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1 nswers: ( HKO Heat vents) reated by: r. Francis Hung Last updated: June Individual Group Individual vents I Given that is divisible by ( a) and ( a), where a is an integer, find the value of a. Let f () = f ( ) = + 8 = 0 + is a factor f () = = 0 is a factor f () = ( + )( )( ) a = I Given that 8, a, b form an.p. and a, b, 6 form a G.P. If a and b are both positive numbers, find the sum of a and b. 8 b a (); b = 6a () Sub. () into (): b = 8(8 + b) b 8b = 0 (b + 6)(b ) = 0 b = 6 (rejected) or b = 8 a = = 6 a + b = 0 I Find the smallest real root of the following equation: 6 Reference: 99 FI. ( + )( 6) = ( )( + ) ( ) = ( ) 0 = ( + ) = 0 or The smallest root =. I In figure, is a square. is a point on such that and. Find the area of the square. = (Pythagoras theorem on ) rea of the square = = Page
2 nswers: ( HKO Heat vents) reated by: r. Francis Hung Last updated: June 08 I If + =, find the value of. I6 I = = = 0 ( + )( + ) = 0 = or + = > 0, = only Given that n is a positive integer which is less than 000. If n is divisible by or, find the number of possible values of n. (Reference: 99 FG8.-, 99 FG8.-, 0 FI.) Number of multiples of = Number of multiples of = 99 Number of multiples of = 66 Number of possible n = = 66 In figure, is a rectangle with. is a point on such that. is the mid-point of and P, are points on and respectively such that P is a straight line. If P : P : k, find the value of k. raw a straight line HG // (H lies on, G lies on ) H = H (Intercept theorem) PH ~ G (equiangular) H ~ (equiangular) P : = H : G (ratio of sides, ~'s) P = : (HG H) (ratio of sides, ~'s) H =. : (.) = : 9 (opp. sides, rectangle) k = 9 I8 Find the last digit of the value of = 6, 6 = 6,, the last digit of 6 0 is 6, the last digit of is. The last digit of the number is 6 8 (mod 0) = 7 = (mod 0) I9 Let a be the positive root of the equation ross multiplying: ( + + ) = ( + + ) = ( + ) = 0 9( )( + ) = 0 a = = 7, find the value of a. I0 Find the sum of all positive factors of 0. Reference 99 HI8, 99 FI., 997 HI, 998 FI., 00 FG., 00 FI. 0 = Positive factors are in the form a b c, 0 a, 0 b, c, a, b, c are integers. Sum of positive factors = ( )( + )( + ) = 6 = 7 G Page
3 nswers: ( HKO Heat vents) reated by: r. Francis Hung Last updated: June 08 Group vents G If, find the value of. Reference: 98 FG0. = G = = ( ) = In Figure, is a triangle. and are the bisectors of the eterior angles and respectively meeting and produced at and. Let and a. Find the value of a. Reference: 986 上海市初中數學競賽 80 a a = = 90 (adj. s on st. line, bisector) 80 a = = (s sum of ) a = =. (vert. opp. s, bisector) 8 = + = 80 + a a a =. = = a (base s isosceles ) a a + a + 7. = 80 (s sum of ) 8 a = G If 6 a and b 6, find the greatest value of a b. 0 a 6 and 9 b 6 6 a b 7 The greatest value = 7. G Let a, b, c be integers such that a = b = c. If c >, find the smallest value of c. Reference: 999 FG. Let a = k, b = k, c = k 6 c > k > The smallest k = The smallest c = 6 = 6 Page
4 nswers: ( HKO Heat vents) reated by: r. Francis Hung Last updated: June 08 G In figure, the area of the parallelogram is 0. and N are the mid-points of and respectively. N intersects and at points P and respectively. Find the area of P. (Reference: 06HI, 08 FG.) Produce and to meet at R. Let = a. Then N = N = a (mid-point) ~ N (equiangular) N (ratio of sides, ~ s) (N = mid-point, opp. sides of //-gram) R P N P t a a N a t rea of = 0 = 60 rea of = = 0 rea of N N rea of rea of N = 0 = 0..() s is the mid-point, R (S) R = (corr. sides s) () lso P ~ NPR (equiangular) P (ratio of sides, ~ s) PR NR a (opp. sides of //-gram, corr. sides s)..() a ombine () and () P R ; R P P P P R 0 R R R 0..() rea of = 0 0 y (): rea of P = rea of 0 6 () rea of N = 0 0 rea of P = rea of N rea of P rea of N = = (by () and ()) R Page
5 nswers: ( HKO Heat vents) reated by: r. Francis Hung Last updated: June 08 G6 In figure, find the number of possible paths from point to point following the direction of arrow heads. Reference 98 FI., 000 HI, 007 HG The numbers at each of the vertices of in the following figure show the number of possible ways. 6 0 So the total number of ways = Figure G7 Find the smallest real root of the equation ( )( ) =. = 0 ( + )( ) = 0 = or G8 The smallest real root is. In figure, four circles with radius touch each other inside a square. Find the shaded area. (orrect your answer to the nearest integer.) The line segments joining the four centres form a square of sides = Shaded area = G9 In figure, is a square and points, F, G, H are the mid-points of sides,,, respectively, find the number of right-angled triangles in the figure. (Reference: 99 HG8) Let the shortest side of the smallest right-angled triangle be. Then =, H =, =, = We count the number of right-angled triangles with different hypotenuses. Hypotenuse Number of triangles 8 Total number of triangles = 0 H F G G0 test is composed of multiple-choice questions. marks will be awarded for each correct answer and mark will be deducted for each incorrect answer. pupil answered all questions and got 70 marks. How many questions did the pupil answer correctly? Reference: 99 FI. Suppose he answer questions correctly and question wrongly. ( ) = 70 = 9 Page
1983 FG8.1, 1991 HG9, 1996 HG9
nswers: (1- HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 6 February 017 - Individual 1 11 70 6 1160 7 11 8 80 1 10 1 km 6 11-1 Group 6 7 7 6 8 70 10 Individual Events I1 X is a point on
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nswers: (997-9 HKMO Het vents) reted by: Mr. Frncis Hung Lst updted: 0 ecember 0 97-9 Individul 0 6 66 7 9 9 0 7 7 6 97-9 Grup 6 7 9 0 0 9 Individul vents I Given tht + + is divisible by ( ) nd ( ), where
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Answers: (99-9 HKMO Final Events) reated by: Mr. Francis Hung Last updated: 7 December 05 Individual Events I a I a 6 I a 4 I4 a 8 I5 a 0 b b 60 b 4 b 9 b c c 00 c 50 c 4 c 57 d d 50 d 500 d 54 d 7 Group
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