a) The average (mean) of the two fractions is halfway between them: b) The answer is yes. Assume without loss of generality that p < r.

Transcription

1 Solutios to MAML Olympiad Level 00. Factioated a) The aveage (mea) of the two factios is halfway betwee them: p ps+ q ps+ q + q s qs qs b) The aswe is yes. Assume without loss of geeality that p <. The, q s p p+ ps< q pq+ ps< pq+ q pq ( + s) < q( p+ ) < ad q q+ s p+ ps< q ps+ s< q+ s s( p+ ) < ( q+ s) < p p + < < q+ s s q q+ s s c) It is possible as the followig table shows ad this is a example of Simpso s Paadox.. ivide ad Coque ecembe FT Made FT Missed Totals Pecet Shaq % Macus % Jauay Shaq % Macus % ec. & Ja. Combied Shaq % Macus % a) b) a c ad+ bc ad + bc ad + ; assume this sum equals the itege e de + c de b d bd b b ad is a itege; sice a ad b ae elatively pime, the b must be a facto of d; b ad + bc bc bc also, be a+ be is a itege; sice c ad d ae elatively pime, the d d d d must be a facto of b; the oly way b ca be a facto of d ad d ca be a facto of b is if b d , ; ; sice the geatest commo facto of these two umbes 7, the both umbes sought must be multiples of 8; theefoe we ca ow coside both sought umbes divided by 8; the least commo multiple of the two ew umbes is ow ad thei sum is 97 ; oe of these ew umbes is a multiple of 7, oe of these ew umbes is a multiple of 5, ad oe of these ew umbes is a multiple of 59; sice , the two ew umbes must be 35 ad 59; multiplyig each by 8 gives the two sought umbes: 65 ad 3780.

2 c) Assume k m, a itege ad let L be the least commo multiple of the set of umbes k S {,,3,..., }. Multiplyig both sides of the equatio by L gives: L Lm. k k By the atue of L, the left-had side, whe expaded will cosist of itege tems. Let be the highest powe of i the set S ad coside the tem o the left side fomed by L. Sice L L is the LCM of S, must be odd sice L does ot cotai ay highe powe of tha. Howeve evey othe tem i the equatio, both o the left-had ad ight-had sides must be eve. The easo why this is tue is because if thee wee othe odd tems, the they would have to be of the fom L k whee k is a itege geate tha. This would imply that + k would also be i S. This cotadicts that is the highest powe of i S. So thee would be a equatio such that a sum of all eve iteges except fo oe is a eve itege. This is clealy impossible. Theefoe the assumptio that the oigial sum is a itege is false. 3. Roud ad Roud a) Let adius of the smalle cicle ad R adius of the lage oe. H By the Pythagoea Theoem R 0. The aea of the aulus A R ( R ) π π π 00π C 0 R b) The aea of the segmet aea of the secto aea of the tiagle R L L + R + 4 (i adias) ; aea of secto R ( θ ) L ; ta L θ ta + L ta L 4 ; L θ R aea of tiagle L + L ; aea of segmet ta L L 4 4 4

3 c) If L 3 ad the aea of the segmet 4π 3 3, usig the fomula i pat b, the ( 3 ) 3 ( 3) 4π 3 3 ta ( 3) + 4π 3 3 ta π 3 4π 3 3 4π 3 3 4π ad L Pogessively Bette a) Istead of the stadad text book appoach, you could use Mathematical Iductio: ak a + a ( a+ a) ; assume it is tue fo, pove it is tue fo k Usig the fomula a a + ( ) d ( ) ( ) + : a a d a + d : + k k + ( ) + ( ( ) ) k k a a + a a + a + a a + a + d + a + d ( a + ( ) d) + ( a + d) ( a + d d) + ( a + d ) d ( a+ d) + ( a+ d) ( a + d) + ( a+ d d ) ( a+ d) + ( a+ d ) ( a + d ) ( a + ( a + d) ) ( a + a + ) b) The th set has elemets. If you ecogize the subscipts of the last tems i each set ae, 3, 6, 0,, which ae the tiagula umbes, you ca coside the sets i evese ode i which case the costat diffeece betwee tems is ow ( ) d. The fist elemet i the th + + evesed set is a ( + ) a + d a+ d ; ow use the fomulas fom ( ) pat a: S a d ( )( d) a ( ) d ( ) ( d) ( a+ ( ) d)

4 c) Usig the fomula fom pat a, if the fist itege is k, the the sum of cosecutive iteges statig with k would be: ( k ) + ; if is odd the the sum ca ot be a powe of ; if is eve, the k+ would be odd ad so the sum is eve a powe of. 4 d) If c, the c a ( b ) Thee ae two cases to coside: (i) + whee a is a o-egative itege ad b is ay positive itege. b < a, the if the fist itege is a b ad the last itege is a + b, thee would be b + b a a a iteges added ad thei sum would equal ( b b) ( b ) (ii) b a, the if the fist itege is a b + ad the last itege is b + a a a a iteges added ad thei sum would equal ( b b ) ( b ) a , thee would be a 5. oes it Really Make a iffeece? a) If a b ( a b) ( a b) ; to get itege solutios fo a ad b, let a+ b 4 ad a b a 8, b 6; If a b ( a b) ( a b) ; to get itege solutios fo a ad b, let a+ b 4 ad a b a 3, b ; let a+ b ad a b 4 a 8, b 4; let a+ b 8 ad a b 6 a 7, b ; 3 48, , ad b) 98 3 ; if a b ( a b) ( a b) ; sice 98 has oly oe facto of, if a+ b is eve the a b is odd, o if a+ b is odd the a b is eve; i eithe case if you added the equatios a would ot be a itege ad so 98 caot be witte as a diffeece of squaes.

5 c) Based o pat b, set S caot cotai positive iteges of the fom ( k + ) whee k is ay 5 o-egative itege. Futhemoe these ae the oly positive iteges ot i set S. If the itege is odd o divisible by 4, it is always possible to expess it as a diffeece of two squaes of oegative iteges. I the case of the itege k beig odd, the a+ b k ad a b, which k + k esults i a ad b. I the case that k 4l, whee l is a positive itege, the a+ b l ad a b, which esults i a l+ ad b l s s d) Coside 3 types of iteges i set S: ( i),whee, ( ii) p p s s ae odd pimes, ad ( iii) p p >, whee the p i s, whee > ad the p i s ae odd pimes. Let f ( ) be the umbe of ways a umbe i set S ca be witte as a diffeece of squaes of positive iteges. ( ),whee i > : I this case you ca split t t ito whee 0 < t t 0 < t. If you let a+ b t ad a b t, you will get a diffeece of squaes of o- egative iteges fo each pai of values of t ad t. If t whee [ ] itege less tha o equal to x, this will give you the umbe of cases. f ( ) s s ( ii) p p : I this case has ( s ) ( s ) will be ( s ) ( ) + s+... x is the geatest. + + factos. If is ot a pefect squae, thee ( ) ( ) pais of factos, each pai poducig a diffeece of two squaes of o- + s+ s+... egative iteges. If is a pefect squae thee will be pai of factos ( s )( s ) will hadle both situatios. f ( p p...) s s ( iii) p p : I this case has ( ) ( s ) ( s ) ( ) ( ) s s + s + s factos. You wat to elimiate all pais of facto i which is cotaied i oe of them. Fom case ( ii) thee ae ( s ) ( s ) + + factos that ae odd. The facto could be multiplied by ay oe of these, which would the be paied with a secod odd facto so that both multiply to. Theefoe you

6 wat to elimiate ( s ) ( s ) + + pais of factos. The total umbe of pais ca be aived at i a simila way to case ( ii ). s s ( ) ( ) ( ) ( ) + + s + s +... f p p... ( s + ) ( s +... ) A secod, moe geeal, fomula fo f ( ) is peseted hee without explaatio: 6 s s Let p p... with, s, s o-egative iteges ad p, p odd pimes, the ( ) ( ) (... ) if is odd s + s + ( ) ( s + ) ( s +... ) if ad ( + ) ( +...ae ) both eve ( ) ( s + ) ( s +... ) + if is eve ad ( s + ) ( s +...is ) odd f s s 6. I m Lost ad You R a) Begiig at the oigi, ( 0,0 ), R poceeds to the poit ( 0,0) ad the chages diectio by 45. Fo the ext 5 miles his x ad y coodiates chage by the amouts ( 5cos45,5si45 ) makig his ew locatio ( 0+ 5cos45,0+ 5si45 ). O the ext leg of his tip the chage i the x ad y coodiates is (.5cos90,.5si90 ) bigig him to the poit ( 0+ 5cos45 +.5cos90,0+ 5si45 +.5si90 ), ad so o. You wat to fid the limitig values fo the the x ad y coodiates of the positio. You should teat each coodiate sepaately. Fo the x coodiate, the temial value 0+ 5cos45 +.5cos90 +.5cos cos cos cos70 + This seies ca be split ito 4 ifiite geometic seies a S with the fist seies cosistig of the st tem, 5th tem, 9th tem,, the secod seies beig the d tem, 6th tem, 0th tem, etc. Fo each of these seies x temial. Note the 3d seies of tems ae all cos45.5cos cos45 +.5cos

7 X.907 miles (Stoig the value i X to use i pat b.) 7 Fo the y coodiate, the temial value 0+ 5si45 +.5si90 +.5si si si si70 + Note the st seies of tems ae all 0. y temial 5si45.5si90.5si35 5si45 +.5si90 +.5si Y 6.5 miles (Stoig the value i Y to use i pat b.) 7 0 b) The total distace that R tavels is miles. At a costat ate of 4mph, R should aive at the temial poit ( XY, ) i 5 hous. Sice R left hous ago he has 3 moe hous befoe he eaches the temial poit. You distace fom the oigi to the temial poit X + Y. Sice you eed to each the poit the same time as R, you ate should be X + Y 3 mph 4.54mph c) Note that i the complex plae 0 i i e 0+ 0,5 i e 5cos45 + 5si45 i, etc. To fid the temial poit i the complex plae with the diectio alteed by θ degees each time 0i θi θi 3θi would amout to addig the ifiite seies: 0e + 5e +.5e +.5e +. This is a ifiite geometic seies with 0.5e θi. Usig the same fomula as pat a: Temial poit i the complex plae θ i ( e θ i ) e (.5 ) (.5 ).5.5 ( ) ( θ i θ) cos si θi θi θi θi θi.5e e e e e.5cosθ 0 5cosθ siθ + i..5cosθ.5cosθ 0 5cos si The coodiates of the temial poit i the Catesia plae ae θ θ,.5cosθ.5cosθ

8 7. What the Hex? 8 a) Let s side of the ie hexago. Use the Law of Cosies: A s x + x x x cos0 L ( ) ( ) s x + x x + x ( x ) s x x x x x s x x Use the fomula fo aea of a egula hexago i tems of its side s: aea ( x x ) 3s F K E J G x B -x H C I b) If the aea 3 3 <, the 4 ( x x ) < < + < x x x x 5 x < x < < x < < x<. The pobability that x is selected fom this iteval 3 c) If pobability that the aea of the hexago is geate tha a squae uits ( + ) 3 x x 3 a 3a > a x x+ > x x+ > + 3a 3 x > a 7 x >. 6 This iequality should epeset itevals which ae of the iteval [ 0, ]. Theefoe 8 3a a a a

9 8. Lowe Left Coe y c... c c c3 c4 c5 c6 7 x a) Coside the followig patitio of the atual umbes: {,,3 } { },4,5,6 { },7,8,9,0 { }, etc. These coespod to the cell umbes of the diagoals i the lattice as you move away fom the oigi. Notice that, as such, the sum of the coodiates of the lowe left coe poit of each of the cells alog ay diagoal is costat. Fo example the thid diagoal cosists of cells umbeed 4, 5, ad 6. The coodiates of the lowe left coe of these cells ae espectively (,0 ),, ( ), ad ( 0, ). Each odeed pai s coodiates add to. I geeal the th diagoal will have lowe left coe poits each of whose coodiates add to. Next ote that the lagest umbe o each of the diagoals is, atually, a tiagula umbe. This meas they have the fom ( ) k k+, k,,3, Cell 00 falls betwee 9 ad 05, the 3th ad 4th tiagula umbes. Theefoe cell 00 is o the 4th diagoal. The sum of the coodiates of the lowe left coe of cell 00 must add to 3. Cell 00 is the 9th oe i the list 9, 93,,05. Sice eve umbeed diagoals have cell umbes that icease with iceasig x coodiates, begiig with 0, the x coodiate of the cell 00 must be 8. Theefoe the lowe left had coodiates of cell 00 ae ( 8,5 ). b) Suppose we set ( + ) k k N ad solve fo k. This gives the positive oot fom the quadatic + + 8N fomula: k. Clealy this will be a itege oly if N is a tiagula umbe. Howeve fom the value of k, you ca detemie betwee which two tiagula umbes N falls.

10 I paticula, N lies betwee [ k] th ad the ([ k ] + ) st tiagula umbes, whee [ ] k is the geatest itege less tha o equal to k. Based o the agumet i pat a, the sum of the coodiates of the lowe left coe poit of cell N must be[ k ]. As with pat a, the ext cosideatio is the paity of [ k ] ad the elative positio of cell N o the [ k] th diagoal. I geeal, the umbe N is the N [ k] ([ k] + ) th oe o the diagoal. If [ k] is odd, the, as i pat a, the x coodiates icease with the cell umbes ad the lowe left coe s x coodiate would be N [ k] ([ k] + ). O the othe had, if [ ] ad the x coodiate of the lowe left coe of the cell would be [ ] [ ] ([ ] ) [ k ] [ ]([ ] ) k is eve, the the situatio is evesed [ ]([ ] ) k k + k k + k k + 3 k N N + + ( N) N Puttig all of this togethe: Give N, detemie k ; if k is odd, the coodiates of the lowe left coe of cell N ( ) [ k] ([ k] + ) [ k] ([ k] + ) [ ] ( ) N, k N if k is eve, the coodiates of the lowe left coe of cell N [ k] ([ k] + 3) [ k] ([ k] + 3) [ ] ( ) ( N) +, k N + c) I the fist quadat the oly itege coodiate poits though which the cicle {( x, y) x + y 44 } passes ae ( ) ( ) ( ) ( ),,,, 9,9,ad 9,9. The fist two poits have a coodiate sum equal to ad theefoe [ k ] fo the umbe N. Sice is eve, the x 5 coodiate equals N + 76 N, usig the fomula i pat b. Theefoe N fo the fist poit, ad N fo the secod. Similaly, fo the secod pai of poits [ k ] 8 ad the x coodiate equals 8 3 N N. Thus, the cell umbes coespodig to these would be N ad N The cell umbes ae 55, 75, 46, 46.