Lily Yen and Mogens Hansen

Transcription

1 SKOLID / SKOLID No. 8 Lily Yen nd Mogens Hnsen Skolid hs joined Mthemticl Myhem which is eing reformtted s stnd-lone mthemtics journl for high school students. Solutions to prolems tht ppered in the lst volume of rux will pper in this volume, fter which time Skolid will e discontinued in rux. New Skolid prolems, nd their solutions, will pper in Mthemticl Myhem when it is relunched in 0. In this issue we present the solutions to the Mritime Mthemtics ompetition, 00, given in Skolid t [0:0 ].. The Vlhll Winter Gmes re held in Ferury, nd the closing ceremonies re on the lst dy of the month. The first Vlhll Winter Gmes were held in the yer 750, nd since tht yer, they hve een held every five yers. How mny times hve the closing ceremonies een held on Ferury 9 th? Note tht yer Y is lep yer if exctly one of the following conditions is true: () Y is divisile y 4 ut Y is not divisile y 00. () Y is divisile y 400. Solution y Gesine Geupel, student, Mx Ernst Gymnsium, rühl, NRW, Germny. The Vlhll Winter Gmes hve een held in yers tht end in 0 or 5. Lep yers re divisile y 4, so the numer formed y the lst two digits of lep yers must e divisile y 4. Therefore you only need to consider yers tht end in 00, 0, 40, 60, or 80. Now list these: Scrtch the full centuries tht re not lep yers fter ll. Then 54 yers remin in which the closing ceremonies were held on Ferury 9th. lso solved y WEN-TING FN, student, urny North Secondry School, urny, ; ROWEN HO, student, École Dr. hrles est Secondry School, oquitlm, ; SZER PINTER, student, Moscrop Secondry School, urny, ; nd LIS WNG, student, Port Moody Secondry School, Port Moody,. Since = 6.05, you do not need to list the yers in the tle to see tht the tle 0 must contin 6 yers. Then sutrct the whole centuries tht re not lep yers. opyright c ndin Mthemticl Society, 0

2 4/ SKOLID. tringle with vertices (0, 0), (, 4), nd (, c) hs re 5. Find ll possile vlues of the numer c. Solution y Rowen Ho, student, École Dr. hrles est Secondry School, The line through nd hs the eqution y = 4 x, so if were on tht line, c = 8. onsider four cses for : ove (tht is, c 4), etween nd the line through nd (tht is, 8 c < 4), etween the x-xis nd the line through nd (tht is, 0 c < 8 ), nd elow the x-xis (tht is, c < 0). c c 4 c c If c 4, then the re of (tht is, 5) cn e found y sutrcting the res of the three other tringles in the leftmost figure from the re of the rectngle. Tht is, 5 = c 4 c (c 4) = c 6 c c, so 9 = c, so c = 6. If 8 c < 4, then the re of cn e found y sutrcting the res of the three other tringles in the second figure from the re of the rectngle. Tht is, 5 = 4 4 (4 c) 4 = 6 6 c 4, so 9 = c, so c = 6, which contrdicts the ssumption tht c < 4. If 0 c < 8, then the re of cn e found y sutrcting the res of the three other tringles in the third figure from the re of the rectngle. Tht is, 5 = 4 c 4 4 = c 6, so c =, which contrdicts the ssumption tht c 0. Finlly, if c < 0, then the re of cn e found y sutrcting the res of the three other tringles in the rightmost figure from the re of the rectngle. Tht is, 5 = (4 c) (4 c) 4 ( c) = c c 6 c, so c =, so c =. Thus c = 6 or c =. lso solved y LEN HOI, student, École Dr. hrles est Secondry School, rux Mthemticorum, Vol. 8(), Jnury 0

3 SKOLID / 5. Let f(x) = x x c, where,, nd c re rel numers. ssume tht f(0), f(), nd f() re ll integers. () Prove tht f(00) is lso n integer. () Decide if f(0) is n integer. Solution sed on contriutions from Len hoi, student, École Dr. hrles est Secondry School, oquitlm, ; Rowen Ho, student, École Dr. hrles est Secondry School, oquitlm, ; nd Lis Wng, student, Port Moody Secondry School, Port Moody,. Since f(0), f(), nd f() re ll integers, f() f() f(0) is lso n integer. Now, f() f() f(0) = (4 c) ( c) (c) =, so is n integer. Since f(0) nd f() re oth integers, f() f(0) is lso n integer. Now, f() f(0) = ( c) (c) =, so is n integer. Finlly, f(0) = c, so c is n integer. Now f(00) = c = c = 09045() 00( ) c, which is n integer ecuse,, nd c re ll integers. lso f(0) = 0 0 c = c = 0055() 0( ) c, which is n integer ecuse,, nd c re ll integers. 4. If x is rel numer, let x denote the lrgest integer which is less thn or equl to x. For exmple, 7.0 = 7. If n is ny positive integer, find (simple) formul for n (n ) (n ). Solution y Rowen Ho, student, École Dr. hrles est Secondry School, oquitlm,. When you divide the integer n y, the reminder is either 0,, or. Tht is, n =, n =, or n = for some integer. If n =, then n (n ) (n ) 6 ( ) ( ) = = 4 = () () ( ) = 6 = n. opyright c ndin Mthemticl Society, 0

4 6/ SKOLID n If n =, then (n ) (n ) ( ) ( ) = = 4 ( ) = () ( ) ( ) = 6 = ( ) = n. n If n =, then (n ) (n ) ( ) = = 4 ( ) 8 ( 4) Thus n = ( ) ( ) ( ) = 6 5 = ( ) = n. ) (n ) (n = n for ll integers n. lso solved y LEN HOI, student, École Dr. hrles est Secondry School, oquitlm, ; SZER PINTER, student, Moscrop Secondry School, urny, ; nd LIS WNG, student, Port Moody Secondry School, Port Moody,. 5. () If is positive numer, prove tht. () If nd re oth positive numers, prove tht 4.5. You my ssume without proof tht f(x) = x x is n incresing function for x. Solution sed on work y Szer Pinter, student, Moscrop Secondry School, urny,. Let f e the function defined y f(x) = x x. Then f is incresing on the intervl [, ) ccording to the hint. Therefore, if, f() f() =. If 0 < <, then >, so f( ) f() =, ut f( ) = f(). Thus f() for ll positive. To prove tht f() f() 4.5 one must consider severl (prtilly overlpping) cses. rux Mthemticorum, Vol. 8(), Jnury 0

5 SKOLID / 7 If, then f() f() =.5. Since f() y Prt (), f() f().5 0 = 4.5. Similrly, if, then f() f().5 0 = 4.5. This proves the inequlity for ll (, ) in the shded region in the figure. If, then. Since f() nd f() y Prt (), f() f() 0.5 = 4.5. This proves the inequlity for ll (, ) in the drker region in the figure, where the curve is given y = (or = ). If nd, then f() f( ) = 6 (since f is incresing) nd f() f( ) = 6. Moreover, = 4, so 4. Therefore f() f() = 55 > 4.5. This proves the inequlity in the drker region of the new figure. If nd, then f() nd f() f( ) = 6. Moreover, =, so. Therefore f() f() 6 = 4.5. Similrly, if nd, then f() f() 6 = 4.5. This proves the inequlity in the drker region of the newest figure. Note tht smll white region still remins. The curve given y = psses through (, ), (, 4 ), ( 4, ), nd (, ). The squre given y 4 nd 4 therefore completely covers the outstnding region. In tht squre, f() f( 4 ) = 5, nd f() f( 4 ) = 5. Moreover, = 9 4, so 4 9. Therefore f()f() = 8 8 > 4.5. This proves the inequlity in the drker region in the figure nd, thus, completes the proof for ll positive nd. 4 One cn esily solve prt () without using the hint: for ny positive numer, ( ) 0, so 0, so. Dividing y the positive numer yields tht s desired. similr rgument proves the hint: / 4/ f(x) f(y) = x x y y = x y y xy x xy xy(x y) (x y) = xy (xy )(x y) =. xy Therefore, if x y, then f(x) f(y) 0, so f(x) f(y). opyright c ndin Mthemticl Society, 0

6 8/ SKOLID 6. hole in concrete wll hs the shpe of semi-circle with rdius of metres. utility compny wnts to plce one lrge circulr pipe or two smller circulr pipes of equl rdius through the hole to supply wter to Wtertown. If they wnt to mximize the mount of wter tht could flow to Wtertown, should they use one pipe or two pipes, nd wht size pipe(s) should they use? Solution y Len hoi, student, École Dr. hrles est Secondry School, If you use single pipe, the rdius is t most. The re of cross section of the pipe is therefore π Š = π. If you use two pipes of rdius r s in the right-hnd figure, the side length of the dotted squre in figure is r. y the Pythgoren Theorem, the digonl of the squre is r. The rdius of the semi-circle is, thus, r r. Tht is, ( )r =, so r = = = Š =. cross section of the two pipes therefore hs re πr = π( ) = π(4 4 ) = ( 8 )π 0.69π > π. Hence two pipes would llow more wter to flow. This issue s prize of one copy of rux Mthemticorum for the est solutions goes to Rowen Ho, student, École Dr. hrles est Secondry School, We hope tht more student reders will sumit their solutions to one or more of the prolems on the fetured contest. rux Mthemticorum, Vol. 8(), Jnury 0