Answers: ( HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 21 September 2018

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1 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember Individual Spare Spare Grup Individual Evens I In hw many pssible ways can 8 idenical balls be disribued disinc bes s ha every b cnains a leas ne ball? Reference: 00 HG, 006 HI6, 0 HI lign he 8 balls in a rw. There are 7 gaps beween he 8 balls. u sicks in w f hese gaps, s as divide he balls in grups. The fllwing diagrams shw ne pssible divisin. The hree bes cnain balls, balls and ball. The number f ways is equivalen he number f chsing gaps as sicks frm 7 gaps The number f ways is I If α and β are he w real rs f he quadraic equain 0, find he value f α 6 + 8β. Reference 99 HG, 0 HG If α, β are he rs f 0, find α + β. α + β, αβ α α + α 6 (α ) (α + ) α + α + α + α(α ) + (α + ) + α + α(α + ) + 6α + α + 7α + (α + ) + 7α + 8α + α 6 + 8β 8(α + β) I If a L +, find he value f a. (Reference: 0 HG) a L L a 0 I Given ha + y + z and + y + z, where, y, z are inegers. If < 0, find he value f y. Le a, where a > 0, hen y + z a + (), y + z a + () Frm (): (y + z) yz(y + z) a + (a + ) a yz(a + ) a + 9a + 7a + 7 a 9a + 7a + a + 9a yz a + () ( a + ) ( a + ) a + a + yz is an ineger a r (y z) (y + z) yz When a,, y + z frm () and yz frm () (y z) < 0, impssible. Rejeced. When a, y + z 8 and yz 6 Slving fr y and z gives, y, z hp:// age

2 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08 I Given ha a, b, c, d are psiive inegers saisfying lg a b and lgc d. If a c 9, find he value f b d. a b and c d a b and c d Sub. hem in a c 9. b d 9 b + d b d 9 d b +, b d (n sluin, rejeced) r b + d 9, b d b, d b, d 8 b d I6 If y + y, where 0, y, find he value f + y. Mehd Le sin, y sin, hen y cs, cs The equain becmes sin cs + cs sin sin ( + ) y sin + sin sin + sin (90 ) sin + cs Mehd ( y ) y y y + y ( ) y + y y ( ) y y + + y + + y + y y + 0 ( + y ) ( + y ) + 0 ( + y ) 0 + y I7. In figure, D is a rapezium. The lenghs f segmens D, and D are, 7 and respecively. If segmens D and are sin α bh perpendicular D, find he value f. sin β Mehd Draw a perpendicular line frm n D. an β ; an(α + β) 7 an( α + β) anβ 7 an α an[(α + β) β] + an( α + β) anβ sin α ; sin β sin α 7 sin β Mehd β (al. s, D // ) sin α 7 7 (Sine law n ) sin β hp:// age

3 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08 I8. In Figure, is a riangle saisfying y z and 7z. If he maimum value f is m and he minimum value f is n, find he value f m + n. 7k, z k, + y + z 80 y 80 k y z 7k 80 k k 8k 80 and 80 k k 0 8 7k 70 m 8, n 70 m + n I9 rrange he numbers,,, n (n ) in a circle s ha adjacen numbers always differ by r. Find he number f pssible such circular arrangemens. When n, here are w pssible arrangemens:,, r,,. When n, here are w pssible arrangemens:,,, r,,,. Deducively, fr any n, here are w pssible arrangemens:,,, 6, 8,, larges even ineger, larges dd ineger,, 7,, r,,, 7,, larges dd ineger, larges even ineger,, 6,,. I0 If is he larges ineger less han r equal, find he number f disinc values in he 00 fllwing 00 numbers:,,, Reference: IMO reliminary Selecin nes - Hng Kng 006. n Le f(n), where n is an ineger frm n + f(n + ) f(n) 00 f(n + ) f(n) < n + < n < f(00) f() 0, f() 0,, f(00) 0, he sequence cnain 0 differen inegers. On he her hand, when n > 00, f(n + ) f(n) > ll numbers in he sequence f(006),, f(00) are differen, al 00 numbers The number f disinc values is 08. hp:// age

4 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08 Spare individual IS In Figure, is an issceles riangle and is a pin n. If + : k :, find he value f k. Reference: 00 FI. Le a, a,, y, Le θ, 80 θ (adj. s n s. line) pply csine rule n and + y a + a cs θ (); cs θ () y Figure + y a + a () + (): + 0 y ( + y a ) + y( + a ) 0 ( + y) + y( + y) a ( + y) 0 ( + y)( + y a ) 0 + y 0 (rejeced, > 0, y > 0) r + y a 0 + y a (*) + : + y : [( + y) y] : [ y] : (a y) : (a y) : : by (*) k Mehd (rvided by hiu Lu Sau Memrial Secndary Schl Ip Ka H) (base s issceles riangle) ( s sum f ) Rae aniclckwise 90 abu he cenre a. and 90 (prpery f rain) 90 (given) (S..S.) (crr. s s) (crr. sides s) : ( + ) : : (yhagras herem) : cs : k hp:// age

5 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08 Grup Evens G Given ha he si-digi number 0yz is divisible by 7, 9,. Find he minimum value f he hree-digi number yz. Reference: 000 FG. There is n cmmn facr fr 7, 9, and he L..M. f hem are is divisible by 7 and is divisible by , The hree-digi number is 8. G Find he smalles psiive ineger n s ha L 009 is divisible by. G G n cpies f 009 Reference: 008 FI. Sum f dd digis sum f even digis muliples f n(0 + 9) n( + 0) m, where m is an ineger. 7n m Smalles n. In figure, is a riangle. D is a pin n such ha D. If 0, find he value f. Reference: 98 FI. Le y, hen y y y 0 0 y ( s sum f ) 80 ( 0 y ) D D 0 + y (base s iss. ) D D 0 + y y 0 (e. f D) 0 Mehd Le y D + y (e. f D) D + y (base s issceles D) + + y + y 0 + y y 0 0 In figure, given ha he area f he shaded regin is cm. If he area f he rapezium D is z cm, find he value f z. Reference 99 HI, 997 HG, 000 FI., 00 FI., 00 HG7, 0 HG Suppse and D inersec a K. 0 S D 60 S DK + S K + S K S K K and DK have he same heigh bu differen bases. K : KD S K : S DK : : 7 K, KD 7 K ~ DK (equiangular) S K : S DK K : DK 7 : 9 : K and DK have he same heigh bu differen bases. S K : S DK K : KD : 7 z S D Figure G Three numbers are drawn frm,,,,, 6. Find he prbabiliy ha he numbers drawn cnain a leas w cnsecuive numbers. Mehd Favurable ucmes {,,, 6,,, 6,,, 6,,, 6, 6, 6, 6}, 6 ucmes 6 rbabiliy Mehd rbabiliy (, 6, 6 r 6) hp:// age

6 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08 G6 Find he minimum value f he fllwing funcin: f() + + L + 000, where is a real number. Reference: 99 HG, 00 HG9, 00 FG., 008 HI8, 008 FI., 0 FGS., 0 FG. Mehd f(00) L ( L + ) Le n be an ineger, fr n 00 and n, n + (00 n) n + 00 n n 00 n + 00 (00 n) 00 n + 0 n 00 n Fr n < 00, n + (00 n) n + 00 n 00 n If 00, f () f (00) 000 n 00 n 000 n 00 n 00 [ ] n + ( 00 n) 00 n + 00 ( 00 n) n 00 [ 00 ( 00 n) ] n n 0 f(00 ) L L L f() f() f(00) 0000 fr all real values f. Mehd We use he fllwing resuls: () a b b a and () a + b a + b LLLLLLLLLLLLLLLLLLLL dd up hese 00 inequaliies: f() + + L ( + 999) n hp:// age 6

7 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08 G7 m Le m, n be psiive inegers such ha < <. Find he minimum value f n. 00 n 009 Reference: 996 FG0., 00 HI Mehd 009 n m 008 > > n n < < n m n m < < 009 n m n m m n m 007 > > n m n m n m < < n m n m m < < 008 n m n m 007 m laim: < < fr a 0,,, L, a n am 009 a rf: Inducin n a. When a 0,, ; prved abve. m Suppse < < fr sme ineger k, where 0 k < k n km 009 k 009 k m n ( k + ) m 008 k > > 00 k 00 k n km n km 009 k 009 k 00 k n km 009 k + < < k 009 k n ( k + ) m 008 k 008 k n km m < < 009 k n ( k + ) m n ( k + ) m 008 k m < < 00 ( k + ) n ( k + ) m 009 ( k + ) y MI, he saemen is rue fr a 0,,,, 008 m u a 008: < < n 008m m < < n 008m m The smalles pssible n is fund by n 008m m, n 008 n 09 m n Mehd < < 00 > > m > n > 009m 00 n 009 m m, n are psiive inegers. We wish find he leas value f n I is equivalen find he leas value f m. When m, 00 > n > 009, n sluin fr n. When m, 00 > n > 08 n 09 hp:// age 7

8 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08 G8 G9 Le a be a psiive ineger. If he sum f all digis f a is equal 7, hen a is called a lucky number. Fr eample, 7, 6, 0 are lucky numbers. Lis all lucky numbers in ascending rder a, a, a,. If a n 600, find he value f a n. Number f digis smalles, L, larges Number f lucky numbers subal 7 6,, L, 6, ,, L, ,, L, 0 6 0,, L, 0 LLLLL L , 0, L, ,, L, 0 6 0, L, 0 LLLL L 600 a ,, 00 6 LLLL L 00 00, L, 00 LLLL L 00 XYZ +++ XYZ ++ 6XYZ XY 7 00 a 8 00 If lg ( + y) + lg ( y), find he minimum value f y. ( + y)( y) y y + T y ( y +) y T + y ( y +) T + y + y T (y + ) y y T + ( T ) 0 [T ()( T )] 0 T 0 T The minimum value f y is. hp:// age 8

9 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08 G0 In Figure, in,,. If and are w pins n and respecively, and, find he value f. Reference:00 HG9, HKEE 00 0 Mehd Jin. (base s iss. ) (base s iss. ) (e. f ) Le R be he mid pin f. Jin R and prduce is wn lengh S s ha R RS. Jin S, S and S. S is a //-gram (diagnals bisec each her) S S (pp. sides f //-gram) S // S (crr. s, S // ) S (S..S.) S (crr. sides, 's) S S S is an equilaeral riangle. S S 60 S (crr. s, 's) S + (crr. s, 's) S R In, ( sum f ) 0 Mehd Le, le y (base s iss. ) y cs y + (base s iss. ) (e. f ) (e. f ) 90 ( sum f iss. ) sin sin cs sin( 90 ) sin sin 0 (Sine law n ) hp:// age 9

10 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08 Mehd Reflec alng R R (by cnsrucin) Jin R, R (crr. sides, 's) (base s iss. ) (base s iss. ) (e. f ) R (crr. s, 's) R R + R + R (given) R (crr. sides, 's) R R (S..S.) R (given) (crr. sides, 's) R is an equilaeral riangle. ( sides equal) R 60 R ( f an equilaeral riangle) ( sum f iss. ) R (crr. s, 's) 0 Mehd Le Use as cenre, as radius draw an arc, cuing a R. R (radius f he arc) (base s iss. ) R (e. f ) R (base s iss. ) R (e. f R) R 90 ( sum f iss. R) R R R (e. f R) ( sum f iss. ) R (sides pp. eq. s) R is an equilaeral riangle. ( sides equal) R 60 0 hp:// age 0

11 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08 Mehd Le, y (base s iss. ) (base s iss. ) (e. f ) (e. f ) E D 90 ( s sum f ) s shwn, cnsruc w riangles s ha D DE Jin E, D,., D (crr. sides 's) D 80 D D () D ; D y (adj. s n s. line) (S..S.) (crr. sides 's) (crr. s 's) DE DE + D D (adj. s n s. line) DE () DE () (by cnsrucin, crr. sides 's) y (), () and (), DE (S..S.) E y + E (crr. sides 's) E is an equilaeral riangle E (angle f an equilaeral riangle) 0 Mehd 6 The mehd is prvided by Ms. Li Wai Man nsruc anher idenical riangle D s ha D, E E D D and D D is a parallelgram (pp. sides equal) E and E // (prpery f //-gram) E is a parallelgram (pp. sides equal and //) E E (prpery f //-gram) E is an equilaeral riangle E D E hp:// age

12 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08 Mehd 7 Le, y (base s iss. ) (base s iss. ) (e. f ) D (e. f ) 90 ( sum f ) y s shwn, reflec alng D DD Jin D,, D., D (crr. sides 's) D 80 D D () D D is an equilaeral riangle. D 60 0 (adj. s n s. line) (S..S.) (crr. sides 's) (crr. sides 's) Spare Grup GS In Figure, D is a recangle. Le E and F be w pins n D and respecively, s ha FE is a rhmbus. If 6 and, find he value f EF. Le F F E E DE 6 F In DE, + (6 ) (yhagras Therem) In D, + 6 (yhagras Therem) 0 G cenre f recangle cenre f he rhmbus G G 0 (Diagnal f a recangle) Le EG FG (Diagnal f a rhmbus) In EG, + G (yhagras Therem) EF G D 6 - E F6 - hp:// age

13 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08 Gemerical nsrucin. Figure shws a line segmen f lengh uni. nsruc a line segmen f lengh 7 unis. Mehd () Draw a line segmen. () Draw he perpendicular bisecr f, O is he mid-pin f. O O. () Use O as cenre, O O as radius draw a semi-circle. () Use as cenre, radius draw an arc, which inersecs he semi-circle a. 90 ( in semi-circle) () Jin. Figure 7 (yhagras Therem) Mehd () Draw a line segmen, wih, 7. () Draw he perpendicular bisecr f, O is he mid-pin f. O O. () Use O as cenre, O O as radius draw a circle wih as diameer. () Draw a perpendicular line. ( frm cenre bisec chrd) y inersecin chrd herem, O 7 O hp:// age

14 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08. Given ha is equilaeral., and R are disinc pins lying n he lines, and such ha O, O, OR and O O OR. Figure shws he line segmen O. nsruc. Figure nsrucin seps () Use O as cenre, O as radius cnsruc an arc; use as cenre, O as radius cnsruc anher arc. The w arcs inersec a H and I. OH and OI are equilaeral. () Use H as cenre, H as radius cnsruc an arc; use as cenre, H as radius cnsruc anher arc. The w arcs inersec a O and J. HJ is equilaeral. () Use I as cenre, I as radius cnsruc an arc; use as cenre, I as radius cnsruc anher arc. The w arcs inersec a O and K. IK is equilaeral. () Use H as cenre, HJ as radius cnsruc an arc; use J as cenre, JH as radius cnsruc anher arc. The w arcs inersec a and. HJ is equilaeral. () Use I as cenre, IK as radius cnsruc an arc; use K as cenre, KI as radius cnsruc anher arc. The w arcs inersec a and. IJ is equilaeral. is he angle bisecr f HJ. O. (6) Use O as cenre, OH as radius cnsruc an arc; use H as cenre, HO as radius cnsruc anher arc. The w arcs inersec a and. OH is equilaeral. (7) Use O as cenre, OI as radius cnsruc an arc; use I as cenre, IO as radius cnsruc anher arc. The w arcs inersec a and R. OIR is equilaeral. (8) Jin, and. Then is he required equilaeral riangle. 6 8 H J O 7 R I 8 K 8 hp:// age

15 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08. Figure shws a line segmen. nsruc a riangle such ha : : and 60. Mehd Sep nsruc an equilaeral riangle D. Sep nsruc he perpendicular bisecrs f and D respecively inersec a he circumcenre O. Sep Use O as cenre, O as radius draw he circumscribed circle D. Sep Lcae M n s ha M : M : (inercep herem) Sep The perpendicular bisecr f inersec he minr arc a X and a. rduce XM mee he circle again a. Le M θ, M α. X X (S..S.) X X (crr. sides s) X X θ (eq. chrds eq. angles) M α, M 80 α (adj. s n s. line) k : sin θ : sin α () (sine rule n M) k : sin θ : sin (80 α) () ( M) Use he fac ha sin (80 α) sin α; () (): : : D 60 ( s in he same segmen) is he required riangle. Mehd Sep Use as cenre, as radius draw an arc H. Sep Draw an equilaeral riangle H (H is any pin n he arc) H 60 Sep Lcae M n H s ha M H (inercep herem) Sep rduce M mee he arc a. Sep Draw a line // H mee prduced a. 60 (crr. s, H // ) ~ M (equiangular) : : M (rai f sides, ~ 's) k O α D X θ M k M H : : is he required riangle. hp:// age

16 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08 Mehd (rvided by Mr. Lee hun Yu, James frm S. aul s -educainal llege) D 7 k 8 O M k 8 6 L 0k N Sep nsruc an equilaeral riangle D. Sep nsruc he perpendicular bisecrs f, D and D respecively inersec a he circumcenre O. Sep Use O as cenre, O as radius draw he circumscribed circle D. Sep Lcae M n s ha M : M : (inercep herem) Sep rduce N s ha N. Le M k, M k, N 0k, hen N : N k : 0k : (signed disance) N divides eernally in he rai :. Sep 6 nsruc he perpendicular bisecrs f MN lcae he mid-pin L. Sep 7 Use L as cenre, LM as radius draw a semi-circle MN which inersecs he circle D a. Sep 8 Jin and, hen is he required riangle. rf: Mehd. D (-,0) M(-,0) (0,0) L(,0) 6 N(0,0) Fr ease f reference, assume M, M Inrduce a recangular c-rdinae sysem wih as he rigin, MN as he -ais. The crdinaes f, M,, L, N are (, 0), (, 0), (0, 0), (, 0) and (0, 0) respecively. Equain f circle MN: ( + )( 0) + y 0 y () Le (, y). ( ) + + y + y + + D 60 is he required riangle by () ( s in he same segmen) + by () hp:// age 6

17 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08 rf: (mehd.) MN k ML LN 6k L k Jin M, N. Draw TL // N TL inersecs, M and a S, R and respecively. MN 90 in semi-circle T k D S R M k k L 6k N T and are he fee f perpendiculars frm and n TL respecively. MRL 90 (crr. s TL//N) Le R RM ( frm cenre bisecs chrd) SR MSR (S..S.) and R MR (S..S.) S MS and M (*) (crr. sides, s) LMR ~ LT (T // MR, equiangular) T : MR L : ML (rai f sides, ~ s) 9k T. 6k TS ~ RS (T // R, equiangular) S : S T : R (rai f sides, ~ s). : : () LMR ~ L ( // MR, equiangular) : MR L : ML (rai f sides, ~ s) k 6k ~ R ( // R, equiangular) : : R (rai f sides, ~ s) : : () y (): S : S : M : M SM // (cnverse, herem f equal rai) y (): : : M : M // M (cnverse, herem f equal rai) SM is a parallelgram frmed by pairs f parallel lines y (*), S MS and M SM is a rhmbus Le SM θ M (rpery f a rhmbus) Le M α, M 80 α (adj. s n s. line) k : sin θ : sin α () (sine rule n M) k : sin θ : sin (80 α) () (sine rule n M) Use he fac ha sin (80 α) sin α; () (): : : D 60 ( s in he same segmen) is he required riangle. hp:// age 7

18 nswers: (009-0 HKMO Hea Evens) reaed by: Mr. Francis Hung Las updaed: Sepember 08 Mehd (rvided by hiu Lu Sau Memrial Secndary Schl Ip Ka H) U T S V R W Sep nsruc an equilaeral riangle R. (R is any lengh) Sep rduce and R lnger. On prduced and W prduced, mark he pins S, T, U, V and W such ha S ST TU V VW, where S is any disance. Sep Jin UW. Sep py UW. Sep py WU. and inersec a. is he required riangle. rf: y sep, R 60 (rpery f equilaeral riangle) y sep, U : W : y sep and sep, UW and WU UW ~ (equiangular) : U : W : (crr. sides, ~ s) The prf is cmpleed. hp:// age 8