Alpha Individual Solutions MAΘ National Convention 2013

Transcription

1 Alpha Idividual Solutios MAΘ Natioal Covetio 0 Aswers:. D. A. C 4. D 5. C 6. B 7. A 8. C 9. D 0. B. B. A. D 4. C 5. A 6. C 7. B 8. A 9. A 0. C. E. B. D 4. C 5. A 6. D 7. B 8. C 9. D 0. B TB. 570 TB. 5 TB. 49

2 Alpha Idividual Solutios MAΘ Natioal Covetio 0 Solutios:. v 8,, si 5 si0 si5 si0 si45 si60 si ta correspods to 60 or 40 i the first rotatio aroud the circle. 60 correspods to r r 4cos 4, which has o solutios. 40 correspods to 4cos 4, so r, givig two poits of itersectio. However, both graphs go through the origi as well, so there are a total of three poits of itersectio. 4. Label the edges of the kite as i the picture. Therefore, we have a x c ad a y b. Subtract the secod equatio from the first to get x y c b, which equals 5 by assumptio. Sice xy 5 also, we must have xy sice 5 x y x y x y. Solvig this system yields x, which is the sought legth. 5. First, we will fid all ordered triples a, b, c with oegative coordiates with a b c. If a 0, the b c 5, ad the oly solutios are 0,0,5 ad 0,9,. A quick search yields o solutios for a (the same is the case for a, 4, 6, 7, or 8 ; it is t ecessary to check a 9 or larger values sice i those cases we would have a, the a b c ). If b c, ad the oly solutios are,5,4 ad,0,. If a 5, the b c 00, ad the oly solutio is 5,0,0. For,5,4 ad,0,, there are six differet arragemets, ad each umber could be positive or egative, so for the two poits, there are 6 96 poits. For umber could be positive or egative, so for that poit there are 5,0,0, there are three differet arragemets, ad each 4 poits. For 0,9,, there are six differet arragemets, ad each ozero umber could be positive or egative, so for that poit there are 6 4 poits. Fially, for 0,0,5, there are three

3 Alpha Idividual Solutios MAΘ Natioal Covetio 0 differet arragemets, ad each ozero umber could be positive or egative, so for that poit there are 6 poits. This gives a total of distict poits. 6. Each factor of the equatio correspods to a sigle poit, ad these ca be foud by completig the square. x y x 4y 5 x y, so if this equals 0, the poit is. x y 4x y 5 x y, so if this equals 0, the poit is,, distace betwee these poits is The 7. Usig the method of fiite differeces, we have: The give iformatio is coected with solid lies, so the miimum degree of f is 4. Fiishig out the rest of the differeces (dashed lies) to get two more values, workig backwards from s alog the bottom row, yields f Sice cos5 si7, the agles beig complemetary, we must solve cos7 si. Agai, the complemetary agle would be the smallest positive agle that works, so the aswer is I the picture, based o the give iformatio, x 5, ad usig the Pythagorea Theorem, y 5. This makes the altitude to the hypoteuse

4 Alpha Idividual Solutios MAΘ Natioal Covetio x x 60 0 x x x 0, so the sum of the solutios is 4 x 6 4 sice the two solutios are differet. c c c c. f f c f c c ad fc, so settig c c 6c c c c these equal mea thatc 0 c or 6c c c 0. Therefore, the ozero solutio is c.5.. Reorderig the -elemet data set, we get,,,,, 5, 5, 7, 7, 7, 7,, ad the sum of the elemets is 60, so A 5. The media would be the average of 5 ad 5, so B 5. Clearly, C 7, so C AB The side legth of the hexago is 4 8, so the eclosed area is Let x The x 5 x ad x 0. Thus 0 x x 5 x. Sice x0, x. x 5 x 5. Coutig up the white area as fractios of the total, we get which is a ifiite geometric series with sum 4. So, actually, oe of the triagle s 4 area is black! 89 si 0 0 si00 8 8, ad usig the fact that 6. 4

5 Alpha Idividual Solutios MAΘ Natioal Covetio 0 si0 si50 si , we get 7 si x must be cotermial with 5 is or. Sice 0, the largest cotermial agle The five least oegative eve itegers are 0,, 4, 6, ad 8. The four greatest odd egative itegers are 7, 5,, ad. The sum of these umbers is The sum of the reciprocals of the positive itegral factors is equal to the sum of the positive itegral factors divided by the umber itself. Sice 8 is a perfect umber, the sum of the positive itegral factors is twice 8, ad the sought sum is 56 (it does t take much to verify 8 this by writig out the positive itegral factors of 8, ad it turs out that the sum of the reciprocals of the positive itegral factors of ay perfect umbers is always ). 0. Let x be the legth of the side of the octago, as i the diagram. Cosiderig the octago as a square with four isosceles right triagles removed, the eclosed area of the x x octago is x 4 x. The square ecloses a area of x x x, so the sought ratio is. 4 csc cot csc cot csc cot, which is oe of the aswer choices Multiply this. Let S 5

6 Alpha Idividual Solutios MAΘ Natioal Covetio 0 equatio by 4 to get S Now, subtract this secod equatio from the first equatio to get 4 S Now multiply this equatio by 4 to get S , ad subtract that from the previous 9 equatio to get Therefore, S S x x cos cos x x x x 6. Squarig both sides yields 9x x x x. However, the positive solutio does ot work because it gives a positive umber o the left side of the equatio just before squarig ad a egative umber o the right side. The egative umber does work, so x is the oly solutio. 4. six six cos x cos x cos x x cos x cos x x cos x, the cosie fuctio is eve ad thus yields 5. For x y (eve if you chose to go with cos x also). 4, usig the stadard coic sectio otatio, ab, ad sice the latus rectum has legth 6. i a, the sought legth is b. 4 4m cis45 cis45 m cis45 m 4 4 i cis00 cis700 m m m. To make this a iteger, we must have 0 m 4 ad m must be a multiple of 4 so that 45 m is cotermial with either 0 or 80. Thus, the valid values of m are 4, 8,, 6, 0, ad 4, ad their sum is Let x be the umber of ways to get to the th step. Clearly x 0 ad x x, but

7 Alpha Idividual Solutios MAΘ Natioal Covetio 0 also x xx x, so the sequece is 0,,,,,,, 4, 5, 7, 9,, 6,, 8, 7, 49, ad the 7th term is Let s be the sum of the th powers of the solutios of the equatio. Therefore, 0 s 0. Usig the method of Newto s sums, we have 0 s 0s s, ad fially, 0 s 0s s s Imagie the 8 8 grid i the plae, where the x-coordiate represets the time i hours after 0 pm whe Sydey arrives ad the y-coordiate represets the time i hours after 0 pm whe Everett arrives. By the give iformatio about the latest each could arrive at the rave, we reduce the possible area dow to the 4rectagle. If Everett arrives at am, the latest he could arrive, Sydey would have to arrive o earlier tha pm i order to have hours of overlap (this poit is represeted by the poit, ). If Sydey arrives at am, the latest she could arrive, Everett would have to arrive o earlier tha pm i order to have hours of overlap (this poit is represeted by the poit 4, ). From those poits, draw lie segmets with a slope of to the edge sice if each back their arrival time up the same amout from those times, they would still overlap for at least hours. The favorable area is betwee the lie segmets, ad that area is 4 7, ad the total possible area is 8, so the probability of their overlap lastig at least hours is Let d be the legth of a diagoal of the larger petago, ad let x be the legth of a side of the larger petago. The, by the Law of Cosies, d x x cos08 x cos08 cos08 4x 4x si 54 d xsi54 Let y be the legth of the part of a diagoal from a vertex of the smaller petago to oe of its closest vertices of the larger petago, ad drop a altitude from the vertex of the smaller petago to oe of the sides of the larger petago closest to it. 7.

8 Alpha Idividual Solutios MAΘ Natioal Covetio 0 x The cos6 x x x y y y cos6 si54. Therefore, the side legth of the smaller x si 54 cos 6 cos7 petago would be xsi54 x x x si54 si54 si54 si54 cos7 x, makig the ratio of the smaller petago s side legth to the larger petago s cos6 cos7 x cos6 cos7 side legth. Sice the larger petago s side legth is cos6, the legth x cos6 of the side of the smaller petago is cos7. Tiebreakers TB TB (sice 0 ). m m m TB m m m m m m m sice m 0, so we are lookig for the least 84 such that 8 8 is a perfect square. Settig this perfect square as p, we are tryig to solve the Pell Equatio k k p. I tryig to solve k ad settig p, fid the smallest solutio i positive itegers, which is kp. All solutios are foud recursively i the followig way:. The ext p is the sum of the previous k ad p.. The ext k is the sum of the previous k ad twice the previous p. 8

9 Alpha Idividual Solutios MAΘ Natioal Covetio 0 The table looks like this: k p k p (oly whe p is odd) It ca be show also that the ad will always alterate, so the ext value of will be 49. I fact, were the last colum to be cotiued, the umbers i the last colum would be the oly umbers that have the property defied i this questio. 9