2001 HG2, 2006 HI6, 2010 HI1

Size: px

Start display at page:

Download "2001 HG2, 2006 HI6, 2010 HI1"

Ruby Clarissa Bryant
5 years ago
Views:

1 - Individual ( 8 ) x, y 0 ( ) Group Individual Events I * see the remark Find the value of the unit digit of (Reference: 996 HG0) (mod 0) 5 (mod 0) ( ) + + ( ) (mod 0) (mod 0) 9 (mod 0) I Given that a, b and c are positive even integers which satisfy the equation a + b + c 0. I How many solutions does the equation have? Reference: 00 HG, 006 HI6, 00 HI Let a p, b q, c r, where p, q, r are positive integers. a + b + c 0 (p + q + r) 0 p + q + r 006 The question is equivalent to find the number of ways to put 006 identical balls into different boxes, and each box must contain at least one ball. Align the 006 balls in a row. There are 005 gaps between these balls. Put sticks into three of these gaps, so as to divide the balls into groups. The following diagrams show one possible division. The three boxes contain balls, 00 balls and ball. p, q 00, r. The number of ways is equivalent to the number of choosing gaps as sticks from 005 gaps The number of ways is In Figure, ABD is a square. The coordinates of B and D are (5, ) and (, ) respectively. If A(a, b) lies in the first quadrant, find the value of a + b. Mid-point of BD M(, ) MB MA (a ) + (b ) (5 ) + (+ ) 0 a + b a b 8 0 () b MA MB a 5 b a () Sub. () into (): a + (a ) a (a ) 8 0 5a 0a 5 0 a a 0 (a )(a + ) 0 a or (rejected) b () 5 a + b D O y A Figure 圖圖 B x Page

2 I4 ( ) Method mbd 5 BDA 45 tan 45 mad b a + m + tan 45 BD mbd tan 45 ; mad + ( ) b a + b 9 a + b a () Mid-point of BD M(, ) b a () (similar to method ) Solving () and () gives a, b 5 a + b 8 Method by Mr. Jimmy Pang from Lee Shing Pik ollege Mid-point of BD M(, ) Translate the coordinate system by x x, y y The new coordinate of M is M (, ) (0, 0) The new coordinate of B is B (5, ) (4, ) Rotate B about M in anticlockwise direction through 90 The new coordinate of A (, 4) Translate the coordinate system by x x +, y y + The old coordinate of A ( +, 4 + ) (, 5) (a, b) a + b 8 D(-, ) D' ' M(, )M'(0,0) O y y' A A' B(5, -) B' Find the number of places of the number 0 5. (Reference: 98 FG0., 99 HI7) The number of places I5 Given that log4 N , find the value of N. (Reference: 994 HI) 9 7 I6 I7 log4 N (sum to infinity of a geometric series, a, r.) N 4 8 Given that a and b are distinct prime numbers, a 9a + m 0 and b 9b + m 0. Find a b the value of +. (Reference: 996 HG8, 996FG7., 00 FG4.4, 005 FG.) b a a and b are prime distinct roots of x 9x + m 0 a + b sum of roots 9 (odd) a and b are prime number and all prime number except, are odd. a, b 7 (or a 7, b ) a b ( 8 ) b a Given that a, b and c are positive numbers, and a + b + c 9. Suppose the maximum value among a + b, a + c and b + c is P, find the minimum value of P. WLOG assume that a + b P, a + c P, c + a P. 8 (a + b + c) (a + b) + (b + c) + (c + a) P 6 P The minimum value of P is 6. x' x Page

3 I8 If the quadratic equation (k 4)x (4k + 4)x has two distinct positive integral roots, find the value(s) of k. learly k 4 0; otherwise, the equation cannot have two real roots. Let the roots be α, β. (4k + 4) 4(48) (k 4) [(7k + ) 48k + 9] (k + 8k + 96) [(k + 4)] 4k α k [ ( k + 4) ] ( 4) 7k + + k + 4 8k k 6, β. k 4 k 4 k k 4 k + For positive integral roots, k is a positive factor of 8 and k + is a positive factor of 6. k,, 4, 8 and k +,,, 6 k, 4, 6, 0 and k, 0,, 4 k 4 only Method provided by Mr. Jimmy Pang from Po Leung Kuk Lee Shing Pik ollege The quadratic equation can be factorised as: [(k )x 8][(k + )x 6] k and k x or k k + By similar argument as before, for positive integral root, k 4 only. I9 Given that x, y are positive integers and x > y, solve x 89 + y. x y (x y)(x + xy + y ) and both and 99 are primes. x + xy + y (x y) + xy x y (x y) + xy xy x y 89 + xy 79. (rejected) 99 + xy xy (rejected) xy (rejected) x, y Page

4 I0 In figure, AE 4, EB 7, A 9 and BD D 0. Find the value of BF. Reference: 005 HI5, 009 HG8 AB 4 + 7, B AB + B A AB 90 (converse, Pythagoras theorem) Let BF a, BF θ, ABF 90 θ Area of BEF + area of BF area of BE a sin θ + a 7cos θ A E 圖圖 Figure F B D 0a sin θ + 7a cos θ 40 () Area of BDF + area of ABF area of ABD 0 a cosθ + a 0sin θ a cos θ + 0a sin θ 0 () () (): 5 a cos θ 80 a cos θ 8 () () (): 50 a sin θ 0 a sin θ 5 (4) () + (4) : BF a ( ) Method AB 90 (similar to method ) A 0 Regard B as the origin, B as the x-axis, BA as the y-axis, then d 0i, c 0i, e 7j, a j Suppose F divides AD in the ratio p and p. 5 Also, F divides E in the ratio t : t. f p d + ( p) a 0pi + ( p)j () f t c + ( t) e 0ti + 7( t)j () ompare coefficients: 0p 0t and ( p) 7( t) p t () and ( p) t (4) () + (4): 5t + t 5 0 E 5 t p F 9 BF f 0( 5 )i + 7( 5 )j 8i + )j B - p - t 0D 0 Page 4

5 Method AB 90 (converse, Pythagoras theorem, similar to method ) Regard B as the origin, B as the x-axis, BA as the y-axis, then Equation of AD: x + y 0 (); equation of E: x + y 0 7 () 5 y x () (): y ; () (): 5 4 x 8 BF x + y ( ) Method 4 AB 90 (similar to method ) Regard B as the origin, B as the x-axis, BA as the y-axis, then A(0, ), E(0, 7), (0, 0), D(0, 0). Let AF : FD r : s Apply Menelaus theorem on ABD with EF. AE B DF EB D FA 4 0 s s 7 0 r r 4 By the point of division formula, F, 5 5 8, 5 BF A(0, ) r E(0,7) B(0,0) F s D(0,0) (0,0) Method 5 (Provided by Mr. Lee hun Yu, James from St. Paul s o-educational ollege) Area of AB 0 0 Area of ABD : area of AD : Area of BDF : area of DF Area of ABF : area of AF : Area of EB : area of EA : Area of FEB : area of FEA Area of FB : area of FA : Area of ABF : area of AF : area of BF : : Area of BF Distance from F to B 4. 0 Area of ABF Distance from F to AB BF (Pythagoras theorem) Page 5

6 Group Events G Given that x, y and z are three consecutive positive integers, and integer. Find the value of x + y + z. x y, z y + y z x z x y y + y + y + y + y + y x x y y z z y y y + ( ) ( ) ( y + y y + + y y ) + ( y ) y ( y ) ( y ) y( y + ) y + y + y + y y + y y + y 6 y 6y ( y ) y( y + ) ( y ) y ( y +) y learly y does not divide y, so y + and y are factors of 6. y y, y + x + y + z 6 y y, but y + 4 which is not a factor of 6, rejected. y or 6 are similarly rejected. Method x y, z y + y z x z x y y + y + y + y + y + y x x y y z z y y y + y y + is an integer y > (); is an integer y () y y + Solving () and () gives y, x, z x + y + z 6 G Given that x is a real number and x + ( 5 x) x G G4 If 5 x, the equation is equivalent to y z x z x y is an x x y y z z 0. Find the value of x. x x x x 0 x 5 x 0 4x 0x + 5 4x x (4)(07) < 0 no real solution, rejected If 5 < x, then the equation becomes x 0 + x 5 x x 0 5 x 07 Evaluate (Answer can be expressed in index form.) ( 005 ) 005 ( + ) Evaluate (Answer can be expressed in surd form.) Page 6

7 G5 Find the minimum value of x + y 0x 6y Reference: 999 HG7, 00 HI, 08 HI x + y 0x 6y (x 5) + (y ) G6 In Figure, AB is an isosceles triangle. Suppose AB A. If D is a point on B produced such that DAB 90 and D, find the length of B. Let AB θ AB (base, isos. ) AD 80 θ (adj. s on st. line) BD sec θ B cos θ BD sec θ cos θ + cos θ 6 0 ( cos θ )(4 cos θ + ) 0 cos θ or (rejected) 4 B cos θ 6 Method Draw AE BD. A ABE AE (R.H.S.) Let BE x E. x x 6 cos B x + x + x + x 7 0 (x 8)(x + 9) 0 B x E x D x 8 B x 6 G7 Given that a x b y c z 0 w and + +, where a, b, c are positive integers (a b c) x y z w and x, y, z, w are real numbers, find the value of a + b + c. log a x log b y log c z log 0 w x log a y log b z log c w log 0 log a logb logc log abc w x y z wlog0 wlog0 wlog0 wlog0 abc 0 a and b (otherwise x log a y log b z log c w log 0 0 w log 0 w 0) a, b, c 5 a + b + c 0 G8 Given that the roots of the equation x + px + q 0 are integers and q > 0. If p + q 60, find the value of q. Let the roots be α and β. α + β p () αβ q > 0 () p + q 60 (α + β) + αβ 60 α β( α) 6 (α )(β ) 6, which is a prime α, β 6 or α, β 6 When α, β 6 α 0, β 60 αβ q 0 (contradicting to the given condition q > 0, rejected) α, β 6 Remark the original question is: Given that the roots of the equation x + px + q 0 are integers and p, q > 0. If p + q 60, find the value of q. α + β p < 0 (), αβ q > 0 () α < 0 and β < 0 and (α )(β ) 6 α, β 6 α 0, β 60 αβ q 0 (rejected), no solution Page 7

8 α, β 6, q αβ 4 G9 Evaluate sin + sin + sin + + sin 59 + sin 60. (Reference 00 FG.) sin + sin 89 sin + cos sin + sin 88 sin + cos... sin 44 + sin 46 sin 44 + cos 44 sin + sin + + sin 89 (sin + sin 89 ) + + (sin 44 + sin 46 ) + sin sin + sin + sin + + sin 59 + sin sin sin G0 In a gathering, originally each guest will shake hands with every other guest, but Steven only shakes hands with people whom he knows. If the total number of handshakes in the gathering is 60, how many people in the gathering does Steven know? (Note: when two persons shake hands with each other, the total number of handshakes will be one (not two).) Suppose there are n persons and Steven knows m persons (where n > m). If everyone shakes hands with each other, then the total number of hand-shaking n In this case, Steven shakes hands with n persons. However, he had made only m hand-shaking. Method n < 60 n By trial and error, n m Method n + n n m 59 + n n ( ) m 60 ( ) 0 55 < m < n 0 ( 8 + n n ) < n n n 8 0 and n n 8 > 0 (n.5) and (n 0.5) 8.5 > n n n > ( n )( ) 0 and ( )( ) 0 ( n ) and ( n < or n > ) < n < 8. 5, 0.5 < < n.5 For integral value, n ( ) + m 60 m 5 Page 8

9 Geometrical onstruction. In the space provided, construct an equilateral triangle AB with sides equal to the length of MN below. () 作綫段 AB, 使得 AB MN () 以 A 為圓心,AB 為半徑作一弧 ; 以 B 為圓心,BA 為半徑作一弧 ; 兩弧相交於 () 連接 A 及 B AB 為等邊三角形 A B. As shown in Figure, construct a circle inside the triangle AB, so that AB, B and A are tangents to the circle. Reference: 009 HS, 04 H, 09 H () 作 AB 的角平分綫 A () 作 AB 的角平分綫 P 為兩條角平分綫的交點 T () 連接 AP 4 4 S (4) 分別過 P 作垂直綫至 B A 及 AB,R S T 為 5 對應的垂足 P BPR BPT (A.A.S.) PR PS (A.A.S.) B R PT PR PS ( 全等三角形的對應邊 ) (5) 若 P 至 B 的垂足為 R, 以 P 為圓心,PR 為半徑作一圓, 此圓內切於三角形的三邊, 稱為內切圓 (inscribed circle) ( 切綫半徑的逆定理 ). Figure shows a triangle PQR. onstruct a line MN parallel to QR so that (i) M and N lie on PQ and PR respectively; and (ii) the area of PMN the area of PQR. 首先, 我們計算 MN 和 QR 的關係 : QPR MPN ( 公共角 ) PQR PMN (QR // MN, 對應角 ) PRQ PNM (QR // MN, 對應角 ) PQR ~ PMN ( 等角 ) PMN 的的的 PQR 的的的 PM PQ PM PQ PM PQ 作圖步驟 : () 利用垂直平分綫, 找 PQ 之中點 O () 以 O 為圓心 OP OQ 為半徑, 向外作一半圓, 與剛才的垂直平分綫相交於 F PFQ 90 ( 半圓上的圓周角 ) PFQ 為一個直角等腰三角形 QPF 45 PQ PF PQ sin 45 () PQ 以 P 為圓心,PF 為半徑, 作一圓弧, 交 PQ 於 M PM (4) 自 M 作一綫段平行於 QR, 交 PR 於 N, 則 PMN 平分 PQR 的面積 F M Q P M Q O P N N R R Page 9

10 Percentage of correct questions 4.48%.58%.66% % 5 0.7% 6 0.8% 7 4.9% 8.08% % 0 4.6% 57.49% 5.% 8.06% % % 6 8.6% % % % 0 4.5% Page 0

Answers: ( HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 23 November see the remark

Answers: ( HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 23 November see the remark 9 50450 8 4 5 8-9 04 6 Individual 6 (= 8 ) 7 6 8 4 9 x =, y = 0 (= 8 = 8.64) 4 4 5 5-6 07 + 006 4 50 5 0 Group 6 6 7 0 8 *4 9 80 0 5 see the remark Individual Events I Find the value of the unit digit