Taiwan International Mathematics Competition 2012 (TAIMC 2012)
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1 Section. (5 points each) orrect nswers: Taiwan International Mathematics ompetition 01 (TIM 01) World onference on the Mathematically Gifted Students ---- the Role of Educators and Parents Taipei, Taiwan, 3rd~8th July 01 Invitational World Youth Mathematics Intercity ompetition Individual ontest , 9361, cm cm Determine the maximum value of the difference of two positive integers whose sum is 034 and whose product is a multiple of 034. Let the two numbers be x and y. From y = 034 x, we have xy = 034x x. If this is divisible by 034, then x is divisible by 034. Now 034 = Hence = 678 must divide x, so that 678 x < 034. It follows that the only possible values for x are 678 and 678 = The corresponding values for y are 1356 and 678 respectively. Hence x y = ±678 and its maximum value is 678. NS:678. The diagram below shows a semicircle sitting on top of a square and tangent to two sides of an equilateral triangle whose base coincides with that of the square. If the length of each side of the equilateral triangle is 1 cm, what is the radius of the semicircle, in cm? Let the triangle be and let O be the centre of the semicircle. Let r be the radius of the semicircle. With the sides of the triangle as bases, the heights of triangles O, O and O are r, r and r respectively. Their total area is equal to the area of triangle. Since the height of triangle r O r is, we have and. NS:
2 3. four-digit number abcd is a multiple of 11, with b + c = a and the two-digit number bc a square number. Find the number abcd. Since the two-digit number bc is a square number, and, we have. Since abcd is a multiple of 11, by trying possible digit d, we have NS:7161, 9361, The area of the equilateral triangle is cm. M is the midpoint of. The bisector of M intersects M at a point N. What is the area of triangle N, in cm? We have N NM M 3 = =. Hence N M N + 3 = and = + 3. Denote the area of the triangle T by [T]. Then [ ] [ ] = = + 3. It follows that [ N] = = = 4 cm. [ N] N NS:4 cm 5. There is a 6 hole on a wall. It is to be filled in using 1 1 tiles which may be red, white or blue. No two tiles of the same colour may share a common side. Determine the number of all possible ways of filling the hole. The top left space can be filled in 3 ways and the bottom left space can be filled in ways, so that the first column from the left can be filled in 3 =6 ways. In moving from column to column, we must retain at least one colour used in the preceding column. If we retain both colours, the only ways is to reverse the positions of the two tiles. If we retain just one colour, the tile with the repeated colour must be placed in a non-adjacent position, and the remaining space is filled with a tile of the third colour. Hence there are 3 ways to fill each subsequent column. It follows that the total number of ways is =1458. NS: Let N = How many perfect squares divide N? The prime factorization of N is Its largest square factor is Its square factors are the squares of the factors of Their number is (15+1)(6+1)(+1)(1+1)=67. NS:67 7. How many positive integers not greater than use only the digits 0, 1 or? N M
3 The first few numbers are 1,, 10, 11, 1, 0, 1,, 100, 101 and so on. These are just numbers in base 3. The base 3 number can be coverted to base 10 as follows Including the number itself, there are 4757 positive integers which use only the digits 0, 1 and. NS: The diagram below shows four points,, and D on a circle. E is a point on the extension of and D is the bisector of E. F is the point on such that DF is perpendicular to. If = F = cm, determine the length of, in cm. Let G be the point on such that FG = F = cm. Then GD = D and DG = DG. Sine D is a cyclic quadrilateral, DG = D. Moreover, D E DG =180 DG =180 DG F G =180 DE = D. It follows that triangles DG and D are congruent, so that G = = cm. Hence = D + DG + G + ++ = 6 cm. NS:6 cm 9. There are 56 different four-digit numbers abcd where each of a, b, c and d is 1,, 3 or 4. For how many of these numbers will ad bc be even? Note that ad bc is even if ad and bc are either both odd or both even. The former occurs when all four numbers are odd. The number of this case is. The latter occurs when a and d are not both odd, and b and c are not both odd. The number of this case is. Hence there are possible numbers. NS: In a plane, given 4 evenly spaced points on a circle, how many equilateral triangles have at least two vertices among the given points? There are 4 3 = 76 pairs of given points. For each pair, we can have an equilateral triangle on each side of the line joining them. However, some of these 76 = 55 triangles have been counted 3 times, because all three vertices are
4 among the given points. There are 4 3 = 8 such triangles. Hence the final count is 55 8 = 536. NS: The diagram below shows a circular sector O which is one-sixth of a circle, and a circle which is tangent to O, O and the arc. What fraction of the area of the circular sector O is the area of this circle? Let be the centre of the circle and let the extension of O cut arc at a point P. Let D be the point on O such that D is perpendicular to O. Let D = r. Then O = r and P = r, so that OP = 3r. Hence the area of the sector is P 1 3 π (3 r) = πr while the area of the circle is 6 desired fraction is 3. π r. The NS: 1. n 8 8 chessboard is hung up on the wall as a target, and three identical darts are thrown in its direction. In how many different ways can each dart hit a different square such that any two of these three squares share at least one common vertex? There are 7 pairs of adjacent rows and 7 pairs of adjacent columns, so that the number of subboards is 7 7 = 49. The three darts must all hit a different square of some subboard, and the square they miss can be any of the 4 squares in the subboard. Hence the total number of ways is 4 49 = 196. NS:196 Section. (0 points each) 1. What is the integral part of M, if O D M = (01 3) (01 ) (01 1) 01 The required answer is 01. y using the inequality ( N 1)( N + 1) < N, we arrived that It follows that (01 1) 01 < (01 1)(01 + 1) < 01, (01 ) (01 1) 01 < (01 ) (01 ) < 01 1,
5 (01 3) (01 ) (01 1) 01 < (01 3) (01 1) Repeating the same process, we conclude < 01 M = (01 ) (01 1) 01 < < 013 This implies the integral part of M is less than 013. nd onversely, 01 01, (01 1) 01 > = 01, (01 ) (01 1) 01 > = 01, (01 3) (01 ) (01 1) 01 > = 01, ontinuing the same process, we have M = (01 ) (01 1) 01 > = 01 In summary, the integral part of M is 01. NS:01 Marking Scheme... Showing M is at least points... Showing M is less than points... orrect answer... points. Let m and n be positive integers such that n < 8m < n + 60( n + 1 n) Determine the maximum possible value of n. When divided by 8, n leaves a remainder no greater than 4. Hence if 60( n + 1 n) < 4, then there will not be a multiple of 8 between n and n n n + 60( + 1 ). It follows that we must have 60( n + 1 n) 4.
6 1 Hence 15 = n n > n, so that n 56. n + 1 n When n = 55 or 56, 60( n + 1 n) < 5, and the remainder when 55 or 56 is divided by 8 is no greater than 1. Hence the desired multiple of 8 cannot exist either For n = 54, 60( 55 54) = > 4 since = 900 > 880 = Now 54 = 916 so that ( 55 54) > 90. Since 90 8 = 365, we can take m = 365. It follows that the maximum value of n we seek is 54. NS:54 Marking Scheme... Show that 60( n + 1 n) points... Solve the inequality...7 points... orrect answer... points 3. Let be a triangle with = 90 and = 0. Let E and F be points on and respectively such that E = 10 and F = 30. Determine FE. Note that F = F. Let D be the midpoint of and let G be the point on such F that GD is perpendicular to. Then G E triangles and DG are similar, so that D =. y symmetry, G D G= G= 0, so that GF=0 also. Hence G bisects F so that F E =. Since E bisects, =. Now FG G E 1 1 F F D E = = = = =. FG FG G G E It follows that G is parallel to EF, so that FE = GF =0. NS: 0 Marking Scheme... Draw the correct auxilary line G...3 points... List equations of ratios of lengths...6 points... Prove that G is parallel to EF...9 points
7 ... orrect answer... points
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