Volume 10, Number 2 May 2005 – July 2005
ٽݙኵᏰᝯᗉϛޟॎኵ୰ᚠ)έ*!
!
າਏᝉ!!Ȟឃࣸᢄ࠲ѿ࠲ୢҗᙴϛᏰ!224054 )
Olympiad Corner
Following are the problems of 2005 Chinese Mathematical Olympiad. Problem 1. Let θi∈(−π /2,π /2), i = 1, 2, 3, 4. Prove that there exists x∈ℝ satisfying the two inequalities
0 ) sin (sin cos cos2θ1 2θ2− θ1 θ2−x 2≥ 0 ) sin (sin cos cos2θ3 2θ4− θ3 θ4−x 2≥ if and only if
∑
∏
∏
= = = + + ≤ 4 1 4 1 4 1 2 2(1 sin cos ). sin i i i i i i θ θ θProblem 2. A circle meets the three sides BC, CA, AB of triangle ABC at points D1, D2; E1, E2 and F1, F2 in turn.
The line segments D1E1 and D2F2
intersect at point L, line segments E1F1
and E2D2 intersect at point M, line
segments F1D1 and F2E2 intersect at
point N. Prove that the three lines AL, BM and CN are concurrent.
Problem 3. As in the figure, a pond is divided into 2n (n ≥ 5) parts. Two parts are called neighbors if they have a common side or arc. Thus every part has
(continued on page 4)
Editors: Ի ஶ(CHEUNG Pak-Hong), Munsang College, HK ଽ υ ࣻ (KO Tsz-Mei)
గ ႀ ᄸ (LEUNG Tat-Wing)
፱ (LI Kin-Yin), Dept. of Math., HKUST ֔ ᜢ ݰ (NG Keng-Po Roger), ITC, HKPU Artist: ྆ ؾ ़ (YEUNG Sau-Ying Camille), MFA, CU Acknowledgment: Thanks to Elina Chiu, Dept. of Math., HKUST for general assistance.
On-line: http://www.math.ust.hk/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 10, 2005.
For individual subscription for the next issue, please send us a stamped self-addressed envelope. 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] ٽ12!!֏џғᐌኵ 1Ȃ2Ȃ3ȂȌȂ 64 Ϸտ༲Σ 8×8 ޟғПלޟ 64 এω ПਿϱȂٺுלԃყ 1ȞПөџоӈ ཎᙽညȟޟӈཎѲএωПਿϱኵϞڷ ᖂೝ5 ᐌଶȉၐᇳ݂౩ҥȄ! ყ1 ၌๎!!ϚџȄή७ҢІᜌݲᜌ݂Ȉ ೩ყ2 ϛ ൷ಒӫᚠ ೩༲ԁޟኵȄ , , , , , , a b cLk lL ყ2 Ӱ5 | b e+ + +f gȂ5 | j e+ + +f gȂհ ৯Ԥ5 | b− jȂ։b≡ j(mod5)Ȃ೩Ꮇኵ ࣏rȄ ӣ౩Ӱ5 | j+ + +f b gȂ5 | e+ + +f b gȂ ࢈5 | j e− Ȃ։ j e≡ (mod5)ȂᡗณڏᎷ ኵζ࣏rȄ ყϛ 64 এωПਿԙ༃ҩࣺޟ לԒȂџு ։ଶُΰڍ ҩ ਿ ϛ ޟ ڍ ኵ Ѵ Ȃ ڏ Ꮇ ҩ ਿ ϛ ޟ , , , , , , b j e g d l L 64 2 30 2 − = এኵೝ 5 ଶӣᎷrȄ ѪΙП७Ȃҥܪন౩Ȃ1 ~ 64 64 এғᐌኵϛശӻԤ 13 এኵೝ 5 ଶӣ ᎷȂᇄࠉ७ுюޟ๖፣ҭࣽȊӰԪȂ ϚԆӵᅖٗᚠ೩ޟ༲ݲȄ ٽ13 ҁ७ΰۡϤᘈAȃBȃCȃDȃ EȂڏϛӈդέᘈϚӵΙޢጣΰȄၐ ᜌȈӈཎӴҢጣࢲ๖࢚ٲᘈȞٲ ጣࢲᆎ࣏ȟȂषܚுڗޟყלϛϚю ౪оϤᘈϛޟӈդέᘈ࣏ഥᘈޟέ ُלȂࠌএყלϚџԤ7 నܖ ӻనȄ ᜌݲ1 ȞІᜌݲȟ೩ყלԤ 7 నܖ ӻనȂࠌӨᘈ࡙ኵڷՍЍ14Ȅ (1) ष࢚ᘈ࡙ኵ 4ȂࠌڏᎷᘈޟ࡙ኵ ڷՍЍ10Ȃҥܪন౩ޣڏϛ҆Ԥ Ιᘈ࡙ኵՍЍ 10 1 3 4 ⎡ ⎤ + = ⎢ ⎥ ⎣ ⎦ Ȟ࡙ኵ2 ൷ϐٗஊȟȂ࢈Ԫਢ҆ณю౪έُלȄ (2) षؐᘈ࡙ኵՍӻ 3Ȃҥܪন౩ ޣՍЍԤ 4 ᘈޟ࡙ኵ 3Ȃᒵڏϛ 2 ᘈȂϚֹ೩࣏AȃBȂи A ᇄ BȃCȃ D ԤጣȂԪਢՃኌ B ᇄ A ϐԤጣȂ ҥܪন౩ޣB ҆ᇄ CȃD ϛ࢚Ιᘈ ԤጣȂኺζю౪ΟέُלȄ Մ(1)ȃ(2)ܚு๖፣ᇄᚠ೩Ɇყלϛ Ϛю౪оϤᘈϛޟӈդέᘈ࣏ഥᘈ ޟέُלɇࣺҭࣽȂ࢈নڼᚠԙҳȄ ᜌݲ2 ȞІᜌݲȟ೩ყלԤ 7 నܖ ӻనȄ ॶӑרঈᄺഅܪȈؐএܪ裏Ԥέ এࣺᘈȂӓџு 3 এܪ 5 10 C = Ȫݧȫ Ȃ άҥܻӣΙనོӵ5 2 3− = এܪ 裏ю౪Ȃࠌ10 এܪ裏ӓԤ7 3 21× = నܖӻనȄ ҥܪন౩ޣȂՍЍԤΙএܪ裏Ԥ 3 నȂՄؐనӵΙএέُלϛശ ӻю౪ΙԩȄ3 నԁᇄڏϛϚ
Mathematical Excalibur, Vol. 10, No. 2, May 05- Jul. 05 Page 2 ӓጣޟࣺέᘈᄺԙΙএέُלȄՄ ᇄᚠ೩ɆყלϛϚю౪оϤᘈ ϛޟӈդέᘈ࣏ഥᘈޟέُלɇࣺ ҭࣽȂ࢈নڼᚠԙҳȄ ݧ ᄇܻճԑᏰҡॎᆗᄺഅܪޟ এኵȂרঈџоՃኌAȃBȃCȃDȃ EϤᘈϛӈڥޟέএᘈᇄ഻ήޟڍ ᘈΙΙᄇᔖȂՄᒵᐅڍᘈޟלԤȈ ABȃACȃADȃAEȃBCȃBDȃBEȃ CDȃCEȃDEȂӓ 10 ᆍȄ ᄇএ୰ᚠ฿հЕժȂ߯ுή७ޟ୰ ᚠȈ ٽ14 ҁ७ΰۡn এᘈȂڏ ϛӈդέᘈϚӓጣȄӈཎӴҢጣࢲ ๖࢚ٲᘈȞٲጣࢲᆎ࣏ȟȂுڗx నȄ
(
n>3)
(1) षጂ߳ყלϛю౪оۡᘈ࣏ഥ ᘈޟέُלȂؑᜌȈ ) 2 ( 3 3 ) 2 )( 1 ( − + − − ≥ n n n n x Ȅ ࿋n n( −3(1)(nn−−2)2) 3+ ᐌኵਢȂؑܚԤ n Ѕᄇᔖ x ޟശωȇ (2) षጂ߳ყלϛю౪оۡᘈ࣏ഥ ᘈޟm(m n< )ׇӒყ(։mᘈϛӈ դڍᘈԤޟყ)ȂؑᜌȈ 2 2 2 C (C 1) 1 C m n m m n x − − − + ≥ Ȅ ᜌ݂ (1)ᢼ˵ȆᄺഅܪȈؐএܪ 裏 Ԥέ এࣺ ᘈȂ ӓџு এ ܪ ȄάҥܻӣΙనོӵ এܪ 裏 ю ౪ Ȃ ਲ਼ ᐃ ܪ ন ౩ ޣ Ȃ ࿋ ਢȂϗጂ߳ԤΙএ ܪ裏Ԥ3 నȂՄ 3 నԁᇄ ڏϛϚӓጣޟࣺέᘈᄺԙΙএέ ُלȄ 3 Cn 1 2 Cn− 1 3 2 Cn 2Cn x⋅ − ≥ +1 ൷ᇳȂጂ߳ყלϛю౪оۡᘈ࣏ ഥᘈޟέُלȂࠌ 3 1 2 2C 1 C n n x − + ≥ Ȃ։ ( 1)( 2) 3( 2) n n n x n 3 − − + − ≥ Ȅ ˶ȆᡗณnȂn−1Ȃn−2ϛԤиѫԤΙ এ3 ޟॻኵȄ (i) ࿋ n ܖn−13 ޟॻኵਢȂΙП७ ( 1)( 2) 3 ( 1) 1 3( 2) 3 2 n n n n n n n − − + = − + − − ᐌ ኵ Ȃ ࠌ 1 2 n− ᐌ ኵ ȇ Ѫ Ι П ७ Ȃ 3 n> n− >2 1Ȃࠌ 1 2 n− ϷኵȄҭࣽȊ Ԫਢn ฒ၌Ȅ (ii) ࿋n−2 3 ޟ ॻ ኵ ਢ Ȃ Ϛ ֹ ೩ ȂՃኌ 2 3 n− = k ) 2 ( 3 3 ) 2 )( 1 ( − + − − n n n n k k k k 3 3 3 ) 2 3 )( 1 3 ( 3 ⋅ + + + = k k k k k 3 1 ) 1 3 3 ( 3 2 + + + − = k k k k 3 1 1 3 3 2 + + + − = ᐌኵȂࠌ1 3−kkᐌኵȄ х1 3− =kk tȂࠌk t(3 + =1) 1ȄՄk 1= Ȃ ։ 3t+ =1 1 k=1Ȃt=0ȄӰԪ 3 2 n= k+ =5Ȅ Մx≥3k2+3k+1Ȃ։ ȄӰԪ x ޟശω7Ȅ 7 x ≥ ᆣӫ(i)ȃ(ii)џޣȂ࿋ ) 2 ( 3 3 ) 2 )( 1 ( − + − − n n n n ᐌኵਢȂn=5Ȃxmin=7Ȅ (2) ᄺഅܪȈؐএܪ裏Ԥ m এࣺ ᘈȂӓџு এܪȄάҥܻӣ Ιనོӵ এܪ裏ю౪Ȅਲ਼ᐃ ܪন౩ޣȂ࿋ Cm n 2 2 Cm n − − 2 2 2 Cm C (Cm 1) n n m x − − ⋅ ≥ − +1 ਢȂϗጂ߳ԤΙএܪ裏Ԥ న ȂՄ నԁᇄڏϛϚӓጣ ޟࣺm ᘈᄺԙΙএ m ׇӒყȄ 2 Cm 2 Cm ൷ᇳȂጂ߳ყלϛю౪оۡᘈ ࣏ഥᘈޟm ׇӒყȂࠌ 2 2 2 C (C 1) 1 C m n m m n x − − − + ≥ Ȅ ݧ ᚠϛԅҔkȃmȃnȃtȃxࡾᐌ ኵȄ оΰ၌ؚኵᏰᝯᗉᚠޟࡦၯᇄПݲ ֙ວרঈȈَӻᜋኄȂџоቨሴਂ Ψȇി௴ಀߝȂϗЍޡҬܒȄ ၌ᚠϛޟཐए౪ȂྛՌҁਢޟРᑖ ТಣȄѫԤӻᢣंȂӻષȂୈᚠਢ ߯ᓍᐠᔖᡐȂһܖᐿ៦ᗑ৷Ȃоम ߔλՄ၌Ȅ ձឈுɆኵᏰԁޖɇ༝ȉѫौձԤ ᑹ፸ȂኵᏰ൷ོᡐுड़ณϚӣȄձ൷ ོཐڧڗኵᏰฒᅾޟᏄΨȂ൷ོڎԤ ฒϚպޟཎסΨȂ൷ོҡฒϚ ᄩޟᏽାΨȄԃݎձਲ਼ҏ൷ؠདྷΰኵ ᏰȂάࡩቄџኡюശ࣏េޟЬ ߆ڹȉܐߨலȂᕡ։ഁȄ Ԥ࡞ӻӣᏰዥདྷኵᏰȂ࣏ӵኵᏰ ༹ݓϾպޟᗉൟΰΙၐ٘Йȃᄢߜჴ ሚՄᓶᓶᢣंȂववᏭାȄרདᏰಬ ϛ߳ࡻߝδޟኵᏰᑹ፸ڷஉᎴഺഅ ܒޟࡦᆰԙђޟᜰᗤȂζپџ ࡻ៉ีޟ߳ራȄՄأڥಀড়Ϟߝ ഺഅܒࡦᆰޟྛࢶȂᏰོᐿҳࡦՃ ඪଽഺഅܒࡦᆰΨޟي๊ȄProblem 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 10,
2005.
Problem 226. Let z1, z2, …, zn be
complex numbers satisfying |z1| + |z2| + ⋯ + |zn| = 1.
Prove that there is a nonempty subset of {z1, z2, …, zn} the sum of whose
elements has modulus at least 1/4. Problem 227. For every integer n ≥ 6, prove that ∑ ⋅ ≤ − − = − 1 1 1 . 5 16 2 1 n k n k k n
Problem 228. In ɔABC, M is the foot of the perpendicular from A to the angle bisector of ʆBCA. N and L are respectively the feet of perpendiculars from A and C to the bisector of ʆABC. Let F be the intersection of lines MN and AC. Let E be the intersection of lines BF and CL. Let D be the intersection of lines BL and AC. Prove that lines DE and MN are parallel.
Problem 229. For integer n ≥ 2, let a1,
a2, a3, a4 be integers satisfying the
following two conditions:
(1) for i = 1, 2, 3, 4, the greatest common divisor of n and ai is 1 and
(2) for every k = 1, 2, …, n – 1, we have (ka1)n + (ka2)n + (ka3)n + (ka4)n = 2n,
where (a)n denotes the remainder when
a is divided by n.
Prove that (a1)n, (a2)n, (a3)n, (a4)n can be
divided into two pairs, each pair having sum equals n.
(Source: 1992 Japanese Math Olympiad)
Problem 230. Let k be a positive integer. On the two sides of a river, there are in total at least 3 cities. From each of these cities, there are exactly k
routes, each connecting the city to a distinct city on the other side of the river. Via these routes, people in every city can reach any one of the other cities.
Prove that if any one route is removed, people in every city can still reach any one of the other cities via the remaining routes.
(Source: 1996 Iranian Math Olympiad, Round 2)
*****************
Solutions
****************
Due to an editorial mistake in the last issue, solutions to problems 216, 217, 218, 219 by D. Kipp Johnson (teacher, Valley Catholic School, Beaverton, Oregon, USA) were overlooked and his name was not listed among the solvers. We express our apology to him.
Problem 221. (Due to Alfred Eckstein, Arad, Romania) The Fibonacci sequence is defined by F0 = 1, F1 = 1 and Fn = Fn−1 + Fn−2 for n ≥ 2. Prove that is divisible by F 3 3 3 2 7Fn+ − Fn − Fn+1 3 8Fn+ n+3.
Solution. HUDREA Mihail (High
School “Tiberiu Popoviciu” Cluj-Napoca Romania) and Kin-Chit O (STFA Cheng Yu Tung Secondary School).
As 3 is divisible by F 1 3 2 7 7 + + + = Fn Fn a n+2 + Fn+1 = Fn+3 and b = is divisible by 2F 3 1+Fn n+1 + Fn = Fn+2 + Fn+1 = Fn+3, so = a − b is divisible by F 3 1 3 3 2 7Fn+ − Fn − Fn+ n+3.
Other commended solvers: CHAN Pak Woon (Wah Yan College, Kowloon, Form 7), CHAN Tsz Lung, CHAN Yee Ling (Carmel Divine Grace Foundation Secondary School, Form 6), G.R.A. 20 Math Problem Group (Roma, Italy), MA Hoi Sang (Shun Lee Catholic Secondary School, Form 5), Anna Ying PUN (STFA Leung Kau Kui College, Form 6), WONG Kwok Cheung (Carmel Alison Lam Foundation Secondary School, Form 6) and WONG Kwok Kit (Carmel Divine Grace Foundation Secondary School, Form 6).
Problem 222. All vertices of a convex quadrilateral ABCD lie on a circle ω. The rays AD, BC intersect in point K and the rays AB, DC intersect in point L.
Prove that the circumcircle of triangle AKL is tangent to ω if and only if the circumcircle of triangle CKL is tangent to ω.
(Source: 2001-2002 Estonian Math Olympiad, Final Round)
Solution. LEE Kai Seng (HKUST) and
MA Hoi Sang (Shun Lee Catholic Secondary School, Form 5).
Let ω1 and ω2 be the circumcircles of
∆AKL and ∆CKL respectively. For a point P on a circle Ω, let Ω(P) denote the tangent line to Ω at P.
Pick D’ on ω(A) so that D and D’ are on opposite sides of line BL and pick L’ on ω1(A) so that L and L’ are on
opposite sides of line BL.
Next, pick D” on ω(C) so that D and D” are on opposite sides of line BK and pick L” on ω2(C) so that L and L” are
on opposite sides of line BK. Now ω, ω1 both contain A and ω, ω2 both
contain C. So ω(A) = ω1(A) ⇔ ∠D’AB = ∠L’AB ⇔ ∠ADB = ∠ALB ⇔ BD ║ LK ⇔ ∠BDC = ∠KLC ⇔ ∠BCD” = ∠KCL” ⇔ ω(C) = ω2(C).
Other commended solvers: CHAN Tsz Lung and Anna Ying PUN (STFA Leung Kau Kui College, Form 6).
Problem 223. Let n ≥ 3 be an integer and x be a real number such that the numbers x, x2 and xn have the same
fractional parts. Prove that x is an integer.
(Source: 1997 Romanian Math Olympiad, Final Round)
Solution. G.R.A. 20 Math Problem
Group (Roma, Italy).
By hypotheses, there are integers a, b such that x2 = x + a and xn = x + b. Since
x is real, the discriminant ɔ = 1 + 4a of x2 − x − a = 0 is nonnegative. So a ≥ 0.
If a = 0, then x = 0 or 1.
If a > 0, then define integers cj, dj so that
xj = c
jx + dj for j ≥ 2 by c2 = 1, d2 = a > 0,
x3= x2+ ax = (1 + a)x + a
leads to c3 = 1 + a, d3 = a and for j > 3, xj
= (x + a)xj−2 = (c
Mathematical Excalibur, Vol. 10, No. 2, May 05- Jul. 05 Page 4 adj−2) leads to cj = cj−1 + acj−2 > cj−1 > 1
and dj = dj−1 + adj−2.
Now cnx + dn = xn = x + b with cn > 1
implies x = (b − dn)/(cn − 1) is rational.
This along with a being an integer and x2
− x − a = 0 imply x is an integer. Other commended solvers: CHAN Tsz Lung, MA Hoi Sang (Shun Lee Catholic Secondary School, Form 5), and Anna Ying PUN (STFA Leung Kau Kui College, Form 6).
Problem 224. (Due to Abderrahim Ouardini) Let a, b, c be the sides of triangle ABC and I be the incenter of the triangle. Prove that 3 3 abc IC IB IA ⋅ ⋅ ≤
and determine when equality occurs.
Solution. CHAN Tsz Lung and
Kin-Chit O (STFA Cheng Yu Tung Secondary School). I B C A P Q R
Let r be the radius of the incircle and s be the semiperimeter (a + b + c)/2. The area of ɔABC is (a + b + c)r/2 =sr and
) )( )( (s a s b s c s − − − by Heron’s formula. So r2 = (s-a)(s-b)(s-c)/s. (*)
Let P, Q, R be the feet of perpendiculars from I to AB, BC, CA. Now s = AP + BQ + CR = AP + BC, so AP = s-a. Similarly, BQ = s-b and CR = s-c. By the AM-GM inequality,
s/3 = [(s−a)+(s−b)+(s−c)]/3
≥3 (s−a)(s−b)(s−c). (**)
Using Pythagoras’ theorem, (*) and (**), we have IA2·IB2·IC2 = [r2+(s−a)2][r2+(s−b)2][r2+(s−c)2] = [(s−a)bc/s][(s−b)ca/s][(s−c)ab/s] ≤ (abc)2/33
with equality if and only if a = b = c. The result follows.
Other commended solvers: HUDREA Mihail (High School “Tiberiu Popoviciu” Cluj-Napoca Romania), KWOK Lo Yan (Carmel Divine Grace Foundation Secondary School, Form 5), MA Hoi Sang (Shun Lee Catholic Secondary School, Form 5) and Anna Ying PUN (STFA Leung Kau Kui College, Form 6).
Problem 225. A luminous point is in space. Is it possible to prevent its luminosity with a finite number of disjoint spheres of the same size?
(Source: 2003-2004 Iranian Math Olympiad, Second Round)
Official Solution.
Let the luminous point be at the origin. Consider all spheres of radius r = 2/4 centered at (i, j, k), where i, j, k are integers (not all zero) and |i|, |j|, |k| ≤ 64. The spheres are disjoint as the radii are less than 1/2. For any line L through the origin, by the symmetries of the spheres, we may assume L has equations of the form y = ax and z = bx with |a|, |b| ≤ 1. It suffices to show L intersects one of the spheres.
We claim that for every positive integer n and every real number c with |c| ≤ 1, there exists a positive integer m ≤ n such that |{mc}| < 1/n, where {x} = x – [x] is the fractional part of x.
To see this, partition [0,1) into n intervals of length 1/n. If one of {c}, {2c}, …, {nc} is in [0,1/n), then the claim is true. Otherwise, by the pigeonhole principle, there are 0 < m’ < m” ≤ n such that {m’c} and {m”c} are in the same interval. Then |{m’c}–{m”c}| < 1/n implies |{mc}| < 1/n for m = m” – m’ ≤ n.
Since |a| ≤ 1, by the claim, there is a positive integer m ≤ 16 such that |{ma}| < 1/16 and there is a positive integer n ≤ 4 such that |{nmb}| < 1/4. Now |{ma}| < 1/16 and n ≤ 4 imply |{nma}| < 1/4. Then i = nm ≤ 64 and j = [nma], k = [nmb] satisfy |j–nma| < 1/4 and |k–nmb| < 1/4. So the distance between the point (i, ia, ib) on L and the center (i, j, k) is less than r.
Therefore, every line L through the origin will intersect some sphere.
Olympiad Corner
(continued from page 1)
three neighbors. Now there are 4n + 1 frogs at the pond. If there are three or more frogs at one part, then three of the frogs of the part will jump to the three neighbors respectively.
Prove that at some time later, the frogs at the pond will be uniformly distributed. That is, for any part, either there is at least one frog at the part or there is at least one frog at each of its neighbors.
Problem 4. Given a sequence {an}
satisfying a1 = 21/16 and 2an- 3an-1
=
3/2n+1, n ≥ 2. Let m be a positive
integer, m ≥ 2.
Prove that if n ≤ m, then
) ) 3 2 ( ( ) 2 3 ( 1/ ( 1)/ 3 m m n m n n m a + + − − . 1 1 2 + − − < n m m
Problem 5. Inside and including the boundary of a rectangle ABCD with area 1, there are 5 points, no three of which are collinear.
Find (with proof) the least possible number of triangles having vertices among these 5 points with areas not greater than 1/4.
Problem 6.
Find (with proof) all nonnegative integral solutions (x, y, z, w) to the equation
Problem 1. Solving for z in the first two inequalities, we get
a=sinB1sin6’~+cos8~cos8~~z~sin8~sin8~-cosB~cos82=b (1) c=sin8~sin~~+COSe~COSe4~z~sin83sin8~-c0s~3c0se4=d (2) res
li ectively. Using cos2 8 = 1 - sin2 8, we get cos2 e1 cos2 e2 - sin2 e1 sin2 e2 = 1 - sin2 e1 - sin 82. Using this equation and a similar one for f& and 04, the third inequality can be re&_ated as
(~0s e1 cos e2 + cos e3 cos ed)2 - (sin e1 sine2 - sin e3 sin e4)2 1 0 This is the same as (a - d)(c - b) > 0.
(3).
If (1) and (2) hold, then a >_ 2 2 d and c > z 2 b and so (u - d)(c - b) 2 0 holds, which is (3). For the converse, first note ISi1 < r/2 implies a > b and c > d. If (3) holds, then either (a 2 d and c > b) or (d > a and b 2 c). In the former case, each of a,c is greater than or equal to each of b, d. So the intersection of [b, u] and [d, c] is nonempty and there exists z satisfying (1) and (2). In the latter case, d > a > b 2 c > d leads to a contradiction.
Problem 2. (Solution 1) Let L’, L” be the feet of the perpendiculars from L to sides AB,AC,respectively. NoteLDlFzL = LDzElLandLFzDlL= LElDzLimplynLDlFz,
LDPE~ are similar. So LF2fLE1 = D~F~/D~EI. Let aI = LBAL, a2 = LCAL,/31 =
LCBM,& = LABM,rl = LACN and 72 = LBCN. Also, let CY~ = LL’F2L and (~4 = LL”ElL. J3y extended sine law, sinas/sinQq = D2FllDlE2. Then
sinal LL’ L F2 sin CY~ DlFz. 024
-=-=
sin QI~ LL” L El sin (~4 = D2E1. DlE2’
We get similar equations for sin
pl/
sin & and sin711 sin 72. The product of these ratios of the sines is 1. By the trigonometric form of Ceva’s theorem, AL, BM, CN concur.(Solution 2) Let P = DIFI rl DzEz, Q = EIDI f~ EzF2, R = FlEl fl FzDz. Applying
Pascal’s theorem to E2, E1, Dl,Fl, F2, Dz, we get A, L, P are collinear. Applying it to Fz,Fl, El, D1, D2, E2, we get B, M, Q are collinear. Applying it to D2, DI, FI, El, E2, F2, we get. C, N, R are collinear.
Let X = E2EI n DlF2 = CA f~ DlF2, Y = FzFl 17 ElD2 = AB f~ ElD2, Z =
‘D2 D1 n FI E2 = BC n FI E2. Applying Pascal’s theorem to DI ,4, El, Ez, DZ , Fz, we get p, R, X are collinear. Applying it to El, DI, 4, Fz, Ez, Dz, we get Q, P, Y are collinear.
Applying it to FI, E1, D1, D2, F2, E2, we get R, Q, Z are collinear.
For AABC and PQR, we have CAn RP = X, AB n PQ = Y, BC n QR = Z. BY ~the converse of Desargues’ theorem, lines A.?’ = AL, BQ = BM, CR = CN concur. .Problem 3. We will say a part is in equilibrium if there is at least one frog in the part or
each of its three neighbors has at least one frog. We also say an explosion occurs in a part if there are three frogs in the part and they each jump to a different neighboring part.
W eventually, then each of its neighbors will have a frog jumped from W and if these neighbors never have any explosion, then W is in equilibrium forever, otherwise there is an explosion in a neighboring part and a frog wil/ jump back into W. For the problem, thus it suffices to show eventually every part will h+ve a frog for the first time.
Label the sectors of the pond in the clot each sector, there are two parts of the pond, I
‘wise direction with numbers 1,2,. . . , n. In n inner part having the center as a vertex and an outer part having an arc on the edge of he pond as its boundary. Assume first, that no frog will ever be in any of the two parts in
i
ector 1. Since 4n + 1 - > 2, there is always 2n
a part exploding. With our assumption, the explosions can only be in parts of sectors 3,4,. . . , n - 1. For each frog in sector k, sssi n it the value k2. Consider the sum of all these values. After an explosion, this sum incr ases by (k - 1)’ + k2 + (k + 1)2 - 3k2 = 2. Since there are finitely many ways of distributi$g the frogs, this increase of the sum cannot happen forever. So our assumption is wrong. ventually there is a frog X in sector 1 for the first time.
- > 2, explosions will continue. By the
2n - 1 % e reasoning as above, explosions cannot only occur in sectors 3,. . . , n - 1. So there will be a second frog Y in sector 1 later. By the assumption, it will be in the same part, /as X. Next, ignore both X and Y. Since
4n - 1
Assume there is never a frog in the 0th r part of sector 1, then ignore X. Since 4n
- > 2, repeating the reasoning again,
2n - 1 th re will be a third frog in the same part as t
X and Y. Then explosion in that part will occ r and a frog will go into the other part of sector 1, a contradiction. Therefore, eventually every part will be in equilibrium.
Problem 4. we have 2”~~ = 3. 2n-1~,_l J 3 4.
SO b, + 2 = 3(b,_l + i). We have bl = 4
Let. b, = 2”u,, then b,, = 3b,,_l + i.
21
; =
-8 .7 24 I -, 8 b,, + 3”-‘(bl + ;) = 3” and so a, =
(;y - &.
The inequality to be prove/l is thenMultiplying both sides by (m - n f l)/(m + l), it is equivalent. to
P-
-2-)@I”’
(m -@(--l)i-) <
m - 1. ) By the AM-GM inequality,I
(l-- m”,l)~+___E_)” ;.1;..; <(m(l-~)+(mn-m))mn_(~)mn, mn mn--m factors
Also, for m L 2,
2 2n/m
Combining the last two inequalities, we get 1 - & < ($)n 5 (2) . Thus, it
suffices to prove that (i)“/“(m - (~)~(~-“‘~) < m - 1. In terms of CL (2/3)“/“, this is the same as e(m - I”‘-‘) < m - 1. Note 0 < e < 1. So (e - 1)m - (e’” - 1) = (! - l)[m - (P”-i + emV2 +. . . + l)] < 0, which can be rearranged as f?(rn - Lmwl) < m - 1 and we are done.
Problem 5. We will first recall the fact that if three points are taken from the inside or boundary of a rectangle I?,, then the area of the triangle formed by these points is at most half the area of the rectangle. To see this, enclose the triangle in the smallest rectangle ‘R’ (with sides parallel to R). If there are two vertices of the triangle on the same edge of R’, then the area of the triangle is easily calculated to be at most half that of R’. If the three vertices of the triangle are on three edges of ‘R’, then the area of the complement of the triangle in 77,’ can easily be computed to be at least half of the area of I?. Since R’ is in R, the fact follows.
Let the five points be M, N, P, Q, R. If three of them are the vertices of a triangle with 1
area at most -, then we say the three points form a gc& triple. We will show there are 4
at least 2 good triples.
Let E, F, G, H be the midpoints of sides AB, CD, DA, BC respectively and 0 be the intersection of EF with GH. By pigeonhole principle, we may assume M, N are in rectangle
AEOG.
If two of P, Q, R are outside rectangle OHCF, then take M, N with each of these two points and note that these triples are in either AEFD or ABHG, hence they are good by the fact above.
Otherwise, there are at least two of P, Q, R, say P, Q, in or on the boundary of
OHCF. If R is in OFDG or OEBH, then M, N, R and P, Q, R are good by the fact
above. Otherwise, it suffices by symmetry to consider P, Q, R in OHCF.
Consider the convex hull of M, N, P, Q, R. It is inside hexagon AEHCFG with area 3
-4. In the case, the convex hull is a pentagon, say MNPQR. Then the sum of the areas of
AMQR, MPQ, MNP is the area of the pentagon, which is at most i. Hence one of these
triangles has area at most a and the vertices of that triangle form a good triple. Similarly, considering ANRM, NQR, NPQ, we get another good triple.
Next if the convex hull is a quadrilateral, say MNPQ. Consider AMNR, NPR, PQR,
QMR. The sum of their areas is the area of MNPQ, which is at most i. There cannot
be three of these with areas greater than a. So at least two have areas at most a. Again, there are at least two good triples.
If the convex hull is a triangle, say MNF” Then the sum of the areas of AMNQ, NPQ, PMQ is the area of MNP, hence at most 4. So one of these triangles has area at most
1
-4. Similarly, the sum of the areas of AMNR, NPR, PMR is the area of MNP. So at the end, we get at least two good triples.
Finally, we give an example of five points with exactly two good triples. Take M on side
AB,NonsideADwithAM:MB=AN3:ND=2:3.Thent~eP=B,Q=C,R=D.
We have areas of AMBN, MDN equal 25 and the others have areas greater than t. Problem 6. Note 5” ‘7”’ + 1 is even implies z 2 1.
Case 1: (y = 0). Then 2” - 5’ .7”’ = 1. If z # 0, then 2= z l(mod 5) and so 412. Then 312” - 1. We get 1 = 2” - 5” .7” zs 1 - (-l)‘(mod 3), a contradiction.
If z = 0, then 2z - 7”’ = 1. Testing 2 = 1,2,3, we find (5,~) = (l,O), (3,l) are solutions. If z 2 4, then Tu E -l(mod 16) and direct computations show there is no possible w. So if y = 0, then we get solutions (2, y, z, w) = (1, 0, 0, 0), (3,0,0,1).
Case 2: (y > 0, x = 1). Then 2 . 3Y - 5= implies z is odd. Then 2 . 3’ G 1 (mod 5)
. 7’” = 1. So -(-1)’ E l(mod 3), which and hence.3 E 1 (mod 4). Assume w # 0. Then 2 . 3” 3 1 (mod 7) implies y = 4 (mod 6), but this contradicts y E 1 (mod 4). So
w = 0 and 2.38 - 5* = 1. If y = 1, then z = 1. If y 2 2, then 5% = -1 (mod 9) so that z G 3 (mod 6). Then 53 + 115” + 1 implies 715’ + 1 = 2.3v, a contradiction. So only one solution (5, y, z, w) = (1, 1, 1,0) in this case.
Case 3: (y > 0, x > 2). Taking (mod 4) and (mod 3) respectively of the equation, we get -(-1)“’ E 1 (mod 4) and -(-1)’ = 1 (mod3). So z and w are odd. Next 2x~3~=52~7Y+1~35+1~4(mod8),whichimplies~=2.Then4~3~-5z~TU=l
with z, w odd. Taking (mod 5) and (mod 7) respectively, we get -33 z 1 (mod 5) and 4 .3Y E 1 (mod 7), which imply y E 2 (mod 12). Write y = 12m + 2, m 10. Then
52 .7w = 4.3y - 1 = (2.3sm+l - 1)(2.3sm+l + 1).
Now2~36m+‘+1=6~93m+1zs6+l~O(mod 7).Since2.36m+‘-1(> 1)and2.3sm+‘+1 are relatively prime, we get 512. 36mf1 - 1. Then
2 .3sm+i _ 1 = 52, 2.3sm+i f 1
5.7.
For m 2 1, the first equation implies 5’ s -1 (mod 9). As in case 2, this leads to 7152 + 1 = 2.3sm+*, a contradiction. So m = 0, then (I, y, z, w) = (2,2,1,1).
In conclusion, all the nonnegative solutions are (2, y, z, w) = (1, 0, 0, 0), (3,0,0, l), (1, 1, 1,O) and (2,2,1,1).
