-1:·~·~
2)(2c
+.3- c2}. =
b2(c2 -:-2c+ 1)(c2- 2c- 3) ···:i.: b2(c.:..1}2(c+
1)(c~ 3).If c = 1, then the original eqtiatibn becomes
: .. · ,_
''L.
1 1-· -·-· +-·-=--·
·. a.+b a+b a+b'
. whicli has
no sblutiori. If d~-
01
then the original equation becomes· ,, ·. '1' '1 . t. · ..
·:-"-+-·=-'-.:,be b. b.'
which illsohas no solution.
H~.tice, ~ ~
1 and a t-0. In particular, ·(c~ ~)(~+
1
)~ b2(c~
;)2
. ·tritist
b~
the'~q~are
ofai:atiori~{
number ... . .
0
1;,1,:::~t
Q *..A be theintersectio~
ofAD
with the circutricircle ofllABC.
'"'':
'·~hen v.ie haveIAGI
=:-2IAEI
becaU;'le the projection of the cent-re of a. circleon a chord of this circle is the midpoint of that chord. Let M pe the
midpoint of AG. Because IABI =
l4fl
= IAMI and because AD is the anglebisector of LBAM, we have that M is the image of B under reflection
in AD. Now we have LDGG
=
LAGC=
LABC = LABD = LDMA=
180°- LDMC, hence DMCG is a cyclic quadrilateral. Then we have
AM2
=
AM/0 = AD/a =AD. A,E, hence AAME N AADM (SAS).Now we have 180°-LEMC = LEMA
=
LMDA=
LBDA=
LCDF=
LDFG
=
LEFC, hence EMCF is a cyclic quadrilateral. Hence, LEBF=
LEMF
=
LECF, which is exactly what we needed to prove. D5. Number the rows from 1 up to 2014 and also number the columns. Consider
the follmving beginning situation: in row i the light bulbs in columns i,
i + 1, ... , i + 1005 are on and the rest are off; in which we take the column
numbers modulo 2014. In each row .and ift'.each column there are exactly 1006 light bulbs that are on. Hence, there is no move possible. ·It is not
always possible to get
tO
a situation in which less than 2014. 1006 lightbulbs are on: · ·
Now we shall show that it is always possible to get to a situation in which at most 2014 · 1006 are on. Suppose, from the contrary, that in a certain
situation at least 2014.·1006 +.1 light bulbs l}.re Ol}. ~d it is not possible to
. get to a situation in wliich less bulbs are on. If th~re. is a row."or c~~umrj. in
which at least 1008 bulbs are on, theri we. can clJ.~gE) the state13 ·of these
bulbs in this row or column and there. will
be
less bulbs thatare
on. Thisis a contradiction, hence fu each row and c?lumn at mos.t
:iooi
buibs are ·on.·
Let· I be the set of rows in which exactly 1007 bulbs are on and let J1
be the set of columns in which exactly 1QQ.'z: bulbs are. in. W.e wi,ll try to
change the state of the bulbs in all rows iri: I. This w~ call th~ big plan.
If after 'executing the big plan .. there is a:· column in whicl;li at leW?t _1098 bulbs are on, then we get a contradiction> Hence. we shall assume· that tliis
does not happen. Let J2 be the set of coiumn8 that, ·afte:~; executing the .
big plan, contain exactly 1007 bulbs that ate on .. If ti:J.ere eXists a square
( i, j) with i E I and j E Jl containing a bulb that is off,we can switch row
i and then column j gets more than 1007 bulbs that are on, which is a
contradiction. Hence every bulb on ( i, j) with i E I and j E J1 is on. If
there exists a square ( i, j) with i E I and j E J2 containing a bulb that is
off, after executing the big plan, then we get a contradiction fu the same way. Hence every bulb on (i,j) with i E I and j E J2 is off after executing
the big plan. Because the colunuis in h contain exactly 1007 bulbs tha~
are on after executing the big plan, there are 1001 -III bulbs that are on
before executing the big plan. (For a set X, we denote by lXI its number
28
l
r
.·-.
of elements.) In the col~mns of J 1 there are exactly 1007 that are on and
this also me!J.llS that J1 and J 2 are.·disjoint. In the other columns at ~ost
109p b1Jlb:o. are qn. The tota1. nu!llber of bulbs that are on before the big
pl~n is at most . ·.. . . . .
(1007-IIDihi+1D071Jlj+1006(2014-IJli-IJ21) = 1006·2014+1Jli+I!2HIIV21·
This has 'to be at least 1006. 2in4 + 1, hence IJll + IJ21-III·IJ21 :?: 1. This
yields lJ1I > (jij-l)jJ2I· If jij;:>; 2, ·then IJ1l > lhl. By executing the big plan, the number of cohimn8 iii which 1007 bulbs are on decreases. · But
after that, we again have a situation with ljj rows containing 1007 bulbs
that· are on,. and in which J1 and J2·have been interchanged. Then we can
again apply the. big plan, to deerease the number of columns that contain 1007 bulbs that are on again. Tijis is a contradiction, because now we are .
back in the beginning situation. We conclude that we mu~t have III = 1.
We have a." situation in which·there is exactly one row with exadly .1007
· bulbs that are on. Analogously, we can show that there must be exa9tly one ·column in which exactly 1007· bulbs are on. Because there are 1006·2014+1 bulbs that are on, each other row and column must contain exactly 1006 bulbs that are on. Now change the state of the bulbs in the row with 1007
bulbs that are on. fu 1007 col~mns there will be 1006 + 1 = 1007 bulbs that
are on. We have already seen that in this situation we can decrease the
number of bulbs.that are on... '
We conclude that if there are more than 1006 · 2014 bulbs that are ·on
.
.
'
it is always possible to decrease this number. Hence, the smallest k is
1006·2014. 0 29 .· .. ~ ·:~ .'
