Volume 12, Number 1 March 2007 – April 2007
From How to Solve It to
Problem Solving in Geometry
K. K. Kwok
Munsang College (HK Island)
Olympiad Corner
Below are the 2007 Asia Pacific Math Olympiad problems.
Problem 1. Let S be a set of 9 distinct integers all of whose prime factors are at most 3. Prove that S contains 3 distinct integers such that their product is a perfect cube.
Problem 2.
Let ABC be an acute angled triangle with ∠BAC=60oand AB > AC.
Let I be the incenter, and H be the orthocenter of the triangle ABC. Prove that 2∠AHI =3∠ABC.
Problem 3. Consider n disks C1, C2, ⋯,
Cn in a plane such that for each 1 ≤ i < n,
the center Ci is on the circumference of
Ci+1, and the center of Cn is on the
circumference of C1. Define the score of
such an arrangement of n disks to be the number of pairs (i, j) for which Ci
properly contains Cj. Determine the
maximum possible score.
Problem 4. Let x, y and z be positive real numbers such that x+ y+ z = 1. Prove that . 1 ) ( 2 ) ( 2 ) ( 2 2 2 2 2 2 2 ≥ + + + + + + + + y x z xy z x z y zx y z y x yz x (continued on page 4)
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 May 31, 2007.
For individual subscription for the next five issues for the 05-06 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]
Geometry is the science of correct reasoning on incorrect figures.
If you can’t solve a problem, then there is an easier problem you can solve, find it.
George Pölya 幾何是：在靜止中看出動態，從變幻 中覓得永恆
數學愛好者，強
Example 1. In the figure below, C is a
point on AE. ΔABC and ΔCDE are equilateral triangles. F and G are the midpoints of BC and DE respectively. If the area of ΔABC is 24 cm2, the area of
ΔCDE is 60 cm2, find the area of ΔAFG.
Idea and solution outline:
This question is easy enough and can be solved by many different approaches. One of them is to recognize that the extensions of AF and CG are parallel. (Why? At what angles do they intersect line AE?) Thus [AFC] = [AFG].
Example 2. In ΔABC, AB = AC. A
point P on the plane satisfies ∠ABP = ∠ACP. Show that P is either on BC or on the perpendicular bisector of BC. Solution:
Apply the sine law to ΔABP and ΔACP, we have sin ∠APB =AB sin∠ABP AP = AC sin∠ACP AP = sin ∠APC.
Thus, either ∠APB = ∠APC or ∠APB + ∠APC = 180o. The first case implies
ΔABP ≅ ΔACP, so BP = CP and P lies on the perpendicular bisector of BC. The second case implies P lies on BC.
Example 3.[Tournament of Towns1993]
Vertices A, B and C of a triangle are connected to points A′, B′ and C′ lying in their respective opposite sides of the triangle (not at vertices). Can the midpoints of the segments AA′, BB′ and CC′ lie in a straight line?
Solution outline:
Let D, E and F be midpoints of BC, AC, and AB respectively. Given any point A′ on BC, let AA′ intersect EF at A′′. Then it is easy to see that A′′ is indeed the midpoint of AA′.
Therefore, the midpoints of the segments AA′, BB′ and CC′ lie respectively on EF, DF and DE, and cannot be collinear.
Example 4. [Tournament of Towns
1993] Three angles of a non-convex, non-self-intersecting quadrilateral are equal to 45 degrees (i.e. the last equals 225 degrees). Prove that the midpoints of its sides are vertices of a square. Idea:
Do you know a similar, but easier problem? For example, the famous Varignon Theorem: By joining the midpoints of the sides of an arbitrary quadrilateral, a parallelogram is formed. G F D B A C E B C A F E D A' A''
Mathematical Excalibur, Vol. 12, No. 1, Mar. 07 – Apr. 07 Page 2 1 x 1 1 n n m m F A E D B C Solution outline:
Extend BC to cut AD at O. Then ΔOAB
and ΔOCD are both isosceles
right-angled triangle. It follows that a 90o rotation about O will map A into B
and C into D, so that AC = BD and they are perpendicular to each other.
Example 5. [Tournament of Towns
1994] Two circles intersect at the points A and B. Tangent lines drawn to both of the circles at the point A intersect the circles at the points M and N. The lines BM and BN intersect the circles once more at the points P and Q respectively. Prove that the segments MP and NQ are equal.
Idea:
MP and NQ are sides of the triangles ΔAQN and ΔAMP respectively, so it is natural for us to prove that the two triangles are congruent. It is easy to observe that the two triangles are similar, so what remains to prove is either AQ = AM or AP = AN. Note that we can transmit the information between the two circles by using the theorem on alternate segment at A. Solution outline:
(1) Observe that ΔAQN ∼ ΔAMP. (2) AP = AN follows from computing ∠APN = ∠APB + ∠BPN
=∠ANB+∠BAN[∠s in same segment] = ∠ANB + ∠AQN [∠ in alt. segment] = 180o − ∠QAN
= 180o − ∠MAP [by step (1)]
= ∠AMB + ∠APB
= ∠AMB + ∠MAB [∠ in alt. segment] = ∠ABP [ext. ∠ of Δ]
= ∠ANP [∠s in the same segment].
Example 6. ABCD is a trapezium with
AD // BC. It is known that BC = BD = 1, AB = AC, CD < 1 and ∠BAC + ∠BDC = 180o, find CD.
Idea:
The condition ∠BAC + ∠BDC = 180o
leads us to consider a cyclic quadrilateral. If we reflect ΔBDC across BC, a cyclic quadrilateral is formed.
Solution outline:
(1) Let E be the reflection of D across BC. (2) ∠BAC + ∠BDC = 180o ⇒ ∠BAC + ∠BEC = 180o ⇒ ABEC is cyclic, AD // BC ⇒ AF = FE, AB = AC ⇒ ∠BEF = ∠FEC ⇒ EC BE EC BF FC = = . (3) Let AF = FE = m, AB = AC = n and DC = EC = x. It follows from Ptolemy’s theorem that AE×BC = AC×BE + AB×EC, i.e. 2m = n (1 + x). Now BF FC BF BF BC BF BE AF AC m n x + = = = = = + 1 2 x BE EC BF FC + = + = + =1 1 1 , i.e. (1 + x)2 = 2. Therefore,x= 2−1.
Example 7. [Tournament of Towns 1995]
Let P be a point inside a convex quadrilateral ABCD. Let the angle bisector of ∠APB, ∠BPC, ∠CPD and ∠DPA meet AB, BC, CD and DA at K, L, M and N respectively. Find a point P such that KLMN is a parallelogram.
Idea:
The angle bisector theorem enables us to replace the ratios that K, L, N and M divided the sides of the quadrilateral by the ratios of the distance from P to A, B, C and D. For instance, we have
BP AP KB AK = and DP AP ND AN = If BP = DP, we have ND AN KB AK = and
hence KN//BD. Similarly, we have LM//BD and so KN//LM.
Therefore, we shall look for a point P such that BP = DP and AP = CP. Solution outline:
(1) Let P be the point of intersection of the perpendicular bisectors of the diagonals AC and BD. Then AP = CP and BP = DP.
(2) By the angle bisector theorem, we have ND AN DP AP BP AP KB AK = = = and so KN // BD. Similarly, LM//BD, KL//AC and MN//AC.
Hence KN//LM and KL//NM, which means that KLMN is a parallelogram. Remark: Indeed, point P in the solution above is the only point that satisfies the condition given in the problem.
Example 8. [IMO 2001] Let a, b, c, d
be integers with a > b > c > d
> 0.
Suppose thatac + bd = (b + d + a − c)(b + d − a + c). Prove that ab + cd is not prime. Remark: This is a difficult problem in number theory. However, we would like to present a solution aided by geometrical insights! (continued on page 4) Q P R S C A B D P Q N M B A B A D C P K L M N
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 May 31, 2007. Problem 271. There are 6 coins that look the same. Five of them have the same weight, each of these is called a good coin. The remaining one has a different weight from the 5 good coins and it is called a bad coin. Devise a scheme to weigh groups of the coins using a scale (not a balance) three times only to determine the bad coin and its weight.
Problem 272. ∆ABC is equilateral. Find the locus of all point Q inside the triangle such that
. 90o = ∠ + ∠ + ∠QAB QBC QCA
Problem 273. Let R and r be the circumradius and the inradius of triangle ABC. Prove that
. sin cos sin cos sin cos 2 2 2 r R C C B B A A + + ≥
(Source: 2000 Beijing Math Contest) Problem 274. Let n < 11 be a positive integer. Let p1, p2, p3, p be prime
numbers such that p pn 3 1+ is prime. If p1+p2=3p, p2 p3 p1(p1 p3) n + = + and p2>9, then determine p1p2p3n.
(Source: 1997 Hubei Math Contest) Problem 275. There is a group of children coming from 11 countries (at least one child from each of the 11 countries). Their ages are from 7 to 13. Prove that there are 5 children in the group, for each of them, the number of children in the group with the same age is greater than the number of children in the group from the same country.
*****************
Solutions
****************
Problem 266. Let
N = 1+10+102+⋯+101997.
Determine the 1000th digit after the
decimal point of the square root of N in base 10. (Source: 1998 Putnam Exam)
Solution. Jeff CHEN (Virginia, USA),
Irfan GLOGIC (Sarajevo College, 4th grade, Sarajevo, Bosnia and Herzegovina), Salem MALIKIĆ (Sarajevo College, 3rd grade, Sarajevo, Bosnia and Herzegovina), Anna Ying PUN (HKU, Math, Year 1) and Fai YUNG.
The answer is the same as the unit digit of . 101000 N We have . 3 10 10 9 1 10 10 10 2000 3998 1998 1000 1000 N= − = − Since (101999−7)2 <103998−102000 <(101999−4)2,
so it follows that 101000 N is between (101999−7)/3=33⋯31 and (101999−4)/3=
33⋯32. Therefore, the answer is 1. Commended solvers: Simon YAU and YEUNG Wai Kit (STFA Leung Kau Kui College, Form 6).
Problem 267. For any integer a, set na = 101a − 100·2a.
Show that for 0 ≤ a, b, c, d ≤ 99, if na+ nb ≡ nc+ nd (mod 10100),
then {a,b}={c,d}. (Source: 1994 Putnam Exam)
Solution. Jeff CHEN (Virginia, USA),
Irfan GLOGIC (Sarajevo College, 4th grade, Sarajevo, Bosnia and Herzegovina), Salem MALIKIĆ (Sarajevo College, 3rd grade, Sarajevo, Bosnia and Herzegovina), Anna Ying PUN (HKU, Math, Year 1) and YEUNG Wai Kit (STFA Leung Kau Kui College, Form 6).
If na+ nb ≡ nc+ nd (mod 10100), then a+b
≡ na+ nb ≡ nc+ nd ≡ c+d (mod 100) and
2a+2b≡ n
a+ nb ≡ nc+ nd ≡2c+2d(mod 101).
By Fermat’s little theorem, 2100 ≡ 1 (mod
101) and so 2a+b ≡ 2c+d (mod 101). Next
(2a −2c)(2a −2d) = 2a(2a −2c −2d)+2c+d
≡ 2a(−2b)+2a+b = 0 (mod 101).
So 2a ≡ 2c (mod 101) or 2a ≡ 2d (mod 101).
Now we claim that if 0 ≤ s ≤ t ≤ 99 and 2s ≡
2t (mod 101), then s=t. To see this, let k be
the least positive integer such that 2k≡1
(mod 101). Dividing 100 by k, we get 100 = kq+r with 0 ≤ r < k. Since 2r = 2100−kq ≡ 1
(mod 101) too, so r = 0, then k is a divisor of 100.
Clearly, 1 < 21, 22, 24, 25 < 101 and 210 =
1024 ≡ 14 (mod 101), 220 ≡ 142 ≡ −6
(mod 101), 225≡ (−6)32 ≡ 10 (mod 101),
250 ≡ 102 ≡ − 1 (mod 101). Hence
k=100. Finally 2t−s ≡ 1 (mod 101) and
0 ≤ t−s < 100 imply t−s=0, proving the claim.
By the claim, we get a=c or a=d. From a+b ≡ c+d (mod 100) and 0 ≤ a, b, c, d ≤ 99, we get a = c implies b = d and similarly a = d implies b = c. The conclusion follows.
Problem 268. In triangle ABC, ∠ ABC = ∠ ACB = 40˚. Points P and Q are inside the triangle such that
∠ PAB = ∠ QAC = 20˚ and ∠ PCB =∠ QCA = 10˚. Must B, P, Q be collinear? Give a proof. (Source: 1994 Shanghai Math Competition)
Solution. Jeff CHEN (Virginia, USA),
Courtis G. CHRYSSOSTOMOS (Larissa, Greece, teacher), Irfan GLOGIC (Sarajevo College, 4th grade, Sarajevo, Bosnia and Herzegovina), Kelvin LEE (Winchester College, England) Salem MALIKIĆ (Sarajevo College, 3rd grade, Sarajevo, Bosnia
and Herzegovina) and NG Ngai Fung (STFA Leung Kau Kui College). Since lines AP, BP, CP concur, by the trigonometric form of Ceva’s theorem,
, 1 sin sin sin sin sin sin = ∠ ∠ ∠ ∠ ∠ ∠ PCB PAC PBA PCA BAP CBP which implies . 1 2 20 sin 10 sin 10 cos 30 sin 20 sin 10 sin 80 sin sin sin = = = ∠ ∠ o o o o o o o PBA CBP So ∠CBP=∠PBA=20o. Replacing P
by Q above, we similarly have . 1 10 sin 80 sin 30 sin 20 sin sin sin = = ∠ ∠ o o o o QBA CBQ So ∠QBA=∠CBQ=20o. Then B, P,
Q are on the bisector of ∠ABC.
Commended solvers: CHIU Kwok Sing (Belilios Public School), FOK Pak Hei (Pui Ching Middle School), Anna Ying PUN (HKU, Math, Year 1) and Simon YAU.
Problem 269. Let f(x) be a polynomial with integer coefficients. Define a sequence a0, a1, … of integers such that
a0 = 0, an+1 = f (an) for all n ≥ 0. Prove
that if there exists a positive integer m for which am = 0, then either a1 = 0 or
Mathematical Excalibur, Vol. 12, No. 1, Mar. 07 – Apr. 07 Page 4 a2 = 0. (Source: 2000 Putnam Exam)
Solution. Irfan GLOGIC (Sarajevo
College, 4th grade, Sarajevo, Bosnia
and Herzegovina), Salem MALIKIĆ (Sarajevo College, 3rd grade, Sarajevo,
Bosnia and Herzegovina) and Anna Ying PUN (HKU, Math, Year 1). Observe that for any integers m and n, m−n divides f(m)−f(n) since for all nonnegative integer k, mk−nk has m−n
as a factor. For nonnegative integer n, let bn=an+1−an, then by the last
sentence, bn divides bn+1 for all n.
Since a0 = am = 0, a1 = am+1 and so
b0=bm. If b0 = 0, then a1 = am+1 = bm+am
= 0.
If b0 ≠ 0, then using bn divides bn+1 for
all n and b0=bm, we get bn = ± b0 for
n=1,2,⋯,m. Since b0+b1+⋯+bm =
am−a0 = 0, half of the integers b0, ⋯, bm
are positive and half are negative. Then there is k < m such that bk−1 = −bk,
which implies ak−1=ak+1. Then am=am+2
and so 0=am=am+2=f(f(am))=f(f(a0))=a2.
Problem 270. The distance between any two of the points A, B, C, D on a plane is at most 1. Find the minimum of the radius of a circle that can cover these four points. (Source 1998 Tianjin Math Competition)
Solution. Jeff CHEN (Virginia, USA).
Case 1: (one of the point, say D, is inside or on a side of
Δ
ABC
) IfΔ
ABC
is acute, then one of the angle, sayo
60 ≥
∠BAC . By the extended sine law, the circumcircle of
Δ
ABC
covers the four points with diameter. 3 2 sin 2 ≤ ∠ = BAC BC R
(Note equality occurs in case
ABC
Δ
is equilateral.) IfΔ
ABC
is right or obtuse, then the circle using the longest side as diameter covers the four points with R ≤ 1/2.Case 2: (ABCD is a convex quadrilateral) If there is a pair of opposite angles, say angles A and C, are at least 90°, then the circle with BD as diameter will cover the four points with R ≤ 1/2. Otherwise, there is a pair of neighboring angles, say angles A and B, both of which are less than 90°. If ∠ADB≥∠ACB≥90o, then the
circle with AB as diameter covers the four points and radius R ≤ 1/2.
If ∠ADB≥∠ACB and ∠ACB<90o,
then D is in or on the circumcircle of ABC
Δ with radius R≤1 3as in case 1. So summarizing all cases, we see the minimum radius that works for all possible arrangements of A,B,C and D is R = 1 3.
Commended solvers: NG Ngai Fung (STFA Leung Kau Kui College) and Anna Ying PUN (HKU, Math, Year 1).
Olympiad Corner
(continued from page 1) Problem 5. A regular (5×5)-array of lights is defective, so that toggling the switch for one light causes each adjacent light in the same row and in the same column as well as the light itself to change state, from on to off, or from off to on. Initially all the lights are switched off. After a certain number of toggles, exactly one light is switched on. Find all possible positions of this light.
From How to Solve It to
Problem Solving in Geometry
(continued from page 2) Idea:
Observe that
ac + bd = (b + d + a − c)(b + d − a + c) ⇔ a2 + c2 − ac = b2 + d2 + bd
The last equal
ity
suggests one to think about usingthe cosine law as follow: a2+ c2 − 2accos 60o= a2 + c2 − ac
= b2 + d2 + bd
= b2 + d2 − 2bd cos 120o.
Solution:
(1) Lemma: Let x, y, and z be positive integers with z < x and z < y. If xy/z is an integer, then xy/z is composite.
[Can you prove this lemma? Is there any trivial case you can see immediately? How about proving the lemma by mathematical induction in z?]
The case z = 1 is trivial. In case z > 1, inductively suppose the lemma is true for all positive integers z’ less than z. Then z has a prime divisor p, say z = pz’. Since xy/z is an integer, either p divides x or p
divides y, say p divides x. Then x = px’. So xy/z=x’y/z’ with z’ < x’ and z’ < z < y. By the induction hypothesis, xy/z=x’y/z’ is composite.
(2) The equality
ac + bd = (b + d + a − c)(b + d − a + c) is equivalent to
a2 + c2 − ac = b2 + d2 + bd
In view of this, we can construct cyclic quadrilateral ABCD with AB = a, BC = c, CD = d, DA = b, ∠ABC = 60o and
∠ADC = 120o.
(3) Considering the ratios of areas and using Ptolemy’s theorem, we have
bd ac cd ab BD AC + + = and AC×BD = ad + bc. (4) Therefore, BD AC AC BD AC bd ac cd ab × = = + + 2 bc ad ac c a + − + = 2 2 , which implies bc ad ac c a bd ac cd ab + − + + = + ( )( 2 2 ) (*).
(5) To get the conclusion from the lemma, it remains to show
ad + bc < ac +bd and ad + bc < a2 + c2 − ac. Now (ac + bd ) − (ad + bc) = (a − b)(c − d) > 0 ⇒ ad + bc < ac + bd. Also, (ab + cd) − (ac + bd) = (a − d)(b − c) > 0 ⇒ ab + cd > ac + bd ⇒ ad + bc < a2 + c2 − ac (by (*)).
Now the result follows from the lemma. c d b a B D C A
Let
It suffices to show that there are 1 5 Z ~ , i ~ , i 3 5 9 such that
S = {2*3'
I @,bi
E Z, @, bi2
0, 15 i 5 9).oil
+
ai, +air=
bi,+
bi,+
bi,=
0 (mod 3 ) . I(t)
For n = 293' E S, let's call (a (mod 3 ) , b (mod 3 ) ) the type of TI. Then there are 9 possible .
types :
Let N ( i , j ) be the number of integers in S of type (i,j). We obtain 3 distinct integers
whose product is a perfect cube when
(0l0)l(0,1)1(0l2),(1,0), (11 11, (1, 21, (2,0),(2,1)1(2,2).
(1) N ( i , j )
2
3 for some i,j, or( 2 ) N ( i , O)N(i, l) N ( i , 2)
#
0 for some i ='O, 1,2, or( 3 ) N ( O , j ) N ( l , j ) N ( 2 , j )
#
0 for some j = 0,1,2, or(4) N ~ i l , j l ) N ( i 2 , j 2 ) N ( i 3 r j 3 )
#
0, where { i 1 2 i 2 ~ i 3 } = { j l l h l j 3 } = {0,1,2}. .Assume that none of the conditions ( 1 ) - ( 3 ) holds.' Since N ( i , j )
5
2 for all ( i , j ) , there are at least five N(i,j)'s that are nonzero. Furthermore, among those nonzero N(i, j)k, no three have the same i nor the same j. Using these facts, one may easily conclude that 'the condition (4) should hold. (For example, if onephceseach nonzero N ( i , j ) in the (i,j)-th box of a regular 3 x 3 array of boxes whose rows and columns are indexed by 0,l. and 2,then one can always find three box@, occupied by at least one nonzero N ( i , j ) , whose rows and columns are all distinct. This implies (49.)
@
Solution. Let D be the intersection point of the lines AH and BC. Let K be the intersection point of the circumcircle 0 of the triangle ABC and the line AH. Let the line through I perpendicular to BC meet BC and the minor arc BC of the circumcircle 0 at E and N , respectively. We have1 1
2 2
LBIC = 180"
-
(LIBC+
LICB) = 180"-
-(LABC+
LACB) = 90"+
-LBAC = 120" and also LBNC = 180"-
LBAC = 120" = LBIC. Since IN I BC, the quadrilateralBICN is a kite and thus IE = EN.
Now, since H is the orthocenter of the triangle ABC, H D = D K . Also because E D
I
ZN and, ED I HK, we conclude that ZHKN is an isosceles trapezoid withZH = N K .
Hence
LAHZ = 180"
-
LZHK = 180"-
LAKN = LABN.Since I E = EN and BE I I N , the triangles ZJ3E and NBE are congruent. Therefore 1
2
LNPE = LIBE = LIBC = LIBA = -LABC
and thus
3
LAHI = LABN = XLABC.
configuration C = {Cl,
..
.
,
C,,}, let SC = {(i,j)I
Ci properly contains Cj }. So, the score of an n-configuration C isIScl.
We'll show that (i) there is an n-configuration C for which
lScl
= (h-
l ) ( n-
2 ) / 2 , and that (ii)lScl
5
(n-
l)(n-
2 ) / 2 for any n-configuration C.Let C1 be any disk. Then for i = 2 , .
.
. ,
n-
1, take Cj inside Cj-1 so that the cir- cumference of Ci contains the center of Finally, let C,, be a disk whose center is on the circumference ofCl
and whose circumference contains the center of C,,-l. This givesSC
= { ( i , j ) 115
i<
j5
n-
1) of size (n-
l ) ( n-
2 ) / 2 , which proves (i).For any n-configuration C, SC must satisfy the following properties: (1) (ili)
f2
sc
I( 2 ) (i
+
114f2
SCl (1,
n) 9,sc
I.
( 3 ) if (i,j), (j, k) E SC, then (i, k) E Sc
,
(4) if (i, j) E SC, then ( j , i) 9, SC
.
Now we show that a set G of ordered pairs of integers between 1 and n, satisfying the conditions ( 1 ) ~ ( 4 ) , can have no more than (n
-
l)(n-
2 ) / 2 elements. Suppose that thereexists a set G that satisfies the conditions (1)-(4), and has more than (n
-
l ) ( n-
2 ) / 2elements. Let n be the least positive integer with which there exists such a set G. Note that G must have (i, i
+
1) for some 15
i 5 n or (n, l), since otherwise G can have at most n(n-
3) 2 (n-
l)(n-
2 ) 2t)
-n=-<
elements. Without loss of generality we may assume that (n, 1) E
G.
Then (1, n-
1) f! GIsince otherwise the condition (3) yields (n, n- 1) E G contradicting the condition ( 2 ) . Now let G' = {(i, j) E G 11
5
i, j5
n- l } , then G' satisfies the conditions ( 1 ) ~ ( 4 ) , with n-
1.We now claim that IG
-
G'I 5 n-
2 :Supposethat IG-G'I > n - 2 , t h e n IG-G'I=n-landhenceforeachl < i < n - 1 ,
either (i, n) or (n, i) must be in G. We already know that (n, 1) E G and (n
-
1, n) EG
(because (n, n-
1)#
G) and this implies that (n, n-
2)#
G and (n-
2, n ) E G. If we keep doing this process, we obtain (1, n) E GI which is a contradiction.Since IG
-
G'I 5 n-
2, we obtain(n
-
2)(n-
3 )2
-
(n-
2) =(n
-
l)(n-
2 )1
6
1
L
.+
+
,&
1.I
Thus, it suflices to show that(x-Y)(x--z)+(Y-z)(Y-z)+(z-x)(.-Y)
>o.
J i q i T q
J
-
J--
Now, m ewithout loss of generality, that x 2 y 2 z. Then we have
+ +
7
and
Solution. We assign the following first labels to the 25 positions of the lights:
1 0 1 1
For each on-off combination of lights in the array, define its first value to be the sum of the first labels of those positions at which the lights are switched on. It is easy to check that toggling any switch always leads to an on-off combination of lights whose first value has the same parity(the remainder when divided by 2) as that of the previous on-off
The 90" rotation of the first labels gives us another labels (let us call it the second
Zukls) which also makes the parity of the second value(the s u m of the second labels of
those positions at which the lights are switched on) invariant under toggling.
(2) combination.
(3)
Since the parity of the first and the second values of the initial status is 0, after certain number of toggles the parity must remain unchanged with respect t o the first labels and the second labels 8s well. Therefore, if exactly one light is on after some number of toggles, the label of that position must be 0 with respect to both labels. Hence according to the above pictures, the possible positions are the ones marked with * i ' s in the following picture:
(4)
I I I I
Now we demonstrate-that i-five positions are possible :
- ..
Toggling the positions checked by t (the order of toggling is irrelevant) in the first picture makes the center(*o) the only position with light on and the second picture makes the position *1 the only position with light on. The other * i ' s can be obtained by rotating
the second picture appropriately.
(Y
-
4 ( x
- Y) -
(Y-
z)(x-
y) 1= (y
-
z)(z-
y)>J-
J
-The quantity is non-negative due t o the fact that
yZ(z
