Online Mathematics Competition Wednesday, November 30, 2016

Transcription

1 Online Mathematics Competition Wednesday, Novembe 0, 206 SOLUTIONS. Suppose that a bag contains the nine lettes of the wod OXOMOXO. If you take one lette out of the bag at a time and line them up left to ight, what is the pobability that you will spell the wod OXOMOXO? () between 0.0 and 0. (B) between 0.00 and 0.0 (C) between and 0.00 (D) between and nswe: (C) between and 0.00 Solution: Note that we can conside the placement of two s, two Xs and one M only, as those will uniquely detemine whee the fou Os ae placed. The wod is nine lettes long. We can place a pai of s in two locations in ( ) 9 2 ways. We can place the two Xs somewhee in the emaining 7 locations in ( ) 7 2 ways. Finally we can place M in any of the emaining 5 locations. Thus, thee ae ( ( ) 9 2) possible wods we can make. Of those, only will spell OXOMOXO, thus the chance is Note that we can count in diffeent ways: we can place M in any of 9 locations. fte M is placed somewhee, a pai of s can be placed in emaining 8 locations in ( ) 8 2 ways and a pai of Xs in the emaining 6 locations in ( ( ( ) 6 2) ways. gain, 9 8 2)

2 Online Competition Wednesday, Novembe 0, motoist tavels the fist 0 kilometes of a tip at 0km/hou. How fast would he have to dive fo the next 0 kilometes if the total tip has an aveage speed of 50 km/hou? () 70 km/h (B) 80 km/h (C) 0 km/h (D) 50 km/h nswe: (D) 50 km/h Solution: is Let x be the speed ove the second 0km. Then the total time fo the 20 km tip x + 0 x hous. Fo the aveage speed to be 50, we must have that 20 Solving fo x, we obtain x x 50

3 Online Competition Wednesday, Novembe 0, 206. Fo how many integes n does the expession 2n+ 4 n+ +6 n yield a pime numbe? () (B) 2 (C) 4 (D) infinitely many nswe: () Solution: Theideaistofacto: 2n+ 4 n+ +6 n 2n 4 2 2n +2 n n 2n 2 2n 2 2n +2 n n ( 2n 2 2n) +2 n ( n 2 n ) ( n 2 n )( n +2 n )+2 n ( n 2 n ) ( n 2 n ) ( n+ +2 n+2) Fo the expession 2n+ 4 n+ +6 n to be pime, one of the two factos must be. Since n+ +2 n+2 > fo all n, we conclude that n 2 n. If n 2, then n 2 n >, and if n, then n 2 n. Thus, n is the only solution; i.e., 2n+ 4 n+ +6 n is pime only when n (inwhich case it is equal to 7).

4 Online Competition Wednesday, Novembe 0, set C of positive integes is called cool if any two numbes in C ae elatively pime. Bob wants to build a cool set fom numbes between and 0 (inclusive), in such a way that his set contains as many numbes as possible. How many diffeent cool sets can he build? () 2 (B) 6 (C) 24 (D) 0 nswe: (C) 24 Solution: Note that a lagest cool set must contain the numbe (if not, we can add to obtain a lage cool set). s well, it cannot contain a numbe which is the poduct of two pime numbes, as othewise we can eplace that numbe with the two pimes and obtain a lage set. Thus, evey numbe in a cool set must be a pime, o a powe of a pime. In othe wods, evey cool set that Bob can build must be of the fom {, 2 n, n 2, 5 n, 7 n 4, n 5, n 6, 7 n 7, 9 n 8 } whee n i. Since all numbes in C must be smalle than o equal to 0, we conclude that n, 2,, 4 n 2, 2, n, 2 and n 4 n 5 n 8. Thus, thee is a total of cool sets.

5 Online Competition Wednesday, Novembe 0, Caly plots a point, and then stats dawing ays stating at, so that all angles she gets (i.e., between any two ays) ae intege multiples of 0 o. What is the lagest numbe of ays she can daw so that all the angles at between any two ays (not just adjacent ays) ae distinct? () 5 (B) 6 (C) 7 (D) 8 nswe: (B) 6 Solution: Denote by n 2 the numbe of ays dawn by Caly. Then thee ae ( ) n 2 n(n ) 2 pais of ays, and each pai of ays detemines two angles which add up to 60 o. Hence the total numbe of angles between all pais of the n ays is exactly n(n ). Each of these angles is smalle than 60 o. Since all angles ae supposed to be intege multiples of 0 o, thee ae at most 5 values fo the measues of these angles. Since they ae distinct, n(n ) 5. If n 6, then n(n ) 0, and if n 7, then n(n ) 42. Thus, n 6. It is possible to daw 6 ays, detemining 0 distinct angles. Fo instance, UV 60 o, VW 40 o, WX 0 o, XY 20 o, YZ 40 o, ZU 90 o. Y X W V U Z Next, we find WY 0 o, VX 50 o, VY 70 o, UW 00 o, UX 0 o, UY 0 o, ZV 50 o, XZ 60 o, WZ 70 o. ll of these angles ae distinct, and smalle than 80 o. Coesponding to these 5 angles, thee ae 5 angles geate than 80 o (all distinct!), yielding a total of 0 distinct angles.

6 Online Competition Wednesday, Novembe 0, The diagonals of squae BCD meet at the point O. The bisecto of the angle OB meets the segment BO at N, and meets the segment BC at P. The length of NO is x. What is the exact length of PC? () x ( 2+ 2 ) (B) 2x (C) x 5 (D) x ( nswe: (B) 2x 2 2 ) Solution: Look at the tiangle BC: the segments OB and P ae bisectos of the angles BC and BC, espectively. Thus, the point N is the incente of the tiangle BC, and hence equidistant fom all thee sides. We conclude that NX NY NO x. Since BN is the diagonal of the squae NXBY, BN x 2. D O x N Y C P X B Tiangles P C and BN ae simila because thei angles ae equal. Thus, B/C BN/PC. Let B a. then C a 2, and B/C BN/PC implies and so PC x 2 22x. a a 2 x 2 PC,

7 Online Competition Wednesday, Novembe 0, lice walks two-thids acoss a aiload bidge fom point to point B when she sees a tain appoaching at 45 km/h. She does a vey quick calculation and ealizes that if she uns at a cetain speed, she can make it to eithe end of the bidge and avoid the tain. What is the smallest value of, i.e., what is the slowest speed at which she can do it? () km/h (B) 2 km/h (C) 5 km/h (D) 6 km/h nswe: (C) 5 km/h Solution: By x we denote the length of the bidge, and by y the distance between the tain and the bidge. If lice uns to the point, then 45 x+y t and 2x/ t, whee t is the time fo lice and the tain to each. TRIN y B x If lice uns to B, then 45 y t 2 and x/ t 2, whee t 2 is the time fo lice and the tain to each B. Fom 2x/ t we find t 2x/, and so 45 x + y t x + y 2x/ (x + y) 2x Since t 2 x/ y, then 45 x/. Setting these equal, we get y x (x + y) 2x and thus y x+y 2, i.e., x y. Now fom 45 x+y t 2x 45 2x we get 45 2x t and 5(km/h). and t 2x 45. Combining with t 2x/, we obtain

8 ltenative solution: The time that it takes the tain to each point B is y 45. Thetimethat it takes lice to each pont B is x/ x. So we need y 45 x and thus y 5x. Thetimethatittakesthetaintoeachpoint is x+y 45. The time that it takes lice to each pont is 2x/ Substituting y 5x and thus 5. 2x. So we need y + x gives 5x 45 + x 45 2x 2x ltenative solution: If we know that lice and the tain can get to the point B at the same time, then if she uns in the opposite diection at the same speed she will be at C when the tain eaches B. We know that lice and the tain will aive at at the same time, which means that lice can un acoss / of the length of the bidge in the time it takes the tain to coss the entie bidge. Thus, lice uns at / of the speed of the tain, i.e., 5 km/h. TRIN B C / / /

9 Online Competition Wednesday, Novembe 0, Tiangle BC is an isosceles tiangle with two inscibed cicles. The lage cicle has adius 2, and the smalle cicle with adius is tangent to the lage cicle and to the two equal sides of the tiangle. The aea of the tiangle BC is x 2. What is x? () 6 2 (B) 8 2 (C) (D) 8 ( ) 2+ nswe: () 6 2 Solution: Label the cetes of the two cicles by X and Y ; M and N ae the feet of the altitudes to BC, D is the point of tangency of the two cicles. Z is the foot of the altitude fom C. See the figue. C X M D Y 2 N Z B The tiangles ΔXCM and ΔYCN ae simila. Note that YC YX+XC 2++XC + XC. Thus, fom XC/Y C 2/ we obtain + XC 2 XC and so XC. Using the Pythagoean Theoem, we find MC XC 2 XM We know that BZC 90 o, and theefoe the tiangles ΔBZC and ΔXMC ae simila. s well, aea(bc) 2aea(BZC).

10 Using the similaity of tiangles ΔBZC and ΔXMC and the fact that the atio of the aeas of simila tiangles equals the atio of thei side lengths squaed, we find: aea(bc) 2aea(BZC) [ ( ) ZC 2 ] 2 aea(xmc) MC ( ) ( XM MC 2 2 ) ] 2[( ) 2

11 Online Competition Wednesday, Novembe 0, The sum! + 2! +! + 4 2! +! + 4! + 5! + 4! + 5! ! + 205! + 206! is equal to () 206! ! (B) 206! ! (C) 206! 2 206! (D) 206! ! nswe: (D) 206! ! Solution: Obseve that n (n 2)! + (n )! + n! (n 2)! (n 2)! n (n )! n n (n )! n +(n ) + n(n ) ( n (n )! n! Now we ae done:! + 2! +! + 4 2! +! + 4! + 5! + 4! + 5! ! + 205! + 206! 2!! +! 4! + 4! 5! ! 206! 2! 206! 206! ! )

12 Online Competition Wednesday, Novembe 0, Conside a game boad shown below. You ae to move a piece fom to X by moving it to an adjacent squae eithe to the ight o down. In how many diffeent ways can you do it? X () 245 (B) 280 (C) 00 (D) 20 nswe:(c) 00 Solution: Label squaes M and N as shown. Note that a path fom to X must pass though one and only one of M o N. M N X To each N fom, one must make 5 moves, of which ae to the ight and 2 ae down. This can be done in ( ( ) 5 2) 5 ways. Then to each X fom N, one must make 6 moves, thee to the ight amd thee down. This can be done in ( ) 6 ways. Thus, the numbe of paths fom to X that pass thought N is ( ) 5 6 2)( 200 The numbe of paths fom to M is ( ( ) 5 2) 5. ThefistmovefomM must be to the ight, and fom thee ae ( ( ) 5 2) 5 possible paths to X. Thus, the numbe of paths fom

13 to X that pass thought M is ( ) 5 5 2)( Thus the total numbe of paths fom to X is ( )( ) ( )( ) ltenative solution: Bute foce. Stating fom count in how many ways one can aive at each squae. Note that the numbe in each squae is equal to the sum of the numbe in the squae above it and to the left of it (assuming that the the squaes outside the boad cay the value zeo). Thus, the answe is X The End