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Olympiad Corner
The 2003 USA Mathematical Olympiad took place on May 1. Here are the problems.
Problem 1. Prove that for every positive
integer n there exists an n-digit number divisible by 5n all of whose digits are odd.
Problem 2. A convex polygon P in the
plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygons P are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.
Problem 3. Let n ≠ 0. For every sequence of integers A = a0, a1, a2, …, an
satisfying 0 ≤ ai ≤ i, for i = 0, …, n, define another sequence t ( A ) = t ( a0 ), t
( a1 ), t ( a2 ),…, t ( an ) by setting t ( ai ) to
be the number of terms in the sequence A that precede the terms ai and are different from ai. Show that, starting from any sequence A as above, fewer than n applications of the transformation t lead to a sequence B such that t ( B ) = B.
(continued on page 4)
Editors:
(CHEUNG Pak-Hong), Munsang College, HK
(KO Tsz-Mei) (LEUNG Tat-Wing)
(LI Kin-Yin), Dept. of Math., HKUST
(NG Keng-Po Roger), ITC, HKPU
Artist: (YEUNG Sau-Ying Camille), MFA, CU
Acknowledgment: Thanks to Elina Chiu, Math. Dept., HKUST
for general assistance.
On-line: http://www.math.ust.hk/mathematical_excalibur/
The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is February 28, 2004.
For individual subscription for the next five issues for the 03-04 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:
Dr. Kin-Yin LI Department of Mathematics
The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong
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j j x m −x Õ Ö×Ø (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.
The deadline for submitting solutions is February 28, 2004.
Problem 191. Solve the equation . 2 3
3 − x = x+
x
Problem 192. Inside a triangle ABC,
there is a point P satisfies PAB = PBC = PCA = . If the angles of
the triangle are denoted by , and ,
prove that . sin 1 sin 1 sin 1 sin 1 2 2 2 2ϕ = α + β + γ
Problem 193. Is there any perfect
square, which has the same number of positive divisors of the form 3k + 1 as of the form 3k + 2? Give a proof of your answer.
Problem 194. (Due to Achilleas
Pavlos PORFYRIADIS, American
College of Thessaloniki “Anatolia”, Thessaloniki, Greece) A circle with
center O is internally tangent to two circles inside it, with centers O1 and O2,
at points S and T respectively. Suppose the two circles inside intersect at points
M, N with N closer to ST. Show that S, N, T are collinear if and only if SO1/OO1 = OO2/TO2.
Problem 195. (Due to Fei Zhenpeng, Yongfeng High School, Yancheng City, Jiangsu Province, China) Given n (n >
3) points on a plane, no three of them are collinear, x pairs of these points are connected by line segments. Prove that if , ) 2 ( 3 3 ) 2 )( 1 ( − + − − ≥ n n n n x
then there is at least one triangle having
these line segments as edges.
Find all possible values of integers n > 3 such that ) 2 ( 3 3 ) 2 )( 1 ( − + − − n n n n is an
integer and the minimum number of line segments guaranteeing a triangle in the above situation is this integer.
*****************
Solutions
****************Problem 186. (Due to Fei Zhenpeng, Yongfeng High School, Yancheng City, Jiangsu Province, China) Let , , be
complex numbers such that
+ + = 1, 2 + 2 + 2 = 3, 3 + 3 + 3 = 7. Determine the value of
21 + 21 + 21 .
Solution. Helder Oliveira de CASTRO
(Colegio Objetivo, 3rd Grade, Sao Paulo, Brazil), CHEUNG Yun Kuen (Hong Kong Chinese Women’s Club College, Form 6), CHUNG Ho Yin (STFA Leung Kau Kui College, Form 7), FOK Kai
Tung (Yan Chai Hospital No. 2 Secondary
School, Form 7), FUNG Chui Ying (True Light Girls’ College, Form 6), Murray
KLAMKIN (University of Alberta,
Edmonton, Canada), LOK Kin Leung (Tuen Mun Catholic Secondary School, Form 6), SIU Ho Chung (Queen’s College, Form 5), YAU Chi Keung (CNC Memorial College, Form 7) and YIM Wing
Yin (South Tuen Mun Government Secondary School, Form 4).
Using the given equations and the identities ( + + ) 2 = 2 + 2 + 2 + 2( + + ), ( + + )( 2 + 2 + 2 – – – ) = 3 + 3 + 3 –3 , we get + + = –1 and = 1.
These imply , , are the roots of f (x) =
x3 – x2 – x – 1 = 0. Let Sn= n + n + n , then S1 = 1, S2 = 3, S3 = 7 and for n > 0,
Sn+3 – Sn+2 – Sn+1 – Sn = n f () – n f () – n f ( ) = 0.
Using this recurrence relation, we find S4
=11, S5 =21, … , S21=361109. Problem 187. Define f (n) = n!. Let
a = 0. f (1) f (2) f (3) … .
In other words, to obtain the decimal
representation of a write the numbers
f (1), f (2), f (3), … in base 10 in a row.
Is a rational? Give a proof. (Source:
Israeli Math Olympiad)
Solution. Helder Oliveira de CASTRO (Colegio Objetivo, 3rd
Grade, Sao Paulo, Brazil), CHEUNG
Yun Kuen (Hong Kong Chinese
Women’s Club College, Form 6),
Murray KLAMKIN (University of
Alberta, Edmonton, Canada) and
Achilleas Pavlos PORFYRIADIS
(American College of Thessaloniki “Anatolia”, Thessaloniki, Greece).
Assume a is rational. Then its decimal representation will eventually be periodic. Suppose the period has k digits. Then for every n > 10k , f ( n ) is nonzero and ends in at least k zeros, which imply the period cannot have k digits. We got a contradiction.
Problem 188. The line S is tangent to
the circumcircle of acute triangle ABC at B. Let K be the projection of the orthocenter of triangle ABC onto line S (i.e. K is the foot of perpendicular from the orthocenter of triangle ABC to S). Let L be the midpoint of side AC. Show that triangle BKL is isosceles. (Source: 2000 Saint Petersburg City
Math Olympiad)
Solution. SIU Ho Chung (Queen’s
College, Form 5).
Let O, G and H be the circumcenter, centroid and orthocenter of triangle ABC respectively. Let T and R be the projections of G and L onto line S. From the Euler line theorem (cf. Math
Excalibur, vol. 3, no. 1, p.1), we know
that O, G, H are collinear, G is between
O and H and 2 OG = GH. Then T is
between B and K and 2 BT = TK. Also, G is on the median BL and 2 LG =
BG. So T is between B and R and 2 RT = BT. Then 2 BR = 2 ( BT + RT ) = TK + TB = BK. So BR = RK. Since LR is
perpendicular to line S, by Pythagorean theorem, BL=LK.
Other commended solvers: CHEUNG Yun Kuen (Hong Kong Chinese
Women’s Club College, Form 6) and
Achilleas Pavlos PORFYRIADIS
(American College of Thessaloniki “Anatolia”, Thessaloniki, Greece).
Problem 189. 2n + 1 segments are
marked on a line. Each of the segments intersects at least n other segments. Prove that one of these segments
M MM
Maatttthaahhheeeemmmmaaaattttiiiiccccaaaallll EExEExxxccccaaaalllliiiibbbbuuuurrrr, Vol. 8, No. 5, Nov 03- Dec 03
intersect all other segments. (Source
2000 Russian Math Olympiad) Solution. Achilleas Pavlos PORFYRIADIS (American College
of Thessaloniki “Anatolia”, Thessaloniki, Greece).
We imagine the segments on the line as intervals on the real axis. Going from left to right, let Ii be the i-th segment we meet with i = 1, 2, … , 2n + 1. Let
Iil and Iir be the left and right endpoints of Ii respectively. Now I1 contains I2l, … , In+1l. Similarly, I2 which
already intersects I1 must contain I3l, … , In+1l and so on. Therefore the
segments I1, I2, … , In+1 intersect each
other.
Next let Ikr be the rightmost endpoint among I1r, I2r, … , In+1r (1 ≤ k ≤ n+1).
For each of the n remaining intervals
In+2, In+3, … , I2n+1, it must intersect at
least one of I1, I2, … , In+1 since it has to
intersect at least n intervals. This means for every j ≥ n + 2, there is at least one m ≤ n + 1 such that Ijl ≤ Imr
≤ Ikr, then Ik intersects Ij and hence every interval.
Problem 190. (Due to Abderrahim Ouardini) For nonnegative integer n,
let [x] be the greatest integer less than or equal to x and
[
1 2]
) (n = n+ n+ + n+ f[
9 +1]
− n .Find the range of f and for each p in the range, find all nonnegative integers n such that f (n) = p.
Combined Solution by the Proposer and CHEUNG Yun Kuen (Hong Kong
Chinese Women’s Club College, Form 6). For positive integer n, we claim that
9 9 ) ( 8 9n+ < g n < n+ , where 2 1 ) (n = n + n+ + n+ g .
This follows from
) 1 ( ( 2 3 3 ) (n 2 = n+ + n n+ g ) ) 2 ( ) 2 )( 1 (n+ n+ + n+ n +
and the following readily verified inequalities for positive integer n,
(n + 0.4 ) 2 < n(n + 1) < (n + 0.5 )2, (n + 1.4 )2 < (n + 1)(n + 2) < (n + 1.5 )2 and (n + 0.7 )2 < (n + 2) n < (n + 1)2. The
claim implies the range of f is a subset of nonnegative integers.
Suppose there is a positive integer n such that f (n) ≥ 2. Then 1 9 1 )] ( [ 9 9n+ > g n > + n+ . Squaring the two extremes and comparing, we see this is false for n > 1. Since f (0) = 1 and f (1) = 1, we have f (n) = 0 or 1 for all nonnegative integers n.
Next observe that
9 9 )] ( [ 8 9n + < g n < n+ is impossible by squaring all expressions. So [g(n)] =[ 9n +8].
Now f (n) = 1 if and only if p = [ g (n) ] satisfies [ 9n+1]= p−1 , i.e. . 8 9 1 9n+ < p ≤ n+
Considering squares (mod 9), we see that
p2 = 9n + 4 or 9n + 7.
If p2 = 9n + 4, then p = 9k + 2 or 9k + 7. In the former case, n = 9k2 + 4k and (9k + 1)2
≤ 9n + 1 = 81k2 + 36k + 1< (9k + 2)2 so that . 1 1 9 ] 1 9 [ n+ = k + = p− In the latter case, n = 9k2 + 14k + 5 and (9k + 6)2 ≤ 9n + 1 = 81k2 + 126k + 46 < (9k + 7)2 so that [ 9n +1]= 9k +6 = p −1. If p2 = 9n + 7, then p = 9k + 4 or 9k + 5. In the former case, n = 9k2 + 8k + 1 and (9k + 3)2 ≤ 9n + 1 = 81k2 + 72k + 10 < (9k + 4)2 so that [ 9n+1]=9k +3= p−1. In the latter case, n = 9k2 + 10k + 2 and (9k + 4)2 ≤ 9n + 1 = 81k2 + 90k + 19 < (9k + 5)2 so that [ 9n+1]= 9k + 4 = p −1. Therefore, f (n) = 1 if and only if n is of the form 9k2 + 4k or 9k2 + 14k + 5 or 9k2 + 8k + 1 or 9k2 + 10k + 2.
Olympiad Corner
(continued from page 1)
Problem 4. Let ABC be a triangle. A
circle passing through A and B intersects segments AC and BC at D and E,
respectively. Rays BA and ED intersect at F while lines BD and CF intersect at M. Prove that MF = MC if and only if MB⋅MD = MC2.
Problem 5. Let a, b, c be positive
real numbers. Prove that
2 2 2 2 2 2 ) ( 2 ) 2 ( ) ( 2 ) 2 ( a c b a c b c b a c b a + + + + + + + + + . 8 ) ( 2 ) 2 ( 2 2 2 ≤ + + + + + b a c b a c
Problem 6. At the vertices of a regular
hexagon are written six nonnegative integers whose sum is 2003. Bert is allowed to make moves of the following form: he may pick a vertex and replace the number written there by the absolute value of the difference between the numbers written at the two neighboring vertices. Prove that Bert can make a sequence of moves, after which the number 0 appears at all six vertices.
(continued from page 2)
4 2 ) ( ) ( 2 2 m x m x x m xj j j j = − + ≤ − 4 4 ) ( 2 1 2 1 2 nm m x m x dc n j n j j j m = ≤ − ≤ = = 2 2 d m d n ≤ − n = 7 d = 4 2 8 2 d d −n = 9 7 ! " #$%&' 4 ( ) * (+, -. 8 / 0 ( ) * ) / 1 23456789 : ;<( = > ? @ / ;< :AB Plotkin C (Plotkin Bound) DE 1DF GHIJHK (;< H LMNEO9
an is divisible by 5” and has only odd digits. Consider the numbers _. Nr = lolaa . . . czn = 1.10” + 5nM = 5n(1 * in + M), N2 = 3ara2 . . . a, = 3 * lon + 5”M = 5n(3 - 2* + M), N3 = 5ulQ. ...a,=540n+5nM=5”(5.2n+M), Nq=7~ra2...a,=7.10”+5~M=5~(7.2*+M), Ns = 9ura2.. . a, = 9.. Id” + 5”M = 5n(9 * 2* + M).
The numbers 1.2” + M, 3.2” + M, 5.2* + M, 7.2” + M, 9 .2” + M give-distinct remainders when divided by 5. Othertise the difference of some two of them would be a multiple of 5, which is impo=ible, because 2” is not a multiple of 5, nor is the difference of any two of the numbers 1,3,5,7,9. It follows that one of the numbers Nr, N2, Ns, N4, Ns is divisible by 5” * 5, and the induction is complete.
f@e;t ion: Let P = Al4.. . A,, where n is an integer with n > 3. The problem is trivial
= 3 because there are no diagonals and thus no dissections. We assume that n 1 4. Our proof is based on the following Lemma.
Lemma 1. Let ABCD be o convex quadrilateral such thut all its sides and diagonals have
rational lengths. If segments AC and BD meet at P, then segments Al’, BP, CP, DP d
have Futionul lengths.
It is clear by Lemma 1 that the desired result holds when P is a convex quadrilateral. Let
AiAj (1 < i < j 5 n) be a diagonal of P. Assume that Cl, C2, . . . , C, are the consecutive
division points od diagonal AiAj (where point Cr is the closest to vertex Ai and C, is the closest to Aj). Then the segments C&(+1, 1 5 L < m - 1, are the sides of all polygons in the dissection. Let Ce be the point where diagonal &Aj meets diagonal ASAt. Then quadrilateral
AiA,AjAt satisfies the conditions of Lemma 1. Consequently, segments A&‘( and CcAj have
rational lengths. Therefore, segments AiCl, A&2,. . . , AjCm all have rational lengths. Thus, C&r+1 = AG+r - ACt is rational. Because i, j, L are arbitrarily chosen, we proved that all sides of all polygons in the dissection are also rational numbers.
Now we present & proof. of Lemma 1 to finish our proof.
We show only that segment AP is rational, the others being klar. Introduce Cartesian coordinates with A = (0,O) and C = (c,O). Put B = (a, b) and
D = (d, e). Then by hypothesis, the numbers
AB=dm, AC=c, AD&m,
BC = d(u - c)2 + b2, BD = J(u - d)2 + (b - e)2, CD = J(d - c)~ + e2, are rational. In particular,
is rational. Because c # 0, a is rational. Likewise d is rational.
Now we have that b2 = AB2 - a2, e2 = AD2 - &, (b - e)2 = BD2 - (u - d)2 are rational,
andsothat2be=b2+e2-(b-e)2’ is rational. Because quadrilateral ABCD is convex,
b and e are nonzero and have opposite sign. Hence p = 3 is rational.
We now calculate so P=(F,O), I! .d-a AP= CL -- c 1 is rational. m
Q$+ Solution: Note ‘first that the transformed sequence t(A) also satisfies the inequalities
0 5 t(ai) < i, for i = 0,. . . , n. Call any integer sequence that satisfies these inequalities an index bounded sequence.
We prove now that that ai 5 t(a;), for i = 0,. . . , n. Indeed, this is clear if ai = 0. Otherwise, letz=ai>Oandy= t(ai). None of the first x consecutive terms ac, al,. . . , a,_1 is greater
than x - 1 so they are all different from z and precede x (see the diagram below). Thus y 1 x, that is, ui < t(u;), for i = 0, . . . , n.
‘.A
index 0 1 . . . x-l . . . i (_ L.
A a0 al . . . a,_1 . . . x
t(A) t(ao) t(ar) . . . t(a,-1) . . . y
This already shows that the sequences stabilize after finitely many applications of the trans- formation t, because the value of the index i term in index bounded sequences cannot exceed i. Next we prove that if ui = t(~), for some i = 0,. . . , n, then no further applications of t will ever change fhe index i term. We consider two cases.
l In this case, we assume that ai = t(ai) = 0. This means that no term on the left of ai
is different from 0, that is, they are alI 0. Therefore the first i terms in t(A) will also be 0 and this repeats (see the diagram below).
. In this case, we assume that ai = t(ai) = x > 0. The first z terms are all different from z. Because t(%) = x, the terms a,, a,+~, . . . , ai- must then all be equal to ti. Consequently, t(aj) = X for j = 2,. . . ,i - 1 and further applications oft cannot change
the index i term (see the diagram below).
index 0 1 . . . X-l X' x+1 . . . i
A a0 al . . . a,--1 x 2 . . . x
t(A) t(m) t(m) . . . t(u,_1) x x . . . x
For 0 _< i 5 n, the index i entry satisfies the following properties: (i) it takes integer values; (ii) it is bounded above by i; (iii) its value does not decrease under transformation t; and (iv) once it stabilizes under transformation t, it never changes again. This shows that no more than n applications of t lead to a sequence that is stable under the transformation t.
..- ..,. ,. :, ::
on n.
For n = 1, the two possible index bounded sequences (ac,ai) = (0,O) and (ae,ai) = (0,l) are already fixed by t so we need zero applications oft.
Assume that any index bounded sequences (a~, ai, . , a,,) reach a fixed sequence after no more than n - 1 applications of t. Consider an index bounded sequence A = (Q, al,. . . , a,+~). It suffices to show that A will be stabilized in no more than n applications of t. We ap proach indirectly by assume on the contrary that n + 1 applications of transformations are needed. This can happen only if a,,+1 = 0 and each application of t increased the index n + 1 term by exactly 1. Under transformation t, tbe resulting value of index term i will not the effected by index term j for i < j. Hence by the induction hypothesis, the subsequence
A’= (~,q,... , a,,) will be stabiiized in no more than n- 1 applications oft. Because index ra
term is stabilized at value z < n after no more than min{z, n - 1) applications oft and index n + 1 term obtains value z after z exactIy applications of t under our current assumptions. We conclude that the index n + 1 term would become equal to the index n term after no more than n - 1 applications of t. However, once two consecutive terms in a sequence are equal they stay equal and stabilize together. Because the index n term needs no more than n - 1 transformations to be stabilized, A can be stabilized in no more than n - 1 applications of t, which contradicts our assumption of n + 1 applications needed. Thus our assumption was wrong and we need at most n applications of transformation t to stabilize an (n + l)-term index bounded sequence. This completes our inductive proof.
Solution: qQJ*
Extend segment DM through M to G such that FG 11 CD.
Then MF = MC if and only if quadrilateral CDFG is a parallelogram, or, FD II CG. MC = MF if and only if LGCD = LFDA, that is, LFDA+ LCGF = 180”.
Hence Because quadrilateral ABED is cyclic, LFDA = LABE. It follows that MC = MF if and only if
180” = LFDA + LCGF = LABE + LCGF, that is, quadrilateral CBFG is cyclic, which is equivalent to
LCBM = LCBG = LCFG = LDCF = L DCM.
Because LDMC= LCMB, LCBM= LDCM if and onlyiftriangles BCM and CDM are
similar, that is
CM DM BM=CM’ or MB’ MD = MC2.
f$+ Solution: By multiplying a, b, and c by a suitable factor, we reduce the problem to the case when a -+ b + c = 3. The desired inequality reads
(a + 3)2 (b + 3)2 202 + (3 - a)2 + 2b2 + (3 - b)2 !- (C+3)2 <*_ 2c2 + (3 - c)’ - Set f(x) = 2x2 + (3 - x)2 (z + 3)2
It su5ces to prove that f(a) + f(b) + f(c) 5 8. Note that
f(x) = 3(22 - 22 + 3) =?i’z2-2x+3 2+6x+9 1 z2+6z+9
!
Hence,
f(a) + f(b) + f(c) 2 ;(4cr + 4 + 4b + 4 + 4c + 4) = 8, -r; \ -,’ .:..
f+y
ote: Let
AFE BcD
/ denote a position, where
A, B, C, D, E, F denote the numbers written on the vertices of the hexagon. We write
AfzD (mod2)
if we consider the numbers written modulo 2.
Solution: Deiine the sum and mtimum of a position to be the sum and maximum of the six numbers at the vertices. We will show that from any position in which the sum is odd, it is possible to reach the all-zero position.
Our strategy alternates between two steps:
(a) from a position with odd sum, move to a positio’n with exactly one odd number; (b) from a position with exactly one odd number, move to a position with odd sum and
strictly smaller maximum, or to the all-zero position.
Note that no move will ever increase the maximum, so this strategy is guaranteed to terminate, because each step of type (b) decreases the maximum by at least one, and it can only terminate at the all-zero position. It sufhces to show how each step can be carried out.
First, consider a position
ABCD FE
with odd sum. Then either A + C + E or B + D + F is odd; assume without loss of generality that A + C f E is odd. If exactly one of A, C and E is odd, say A is odd, we can make the sequence of moves
O*Cl 10
10 o-+0 i i 0 (mod 2),
where a letter or number in boldface represents a move at that vertex, and moves that do not affect each other have been written as a single move for brevity. Hence we can reach a position with exactly one odd number. Similarly, if A, C, E are all odd, then the sequence of moves
lB1
F1D4;:040,, Cl ’ 0 (mod 2),brings us to a position with exactly one odd number. Thus we have shown how to carry out step (a).
Now assume that we have a position
ABCD FE
with A odd and all other numbers even. We want to reach a position with smaller maximum. Let M be the maximum. There are two cases, depending on the parity of M.
l In this case, M is even, so one of B, C, D, E, F is the maximum. In particular, -A < M.
We claim after making moves at B, C, D, E, and F in that order, the sum is odd and the maximum is less than M. Indeed, the following sequence .c
loo
00 041 10 o-+1 11 00 00 041 11 00 14 1: : l+ 1: :l (mod2). shows how the numbers change in parity with each move. Call this new position, B’ C’
A F’ E’ D’. The sum is odd, since there are five odd numbers. The numbers A’, B’, C’, D’, E’ are all less than M, since they are odd and M is even, and the maximum can never increase. Also, F’ = IA’ - E’I 5 max(A’, E’} < M. So the maximum has been decreased.
. In this case, M is odd, so M = A and the other numbers are all less than M.
If C > 0, then we make moves at B, F, A, and F, in that order. The sequence of positions is 00 loo O+l 10 o-+1 10 00 10 O*O:~O+O~~O (mod2). , B’ C’
Call this new position A F, E, D’. The sum is odd, since there is exactly one odd number. ks before, the only way the maximum could not decrease is if B’ = A; but this is impossible, since B’ = IA - Cl < A because 0 < C < M = A. Hence we have,reached a position with odd sum and lower maximum.
If E > 0, then we apply a similar argument, interchanging B with F and C with E. If C = E = 0, then we can reach the all-zero position by the following sequence of moves:
ABO FOD-A;;d-O;~O+O;;O.
(Here 0 represents zero, not any even number.)
Hence we have shown how to carry out a step of type (b), proving the desired result. The
problem statement follows since 2003 is odd. -
