Volume 13, Number 2 May-June, 2008
Geometric Transformations I
Kin Y. Li
Olympiad Corner
The following are the four problems of the 2008 Balkan Mathematical Olympiad.
Problem 1. An acute-angled scalene triangle ABC is given, with AC > BC. Let O be its circumcenter, H its orthocenter and F the foot of the altitude from C. Let P be the point (other than A) on the line AB such that
AF=PF and M be the midpoint of AC.
We denote the intersection of PH and
BC by X, the intersection of OM and FX by Y and the intersection of OF and AC by Z. Prove that the points F, M, Y
and Z are concyclic.
Problem 2.
Does there exist a sequence a1, a2, a3, …, an, … of positive real numbers satisfying both of the following conditions:
(i) 2, 1 n a n i i≤
∑
=for every positive integer n; (ii) 1 2008, 1 ≤
∑
= n i aifor every positive integer n ?
(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 August 20, 2008.
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]
Too often we stare at a figure in solving a geometry problem. In this article, we will move parts of the figure to better positions to facilitate the way to a solution.
Below we shall denote the vector from X to Y by the boldface italics XY. On a plane, a translation by a vector v moves every point X to a point Y such that XY = v. We denote this translation by T(v).
Example 1. The opposite sides of a
hexagon ABCDEF are parallel. If
BC−EF = ED−AB = AF−CD > 0, show
that all angles of ABCDEF are equal.
Solution. One idea is to move the side
lengths closer to do the subtractions. Let T(FA) move E to P, T(BC) move A to Q and T(DE) move C to R.
A B C D E F Q P R
Hence, EFAP, ABCQ, CDER are parallelograms. Since the opposite sides of the hexagon are parallel, P is on
AQ, Q is on CR and R is on EP. Then,
we get BC − EF = AQ − AP = PQ. Similarly, ED − AB = QR and AF − CD
= RP. Hence,
Δ
PQR is equilateral.Now,
∠
ABC =∠
AQC = 120°. Also,∠
BCD=∠
BCQ+∠
DCQ = 60° + 60°= 120°. Similarly,
∠
CDE =∠
DEF =∠
EFA =∠
FAB = 120°.Example 2. ABCD is a convex
quadrilateral with AD =BC. Let E, F be midpoints of CD, AB respectively. Suppose rays AD, FE intersect at H and rays BC, FE intersect at G. Show that
∠
AHF =∠
BGF.Solution. One idea is to move BC
closer to AD. Let T(CB) move A to I.
A B C D E F G H I
Then BCAI is a parallelogram. Since F is the midpoint of AB, so F is also the midpoint of CI. Applying the midpoint
theorem to ∆CDI, we get EF||DI.
Using this and CB||AI, we get
∠
BGF =∠
AID. From AI = BC = AD, weget
∠
AID =∠
ADI. Since EF || DI,∠
AHF =∠
ADI =∠
AID =∠
BGF.Example 3. Let M and N be the
midpoints of sides AD and BC of quadrilateral ABCD respectively. If
2MN = AB+CD, then prove that AB||CD.
Solution. One idea is to move AB, CD
closer to MN. Let T(DC) move M to E and T(AB) move M to F.
A B C D E F M N K
Then we can see CDME and BAMF are parallelograms. Since EC = ½AD = BF,
BFCE is a parallelogram. Since N is the
midpoint of BC, so N is also the midpoint of EF.
Next, let T(ME) move F to K. Then
EMFK is a parallelogram and MK = 2MN = AB+CD
= MF+EM = MF+FK.
So F, M, K, N are collinear and AB||MN. Similarly, CD||MN. Therefore, AB||CD.
Mathematical Excalibur, Vol. 13, No. 2, May-Jun. 08 Page 2 On a plane, a rotation about a center
O by angle α moves every point X to a
point Y such that OX = OY and ∠XOY
= α (anticlockwise if α > 0, clockwise
if α < 0). We denote this rotation by
R(O,α).
Example 4. Inside an equilateral
triangle ABC, there is a point P such that PC=3, PA=4 and PB=5. Find the perimeter of ∆ABC.
Solution. One idea is to move PC, PA,
PB to form a triangle. Let R(C,60°)
move ∆CBP to ∆CAQ. 5 4 3 3 3 5 A B C P Q
Now CP=CQ and ∠PCQ = 60° imply ∆PCQ is equilateral. As AQ = BP = 5,
AP = 4 and PQ = PC = 3, so ∠APQ =
90°. Then ∠APC =∠APQ +∠QPC = 90°+60° = 150°. So the perimeter of ∆ABC is o 150 cos 12 4 3 3 3AC= 2+ 2− =3 25+12 3.
For our next example, we will point out a property of rotation, namely
P B1
O
B A1 A
if R(O,α) moves a line AB to the line A1B1 and P is the intersection of the
two lines, then these lines intersect at an angle α.
This is because ∠ OAB= ∠ OA1B1 implies O,A,P,A1 are concyclic so that ∠BPB1=∠AOA1=α.
Example 5. ABCD is a unit square.
Points P,Q,M,N are on sides AB, BC,
CD, DA respectively such that AP + AN + CQ + CM = 2.
Prove that PM⊥QN.
Solution. One idea is to move AP, AN
together and CQ, CM together. Let
R(A,90°) map B→D, C→C1, D→D1, Q→Q1, N→N1 as shown below. A B C D C1 D1 P M N Q Q1 N1
Then AN=AN1 and CQ=C1Q1. So
PN1= AP+AN1=AP+AN = 2−(CM+CQ) = CC1−(CM+C1Q1) = MQ1.
Hence, PMQ1N1 is a parallelogram and
MP||Q1N1. By the property before the example, lines QN and Q1N1 intersect at 90°. Therefore, PM⊥QN.
Example 6. (1989 Chinese National
Senoir High Math Competition) In
∆ABC, AB > AC. An external bisector of ∠BAC intersects the circumcircle of
∆ABC at E. Let F be the foot of perpendicular from E to line AB. Prove that
2AF = AB−AC.
Solution. One idea is to move AC to
coincide with a part of AB. To do that, consider R(E,∠CEB). C B A T E F D
Observe that ∠EBC=∠EAT=∠EAB= ∠ECB implies EC=EB. So R(E,∠CEB)
move C to B. Let R(E,∠CEB) move A to
D. Since ∠CAB =∠CEB, by the property
above and AB > AC, D is on segment AB.
So R(E,∠CEB) moves ∆AEC to ∆DEB.
Then ∠DAE =∠EAT =∠EDA implies ∆AED is isosceles. Since EF⊥AD,
2AF=AD=AB −BD=AB −AC. ****************
On a plane, a reflection across a line moves every point X to a point Y such that the line is the perpendicular bisector of segment XY. We say Y is the mirror image of X with respect to the line.
Example 7. (1985 IMO) A circle with
center O passes through vertices A and C of ∆ABC and cuts sides AB, BC at K, N respectively. The circumcircles of ∆ABC and ∆KBN intersect at B and M. Prove that ∠OMB = 90°.
Solution. Let L be the line through O
perpendicular to line BM. We are done if we can show M is on L. L O C A B K N M C' K'
Let the reflection across L maps C
→C’ and K →K’. Then CC’⊥L and
KK’⊥L, which imply lines CC’, KK’, BM are parallel. We have
∠KC’C=∠KAC = ∠BNK=∠BMK,
which implies C’,K,M collinear. Now ∠C’CK’= ∠CC’K=∠CAK
= ∠CAB=180 ° −∠BMC = ∠C’CM,
which implies C,K’,M collinear. Then lines C’K and CK’ intersect at M. Since lines C’K and CK’ are symmetric with respect to L, so M is on L.
Example 8. Points D and E are on
sides AB and AC of ∆ABC respectively with∠ABD = 20°, ∠DBC = 60°, ∠
ACE = 30° and ∠ECB = 50°. Find
∠EDB.
Solution. Note ∠ ABC= ∠ ACB.
Consider the reflection across the perpendicular bisector of side BC. Let the mirror image of D be F. Let BD intersect CF at G. Since BG = CG, lines BD, CF intersect at 60° so that ∆BGC and ∆DGF are equilateral. Then DF=DG. 20° 30° 60° 50° A B C D E F G We claim EF = EG (which implies ∆EFD ≅ ∆EGD. So ∠EDB
= ½ ∠ FDG = 30°).
For the claim, we have ∠EFG=∠CDG = 40°
and ∠FGB = 120°. Next ∠BEC = 50°. So
BE=BC. As ∆BGC is
equilateral, so BE = BC = BG .This gives ∠EGB = 80 °. Then
∠EGF =∠FGB −∠EGB = 40° =∠EFG ,
which implies the claim.
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 August 20,
2008.
Problem 301. Prove that it is possible to decompose two congruent regular hexagons into a total of six pieces such that they can be rearranged to form an equilateral triangle with no pieces overlapping.
Problem 302. Let ℤ denotes the set of all integers. Determine (with proof) all functions f:ℤ→ℤ such that for all x, y in ℤ, we have f (x+f (y)) = f (x) − y. Problem 303. In base 10, let N be a positive integer with all digits nonzero. Prove that there do not exist two permutations of the digits of N, forming numbers that are different (integral) powers of two.
Problem 304. Let M be a set of 100 distinct lattice points (i.e. coordinates are integers) chosen from the x-y coordinate plane. Prove that there are at most 2025 rectangles whose vertices are in M and whose sides are parallel to the x-axis or the y-axis.
Problem 305. A circle Γ2 is internally tangent to the circumcircle Γ1 of ∆PAB at P and side AB at C. Let E, F be the intersection of Γ2 with sides PA, PB respectively. Let EF intersect PC at D. Lines PD, AD intersect Γ1 again at G, H respectively. Prove that F, G, H are collinear.
*****************
Solutions
****************
Problem 296. Let n > 1 be an integer. From a n×n square, one 1×1 corner square is removed. Determine (with proof) the least positive integer k such that the remaining areas can be partitioned into k triangles with equal areas.
(Source 1992 Shanghai Math Contest)
Solution. Jeff CHEN (Virginia, USA), O
Kin Chit Alex (GT Ellen Yeung College), PUN Ying Anna (HKU Math Year 2), Simon YAU Chi-Keung (City University of Hong Kong) and Fai YUNG.
A B C
The figure above shows the least k is at most 2n+2. Conversely, suppose the required partition is possible for some k. Then one of the triangles must have a side lying in part of segment AB or in part of segment BC. Then the length of that side is at most 1. Next, the altitude perpendicular to that side is at most n − 1. Hence, that triangle has an area at most (n−1)/2. That is (n2 −1)/k ≤ (n−1)/2. So
k ≥ 2n + 2. Therefore, the least k is 2n+2.
Problem 297. Prove that for every pair of positive integers p and q, there exist an integer-coefficient polynomial f(x) and an open interval with length 1/q on the real axis such that for every x in the interval, |f(x) − p/q| < 1/q2.
(Source:1983 Finnish Math Olympiad)
Solution. Jeff CHEN (Virginia, USA) and
PUN Ying Anna (HKU Math Year 2). If q = 1, then take f(x) = p works for any interval of length 1/q. If q > 1, then define
the interval . 2 3 , 2 1 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = q q I
Choosing a positive integer m greater than (log q)/(log 2q/3), we get [3/(2q)]m < 1/q. Let a = 1−[1/(2q)]m. Then for all x in I, we have 0 < 1 − qxm < a < 1.
Choosing a positive integer n greater than −(log pq)/(log a), we get an < 1/(pq). Let
]. ) 1 ( 1 [ ) ( qxm n q p x f = − − Now
∑
− = − − − = 1 0 ) 1 ( )] 1 ( 1 [ ) ( n k k m m qx qx q p x f∑
− = − = 1 0 ) 1 ( n k k m m qx pxhas integer coefficients. For x in I, we have . 1 ) 1 ( ) ( 2 q a q p qx q p q p x f − = − m n < n<
Problem 298. The diagonals of a convex quadrilateral ABCD intersect at
O. Let M1 and M2 be the centroids of ∆AOB and ∆COD respectively. Let
H1 and H2 be the orthocenters of ∆BOC and ∆DOA respectively. Prove that M1M2⊥H1H2.
Solution. Jeff CHEN (Virginia, USA).
A B C D O A1 C1 B1 H1 D1 H2 E F M1 M2
Let A1, C1 be the feet of the
perpendiculars from A, C to line BD respectively. Let B1, D1 be the feet of the perpendiculars from B, D to line AC respectively. Let E, F be the midpoints of sides AB, CD respectively. Since
OM1/OE = 2/3 = OM2/OF, we get EF || M1M2. Thus, it suffices to show H1H2⊥EF.
Now the angles AA1B and BB1A are right angles. So A, A1, B, B1 lie on a circle Γ1 with E as center. Similarly, C, C1, D, D1 lie on a circle Γ2 with F as center. Next, since the angles AA1D and DD1A are right angles, points A,D,A1,D1 are concyclic. By the intersecting chord theorem, AH2·H2A1=DH2·H2D1.
This implies H2 has equal power with respect to Γ1 and Γ2. Similarly, H1 has equal power with respect to Γ1 and Γ2. Hence, line H1H2 is the radical axis of Γ1 and Γ2. Since the radical axis is perpendicular to the line joining the centers of the circles, we get H1H2⊥EF.
Comments: For those who are not
familiar with the concepts of power and radical axis of circles, please see Math.
Excalibur, vol. 4, no. 3, pp. 2,4.
Commended solvers: PUN Ying Anna
(HKU Math Year 2) and Simon YAU Chi-Keung (City University of Hong Kong).
Problem 299. Determine (with proof) the least positive integer n such that in every way of partitioning S = {1,2,…,n} into two subsets, one of the subsets will contain two distinct numbers a and b such that ab is divisible by a+b.
Mathematical Excalibur, Vol. 13, No. 2, May-Jun. 08 Page 4 PUN Ying Anna (HKU Math Year 2).
Call a pair (a,b) of distinct positive integers a good pair if and only if ab is divisible by a+b. Here is a list of good pairs with 1 < a < b < 50 : (3,6)
,
(4,
12), (5,20), (6,12), (6,30), (7,42), (8,24), (9,18), (10,15), (10,40), (12,24), (12,36), (14,35), (15,30), (16,48), (18,36), (20,30), (21,28), (21,42), (24,40), (24,48), (30,45), (36,45).Now we try to put the positive integers from 1 to 39 into one of two sets S1, S2 so that no good pair is in the same set. If a positive integer is not in any good pair, then it does not matter which set it is in, say we put it in S1. Then we get
S1={1, 2, 3, 5, 8, 10, 12, 13, 14, 18, 19, 21, 22, 23, 30, 31, 32, 33, 34, 36} and S2={4, 6, 7, 9, 11,15, 17, 20, 24, 25, 26, 27, 28, 29, 35, 37, 38, 39}.
So 1 to 39 do not have the property. Next, for n = 40, we observe that any two consecutive terms of the sequence 6, 30, 15, 10, 40, 24, 12, 6 forms a good pair. So no matter how we divide the numbers 6, 30, 15, 10, 40, 24, 12 into two sets, there will be a good pair in one of them. So, n = 40 is the least case.
Problem 300. Prove that in base 10, every odd positive integer has a multiple all of whose digits are odd.
Solution. Jeff CHEN (Virginia, USA)
and G.R.A. 20 Problem Solving Group (Roma, Italy), PUN Ying Anna (HKU Math Year 2).
We first show by induction that for every positive integer k, there is a
k-digit number nk whose digits are all odd and nk is a multiple of 5k. We can take n1=5. Suppose this is true for k. We will consider the case k + 1. If nk is a multiple of 5k+1, then take n
k+1 to be nk + 5×10k. Otherwise, n
k is of the form 5k(5i+j), where i is a nonnegative integer and j = 1, 2, 3 or 4. Since gcd(5,2k) = 1, one of the numbers 10k+n
k, 3×10k+nk, 7×10k+nk, 9×10k+nk is a multiple of 5k+1. Hence we may take it to be nk+1, which completes the induction.
Now for the problem, let m be an odd number. Let N(a,b) denote the number whose digits are those of a written b times in a row. For example, N(27,3) = 272727.
Observe that m is of the form 5kM,
where k is a nonnegative integer and gcd(M,5) = 1. Let n0 = 1and for k > 0, let
nk be as in the underlined statement above. Consider the numbers N(nk,1), N(nk,2), …,
N(nk, M + 1). By the pigeonhole principle, two of these numbers, say N(nk, i) and
N(nk, j) with 1 ≤ i < j ≤ M + 1, have the same remainder when dividing by M. Then N(nk, j) − N(nk, i) = N(nk, j−i) × 10ik is a multiple of M and 5k.
Finally, since gcd(M, 10) = 1, N(nk, j−i) is also a multiple of M and 5k. Therefore, it is a multiple of m and it has only odd digits.
Olympiad Corner
(continued from page 1) Problem 3. Let n be a positive integer. The rectangle ABCD with side lengths
AB=90n+1 and BC=90n+5 is partitioned
into unit squares with sides parallel to the sides of ABCD. Let S be the set of all points which are vertices of these unit squares. Prove that the number of lines which pass through at least two points from S is divisible by 4.
Problem 4. Let c be a positive integer. The sequence a1, a2, …, an, … is defined by
a1=c and an+1=an2+an+c for every positive integer n. Find all values of c for which there exist some integers k ≥ 1 and m ≥ 2 such that ak2+c3 is the mth power of some positive integer.
Geometric Transformations I
(continued from page 2)
On a plane, a spiral similarity with center
O, angle α and ratio k moves every point X
to a point Y such that ∠XOY = α and
OY/OX = k, i.e. it is a rotation with a
homothety. We denote it by S(O,α, k).
Example 9. (1996 St. Petersburg Math
Olympiad) In ∆ABC, ∠BAC=60°. A
point O is inside the triangle such that ∠AOB = ∠BOC = ∠COA. Points D
and E are the midpoints of sides AB and
AC, respectively. Prove that A, D, O, E
are concyclic. 120° 120° B A C O D E
Solution. Since ∠AOB=∠COA=120°
and ∠OBA=60°−∠OAB=∠OAC, we
see ∆AOB~∆COA. Then the spiral
similarity S(O,120°,OC/OA) maps ∆AOB→∆COA and also D→E. Then ∠DOE = 120° = 180°−∠BAC, which
implies A, D, O, E concyclic.
Example 10. (1980 All Soviet Math
Olympiad) ∆ABC is equilateral. M is
on side AB and P is on side CB such that MP||AC. D is the centroid of ∆MBP and E is the midpoint of PA. Find the angles of ∆DEC.
A B C M P D E K H
Solution. Let H and K be the
midpoints of PM and PB respectively. Observe that S(D,−60°,1/2) maps
P→H, B→K and so PB→HK. Now H, K, E are collinear as they are midpoints
of PM, PB, PA. Note BC/BP = BA/BM
= KE/KH, which implies S(D,−60°,1/2)
maps C→E. Then ∠EDC = 60° and
DE=½DC. So we have∠DEC = 90°
and ∠DCE = 30°.
Example 11. (1998 IMO Proposal by
Poland) Let ABCDEF be a convex
hexagon such that ∠B+∠D+∠F = 360° and (AB/BC)(CD/DE)(EF/FA)=1. Prove (BC/CA)(AE/EF)(FD/DB)=1. F A B C D E A'' A' Solution. Since ∠B+∠D+∠F =360°,
S(E, ∠ FED,ED/EF) maps ∆FEA→
∆DEA’ and S(C,∠BCD,CD/CB) maps ∆BCA→ ∆DCA’’. So ∆FEA ~∆DEA’
and ∆BCA~∆DCA’’. These yield
BC/CA=DC/CA’’, DE/EF=DA’/FA and
using the given equation, we get , ' '' CD DA EF FA CD DE BC AB DC D A = = =
which implies A’=A’’. Next ∠AEF = ∠A’ED implies ∠DEF = ∠A’EA. As DE/FE=A’E/AE, so ∆DEF~∆A’EA
and AE/FE=AA’/FD. Similarly, we get ∆DCB~∆A’CA and DC/A’C=DB/A’A. Therefore, . 1 ' '' = = DB AA CA DC DB FD EF AE CA BC
PROBLEMS AND SOLUTIONS FOR 25& BALKAN MATHEMATICAL OLYMPLAD
Problem 1
An acute-angled scalene triangle ABC is given, with AC> BC. Let 0 be its circurncentre, Hits orthocentre, and F the foot of the altitude fiom C. Let P be the psint (other than A ) on the line AB such that AF=PF,.and M be the midpoint of AC. We denote
the intersection of P H and BC by - - - X, the intersection of OM and FX by
Y,
and the intersection of OF and AC by 2. ~ r i v e that theboints F, M, Y and Zare concyclic.Solution:
It is enough to show that OF
I
FX.
Let OE I AB , then it is trivial that :CH = 20E. (1)
Since fi-om the hypothesis we have PF*
AF
then we take PB = P F -BF orPB = A F - B F (2)
Also, LXPB =
LFt4P
+
LHAP
= LHCX since AFGC in inscribable (where G is the foot of the altidude fiom A);so LYPB = LHCX and since LBXP = LNXC, the triangles XHC and XBP are similar. If XL and XD are respectively the heights of the triangles 'XHCand XBP we have:
X D
PB
-=- X L CH'
and fiom (1) and. (2) we get: -.
XD - A F
-
BF . FEXD
;E-- --ap=- -
XL
2 0 E OE FD OE . .Therefo~e the triangles XFD, OEF are similar and we get:
LOFX= LOFC+LLFX=LFOE+LFXD=LXFD+LFXD=90° , S O O F I F X . --
Does there exist a sequence al, a2,,
..,
an,...
o f positive.real numbers satisfying both of the. .
following conditions: . .
. .
"
1 . .(ii)
2
-
<
2008, for every positive.;integer n ?, . i-1 aiSolution.
The answer is no. It is enough to show that
n ' 1 n
if a, S nZ for any n
,
thenz.-
>-
.
(or any other precise estimate)1=1 1-2 a, 4
I
For this, we use that
2
o,2
-
>
2"
for anyk
50
by the arithmetic-harmonic meanl=zk t1 l=zk +I a,
zkH P+' : z M 1 I
Since a,
<
x a , 2z ' ~ + ' ,
it follows thatC
-
>-
,=7?+1- 1-1 .r=2*+, a, 4 Z" n-I 2" 1
and hence
xT
>2
-
>-
.
(it can be stated b,words)a tnO lCzk+l ai 4
Problem 3
Let n be a positive integer. The rectangle ABCD with side lengths AB= 90n,+l and BC = 90n
+
5 is partitioned into 'unit squares with sides parallel to the sides of ABCD. LetS be the set of all points which are vertices of these unit squares. Prove that the number of lines which pass through at least two points fi-om S is divisible by 4 .
Solution. . .
Denote 90n
+
1 = m.
We investigat& the -n&nber . . of the '1ines"~odulo 4 .consecutively ,reducing different types of lines. .
. . :...
The vertical and horizontal lines are
(m f5)
+
(in+
1) = 2(m+
3) which is di~isible~to 4.
Moreover, every line which.'makes an acute >mgle to the axe Ox (i.e. that line has a positive angular coefficient) corresponds [email protected] line with an obtuse angle (conqider the symmetry with respect to the line through the midpoints of
AB
.and CD). Therefore it is enough to prove that the lines with acute angles are an even number. 'Exery line which does not pass through the center 0 of the rectangle corresponds to another line with the same angular coefficent(consider the symmetry with respect to 0).
Therefore it is enough to-consider the lines through 0 .
-P
Every line through 0 has an angular coefficient -, where ( p , q ) = 1, p and q are odd - 9 ' . ' .
positive integers. of the line).
(To see this, consider the two nearest,,fiom the two sides, to 0 points
. . . .. . . . . ,.:.:
-
. - .2 and 5 m , d;js2n&:&guis ckfiCi=nt:;
P,
:uniquely. . . 4'
- . . .
'...
4
correspands to the line with angular coeffici*
-.
It remains to prove that the number of' . P
the remaining lines is even. The last number is
p(rn+2)
+pOv1=
9(nr+2)+~(01+4) 1+-
2 2 . : . . 2 because we have: 1)onelinewith p = q = l ; , p < m is odd and (p,m+2)=1; 2 ) lines with angular coefficient-rn+2
3 )
Q'o
-1 lines with angular coefficied't2,
p L m is odd and2 m
+
4.'(p,yr+4)=1..
NOW the assertion follows from the fact that the number
p(m
+
2)+
p(m+
4 ) = p(90n+
3)+
p(90n+
5 ) is divisible to 4.
Problem 4
Let c be a positive integer. The sequence al,a2
,...,
an,...
is defined by a1 = c , and2
an+^ =an +an +c3, for every positive integer n . Find all values of c for which there exist some integers k 2 1 and m 2 2 , such that a;
+
c3 is the m f h power of some positiveinteger. Solution. First, notice:
a2,+c3 = ( a , 2 + a , + ~ ' ) ~ + c ~ =(a,2+c3)(a:+2an+1+c3)
We first prove that a:
+
c3 and a:+
2an+
1+
c3 are coprime.We prove by induction that 4c3
+
1 is qprirne with 2an+
1 , for every n 2 1.Let n = 1 and p be a prime divisor of 42-1-1 and 2a,
+ I =
2c+l. Then p divides 2(4c3+
1) = (2c+
1)(4cZ-
2c+
1)+
l,
hence p divides 1,a contradiction. Assume now
that (4c3+
1, 2an+
1) = 1 for some nr
1 and the prime p divides 4c3+
1 and 2an+,+
1 .Then p divides 4a,,
+
2 = (2a,+ +
4c3+
1
,
which gives a contradiction.Assume that for some n
r
1 the numbera:+, +c3 =(a:+=, +c3)'+c3 =(a: +c3)(a,2 +2a, +1+c3)
is a power. Since a:
+
c3 and a:+
2an+
1+
c3 are coprime;than a:
+
c3 is a power as well.The same argument can be further applied giving that a:
+
"c c2 +c3 = c2 (c+
1) is apower.
If a2(a+l) = tm with odd m 2 3 , then a = t; and a
+
1 = t," , which is impossible. Ifa2(a +I) = tZq with m,
