[PDF] MOEMS Division M Problems 2010-2011 with Solutions and Answers.pdf | Plain Text

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 1/16   O L Y M PIA D   P R O BLE M S   2 010-2 011   D IV IS IO N  M M ath em atic a l O ly m pia d s f o r E le m en ta ry a n d M id dle S ch ools  M ath em atic a l O ly m pia d s f o r E le m en ta ry a n d M id dle S ch ools  M ath em atic a l O ly m pia d s f o r E le m en ta ry a n d M id dle S ch ools  M ath em atic a l O ly m pia d s f o r E le m en ta ry a n d M id dle S ch ools  M ath em atic a l O ly m pia d s f o r E le m en ta ry a n d M id dle S ch ools    A N onpro fit P ublic F oundatio n 2 1 54 B ellm or e A ve nue B el lm o r e, N Y 11 710-5 6 45 P H O N E: (5 1 6 ) 7 8 1- 24 00   F A X : (5 1 6 ) 7 8 5 -6 6 4 0   E - M A IL : o f fi c e @ m oem s. o r g W E BS IT E : w w w .m o e m s .o r g O ur T hir ty -S eco n dY ea r Sin ce 1 979 M ATH O LY M PIA D S O L Y M PIA D   P R O BLE M S 2 010-2 011 D IV IS IO N  M W IT H A N SW ER S A N D S O LU TIO NS

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 2/16   C opyrig ht © 2 011 b y M ath em atic a l O ly m pia ds f o r E le m enta ry a nd M id dle S ch ools , I n c. A ll r ig hts r e se rv e d.

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 3/16   C opyrig ht © 2 011 by M ath em atic a l O ly m pia ds fo r E le m en ta ry a nd M id dle S ch ools , In c. A ll rig hts re se rv ed .   P age 3  o n C on te st D iv is io n OLY M PIA D S M ATH   f o r E le m en ta ry a n d M i ddle S ch ools   M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s  Div is io n M 1 A T im e: 3 m in ute s 2 5 d ig it s a re s h ow n. F in d t h e s u m o f t h e d ig it s . 1 B   T im e: 4 m in ute s H ow m any q uarte rs ( w orth 2 5 c e nts e ach ) m ust b e a dded t o 1 2 n ic ke ls ( w orth 5 c e nts e ach ), s o t h at t h e a ve ra ge v a lu e o f a c o in in t h e n ew e nla rg ed c o lle ctio n is 1 0 ce nts ? 1 C   Tim e: 5 m in ute s H ow m any d if fe re nt s u m s c a n b e o bta in ed b y a ddin g t w o d if fe re nt i n te gers c h ose n fr o m t h e s e t b elo w ? {– 12, – 11, – 10, … , + 6, + 7, + 8} 1 D   Tim e: 5 m in ute s 5 61 is t h e p ro duct o f 3 d if fe re nt p rim e n um bers . H ow m any f a cto rs o f 5 61 a re n ot prim e? 1 E   T im e: 7 m in ute s In r e cta ngle   A BC D ,   P  i s t h e m id poin t o f s id e B C  a nd Q  i s t h e m id poin t o f C D . T he a re a o f ∆∆∆∆∆  A PQ  is w hat f r a ctio nal p art o f  th e a re a o f r e cta ngle   A BC D ? N   OVEM BER    1 7, 2 010  2 222 2262 2662 6662 4444 44444  A B D C P Q 1  

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 4/16   C on te st OLY M PIA D S M ATH   f o r E le m en ta r y a n d M id d le S ch ools   M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s  D iv is io n  P age 4   C opyrig ht © 2 011 by M ath em atic a l O ly m pia ds fo r E le m en ta ry a nd M id dle S ch ools , In c. A ll rig hts re se rv ed .  o n D iv is io n M D   EC EM BER    1 6, 2 00 9  D  EC EM BER    1 5, 2 01 0  2   2 A Tim e: 3 m in ute s W rit e a s a s in gle d ecim al: 1 – 2 1 0  + 3 1 00  – 4 1 000 2 B   T im e: 4 m in ute s  A m y p ic k s a w ho le num ber , s q u are s it a nd th e n su b tr a c t s 1 . S he giv e s h er fin a l n um ber  to B ria n. B ria n a dds 3 t o t h e n um ber A m y g ave h im a nd t h en d ouble s t h at r e su lt . B ria n’s f in al r e su lt i s 5 4. W it h w hat n um ber d id A m y s ta rt? 2 C   Tim e: 6 m in ute s  A le x, B ru no, a nd C harle s e ach a dd th e le ngth s o f tw o s id es o f th e s a m e tr ia ngle c o rre ct ly . T hey g et 2 7 c m , 3 5 c m , a nd 3 2 c m , r e sp ectiv e ly . F in d t h e p erim ete r o f t h e tr ia ngle , in c m . 2 D  Tim e: 6 m in ute s T he f ir s t t h re e t e rm s i n a s e quence are : 1 , 2 , 3 . E ach t e rm a fte r t h at i s t h e o pp osit e o f  t h e s u m o f t h e t h re e p re vio us te rm s. F or e xa m ple , t h e 4 th   t e rm i s  –   6 ( th e o pposit e o f  1 + 2+ 3), a nd t h e 5 th   t e rm i s 1 . W hat i s t h e 9 9 th   t e rm ? 2 E  T im e: 7 m in ute s F in d t h e w hole n um ber v a lu e o f  ... + + + + + + 1 3 5 4 5 4 7 4 9

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 5/16   C opyrig ht © 2 011 by M ath em atic a l O ly m pia ds fo r E le m en ta ry a nd M id dle S ch ools , In c. A ll rig hts re se rv ed .   P age 5  o n C on te st D iv is io n OLY M PIA D S M ATH   f o r E le m en ta ry a n d M i ddle S ch ools   M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s  Div is io n M 3   3 A Tim e: 4 m in ute s W hen I s a ac o pens a b ook, t h e p ro duct o f t h e p age n um bers o n t h e o pen p ages i s 4 20. F in d t h e s u m o f t h e t w o p age n um bers . 3 B   T im e: 5 m in ute s  A sq uir re l b urie s a to ta l o f 8 0 aco rn s in N  h ole s. F in d t h e g re ate st p ossib le v a lu e o f N , pro vid ed: (1 ) N o h ole is e m pty , a nd (2 ) N o t w o h ole s c o nta in t h e s a m e n um ber o f a co rn s. 3 C   Tim e: 5 m in ute s T he s u m o f a p ro per f r a ctio n in lo w est t e rm s a nd it s r e cip ro ca l e quals 2 4 1 5 . F in d th e orig in al p ro per f r a ctio n. 3 D   Tim e: 7 m in ute s T he p ic tu re s h ow s a “ s p ir a l” t h at b egin s a t t h e o rig in ( 0 ,0 )  a nd p asse s th ro ugh e ve ry la ttic e p oin t in t h e p la ne. E ach s m all a rro w is 1 u nit in le ngth . F ollo w in g t h e “ s p ir a l” , w hat i s t h e l e ngth o f t h e p ath f r o m th e o rig in t o t h e poin t (5 ,3 ) ? 3 E   T im e: 7 m in ute s H ow m any d egre es a re i n t h e a ngle f o rm ed b y th e h ands o f a c lo ck a t 8 :2 4?  x  y 3   9  6  1 2  J   A N UAR Y   1 2, 2 011 

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 6/16   C on te st OLY M PIA D S M ATH   f o r E le m en ta r y a n d M id d le S ch ools   M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s  D iv is io n  P age 6     C opyrig ht © 2 011 by M ath em atic a l O ly m pia ds fo r E le m en ta ry a nd M id dle S ch ools , In c. A ll rig hts re se rv ed .  o n D iv is io n M 4 A T im e: 4 m in ute s S om e s tu dents a re i n a l in e. A bby i s i n t h e c e nte r o f t h e l in e. S ara i s 3 p la ce s i n f r o nt o f h er, E l i i s 4 p la ce s b ehin d S ara , a nd K ayla i s 2 p la ce s i n f r o nt o f E li. K ayla i s t h e th ir d p ers o n in lin e. H ow m any s tu dents a re in t h e lin e? 4 B   T im e: 4 m in ute s G iv e n t h e d ata : 3 , 6 , 6 , 8 , 1 0, 1 2. E xp re ss i n l o w est t e rm s: × × 3 m e d i a n – m o d e 6 m e a n 4 C   Tim e: 4 m in ute s F in d t h e i n te ger t h at e xce eds  –   5 b y t h e s a m e am ount t h at + 13 e xce eds  –   1 . 4 D   Tim e: 6 m in ute s  A c ir c le w it h r a diu s 5 c m in te rs e cts a cir c le w it h r a diu s 3 c m a s sh ow n. T he a re a o f t h e s h aded r e gio n i s π 7 2   s q uare cm . F in d t h e to ta l c o m bin ed a re a in sid e th e c ir c le s, b ut o uts id e th e s h aded re gio n. L eave y o ur a nsw er in t e rm s o f π . 4 E   T im e: 8 m in ute s H ow m any d if fe re nt t r ia ngle s c a n b e f o rm ed w hose 3 v e rtic e s a re c h ose n f r o m th e re cta ngula r a rra y o f 8 p oin ts s h ow n? F   EBR UAR Y   9 , 2 01 1  4  

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 7/16   C opyrig ht © 2 011 by M ath em atic a l O ly m pia ds fo r E le m en ta ry a nd M id dle S ch ools , In c. A ll rig hts re se rv ed .   P age 7    o n C on te st D iv is io n OLY M PIA D S M ATH   f o r E le m en ta ry a n d M i ddle S ch ools   M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s M ath em atic a l O ly m pia d s  Div is io n M 5 A T im e: 4 m in ute s H ow m any 2 -d ig it n um bers a re t h ere in w hic h t h e o nes d ig it is g re ate r t h an t h e t e ns d ig it ? 5 B   T im e: 5 m in ute s  A bank h as tw o pla ns fo r c h eckin g a cco unt s. In p la n A , th e c h a rg e is $ 7.5 0 a m onth w it h n o f e e f o r e ach ch eck. In p la n B , t h e ch arg e i s $ 3 a m onth p lu s a n a d dit io nal 2 0 ce nts f o r e ach c h eck w rit te n. W hat is t h e le ast n um ber o f c h ecks a c u sto m er m ust w rit e e ach m onth s o t h at p la n A c o sts le ss t h an p la n B ? 5 C  Tim e: 5 m in ute s S uppose t h e b ase o f a t r ia ngle i s in cre ase d b y 2 0% , a nd i t s h eig ht i s i n cre ase d by 30% . B y w hat p erc e nt i s t h e a re a o f th e t r ia ngle i n cre ase d? 5 D   Tim e: 6 m in ute s S ta rtin g w it h 1 , S ara lis ts t h e c o untin g n um bers in o rd er b ut o m it s a ll t h ose t h at u se th e d ig it 9 . W hat is th e 3 00 th   n um ber o n h er l is t? 5 E   T im e: 8 m in ute s L in e s e gm ents f o rm a p ath t h at s ta rts a t ( 0 ,0 ), is d ra w n to ( 1 ,0 ), a nd t h en t o ( 1 ,2 ). E ach n ew s e gm ent fo rm s a r ig ht a ngle w it h th e s e gm ent b efo re it a nd is 1 u nit lo nger t h an t h at s e gm ent. T he p ath e nds a t ( 0 ,0 ). H ow m any s e gm ents a re in th e sh orte st p ossib le p ath ? ( H in t: C on sid er h o riz o n ta l a nd v er tic a l s e g m en ts s e p ara te ly . ) M   AR CH   9 , 2 01 1  5  

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 8/16   D i v is io n   M   M M M M  Copyrig ht © 2 011 b y M ath em atic a l O ly m pia ds fo r E le m en ta ry a nd M id dle S ch ools , In c. A ll r ig hts r e se rv ed .  P age 8 1 A M E T H O D 1 :  Str a te g y: C ount th e n um ber o f tim es e a ch d ig it a ppea rs. M ult ip ly e ach v a lu e b y t h e n um ber o f tim es it a ppears . A dd t h ose p ro ducts . 1 0× 2 + 9× 4 + 6× 6 = 92. T he s u m is 9 2 . M ETH O D 2 :  S tr a te g y: S ep ara te in to a re cta ngle o f 2 s a nd a tr ia ngle o f 4 s. 2 222 2262 2662 6662 4444 44444   = 2222 2222 2222 2222   + 4 4 4 4 4 4 4444 44444 The su m is 1 6 × 2 + 15 × 4 = 92. 1 B M ETH O D 1:  S tr a te g y: C om pare e a ch c o in to th e a vera ge v a lu e. E ach n ic ke l is w orth 5 c e nts le ss t h an t h e a ve ra ge; e ach q uarte r is w orth 1 5 c e nts m ore . T hen e ach q uarte r, c o m bin ed w it h 3 n ic ke ls , h as a n a ve ra ge v a lu e of 1 0 c e nts . T here a re 1 2 n ic ke ls , s o 4 q uarte rs m ust b e a dded t o t h e c o lle ctio n . M ETH O D 2 :  S tr a te g y: U se a lg eb ra . L et Q r e pre se nt t h e n um ber o f q uarte rs t o b e a dded. T he t o ta l v a lu e o f t h e c o in s is 2 5Q + 6 0. T he to ta l n um ber o f c o in s is Q + 1 2. T hen + + 2 5 6 0 12 Q Q  = 1 0. S olv in g, Q = 4 . T hus, 4 q uarte rs m ust b e a dded.  F O LLO W   -U    P :   H ow m any $ 5 b ill s m ust b e a dded to t w en ty $ 100 b ills so th at th e a vera ge va lu e o f a ll th e b ills is $ 10? [3 60] 1 C   Str a te g y: F in d th e le a st a nd g re a te st p ossib le s u m s. T he l e ast p ossib le s u m i s o bta in ed b y a ddin g –12 a nd – 11. T he g re ate st p ossib le s u m i s o bta in ed b y a ddin g 7 a nd 8 . E ve ry in te ger b etw een th ese e xtr e m es is a ls o a p ossib le s u m . B y e xa m in in g a n um ber lin e f r o m – 23 t o + 15 in clu siv e , y o u s h ould s e e 2 3 negativ e su m s, 1 5 p osit iv e s u m s, a nd z e ro . T here a r e 3 9 d if fe re nt s u m s th at c a n b e o b ta in ed.  F O LLO W   -U    P :   H ow m any d is tin c t s u m s c a n b e o bta in e d b y a ddin g tw o d if fe re n t in t eg ers c h ose n fr o m th e c o nse cu tiv e e ven in te g ers fr o m -1 2 to + 8, in clu siv e?   [1 9] MATH O LY M PIA D S M ATH O LY M PIA D S   A N SW ER S  A N D  S O L UTIO N S  N ote : N um ber i n p are n th ese s in dic a te s p erc en t o f a ll c o m peti to rs w ith a c o rr e ct a n sw er . O L Y M P I A D 1 N O V E M B E R 1 7 , 2 0 1 0    A n sw e r s: [1 A ] 92   [1 B ] 4   [1 C ] 39   [1 D ] 5   [1 E ] 3/8 6 9% c o rre ct 5 8% 10%

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 9/16     P age 9 C opyrig ht © 2 011 b y M ath em atic a l O ly m pia ds fo r E le m en ta ry a nd M id dle S ch ools , In c. A ll r ig hts r e se rv ed . D iv is io n   M   M M M M  1 D M E T H O D 1 :  Str a te g y: U se t h e d iv is ib ili ty r u le s t o f in d f a cto r s. T he s u m o f t h e d ig it s o f 5 61 is 1 2, s o 3 is a f a cto r a nd 5 61 ÷ 3 = 1 87. T hen 187 s a tis fie s t h e t e st o f d iv i sib ilit y f o r m ult i p le s o f 1 1 ( th at is , 1 – 8 + 7 = 0 ), s o 1 1 is a ls o a f a cto r a nd 1 8 7 ÷ 1 1 = 1 7. T hus 5 61 f a cto rs in to 3 × 1 1 × 1 7. T he ta ble s h ow s a ll f a cto r p air s o f 5 61. O f t h e 8 fa cto rs , 3 a re p rim e. T here fo re , th ere a re 5 f a cto rs o f 5 61 w hic h a re n ot p r im e. M ETH O D 2 :  S tr a te g y: F in d t h e t o ta l n um ber o f f a cto rs w ith out f a cto rin g. C all th e p rim e fa cto rs P , Q , a nd R . T heir p ro duct is P Q R  a nd th e ta ble s h ow s a ll i t s f a cto r p air s . I n a ll, 3 o f t h e 8 fa cto rs a re p rim e. T hus, t h ere a re 5 f a cto rs o f 5 61 w hic h a re n ot p rim e.  F O LLO W   -U    P S :   (1 ) N i s t h e p ro duct o f 4 diffe re n t p rim e n um be rs. H ow m any f a cto rs o f N a re n ot p rim e?  [ 1 2] (2 ) N i s t h e p ro duct o f 5 d iffe re n t p rim e n um bers. H ow m any f a cto rs d oes it h ave a lto geth er?  [ 3 2] (3 ) W hat i s t h e l e a st n um ber w ith e xa ctly 8 f a cto rs?  [ 3 0] (4 ) W hy i s 1 neith er p rim e n or c o m posite ?  [ A p rim e h as e x actly 2 f a cto rs a n d a co m posite h as a t le ast 3 f a cto rs . 1 h as o nly o ne fa cto r. ] 1 E   Str a te g y: F in d th e u nsh aded a re a .   M ETH O D 1 :  S tr a te g y: A ssig n c o nven ie n t n um eric a l l e n gth s t o th e s id es. S uppose f o r e xa m ple t h at B C  = 6 a nd C D  = 4 . T hen B P  a nd P C each a re 3 , a nd C Q  a nd Q D  e ach a re 2 , a nd t h e a re a o f r e cta ngle   A BC D  is 2 4. T he a re a o f ∆∆∆∆∆  A BP  is 1 2   .   4 .  3 = 6 , o f ∆∆∆∆∆ P C Q  is 1 2   .   3 .   2 = 3 , and o f ∆∆∆∆∆ Q DA  is 1 2   .   6 .  2 = 6. T he to ta l u nsh aded are a is 6 + 3 + 6 = 1 5, a nd t h e a re a o f ∆∆∆∆∆  A PQ  i s 9 . T he a re a o f ∆∆∆∆∆  A PQ   is   9 2 4  = 3 8  o f th e a re a o f r e cta n gle   A B C D . M ETH O D 2 :  S tr a te g y: S plit th e fig ure in to m ore c o nven ie n t  s h apes. D ra w P R  a nd Q S  p ara lle l t o t h e s id es o f t h e r e cta ng le a s s h ow n. T he a re a o f ∆∆∆∆∆  A BP  is 1 2   t h e a re a o f r e cta ngle   A BPR , w hic h is 1 2   t h e a re a o f  A BC D . T hen ∆∆∆∆∆  A BP  is 1 4  t h e t h e a re a of A BC D. The a re a o f ∆∆∆∆∆ Q DA  is 1 2   t h e a re a o f r e cta ngle S Q DA  w hic h is 1 2   t h e a re a o f  A BC D . S o ∆∆∆∆∆ Q DA  is 1 4  t h e a re a o f  A BC D . The a re a o f ∆∆∆∆∆ P C Q  is 1 2  t h e a re a o f r e cta ngle P C Q T whic h is 1 4   t h e a re a o f  A BC D . So ∆∆∆∆∆ P C Q  is 1 8   t h e a re a o f  A BC D . The u nsh aded a re a t h en is 1 4   + 1 4  + 1 8  = 5 8  o f t h e a re a o f  A BC D . Thus, t h e a re a o f ∆∆∆∆∆  A PQ  is 3 8   o f t h e a re a o f r e cta ngle   A BC D . 19%  A B D C P Q 4   3 3  2  2  6   A B D C P Q R S   T 5 6 1     1 × 56 1    3  × 1 8 7  11   × 5 1  1 7   × 3 3  P Q R     1 ×     P Q R   P   × Q R  Q   × P R  R  × P Q  10%

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 10/16   D i v is io n   M   M M M M  Copyrig ht © 2 011 b y M ath em atic a l O ly m pia ds fo r E le m en ta ry a nd M id dle S ch ools , In c. A ll r ig hts r e se rv ed .  P age 1 0 2 A M E T H O D 1 :  Str a te g y: A dd d ecim als in a c o nven ie n t o rd er. R ew rit e e ach te rm a s a d ecim al a nd th en c o m bin e te rm s o f th e s a m e s ig n. 1 – 2 1 0  + 3 1 00  – 4 1 000 = 1 – .2 + .0 3 –.0 04 = (1 + .0 3) – (.2 + .0 04) = 1.0 3 – .2 04 = .8 26 M ETH O D 2 :  S tr a te g y: E lim in ate th e d en om in ato rs (te m pora rily ). M ult ip ly e ach t e rm b y 1 000 a nd r e duce . T hen d iv id e b y 1 000 t o u ndo s te p 1 .   1000 – 2000 10  + 3 000 100  – 4 000 1000 = 1000 – 200 + 30 – 4 = 826 N ow d iv id e b y 1 000 t o u ndo t h e fir s t s te p. T he r e su lt is 0 .8 26. 2 B   Str a te g y: W ork b ackw ard s.   A m y’s n um ber     A m y p ic ke d a w hole n um ber , s o s h e s ta rte d w it h 5 . 2 C M E T H O D 1 :  Str a te g y: U se th e s y m metr y o f th e g iv en in fo rm atio n. 2 7, 3 5, a nd 3 2 a re e ach th e s u m o f a d if fe re nt p air o f s id es o f th e t r ia ngle . T hen 2 7 + 3 5 + 3 2 is th e s u m o f th e th re e s id es, e ach c o unte d tw ic e . T hus 9 4 is tw ic e th e p erim ete r  a nd th e p erim ete r o f t h e t r i a ngle is 4 7 c m . M ETH O D 2 :  S tr a te g y: U se a lg eb ra to fin d th e le n gth o f e a ch s id e. L et a , b , a nd c r e pre se nt t h e s id es o f t h e t r ia ngle . T hen t h e t h re e e quatio ns a re : (1 ) a + b = 27, ( 2 ) a + c = 35, a nd (3 ) b + c = 32. O ne w ay t o s o lv e t h is s yste m is t o a dd e quatio ns ( 1 ) a nd (2 ), a nd t h en s u btr a ct ( 3 ) f r o m th e r e su lt . ( 1 ) a + b = 2 7 ( 2 ) a + c = 3 5   2a + b + c = 62 ( 3 ) b + c = 3 2 2 a = 3 0 ( 4 ) a = 1 5 T h e p e r i m e t e r o f t h e t r i a n g l e i s 4 7 c m .  F O LLO W   -U    P S :   (1 ) S uppose th e 3 n um bers g i ven in th e p ro ble m a r e e a ch th e s u m o f 2 s id es o f a p ara lle lo gra m . W hat is th e p erim ete r o f th e p ara lle lo gra m ?  [6 2 2 /3 c m ]   (2 ) T he a re a o f 3 d iffe re n t fa ces o f a b ox (r e cta ngula r so lid ) a re 2 0, 2 8, a nd 3 5 sq c m . W hat is th e v o lu m e o f th e b ox? [1 40 c m 3 ] 4 7% 78% O L Y M P I A D 2 D E C E M B E R 1 5 , 2 0 1 0    A n sw e r s: [2 A ] .8 26   [2 B ] 5   [2 C ] 47   [2 D ] 3   [2 E ] 25 53% c o rre ct 24 N ow a dd ( 4 ) a nd ( 3 ): ( 4 ) a = 1 5 ( 3 ) b + c = 3 2 a + b + c = 4 7 s q u a r e d – 1 + 3 x 2 s q u a r e r o o t + 1 ÷ 2 2 7 ± 5 - 3 5 4

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 11/16     P age 1 1 C opyrig ht © 2 011 b y M ath em atic a l O ly m pia ds fo r E le m en ta ry a nd M id dle S ch ools , In c. A ll r ig hts r e se rv ed . D iv is io n   M   M M M M  41% 38% O L Y M P I A D 3 J A N U A R Y 1 2 , 2 0 1 1    A n s w er s: [3 A ] 41   [3 B ] 12   [3 C ] 3/5   [3 D ] 82   [3 E ] 108 66% c o rre ct 2 D   Str a te g y: C ontin ue th e s e q uen ce a nd lo ok fo r a p atte rn . T he s ix th t e rm is  –   ( 3 + - 6 + 1 ) = - ( - 2 ) = 2 . T he s e quence is 1 , 2 , 3 , - 6 ,  1 , 2 , 3 , - 6 , 1, 2 , 3 , - 6 , and s o o n. T he t e rm s r e peat in g ro up s o f f o ur . T hus, e ve ry f o urth t e rm is - 6 . T here fo re t h e 100 th   te rm is - 6 , a nd s o th e 9 9 th   t e rm is 3 .  F O LLO W   -U    P :   A s e q uen ce b eg in s 1 , 2 , 3 , … The f o urth te rm i s th e t h ir d l e ss th e s e co nd; th e  f ifth te rm is th e fo urth le ss th e th ir d , a nd so on. W hat is th e 4 9 th   te rm ? [-1 ] 2 E M E T H O D 1 :  Str a te g y: F in d th e s u m o f th e n um bers. In th e s e rie s 1 + 3 + 5 … + 4 9, w e g et fr o m 1 to 4 9 b y a ddin g 2 tw enty fo ur tim es, s o th e se rie s h as 2 5 t e rm s. P air t h ese t e rm s a s f o llo w s, w ork in g f r o m t h e o uts id e in w ard ( 1 + 4 9) + ( 3 + 4 7) + ( 5 + 4 5), a nd s o o n. T he s u m of e ach p air is 5 0 a nd th ere a re 1 2 p air s . T he n um ber w it h out a p air i s t h e m id dle o ne, 2 5. T he s u m i s t h en 1 2 × 5 0 + 2 5 = 625. T hen, 6 25  is 2 5. M ETH O D 2 :  S tr a te g y: L ook fo r a p atte rn in th e p artia l s u m s. T he t a ble b elo w e xa m in es t h e s q uare r o ot o f t h e s u m s o f th e f ir s t f e w t e rm s. I n e ach c a se , th e s q uare r o ot is e qual t o t h e n um ber o f t e rm s a dded. N u m b e r o f T e r m s S u m o f T e r m s S quare R oot o f t h e S um o f T erm s 1 1 1 2 4 2 3 9 3 4 1 6 4 5 2 5 5 There a re 2 5 te rm s in th e g iv e n s e quence ( 2 5 o dd n um bers fr o m 1 to 4 9). T here fo re th e sq uare r o ot o f t h e s u m o f t h e s e rie s is 2 5. 3 A M E T H O D 1 :  Str a te g y: L ook fo r a p erfe ct sq uare n ea r 4 20. T he p age n um bers d if fe r b y 1 , s o t h e f a cto rs o f 4 20 m ust b e n early e qual. 2 0 2  = 4 00, s o t r y 2 0 a s a fa cto r . 4 20 = 2 0 × 2 1, a nd t h e s u m o f th e t w o p age n um bers i s 4 1. M ETH O D 2 :  S tr a te g y: C om bin e th e p rim e fa cto rs o f 4 20. T he p rim e f a cto riz a tio n o f 4 20 i s 2 × 2 × 3 × 5 × 7 . C om bin in g t h e f a cto rs i n to t w o p ro ducts w hose d if fe re nce is 1 , w e g et 2 0 × 2 1. T heir s u m is 4 1.  F O LLO W   -U    P :   T he s u m o f th e 6 p a ge n um bers in a c h a pte r o f th e b ook is 1 53. W hat is th e num ber o n th e fir st p age o f th e c h apte r?  [2 3]

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 12/16   D i v is io n   M   M M M M  Copyrig ht © 2 011 b y M ath em atic a l O ly m pia ds fo r E le m en ta ry a nd M id dle S ch ools , In c. A ll r ig hts r e se rv ed .  P age 1 2 5 2% 3 B   Str a te g y: M in im iz e th e n um ber o f a co rn s in e a ch h ole . E ach h ole r e quir e s a d if fe re nt n um ber o f a co rn s, s o p ut 1 in t h e f ir s t h ole , 2 in t h e s e co nd, and s o o n. C ontin ue u ntil t h e t o ta l n um ber o f a co rn s is n ear 8 0. 1 + 2 + 3 + … + 1 1 + 1 2 = 78, a nd 1 + 2 + 3 + … + 1 3 = 91. 1 3 hole s r e quir e at le ast 9 1 a co rn s. T he s q uir re l p uts t h e fir s t 7 8 a co rn s in to 1 2 h ole s a s in dic a te d. T he s q uir re l th en c a n p ut th e o th er tw o a co rn s in to t h e 1 2 th   h ole , m akin g 1 4 a co rn s in t h at h ole . T he g re a te st p ossi b le v a lu e o f N is 1 2.  F O LLO W   -U    P S :   (1 ) A sid e fr o m 1 ,2 ,3 , . . .,1 1,1 4 a b ove, h ow m any o th er s e ts o f 1 2 d i ffe re n t  c o untin g n um bers h ave a s u m o f 8 0? [1 : t h e l a st 2 n um bers a re 1 2 a n d 1 3] (2 ) I n h ow m any diffe re n t w ays c a n th e s q uir re l b ury 2 0 a co rn s in tw o h ole s if e a ch h ole h as a d iffe re n t  n um ber o f a co rn s a nd n o h ole is e m pty ? [9 ] (3 ) 2 0 a co rn s in 3 h ole s?  [2 4] 2 9% 3 C   Str a te g y: L is t th e p ossib le n um era to rs a nd d en om in ato rs. D enote t h e f r a ctio n a nd it s r e cip ro ca l b y  A B   a nd B  A . T heir s u m h as a d enom in ato r o f 1 5, s o  A and B m ust b e c h ose n fr o m { 1 , 3 , 5 , 1 5}. N eit h er  A  n or B  c a n b e 1 5, f o r 15 1 , 15 3 , a nd 15 5  a re e ach la rg er t h an 2 4 15 . L ik e w is e , n eit h er  A  n or B  c a n b e 1 , f o r 5 1  a nd 3 1  a re a ls o la rg er t h an 2 4 15 . T he o nly p ossib ilit y is t h at t h e f r a ctio ns a re 3 5  a nd 5 3 . I n f a ct, t h eir s u m is 3 4 15  = 2 4 15  . O f  th ese t w o, th e p ro per f r a ctio n is 3 5 .  F O LLO W   -U    P :   W hat is th e le a st p ossib le s u m o f a p ositiv e fr a ctio n (n ot n ecessa rily p ro per) a nd its r e cip ro ca l? [ 2 ] 3 D   Str a te g y: L ook fo r a p atte rn . C onsid er th e p oin ts w here th e p ath c h anges d ir e ctio n. S in ce (5 ,3 ) is in t h e f ir s t q uadra nt, f in d t h e p ath l e ngth t o e ach o f t h e upper r ig ht “ c o rn ers ”. C o o rd i na te s o f c o rn e r p o i nt P a th le n g th ( 1 , 0 ) 1 ( 2 , 1 ) 1 + 1 + 2 + 2 + 3 = 9 ( 3 , 2 ) 9 + 3 + 4 + 4 + 5 = 2 5 The p ath le ngth s a re c o nse cu tiv e o dd s q uare s. I n f a ct, t h ey a re t h e s q uare s o f t h e su m o f  th e c o ord in ate s o f t h e c o rn er p oin ts . T he c o rn er p oin t c lo se st t o ( 5 ,3 ) is (5 ,4 ) a nd t h e p ath le ngth t o ( 5 ,4 ) is 8 1 u nit s . F ollo w in g t h e sp ir a l, ( 5 ,3 ) is t h e n ext la ttic e p oin t r e ach ed, s o th e le ngth o f t h e p a th is 8 2 u nit s .  F O LLO W   -U    P :   I n th e o th er 3 q uadra nts w h at p atte rn s a r e fo rm ed b y th e p ath ’ s le n gth s to t h e c o rn ers? 3 E M E T H O D 1 :  Str a te g y: F in d th e s p eed o f e a ch h and in d eg re es p er m in ute . T he m in ute h and m ove s 3 60 o  in 1 h our a nd t h us m ove s 6 o  p er m in ute . T he hour h and m ove s 1 12  a s fa r a s th e m in ute h and e ve ry h our a nd th ere fo re m ove s 1 12  a s f a r e ve ry m in ute , i. e . 1 2 º p er m in ute . A t 8 :0 0 t h e h our h and is 2 40º a head o f t h e m in ute h and. I n t h e n ext 2 4 m in ute s t h e h o ur h and m ove s a n a ddit io nal 1 2º a nd th e m in ute h and m ove s 1 44º.  A t 8 :2 4 th e ang le b etw ee n t h e h ands i s  2 40 + 1 2 – 1 44 = 108º.   2 2%  x  y 1 4% 3  9  6  1 2  M ETH O D 2 o n n ext p age 

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 13/16     P age 1 3 C opyrig ht © 2 011 b y M ath em atic a l O ly m pia ds fo r E le m en ta ry a nd M id dle S ch ools , In c. A ll r ig hts r e se rv ed . D iv is io n   M   M M M M  O L Y M P I A D 4 F E B R U A R Y 9 , 2 0 1 1    A n s w er s: [4 A ] 7   [4 B ] 1/3   [4 C ] 9   [4 D ] 27 πππππ   [4 E ] 48 53% c o rre ct 4 5% 3 E M E T H O D 2 :  Str a te g y: S ta rt a t 1 2:0 0 a nd s e e h ow fa r e a ch h and h as r o ta te d . F ro m 8 :0 0 t o 8 :2 4, t h e m in ute h and has r o ta te d 24 60   = 2 5  o f t h e w ay a ro und th e c lo ck. T hat i s , it h as r o ta te d 2 5  o f 3 60º = 1 44º. T hin k o f 1 2:0 0 a s 0 º, 3 :0 0 a t 9 0º, a nd 6 :0 0 a s 1 80º. T hen th e m in ute h and is p oin tin g t o 1 44º. M eanw hile , th e h our h and, w hic h n eeds 1 2 h ours to r o ta te 3 60º, r o ta te s 3 0º e ve ry h our. T hus, a t 8 :0 0, it w as p oi ntin g t o 2 40º. A t 8 :2 4, it h as r o ta te d a n oth er 2 5  o f 3 0º = 1 2º a nd is p oin tin g t o 2 52º. A t 8 :2 4 t h e a ngle b etw een t h e hands is 2 52 – 1 44 = 1 08º. 4 A M E T H O D 1 :  Str a te g y: S ta rt w ith K ayla , w hose p ositio n is k n ow n. K a y l a i s t h e t h i r d p e r s o n . , , K , … Kayla is 2 p la ce s in fr o nt o f E l i . , , K , , E , … E li is 4 p la ce s b ehin d S ara . S , , K , , E , … S ara is 3 p la ce s in fr o nt o f  A b b y . S , , K ,  A , E , …  A b by i s in th e ce nte r o f th e l in e . S , , K , A , E , ,  A bby is t h e 4 th p ers o n in lin e. T here a re 3 people in fr o nt o f h er a nd 3 p eople b ehin d her. T here a re 7 s tu dents in th e lin e. M ETH O D 2 :  S tr a te g y: C re a te th e lin e, k eep in g A bby in th e c en te r. S a ra i s 3 p l a ce s i n f ro n t o f A b b y . … ,  S , , ,  A , , , E l i i s 4 p l a c e s b e h i n d S a r a . … , S , , , A , E , , K a y l a i s 2 p l a c e s i n f r o n t o f E l i . … , S , , K , A , E, , Kayla is t h e 3 rd   p ers o n in lin e, s o n o a ddit io nal s p ace s a re n eeded in f r o nt o f S ara . T here a re 7 s tu dents in t h e lin e. 4 B M E T H O D 1 :  Str a te g y: C alc u la te th e re q uir e d v a lu es. T he m edia n is + 8 6 2   = 7. T he m ode is 6 . T he m ean is + + + + + 3 6 6 8 1 0 1 2 6   = 4 5 6 . Then ( ) × − × 45 6 3 7 6 6   = 1 5 45   = 1 3 . F   OLLO W   -U   PS  : (1 ) T he m ea n a nd m ed ia n o f a s e t o f fiv e d iffe r en t p ositiv e in t eg ers is 1 2. O ne n um ber is 3 le ss th an th e m ed ia n a nd a not her is h alf th e m ea n. W hat is th e g re a te st   p ossib le i n te g er i n th e s e t? [2 0] (2 ) T he m ea n, m ed ia n a nd m ode o f a s e t o f f iv e p ositiv e in te g ers a re a ll e q ual . T hre e o f th e n um bers a re 9 , 1 3, a nd 4 1. F in d th e m is sin g n um bers th at sa tis fy th is c o nditio n. [ 2 1, 2 1] a nd [4 1, 1 01]

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 14/16   D i v is io n   M   M M M M  Copyrig ht © 2 011 b y M ath em atic a l O ly m pia ds fo r E le m en ta ry a nd M id dle S ch ools , In c. A ll r ig hts r e se rv ed .  P age 1 4 1 6% 61% –   5 0  +   5  +   1 0  +   1 5  (  ?    )    e xce eds –  5   1 3 e xce eds –  1   = =  A B 7 % 4 E M E T H O D 1 :  Str a te g y: C ount in a n o rg aniz e d w ay. T o fo rm a tr ia ngle , tw o p oin ts m ust b e c h ose n fr o m 1 r o w and o ne fr o m t h e o th er . S uppose tw o p oin t s a re c h ose n fr o m t h e t o p r o w . L abel t h e f o ur p oin ts i n t h e to p r o w A , B , C , D. T w o p oin t s m ay b e c h ose n in 6 w ays ( A B, A C , A D , B C , BD , C D). F or e ach o f t h ese 6 pair s o f p oin ts , t h e t h ir d v e rte x m ay b e a ny o f t h e 4 poin ts i n t h e b otto m r o w . T h ere a re t h en 6 × 4 = 2 4 tr ia ngle s u sin g 2 p oin ts fr o m th e to p r o w a nd 1 fr o m th e b otto m ro w . Lik e w is e , t h ere a re 2 4 t r ia ngle s u sin g 2 p oin ts f r o m t h e b otto m r o w a nd 1 fr o m t h e t o p. In a ll, 4 8 t r ia ngle s c a n b e f o rm ed M ETH O D 2 :  S tr a te g y: M ake a n o rg aniz e d lis t. L abel a ll 8 p oin ts , a s s h ow n. L is t a ll 1 8 tr ia ngle s c o nta in in g v e rte x  A , a ll 1 4 tr ia ngle s co nta in in g v e rte x B  b ut n ot  A , a ll 1 0 t r ia ngle s c o nta in in g v e rte x C but n ot  A  o r B , a nd a ll 6 t r ia ngle s c o nta in in g v e rte x D  b ut n ot  A   o r B   or C . A to ta l o f 4 8 t r ia n gle s c a n b e f o r m ed. .  A B D C S ∆∆∆∆∆  A C S  i s s h ow n.  A B D C S P Q R Δ  A BP   Δ  A C P   Δ  A D P   Δ  B C P   Δ  B D P   Δ CD P   Δ CPQ   Δ  D PQ Δ  A BQ   Δ  A C Q   Δ  A D Q   Δ  B C Q   Δ  B D Q   Δ CD Q   Δ CPR   Δ  D PR Δ  A BR   Δ  A C R   Δ  A D R   Δ  B C R   Δ  B D R   Δ CD R   Δ CPS    Δ  D PS  Δ  A BS    Δ  A C S    Δ  A D S    Δ  B C S    Δ  B D S    Δ C D S    Δ C Q R   Δ  D QR Δ  A PQ   Δ  A PR   Δ  A PS    Δ  B PQ   Δ  B Q R   Δ CQ S    Δ C RS    Δ  D QS  Δ  A Q R   Δ  A Q S    Δ  A RS    Δ  B PR   Δ  B Q S    Δ  D RS  Δ  B PS    Δ  B RS  M ETH O D 3 on n ext p a ge  4 C M E T H O D 1 :  Str a te g y: D o th e a rith m etic . 1 3 e xce eds  –   1 b y 1 3 –  –   1 = 1 4. T he in te ger t h at e xce eds  – 5 b y 1 4 is    –   5 + 1 4 = 9 . M ETH O D 2 :  S tr a te g y: D ra w a n um ber lin e. T he u pper lin e se gm ent s h ow s th e am ount b y w hic h 1 3 e xce eds  –   1 . T he lo w er lin e se gm ent s h ow s th e am ount b y w hic h t h e d esir e d n um ber e xce eds  –   5 . S lid in g th e u pper s e gm ent in to th e lo w er p osit io n re quir e s m ovin g t h e l e ft e ndpoin t f r o m  –   1 t o  –   5 , 4 u nit s t o th e le ft. T hen t h e r ig ht e ndpoin t a ls o slid es 4 u nit s t o t h e le ft, f r o m 1 3 to 9 . T he in te ger is 9 . 4 D M E T H O D 1 :  Str a te g y: F in d th e u nsh aded a re a in sid e e a ch c ir c le . T he a re as in sid e t h e t w o c ir c le s a re 2 5 π  a nd 9 π   re sp ectiv e ly . T he u nsh aded a re a in sid e c ir c le   A  is 2 5 π  – π 7 2   = π 4 3 2 . T he u nsh aded a re a in sid e c ir c le B is 9 π  – =   . T he t o ta l u ns haded a re a is   + 4 3 1 1 2   π   = 2 7 πππππ   sq c m . M ETH O D 2 :  S tr a te g y: S ta rt w ith th e to ta l a re a o f th e tw o c ir c le s. T he s u m o f t h e a re as o f t h e cir c le s is 2 5 π   + 9 π   = 3 4 π . T his , h ow eve r , c o unts t h e sh aded are a tw ic e , o nce fo r e ach c ir c le . T he to ta l u nsh aded are a is 3 4 π   – 2 (   π 7 2 ) = 2 7 π . 7 π 2 11 π 2

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 15/16     P age 1 5 C opyrig ht © 2 011 b y M ath em atic a l O ly m pia ds fo r E le m en ta ry a nd M id dle S ch ools , In c. A ll r ig hts r e se rv ed . D iv is io n   M   M M M M  O L Y M P I A D 5 M A R C H 9 , 2 0 1 1    A n s w er s: [5 A ] 36   [5 B ] 23   [5 C ] 56   [5 D ] 363   [5 E ] 7 60% c o rre ct 6 3% 4 E M E T H O D 3 :  Str a te g y: U se c o m bin atio ns. S ubtr a ct th e tr ip le s th at d on’t w ork . G iv e n e ig ht p oin ts , t h re e p oin ts m ay b e ch ose n in 8 C 3   = 5 6 w ays. N o tr ia ngle is fo rm ed if a ll th re e p oin t s a re c h ose n fr o m th e s a m e r o w . T here a re 4 C 3   = 4 s e ts o f t h re e p oin ts in t h e to p r o w a nd a noth er 4 s e ts in t h e b otto m ro w . Thus 5 6 – 4 – 4 = 4 8 tr ia ngle s c a n b e f o rm ed.  F O LLO W   -U    P :   H ow m any tr ia ngle s c a n b e fo rm ed w ith v ertic es c h o se n fr o m a 3 x 3 a rra y o f   p oin ts ?  [7 6] 5 A M E T H O D 1 :  Str a te g y: G ro up th e n um bers b y te n s a nd lo ok fo r a p atte rn . In te r v a l S a tis fa c to ry N um bers N um ber o f N u m bers 1 0 - 1 9 12 , 1 3, 1 4 , 1 5, … , 1 9 8 2 0 - 2 9 2 3 , 2 4 , 2 5 , … , 2 9 7 3 0 - 3 9 3 4 , 3 5 , … , 3 9 6  … … … 8 0 - 8 9 8 9 1 T o t a l 3 6 METH O D 2 :  S tr a te g y: E lim in ate a ll u nw ante d n um bers. T here a re 9 0 t w o-d ig it c o untin g n um bers . E lim in ate t h e 9 num bers w it h e qual d ig it s ( 1 1, 2 2, etc .) . T hen e lim in ate th e 9 m ult ip le s o f 1 0. T he re m ain in g 7 2 n um bers h ave 2 u nequal nonze ro d ig it s . I n h alf o f t h em t h e o nes d ig it e xce eds t h e t e ns d ig it . T here a re t h en 3 6 s u ch n um bers . T here a re 3 6 tw o-d ig it n um bers in w hic h th e o nes d ig it is g re ate r  th an th e te ns d ig it . 5 B M E T H O D 1 :  Str a te g y: M ake a ta ble . N um ber o f C h ecks A : F la t F e e = $7 .5 0 B : $ 3 + 20 ce nts / c h eck 1 $ 7 . 5 0 $ 3 . 2 0 1 0 $ 7 . 5 0 $ 5 . 0 0 2 0 $ 7 . 5 0 $ 7 . 0 0 2 1 $ 7 . 5 0 $ 7 . 2 0 2 2 $ 7 . 5 0 $ 7 . 4 0 23 $7.5 0 $7.6 0 T here fo re , th e l e ast n um ber o f c h ecks s u c h t h at P la n A c o sts l e ss t h an P la n B i s 2 3 . M ETH O D 2 :  S tr a te g y: U se A lg eb ra . L et N  r e pre se nt t h e n um ber o f c h ecks f o r w hic h p la n A is c h eaper. 2 0 N   + 300 > 750 (c o st, in ce nts to avo id decim als ) 2 0 N > 4 50   N > 22½  A t le ast 2 3 ch ecks a re n eeded f o r P la n A to c o st le ss th an P la n B .

5/20/ 2018 M O lym piad S am pl e - slide pdf .com http: //s lide pdf .com /r e ade r/f ul l/m -ol ym piad- sa m pl e 16/16   2 7% 5 C M E T H O D 1 :  Str a te g y: A ssig n a rb itr a ry d im en sio ns. In r a tio p ro ble m s, a ssig nin g c o nve nie nt m easu re s w ill n ot a ffe ct t h e a nsw er . F or e ase o f c o m puta tio n, l e t t h e b ase b e 2 0 a nd t h e h eig ht 1 0 s o t h at t h e a re a i s 1 2 ( 2 0)(1 0) = 1 00 sq u nit s . I f t h e b ase is in cre ase d b y 2 0% , t h e n ew b ase is 2 4. If t h e h eig ht is in cre ase d by 3 0% , th e n ew h eig ht is 1 3. T he a re a o f th e n ew tr ia ngle is 1 2 ( 2 4)(1 3)= 156 s q uare u nit s . T he i n cre ase o v er t h e o rig in al 1 00 i s 5 6. T here fo re , th e a re a i s i n cre ase d b y   56 co m pare d t o 1 00 =   5 6% . M ETH O D 2 :  S tr a te g y: C om pare a re a s u sin g th e a re a fo rm ula . T he o rig in al a re a o f t h e t r ia ngle is g iv e n b y  A   = 1 2 b h . T he c h anged a re a o f t h e t r ia ngle i s g iv e n b y 1 2 (1 .2 0 b )(1 .3 0 h ) = 1 2 (1 .5 6) b h , w hic h i n t u rn e quals 1 .5 6 t im es 1 2 b h . T here fo re th e a re a o f th e n ew tr ia ngle is 1 .5 6 tim es a s g re at a s th e a re a o f th e o rig in al. T his is 5 6%   gre ate r t h an t h e a re a o f th e o rig in al. T he in cre ase is 5 6% .  F O LLO W   -U    P :   S uppose th e b ase o f a tr ia ngle is d ecre a se d by 2 0% , a nd its h eig ht is d ecre a se d b y 3 0% . B y w hat p erc en t is th e a re a o f th e tr ia ngle d ecre a se d ?  [4 4] 1 4% 5 D   Str a te g y: D ete rm in e h ow m any o f th e fir st 3 00 n um bers c a n’t b e u se d . In e ach o f t h e s e ts 1 t h ro ugh 1 00, 1 01 t h ro ugh 2 00, a nd 2 01 th ro ugh 3 00, 1 0 n um bers h ave a o nes d ig it o f 9 , a nd 1 0 n um bers h ave a te ns d ig it o f 9 . H ow eve r , in e ach s e t t h e num ber e ndin g i n 9 9 h as b een c o unte d t w ic e , s o e ach s e t h as 1 9 n um bers t h at c o nta in a d ig it o f 9 . In t h e o ve ra ll l is t 1 t h ro ugh 3 00, 5 7 n um bers m ust b e e lim in ate d. R efill t h e lis t w it h th e n ext 5 7 n um bers , 3 01 th ro ugh 3 57. T here a re 5 n um bers in th is s e t th at ca n’t b e u se d ( 3 09, 3 19, 3 29, 3 39, 3 49). A dd o n 5 m ore n um bers , 3 58 th ro ugh 3 62. O ne o f t h ese , 3 59, c a n’t b e use d, s o a dd o n one m ore n um ber. T he 3 00 th   n um ber o n S ara ’s lis t i s 3 63. 5 E   Str a te g y: D ra w possib le p ath s. 1 2% 2 4 5 6 1 7  x  y 2 + 1 B y d ra w in g s o m e p ath s, y o u m ay s e e t w o th in gs: ( 1 ) t h e p ath c a n tu rn e it h er le ft o r r ig ht, a nd ( 2 ) e ach h oriz o nta l s e gm ent has a n o dd le ngth w hile e ach v e rtic a l s e gm ent h as a n e ve n le ngth . C onsid er fir s t th e h oriz o nta l s e gm ents th at c a n e nd b ack a t th e s ta rtin g p oin t. N eit h er { 1 u nit a nd 3 u nit s } n or { 1 , 3 a nd 5 u nit s } c a n e nd a t t h e o rig in , b ut { 1 , 3 , 5 , a nd 7 u nit s } c a n. B y tr a ve lin g 1 a nd 7 u nit s t o t h e r ig ht, a nd 3 and 5 u nit s t o t h e l e ft, th e h oriz o nta l p art o f t h e p ath c a n e nd a t z e ro . N ow c o nsid er 6 a nd 8 , t h e e ve n n um bers o n e it h er s id e o f 7 . S in ce 2 + 4 = 6 , if 2 a nd 4 a re d ir e cte d u p a nd 6 i s d ir e cte d d ow n, th e v e rtic a l p art o f t h e p ath a ls o e nds a t z e ro . T hus t h e sh orte st p ossib le p ath c o nsis ts o f 7 s e gm ents a nd is s h ow n a bove .  F O LLO W   -U    P S :   (1 ) A path co nsis tin g of  N   lin e s e g m en ts is d ra w n in th e c o ord in ate  p la ne. T he fir st s e g m en t s ta rts a t ( 0 ,0 ) a nd is d ra w n to (2 ,0 ). T he se co nd se g m en t  s ta rts a t (2 ,0 ) a nd is d ra w n to (2 ,4 ). E ach o f th e  N   s e g m en ts is d ra w n a t a r ig ht  a ngle to th e s e g m en t b efo re it a nd is 2 u nits lo nger th an th at s e g m en t. T he  N th s e g m en t e n ds a t (0 ,0 ). W hat is th e le a st p ossib le v a lu e o f  N ?   [ 7 ]   (2 ) W hat is th e le a st p ossib le v a lu e o f  N   g re a te r th an 7 ?   [8 ]  (3 ) W hat a re th e n ext 2 p ossib le v a lu es o f  N ?   [ 1 5, 1 6] D iv is io n M C opyrig ht © 2 011 by M ath em atic a l O ly m pia ds fo r E le m en ta ry a nd M id dle S ch ools , In c. A ll rig hts re se rv ed .   P age 1 6   o n