The ENTRE fo EDUATIN in MATHEMATIS and MPUTING cemc.uwateloo.ca 2017 Galois ontest Wednesday, Apil 12, 2017 (in Noth Ameica and South Ameica) Thusday, Apil 13, 2017 (outside of Noth Ameica and South Ameica) Solutions 2017 Univesity of Wateloo
2017 Galois ontest Solutions Page 2 1. (a) In Box E, 6 of the 30 cups wee puple. The pecentage of puple cups in Box E was 6 30 0% = 2 0% = 20%. (b) n Monday, 30% of Daniel s 90 cups o 30% 90 = 30 90 = 27 cups wee puple. 0 Daniel had 9 puple cups in Box D and 6 puple cups in Box E. Theefoe, the numbe of puple cups in Box F was 27 9 6 = 12. (c) Daniel had 27 puple cups and 90 cups in total. n Tuesday, Avil added 9 moe puple cups to Daniel s cups, binging the numbe of puple cups to 27 + 9 = 36, and the total numbe of cups to 90 + 9 = 99. Bay bought some yellow cups and included them with the 99 cups. Let the numbe of yellow cups that Bay bought be y. The total numbe of cups was then 99 + y, while the numbe of puple cups was still 36 (since Bay bought yellow cups only). Since the pecentage of cups that wee puple was again 30% o 30 30, then of 99 + y 0 0 must equal 36. Solving, we get 30 (99 + y) = 36 o 30(99 + y) = 3600 o 99 + y = 120, and so y = 21. 0 Theefoe, Bay bought 21 cups. 2. (a) Abdi aived at 5:02 a.m., and so Abdi paid $5.02. aleigh aived at 5: a.m., and so aleigh paid $5.. In total, Abdi and aleigh paid $5.02 + $5. = $.12. (b) If both Daniel and Emily had aived at the same time, then they each would have paid the same amount, o $12.34 2 = $6.17. In this case, they would have both aived at 6:17 a.m. If Daniel aived 5 minutes ealie, at 6:12 a.m., and Emily aived 5 minutes late, at 6:22 a.m., then they would have aived minutes apat and in total they would have still paid $12.34. (We may check that these aival times ae minutes apat, and that Daniel and Emily s total pice is $6.12 + $6.22 = $12.34, as equied.) (c) To minimize the amount that Kala could have paid, we maximize the amount that Isaac and Jacob pay. Isaac and Jacob aived togethe and Kala aived afte. Since Kala aived at a late time than Isaac and Jacob, then Kala paid moe than Isaac and Jacob. If Isaac and Jacob both aived togethe at 6:18 a.m., then they would each have paid $6.18, and Kala would have paid $18.55 $6.18 $6.18 = $6.19. This is the minimum amount that Kala could have paid. Why? If Isaac and Jacob aived at 6:19 a.m. o late, then Kala would have aived at a time ealie than 6:19 a.m. (since $18.55 $6.19 $6.19 = $6.17). Since Kala aived afte Isaac and Jacob, this is not possible. If Isaac and Jacob aived ealie than 6:18 a.m., then they would have each paid less than $6.18, and so Kala would have paid moe than $6.19. Theefoe, the minimum amount that Kala could have paid is $6.19. (d) If Lay aived ealie than 5:39 a.m., then he would have paid less than $5.39 and so Mio would have paid moe than $11.98 $5.39 = $6.59. Since Mio aived duing the time of the special picing, it is not possible fo Mio to have
2017 Galois ontest Solutions Page 3 paid moe than $6.59, and so Lay must have aived at 5:39 a.m. o late. If Lay aived between 5:39 a.m. and 5:59 a.m. (inclusive), then Lay would have paid the amount between $5.39 and $5.59 coesponding to his aival time. Theefoe, Mio would have paid an amount between $11.98 $5.59 = $6.39 and $11.98 $5.39 = $6.59 (inclusive). Each of the amounts between $6.39 and $6.59 coesponds to an aival time fo Mio between 6:39 a.m. and 6:59 a.m., each of which is a possible time that Mio could have aived duing the special picing peiod. That is, each aival time fo Lay fom 5:39 a.m. to 5:59 a.m. coesponds to an aival time fo Mio fom 6:39 a.m. to 6:59 a.m. Each of these times is duing the peiod of the special picing and each coesponding pai of times gives a total pice of $11.98. To see this, conside that if Lay aived x minutes afte 5:39 a.m. (whee x is an intege and 0 x 20), then Mio aived x minutes befoe 6:59 a.m., and in total they paid $5.39 + x + $6.59 x = $11.98. Since Lay s aival time and Mio s aival time may be switched to give the same total, $11.98, then Lay could also have aived between 6:39 a.m. and 6:59 a.m. The only times left to conside ae those fom 6:00 a.m. to 6:38 a.m. If Lay aived at one of these times, his pice would have been between $6.00 and $6.38, and so Mio s pice would have been between $11.98 $6.38 = $5.60 and $11.98 $6.00 = $5.98. Since thee ae no aival times which coespond to Mio having to pay an amount between $5.60 and $5.98, then it is not possible that Lay aived at any time fom 6:00 a.m. to 6:38 a.m. Theefoe, the anges of times duing which Lay could have aived ae 5:39 a.m. to 5:59 a.m o 6:39 a.m. to 6:59 a.m. 3. (a) Since P Q = 90, then P Q is a ight-angled tiangle. By the Pythagoean Theoem, Q 2 = P 2 + P Q 2 = 18 2 + 24 2 = 900, and so Q = 900 = 30 (since Q > 0). Line segment S is a adius of the cicle and thus has length 18. Theefoe, SQ = Q S = 30 18 = 12. (b) Sides AB, B, D, and DA ae tangent to the cicle at points E, F, G, and H, espectively. 12 20 Theefoe, adii E, F, G, and H ae pependicula to H 12 thei coesponding sides, as shown. 12 In quadilateal DHG, GD = GDH = DH = 90 F 12 15 and so GH = 90. Since H = G = 12 (they ae adii of the cicle), then DHG is a squae with side length 12. A E B Similaly, HAE is also a squae with side length 12. Since G = 90, then G is a ight-angled tiangle. By the Pythagoean Theoem, G 2 = 2 G 2 = 20 2 12 2 = 256, and so G = 256 = 16 (since G > 0). It can be similaly shown that F = 16. Since EB = 90, then EB is a ight-angled tiangle. By the Pythagoean Theoem, EB 2 = B 2 E 2 = 15 2 12 2 = 81, and so EB = 81 = 9 (since EB > 0). It can be similaly shown that F B = 9. D G
2017 Galois ontest Solutions Page 4 Theefoe, the peimete of ABD is GD + DH + HA + AE + EB + BF + F + G o 4 12 + 2 9 + 2 16 = 98. (c) In Figue 1: Figue 1 Since the cicles ae inscibed in thei espective squaes, then T U is a tangent to the lage cicle and UV is a tangent to the smalle cicle. Let T U touch the lage cicle at W, and let UV touch the smalle cicle at X. The adius W is pependicula to T U, and the adius X is pependicula to UV. T W U X V Figue 2 In Figue 2: The diamete of the lage cicle is equal to the side length of the lage squae. To see this, label the points P and R whee the vetical sides P R of the lage squae touch the lage cicle. S Q Join P to and join R to. The adius P is pependicula to P T and the adius R is T W U X V pependicula to RU. In quadilateal P T W, P T = P T W = T W = 90, and so P W = 90. Similaly, in quadilateal RUW, RW = 90. Theefoe, P W + RW = 180 and so P R passes though and is thus a diamete of the lage cicle. In quadilateal P T UR, all 4 inteio angles measue 90, and so P T UR is a ectangle. It can similaly be shown that if S and Q ae the points whee the vetical sides of the smalle squae touch the smalle cicle, then SQ is a diamete of the smalle cicle and SUV Q is a ectangle. In Figue 3: The aea of the lage squae is 289, and so each side of the lage squae has length Figue 3 289 = 17. The diamete of the lage cicle is equal to the side length of the lage squae, o P R = T U = 17. Since is the midpoint of P R, and W is pependicula to T U, then W is the midpoint of T U. P 8.5 Theefoe, W U = R = W = 17 2 = 8.5. T W The aea of the smalle squae is 49, and so each side of the smalle squae has length 49 = 7. Similaly, X is the midpoint of UV and so UX = S = X = 7 2 = 3.5. 8.5 S R U X 3.5 3.5 Q V
2017 Galois ontest Solutions Page 5 In Figue 4: Figue 4 Finally, we constuct the line segment fom, paallel to XW, and meeting W at Y. In quadilateal Y W X, Y is paallel to XW, Y W is pependicula to XW, and X is pependicula to XW, and so Y W X is a ectangle. 5 Thus, X = Y W = 3.5, and Y 3.5 3.5 Y = XW = XU + W U = 3.5 + 8.5 = 12. T W U X V Since Y = 90, then Y is a ight-angled tiangle 12 with Y = 12, and Y = W Y W = 8.5 3.5 = 5. By the Pythagoean Theoem, 2 = Y 2 + Y 2 = 12 2 + 5 2 = 144 + 25 = 169, and so = 169 = 13 (since > 0). 4. (a) The total aea of the m = 14 by n = Koelle-ectangle is m n = 14 = 140. The dimensions of the shaded aea inside a Koelle-ectangle ae (m 2) by (n 2) since the 1 by 1 squaes along the sides ae unshaded, so each dimension is educed by 2. Theefoe, the shaded aea of a 14 by Koelle-ectangle is (14 2) ( 2) = 12 8 = 96. The unshaded aea is the diffeence between the total aea and the shaded aea, o mn (m 2)(n 2) = mn (mn 2m 2n+4) = 2m+2n 4 o 2 14+2 4 = 44. Finally, is the atio of the shaded aea to the unshaded aea, o 96 44 = 24 (o 24 : 11). 11 (b) As we saw in pat (a), the shaded aea of an m by n Koelle-ectangle is (m 2)(n 2), and the unshaded aea is 2m + 2n 4. Theefoe, = (m 2)(n 2) 2(m 2). When n = 4, = 2m + 2n 4 2m + 4 We ewite m 2 m + 2 as m + 2 4 m + 2 = m + 2 m + 2 4 m + 2 = 1 4 m + 2. = 2(m 2) 2(m + 2) = m 2 m + 2. We must detemine all possible intege values of u fo which = 1 4 m + 2 = u 77, fo some intege m 3. Simplifying this equation, we get 1 4 m + 2 = u 77 1 u 77 = 4 m + 2 77 u 4 = 77 m + 2 (m + 2)(77 u) = 4 77 Both u and m ae integes, and so (m + 2)(77 u) is the poduct of two integes. If a and b ae positive integes so that ab = 4 77 = 2 2 7 11, then thee ae 6 possible facto pais (a, b) with a < b. These ae: (1, 308), (2, 154), (4, 77), (7, 44), (11, 28), and (14, 22). Since m 3, then m + 2 5 and so m + 2 cannot equal 1, 2 and 4. Howeve, m + 2 can equal any of the emaining 9 divisos: 7, 11, 14, 22, 28, 44, 77, 154, 308. In the table below, we detemine the possible values fo u given that (m + 2)(77 u) = 2 2 7 11, and m + 2 5.
2017 Galois ontest Solutions Page 6 m + 2 7 11 14 22 28 44 77 154 308 77 u 44 28 22 14 11 7 4 2 1 u 33 49 55 63 66 70 73 75 76 The intege values of u fo which thee exists a Koelle-ectangle with n = 4 and = u 77, ae u = 33, 49, 55, 63, 66, 70, 73, 75, 76. (Fo example, the 5 by 4 Koelle-ectangle has = m 2 m + 2 = 3 7 = 33, and so u = 33.) 77 (m 2)(n 2) (c) As in pat (b), = 2m + 2n 4 Reaanging this equation, we get =, and when n =, = 8(m 2) 2m + 16 4(m 2) 4 = m 2 4 = 4 = 4 = 1 = 1 4 = 1 u (since = u 4p 2 p ) 2 = 4p2 u 4p 2 40p 2 = ()(4p 2 u) = 4(m 2). Since p, u and m ae integes, then ()(4p 2 u) is the poduct of two integes. We must detemine all pime numbes p fo which thee ae exactly 17 positive intege values of u fo Koelle-ectangles satisfying this equation 40p 2 = ()(4p 2 u). Fo p = 2, 3, 5, 7, and then p 11, we poceed with the following stategy: detemine the value of 40p 2 count the numbe of divisos of 40p 2 eliminate possible values of, thus eliminating possible values of 4p 2 u count the numbe of values of u by counting the numbe of values of 4p 2 u If p = 2, then 40p 2 = 40 2 2 = 2 5 5, and so 2 5 5 = ()(16 u). Each diviso of 2 5 5 is of the fom 2 i 5 j, fo integes 0 i 5 and 0 j 1. That is, thee ae 6 choices fo i (each of the integes fom 0 to 5) and 2 choices fo j (0 o 1), and so thee ae 6 2 = 12 diffeent divisos of 2 5 5. Since 2 5 5 = ()(16 u), then thee ae at most 12 diffeent intege values of 16 u (the 12 divisos of 2 5 5), and so thee ae at most 12 diffeent intege values of u. Theefoe, when p = 2, thee cannot be exactly 17 positive intege values of u. If p = 5, then 40p 2 = 40 5 2 = 2 3 5 3, and so 2 3 5 3 = ()(0 u). Each diviso of 2 3 5 3 is of the fom 2 i 5 j, fo integes 0 i 3 and 0 j 3. That is, thee ae 4 choices fo i and 4 choices fo j, and so thee ae 4 4 = 16 diffeent divisos of 2 3 5 3. Since 2 3 5 3 = ()(0 u), then thee ae at most 16 diffeent intege values of
2017 Galois ontest Solutions Page 7 0 u, and so thee ae at most 16 diffeent intege values of u. Theefoe, when p = 5, thee cannot be exactly 17 positive intege values of u. If p = 3, then 40p 2 = 40 3 2 = 2 3 3 2 5, and so 2 3 3 2 5 = ()(36 u). Each diviso of 2 3 3 2 5 is of the fom 2 i 3 j 5 k, fo integes 0 i 3, 0 j 2, and 0 k 1. That is, thee ae 4 3 2 = 24 diffeent divisos of 2 3 3 2 5. Since m 3, then 11, and so the divisos of 2 3 3 2 5 which cannot equal ae: 1, 2, 3, 4, 5, 6, 8, 9, and. Since thee ae 9 divisos which cannot equal, then thee ae 9 divisos that 36 u cannot equal. (These divisos can be detemined by dividing 2 3 3 2 5 by each of the 9 divisos 1, 2, 3, 4, 5, 6, 8, 9, and.) So then thee ae 24 9 = 15 diffeent intege values of 36 u, and so thee ae exactly 15 diffeent intege values of u when p = 3. Theefoe, thee ae not 17 positive intege values of u when p = 3. If p = 7, then 40p 2 = 40 7 2 = 2 3 5 7 2, and so 2 3 5 7 2 = ()(196 u). Each diviso of 2 3 5 7 2 is of the fom 2 i 5 j 7 k, fo integes 0 i 3, 0 j 1, and 0 k 2. That is, thee ae 4 2 3 = 24 diffeent divisos of 2 3 5 7 2. Since m+8 11, then the divisos of 2 3 5 7 2 that m+8 cannot equal ae: 1, 2, 4, 5, 7, 8, and. Since thee ae 7 divisos which cannot equal, then thee ae 7 divisos that 196 u cannot equal. (These divisos can be detemined by dividing 2 3 5 7 2 by each of the 7 divisos 1, 2, 4, 5, 7, 8, and. We also note that each of the emaining divisos that 196 u can equal, is less than 196, giving a positive intege value fo u.) So then thee ae 24 7 = 17 diffeent intege values of 196 u, and so thee ae exactly 17 diffeent intege values of u when p = 7. Fo all emaining pimes p 11, we get 40p 2 = 2 3 5 p 2, and so 2 3 5 p 2 = ()(4p 2 u). Since p 2 and p 5, each diviso of 2 3 5 p 2 is of the fom 2 i 5 j p k, fo integes 0 i 3, 0 j 1, and 0 k 2. That is, thee ae 4 2 3 = 24 diffeent divisos of 2 3 5 p 2. Since 11 and p 11, then the divisos of 2 3 5 p 2 that cannot equal ae: 1, 2, 4, 5, 8, and. Since thee ae 6 divisos which m+8 cannot equal, then thee ae 6 divisos which 4p 2 u cannot equal. (These divisos can be detemined by dividing 2 3 5 p 2 by each of the 6 divisos 1, 2, 4, 5, 8, and. We also note that each of the emaining divisos that 4p 2 u can equal, is less than 4p 2, giving a positive intege value fo u.) So then thee ae 24 6 = 18 diffeent intege values of 4p 2 u, and so thee ae exactly 18 diffeent intege values of u fo all pime numbes p 11. Theefoe, p = 7 is the only pime numbe fo which thee ae exactly 17 positive intege values of u fo Koelle-ectangles with n = and = u p 2.