Download [PDF] 2018 AMC Senior Years 11 and 12 Solutions Australian Mathematics Competition

File Information

Filename:	[PDF] 2018 AMC Senior Years 11 and 12 Solutions Australian Mathematics Competition.pdf
Filesize:	239.15 KB
Uploaded:	18/08/2021 16:18:49
Keywords:	2018 amc senior years 11 and 12 solutions australian mathematics competition pdf
Description:	Download file or read online AMC past exam paper 2018 Senior Years 11 and 12 Solutions - Australian Mathematics Competition.
Downloads:	18

File Preview

Download Urls

Short Page Link

https://www.edufilestorage.com/3u5

Full Page Link

https://www.edufilestorage.com/3u5/PDF_2018_AMC_Senior_Years_11_and_12_Solutions_Australian_Mathematics_Competition.pdf

HTML Code

<a href="https://www.edufilestorage.com/3u5/PDF_2018_AMC_Senior_Years_11_and_12_Solutions_Australian_Mathematics_Competition.pdf" target="_blank" title="Download from eduFileStorage.com"><img src="https://www.edufilestorage.com/cache/plugins/filepreviewer/3397/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg"/></a>

Forum Code

[url=https://www.edufilestorage.com/3u5/PDF_2018_AMC_Senior_Years_11_and_12_Solutions_Australian_Mathematics_Competition.pdf][img]https://www.edufilestorage.com/cache/plugins/filepreviewer/3397/pdf/150x190_middle_46f4e7862b1eb5bd4935adbbba5d79e8.jpg[/img][/url]

[PDF] 2018 AMC Senior Years 11 and 12 Solutions Australian Mathematics Competition.pdf | Plain Text

2018 AMC Senior Solutions Solutions { Senior Division 1. (Also I4) The three angles at Padd to 90  , so \X P Y = 90 50 = 40  , hence (D). 2. Since 8 30 = 240, slightly more than 30 km per day is required, hence (C). 3. The number is 64, so twice the number is 128, hence (D). 4. (Also UP18, J14, I10) There are many subdivisions of the hexagon into equal areas that show that the shaded area is 2 3 of the total: 2 3 4 6 4 6 8 12 hence (B). 5. (Also I9) Using the distributive law, 9 1:2345 9 0:1234 = 9 (1:2345 0:1234) = 9 1:1111 = 9:9999 hence (A). 6. 20 = 1 and 1 8 = 1 so that 2 0 18 = 0, hence (A). 7. (Also J11) 1000% means `10 times', since 1000% = 10 100%. So 10 times the number is 100, and the number is 10, hence (C). 8. (Also J13, I11) To feed 4 dogs for 1 day costs $60 3 = $20, and then to feed 1 dog for 1 day costs $20 4 = $5. To feed 7 dogs for 7 days will cost $5 7 7 = $245, hence (C). c Australian Mathematics Trust www.amt.edu.au 69

2018 AMC Senior Solutions 9. Four of the statements can be eliminated through a well-chosen example. A BC M 1 1 p 5 p 5 2 Suppose AM=M B = 1,M C = 2 and AB?M C as shown. Then AC =B C =p 5 by Pythagoras' theorem. This eliminates (C) and (D). Also AM6=C M so that 4AC M is not isosceles. Then \C AM6= \AC M , so that (A) is false. Also 2\C AM > 2 45 = \C M B , so that (B) is false. In general (E) is always true, since triangles 4AM Cand4M B C have bases of the same length and identical altitude, hence (E). 10. The di erence between 1 + 2 +


+ 100 and 101 + 102 +


+ 200 is 100 100 = 10000. Therefore 101 + 102 +


+ 200 = 5050 + 10000 = 15050, hence (A). 11. Alternative 1 If the edge of the triangular tile is 1 unit, then the row of tiles has perimeter 30 units. To t a single row, this will split 30 = 14 + 1 + 14 + 1 like this: 14 1 14 1 Then there are 28 triangles, 14 pointing up and 14 pointing down, hence (D). Alternative 2 There are 10 3 = 30 triangle edges in the perimeter. If the row has ntriangles, then there are 3 ntriangle edges, but some are used in a `join'. The triangles are in a single row, so there are n 1 joins, each using 2 edges. So the perimeter is 3n 2(n 1) = n+ 2 edges. Solving n+ 2 = 30, we deduce that 28 tiles are needed, hence (D). 12. (Also I16) Since x+ 3 x+ 200 = 360, x= 40. Triangle 4AC Dis isosceles with \AC D= 40 and \DAC =\C DA. Then 180  = 40  + 2 \DAC , so that \DAC= 70 . Similarly in 4AB C,\B AC =\AB C =1 2 (180 120) = 30  . Hence \DAC :\B AC = 70 : 30 = 7 : 3, hence (E). c Australian Mathematics Trust www.amt.edu.au 70

2018 AMC Senior Solutions 13. If the number is x, then 4x330 = x 4 + 330 16x 4 330 = x+ 4 330 15x = 8330 x = 8 330 15 = 8 22 = 176 hence (E). 14. Subdivide the octagon with horizontal and vertical lines, and use the standard ratios of right-isosceles triangles to nd the spacing between horizontal lines. 1 1 1 1 p 2=2 1p 2=2 Then the shaded triangle has base 1, altitude 1 + p 2 and area 1 2 (1 + p 2), hence (E). 15. Alternative 1 If the fares are pand wand the total cost is C, then C = 6 p+ 7 w 4C= 24 p+ 28w C = 8 p+ 4 w 3C= 24 p+ 12w 4C 3C = 24 p+ 28w 24p 12w C = 16w That is, for cost Cshe can take 16 Warriors, hence (C). Alternative 2 From the information, swapping 2 Panthers for 3 Warriors doesn't change the fare. Consequently starting with 6 Panthers and 7 Warriors, she can swap 6 Panthers for 9 Warriors. On this trip she could take 16 Warriors, hence (C). c Australian Mathematics Trust www.amt.edu.au 71

2018 AMC Senior Solutions 16. LetP S=S Q =pand P T=T R =q. P Q RS T q q p p Then tan = 2p q and tan = p 2q . So tan : tan = 2p q : p 2q = 4 : 1 ; hence (B). 17. Distinguishing between the three dice, the sample space consists of 6 3 = 216 equally likely outcomes. There are 4 possible selections of consecutive numbers: 123; 234;345; 456. Each of these will appear 6 times in the sample space. So the required probability is 24 216 = 1 9 , hence (B). 18. (Also I21) We can approximate 20 18 = 2 18  1018 = 2 8  210  1018  200 103  1018 = 2 1023 , which has 24 digits. This indicates that 20 18 has 24 digits. More formally, 2 18 = 2 8  210 > 100 1000 = 10 5 and 2 18 = 2 9  29 < 1000 1000 = 10 6 . Then 10 23 < 2018 < 1024 and 20 18 has 24 digits, hence (A). Note: 2018 = 262 144 000 000 000 000 000 000 19. (Also J25, I22) Alternative 1 Start with some simple cases: 1 1 1 1 1 1 1 1 2 2 2 2 1 1 1 0 8 8 8 9 sum of digits = 3 1 + 3 8 + 9 = 36 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 1 1 1 1 0 8 8 8 8 9 sum of digits = 4 1 + 4 8 + 9 = 45 Clearly this pattern continues, and we can generalise. Then the sum of all digits is 49  1 + 0 + 49 8 + 9 = 450, hence (D). c Australian Mathematics Trust www.amt.edu.au 72

2018 AMC Senior Solutions Alternative 2 111: : :111 | {z } 100 222 : : :222 | {z } 50 =111 : : :111 | {z } 100 111 : : :111 | {z } 50 111 : : :111 | {z } 50 =111 : : :111 | {z } 50 000 : : :000 | {z } 50 111 : : :111 | {z } 50 =111 : : :110 | {z } 50 999 : : :999 | {z } 50 111 : : :110 | {z } 50 =111 : : :110 | {z } 50 888 : : :889 | {z } 50 In the digit sum of this number, we can pair the 1 + 8 terms, giving 50 9 = 450, hence (D). 20. Alternative 1 Let the smaller polygon have nsides, so that the larger one will have n+ 5. The sum of the interior angles in a regular n-sided polygon is ( n 2) 180  and the size of the interior angle is 180(n 2) n  . Then 180[(n + 5)2] n + 5 180(n 2) n = 1 180(n + 3)n 180(n 2)(n + 5) = n(n+ 5) 180(n 2 + 3 n n2 3n + 10) = n2 + 5 n n 2 + 5 n 1800 = 0 (n 40)(n + 45) = 0 Since nis positive, n= 40. The larger polygon has n+ 5 = 45 sides, hence (C). Alternative 2 Let the exterior angle on the larger and smaller polygons be  and ( + 1) , respectively. Then the polygons have 360  and 360  + 1 sides, respectively. Then 5 + 360  + 1 = 360  5 ( + 1) + 360 = 360(+ 1) 5( + 9)( 8) = 0 Since  >0, we have = 8. Therefore the larger polygon has 360 8 = 45 sides, hence (C). c Australian Mathematics Trust www.amt.edu.au 73

2018 AMC Senior Solutions 21. Since n= p 100 m2 , we have 100 m2  0 so that m2  100 and 10m 10. Also 100 m2 is a perfect square. By checking possible values of m2 = 0 2 ; 1 2 ; : : : ; 102 , the possible pairs ( m; n) are (10; 0); (8; 6); (6; 8); (0;10) Therefore there are 7 possible solutions (m; n ), hence (C). 22. The tetrahedron can be made by slicing 4 triangular pyramids o the cube. Each of these triangular pyramids has base area 2, height 2 and volume 4 3 . The volume of the tetrahedron is then 8 4 4 3 = 8 3 , hence (A). 23. The diagonal of the rectangle is p 12 2 + 5 2 = p 169 = 13 cm. 12 13 5 O A B D C P r r r To nd the radius of the circle, note that the area of 4ADBis 30, which is also equal to the combined areas of 4AOB,4DOA , and4B OD , which is 12r 2 + 5r 2 + 13r 2 = 15 r. Thus r = 2. We can then subdivide the rectangle horizontally and vertically: 2 82 2 1 2 O A B D C P Then by Pythagoras' theorem, OP=p 8 2 + 1 2 = p 65, hence (C). c Australian Mathematics Trust www.amt.edu.au 74

2018 AMC Senior Solutions 24. Express each p
as (
) 1 2 , then (


((28 ) 1 2 ) 1 2


)1 2 | {z } 60 = 2 8 x 2 8 ( 1 2 ) 60 = 2 2 3x 8  1 2  60 = 2 3x 2 57 = 23x x = 19 hence (B). 25. Alternative 1 C B A E D Label the original triangle and the rst two smaller triangles as shown. Note that all the triangles are similar, since they have a right angle and a common acute angle. We rst nd the fraction of trapezium B DC E that is shaded. Similar triangles 4AB C,4C B D ,4DC E demon- strate that DE DC = AC AB = 4 5 . So the area of 4DC E is 4 5
2 = 16 25 times the area of 4C B D. That is, 4DC E is16 41 of the area of trapezium BDC E. The same ratio applies to the trapezium formed by any pair where the white triangle is the larger of the two. Furthermore 4AB Cis comprised of the sequence of such trapeziums. Consequently the shaded area overall is 16 41 of the area of 4AB C. That is, the shaded area is 16 41  6 = 96 41 hence (E). Alternative 2 Suppose xis the area shaded, so that 6 xis unshaded. The leftmost unshaded triangle is similar to the whole triangle (area 6) but with hypotenuse 3 instead of 5. So the area of the leftmost triangle is  3 5  2 6 = 54 25 . Removing the leftmost triangle leaves a gure with area 6 54 25 = 96 25 with xshaded and 96 25 xunshaded. This gure is similar to the original, except that the shading is reversed. Consequently the ratio of unshaded to shaded in the original triangle is equal to the ratio c Australian Mathematics Trust www.amt.edu.au 75

2018 AMC Senior Solutions of shaded to unshaded once the leftmost triangle is removed: 6 x x = x 96 25 x x 2 = (6 x)( 96 25 x) 25x 2 = (6 x)(96 25x) = 25x 2 246x + 576 x = 576 246 = 96 41 hence (E). Alternative 3 In the in nite sequence of triangles (white, shaded, white, . . . ) each triangle is similar to the 3:4:5 triangle. By comparing sides, we deduce that each is r= 16 25 the area of the previous one. That is, the areas form a geometric series: 6 =a+ ar +ar2 + ar3 +


= a 1 r For a general geometric series S= a+ ar +ar 2 + ar 3 + ar 4 +


, consider the even exponent and odd exponent terms separately: A= a+ ar2 + ar4 +


and B=ar +ar3 + ar5 +


Then B=rA and A+ B =S. Therefore A= 1 1 + rS and B= r 1 + rS . In the current problem, the shaded triangles are represented by the series Babove, so with S = 6, r= 16 25 we have B= r 1 + rS = 16 25 41 25
6 = 96 41 hence (E). 26. (Also J30) A number is divisible by 9 if and only if the sum of its digits is divisible by 9. Thus, B and Care divisible by 9. Since Ais a 2018-digit number, B

2018 AMC Senior Solutions 27. The area of ABC Dis120 2 (100 + 150) = 15000, so each small quadrilateral has area 3750. A B CD Q PE F GH R 50 50 60 75 75 Let the sides' midpoints be E ; F; G; Has shown and let R be the foot of the perpendicular from Qto AD . The trapezium AE GDand the pentagon AE QGDare both half the area of AB C D|the trapezium because it has half the base and top of AB C Dand the same al- titude, and the pentagon because it is 2 of the 4 equal quadrilaterals. This can only occur if E QGis a straight line as shown. Let Pbe the midpoint of E G. ThenAE P His a trapez- ium with AE= 50, AH= 60 and H P=125 2 = 62:5, so AE P Hhas area 60 2 (50 + 62:5) = 3375. Consequently 4H P Qhas area 3750 3375 = 375. Let h= H R , the altitude of 4H P Q, then 375 = 1 2 125 2 h =) h= 12 Then AR= 60 + 12 = 72 and RQ= 50 + 72 25 120 = 65. By Pythagoras' theorem, AQ =p 65 2 + 72 2 = p 9409. For AQ to be an integer, the last digit of AQwill be 3 or 7. Also AQ2  10000 = ) AQ 100, which suggests that AQ= 97. Checking, 97 2 = (100 3)2 = 10000 600 + 9 = 9409 so that AQ= 97, hence (97). 28. Alternative 1 There are 6 5 4 3 2 1 = 720 ways to arrange the six shoes in a row, if we ignore the fact that we want each pair to appear in the correct order. In half of these con gurations, the blue shoes will be in the correct order. In half of these con gurations again, the red shoes will be in the correct order. And in half of these con gurations again, the white shoes will be in the correct order. Therefore, the number of ways to do this is 720 1 2  1 2  1 2 = 90, hence (90). Alternative 2 Start with the two blue shoes in a row B L and B R. Next place the red shoes R L and R R to make a row of 4 shoes, two of which are red. There are 4 2
= 6 ways of doing this. Finally place the two white shoes W Land W R, which can be done in 6 2
= 15 ways. Therefore the number of ways of placing the shoes is 6 15 = 90, hence (90). c Australian Mathematics Trust www.amt.edu.au 77

2018 AMC Senior Solutions 29. (Also I30) A O B C Let Abe the centre of the pattern, then for each circle there is a centre Oand an isosceles triangle 4AB Cas pictured. Then \B AC =360 n \AB C =\AC B =1 2  180 360 n  = 90 180 n \AOC = 2\AB C = 180360 n Then re ex angle AOC= 180 + 360 n = 1 2 + 1 n   360. Consequently the visible arc is 1 2 + 1 n of the circle and it has arc length 1 2 + 1 n . The total of all visible arcs is then n 1 2 + 1 n  = n 2 + 1 = 60. Therefore n 2 = 59 and n= 118, hence (118). 30. Consider the number 1 in the top-left cell of the n ngrid. After one shue it has moved one cell to the right and one cell down. Similarly, after a second shue it has moved a further cell to the right and down. This continues until it reaches the bottom-right cell after n 1 shues in total. The next shue sends it to the cell below the top-left cell. The next n 2 shues continue to move it right and down until it reaches the bottom row immediately to the left of the bottom-right cell. The next shue sends the number 1 back to top-left cell. Hence it takes (n1) + 1 + ( n 2) + 1 = 2 n 1 shues for the number 1 to be returned to its original position, as illustrated in the path on the left. Every other number that starts on this path will also follow the same cycle, so they too will return to their starting positions after 2n 1 shues. 1 path of 1 after each shue 2 path of 2 after each shue Similarly, the path taken by the number 2 can be traced as illustrated on the right. After n 2 shues it reaches the nal column, immediately above the bottom-right cell. The next shue sends it immediately back to its starting position, via the bottom-left cell. Hence it takes (n2) + 1 = n 1 c Australian Mathematics Trust www.amt.edu.au 78

2018 AMC Senior Solutions shues for the number 2 to return to its original position. Again this applies to every number that starts on this path. A similar analysis shows that all other numbers in the grid will also take n 1 shues to be returned. Therefore lcm(2n 1; n 1) shues are required to return all of the numbers in the grid to their original position. Since 2n 1 and n 1 have no common factors, other than 1, the number of shues is in fact given by S= (2n 1)(n 1). To nd when S rst exceeds 20000, it is convenient to approximate it by the simpler expression 2n 2 . Hence 2n2  20000 n 2  10000 n  100 If n= 100 then S= 199 99