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nswers: (008-09 HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 8 ugust 08 08-09 Individual 00 980 007 008 0 7 9 8 7 9 0 00 (= 8.) Spare 0 9 7 Spare 08-09 Group 8 7 8 9 0 0 (=.) Individual Events I Let x = 0.+ 0.00 + 0.0000 + 0.000000 + L, find the value of x. I eference: 000 HI Let x = 0.7+ 0.07+ 0.007+... 0. = ; 0.00 = ; 0.0000 = ; 99 9900 990000 00 00 x = + + + L= + + + L = = = 99 9900 990000 99 00 0000 99 99 99 00 980 In Figure, a regular hexagon and a rectangle are given. The vertices of the rectangle are the midpoints of four sides of the hexagon. If the ratio of the area of the rectangle to the area of the hexagon is : q, find the value of q. Let one side of the hexagon be a, the height of the rectangle be x, the length be y. x = a cos 0 = a, y = a cos 0 + a = a o atio of area = a a : ( a) sin 0 = : = : q = The following method is provided by Mr. Jimmy ang from Sai Kung Sung Tsun atholic School (Secondary). Let the hexagon be EF, the rectangle be S as shown. Fold F along to F'. Fold S along S to 'S. Fold ES along S to F''S. Fold along to F''. Then the folded figure covers the rectangle completely. atio of area = : ; q = I Let sin θ = + cos θ and 0 θ 90, find the value of θ. sin θ = + ( sin θ ) sin θ + sin θ = 0 ( sin θ )( sin θ + 7) = 0 sin 7 θ = or (rejected) sin θ = or (rejected) F E F' ' S I θ = 0 Let m be the number of positive factors of gcd(008, 8), where gcd(008, 8) is the greatest common divisor of 008 and 8. Find the value of m. 008 = 8 ; 8 = 9 gcd = = 0 The positive factors are,,, 0; m = http://www.hkedcity.net/ihouse/fh7878/ age

nswers: (008-09 HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 8 ugust 08 I Given that x + (y ) = 7, where x and y are real numbers. If the maximum value of y + x is k, find the value of k. (eference: 00 FG., 0 HI) x = 7 (y ) ; sub. into y + x = y + 7 (y ) = y + 7 y + y 9 = y + y = (y y +.. ) = (y.) + 0. = = (y.) + 8. Maximum value = k = 8. Method z = y + x = y + y y y + (z + ) = 0 iscriminant 0 for all real value of y. ( ) ()(z + ) 0 z 8 0 z ; k = I Let f (x) = x eference: 999 FI. f (x) = f (f (x)) = f x = I7 and f n (x) = f (f n (x)), where n =,,,. Find the value of f 009 (008). x = x x = x x x f (x) = f (f (x)) = f x = = x, f n (x) = x for all positive integer n. ( x ) 008 007 f 009 (008) = f (f 007 (008)) = f (f (9) (008)) = f (008) = = 008 008 In Figure, EF is a regular hexagon centred at the point. ST is an equilateral triangle. It is given that = cm, = cm and T = cm. If the area of the common part of the hexagon and triangle is c cm, find the value of c. eference: 0 FI. Join and. is an equilateral triangle, side = cm + = 0 + 0 = 80 is a cyclic quadrilateral (opp. sides supp.) = (ext., cyclic quad.) = 0 = = sides of an equilateral triangle (S) Shaded area = area of + area of = area of + area of = area of = sin 0 = 9 I8 Find the unit digit of 7 009. I9 o cm ; c = 9 eference: 00 HI9: Given that the units digit of 7 = 7, 7 9 (mod 0), 7 (mod 0), 7 (mode 0) 7 009 = (7 ) 0 7 7 mod 0 00 7 is,... Given that a and b are integers. Let a 7b = and log b a =, find the value of a b. (b) = a b = 7b + b 7b = 0 (b )(b + ) = 0 b = or (rejected) a = 7b + = ; a b = http://www.hkedcity.net/ihouse/fh7878/ age

nswers: (008-09 HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 8 ugust 08 I0 In Figure, is a rectangle. oints E and F lie on and respectively, such that F = 8 cm and E = cm. Given that the area of the shaded region is 80 cm. Let the area of the rectangle be g cm, find the value of g. Let F = x cm, E = y cm g = xy + ( 8 + x) + ( y + ) 8 + 80 = ( xy + x + 8y + 0 + 00) = [( x + 8)( y + ) + 00] = g + 00 g = 00 IS Given that a is a negative real number. If =, find the value of a. a = a + a = a + a + Group Events G + a+ a a = a = 008 If a is a positive integer and + + L+ =, find the value of a. a( a + ) ( a + )( a + ) 009 0 7 + + L + = a a + a + a + 008 009 0 7 7 7 + 87 = = + = = = ; a = 0 a 009 0 a 009 009 009 7 7 G Let x = +, find the value of x x + x x 0x. eference: 99 HI9, 000 HG, 00 FG., 007 HG Method x =, (x ) = x x = 0 y division, x x + x x 0x = (x x )(x + x + ) + x = x = ( + ) = Method ivide x x + x x 0x by (x ) successively: 0 Let y = x, the expression in terms of y becomes 8 y + y + y + y y 8 0 0 = + ( ) + ( ) + ( ) 0 0 = + + 0 + 8 8 = 7 http://www.hkedcity.net/ihouse/fh7878/ age

nswers: (008-09 HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 8 ugust 08 G Given that p and q are integers. If + =, find the maximum value of p q. p q eference 008 HI + = x y. If < y 0 < 0 and x 0 + y 0 = z 0 q + p = pq pq p (q ) = (p )(q ) = (p, q ) = (, ), (, ), (, ), (, ) (p, q) = (, ), (, ), (, ), (0, 0) maximum p q = = 9 G Given that 0 x 80. If the equation cos 7x = cos x has r distinct roots, find the value of r. 7x = x + 0n or 7x = 0 x + 0n x = 80n or x = 0(n + ) n = 0, x = 0 or 0 n =, x = 80 or 0 n = x = 90 n = x = 0 n = x = 0 r = 7 G Let x, y and z be positive integers and satisfy z 8 = x y. Find the value of x + y + z. G ( x ) z 8 = y z 7 = x + y xy x + y = z; xy = 7 x = 7, y = ; z = + 7 = 8; x + y + z = In Figure, is a square and M = N = E = F = cm and MN = cm. Let the area of quadrilateral S be c cm, find the value of c. (eference: 07 HG) F = cm, = = cm, F = + cm = cm = N F NF (SS) F = NF (corr. s 's) S = FS (sides opp. eq. s) ut SF NS (S) S = SF = F = cm Let H be the mid point of EF. EH = HF = cm Suppose S intersects at G. It is easy to show that F ~ HF ~ GS F FH F = = F cm S = SF F = cm cm = cm rea GS = rea F S F F = cm (ratio of sides, ~ 's) rea GS = cm rea of S = area of SG = cm, c = rea SG = cm Method Let FH = θ = H = M (corr. s and alt. s, //-lines) tan θ = H = H = cm cm 8 = H H = H = cm; H = 8 cm cm = cm Further, FS is a //-gram. S = F = cm (opp. sides, //-gram) rea of S = S = cm ; c = cm cmh cm cm cm E S M G F N http://www.hkedcity.net/ihouse/fh7878/ age

nswers: (008-09 HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 8 ugust 08 G7 Given that x is a real number and satisfies x+8 + = x. Find the value of x. Let y = x, y = x, the equation becomes: 8 y + = y y y + = 0 (y ) = 0 y = x = ; x = G8 In Figure, is a right angle, = = cm and E = F = cm. If = d cm, find the value of d. eference: 00 HI, 0 HI0 = = We find the area of in two different ways. S = S + S = S F + S E S E S F o o d sin = d sin G9 0 d = 8 d = Method Set up a coordinate system with as x-axis, as y-axis, and as the origin. Equation of F in intercept form: x + y = () Equation of is y = x () Sub. () into (): x + y = x = d = sin = If there are different values of real number x that satisfies x x 0 = f, find the value of f. (eference: 00 FG., 00 FG., 0 FG.) emark: The original solution is wrong. Thanks for Mr. Ng Ka Lok s (from E) comment. bsolute values must be non-negative, f 0 (*) x x 0 = ±f x x = 0 + f or x x = 0 f (**) x x = 0 + f, x x = 0 f, x x = 0 + f, x x = 0 f x x f = 0 (), x x +f = 0 (), x x f = 0 (), x x +f = 0 () The respective discriminants are: = ( + f), = ( f), = ( + f), = ( f) onsider the different values of f ( 0) by table: f = 0 0 < f < f = < f < f = < f + + + + + + + + + + 0 + + + + + + + + 0 onsider the following cases: (I) the equation has real roots and complex roots From the table, the range of possible values of f is < f <. Sub. < f < into (**): x x = 0 + f or x x = 0 f (no solution, HS<0) x x = 0 + f or x x = 0 f x x f = 0 () or x x + f = 0 () For < f <, > 0, < 0 The equation has only real roots,contradicting to our assumption that there are real roots (II) The equation has 8 real roots. of these roots are distinct and other roots are the same as the first roots. If () (), then f = + f f = 0, if () (), then f = + f f = 0 If () (), then f = f no solution, if () (), then +f = + f no solution If () (), then f = + f f = 0 (contradict with (*), rejected) If () (), then +f = f f = 0 If f = 0 the equations are x x = 0 and x x = 0, which have only roots,rejected If f = 0, the equations are x x = 0, x x = 0 and x x = 0, which have distinct real roots, accepted. http://www.hkedcity.net/ihouse/fh7878/ age

nswers: (008-09 HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 8 ugust 08 Method The following method is provided by Mr. Jimmy ang from Sai Kung Sung Tsun atholic School (Secondary). First sketch the graph y = x x 0. From the graph, draw different horizontal lines. The line y = 0 cuts y = x x 0 at different points. f = 0 G0 In Figure, is a triangle, E is the midpoint of, F is a point on E where E = F. The extension segment of F meets at. Given that the area of is 8 cm. Let the area of F be g cm, find the value of g. From E, draw a line EG // which cuts at G. E = F F : FE = : ; let E = k, FE = k E is the midpoint of E = E = t S E = S E = 8 cm = cm (same base, same height) : G = F : FE = : (theorem of equal ratio) G : G = E : E = : (theorem of equal ratio) : G : G = : : S EG : S EG = : (same height, ratio of base = : ) S EG = cm 7 = cm + 7 F ~ GE S F = SEG = cm 8 = cm 8 ; g = 9 9 t k k F E t G http://www.hkedcity.net/ihouse/fh7878/ age

nswers: (008-09 HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 8 ugust 08 GS Let be the remainder of 88 009 divided by 97. Find the value of. eference: 008 FIS. the remainder of 88 008 divided by 97 88 009 = (97 +) 009 = (97 ) 009 + 009 (97 ) 009 + + 009 = 97m+ 009, where m is an integer. Note that = 9 = 97 (mod 97); = 8 = 97 + (mod 97); = ( ) ( ) (mod 97) 009 = ( ) ( ) ( ) (mod 97); d = http://www.hkedcity.net/ihouse/fh7878/ age 7

nswers: (008-09 HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 8 ugust 08 HKMO 009 Geometrical onstruction Sample aper solution. 在下列三角形中, 試作出一點使它與該三角形各邊的距離相等 First, we find the locus of a point which is equidistance to a given angle. (Figure ) E F Figure Figure () From, let E and F be the feet of perpendiculars onto and respectively. (Fig. ) () Join. = (common side) E = F = 90 (y construction) E = F (given that is equidistance to and ) E F (HS) E = F (corr. s 's) lies on the angle bisector of. In, if is equidistance to, then must lie on the intersection of the three angle bisectors. (i.e. the incentre of.) The three angle bisectors must concurrent at one point We need to find the intersection of any two angle bisectors. The construction is as follows: () raw the bisector of. () raw the bisector of. is the intersection of the two angle bisectors. () Join. () Let, S, T be the feet of perpendiculars from onto, and respectively. T S T = = S T S T = S (..S.) (..S.) (orr. sides, 's) (.H.S.) (orr. s, 's) is the angle bisector of. The three angle bisectors are concurrent at one point. () Using as centre, as radius to draw a circle. This circle touches internally at, S, and T. It is called the inscribed circle. T S http://www.hkedcity.net/ihouse/fh7878/ age 8

nswers: (008-09 HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 8 ugust 08 Note that the ex-centres are equidistance from,,. I shall draw one ex-centre as demonstration. () Extend and to S and T respectively. () raw the exterior angle bisectors of and respectively. The two exterior angle bisectors intersect at I. () Let,, be the feet of perpendiculars drawn from I onto to, and respectively. () Join I. () I I (S); I I (S) I = I = I (corr. sides, 's) I is equidistance from, and. I I I I = I (.H.S.) (corr. sides, 's) S T I is the interior angle bisector of Two exterior angle bisectors and one interior angle bisector of a triangle are concurrent. Use I as centre, I as radius to draw a circle. This circle touches triangle externally. It is called the escribed circle or ex-circle. I http://www.hkedcity.net/ihouse/fh7878/ age 9

nswers: (008-09 HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 8 ugust 08. 已知一直綫 L, 及兩點 位於 L 的同一方 試在 L 上作一點 T 使得 T 及 T 的長 度之和最小 (Figure ) L Figure L S T Figure Let be one of the end point of L nearer to. () Use as centre, as radius to draw an arc, which intersects L at and. () Use as centre, as radius to draw an arc. Use as centre, as radius to draw an arc. The two arcs intersect at. () Join which intersects L at S. () Join which intersects L at T = (same radii) = (common side) = (same radii) (S.S.S.) = (corr. sides s) S = S (common side) S S (S..S.) S = S (corr. sides s) = 90 (adj. s on st. line) S = S (corr. sides s) ST = ST (common side) ST ST (S..S.) T = T (corr. sides 's) T + T = T + T It is known that T + T is a minimum when, T, are collinear. T is the required point. http://www.hkedcity.net/ihouse/fh7878/ age 0

nswers: (008-09 HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 8 ugust 08. 試繪畫一固定圓的切綫, 且該切綫通過一固定點 ( 註 : 在該圓形外 ) O Figure First, we locate the centre of the circle. (Figure ) () Let and be two non-parallel chords. () raw the perpendicular bisectors of and respectively which intersect at O. () Join O. () raw the perpendicular bisector of O, N K is the mid-point of O. () Using K as centre, K as radius to draw a circle, which intersects the given circle at M and N. O K () Join OM, ON, M, N. ON = 90 = OM ( in semi-circle) M, N are the required tangents. (converse, tangent radius) M Figure Method () From, draw a line segment cutting the T circle at and. ( lies between and.) () raw the perpendicular bisector of, M which intersects at O. 7 () Use O as centre, O = O as radius to 8 draw a semi-circle T. O () From (the point between ), draw a line T perpendicular to, cutting the 8 semi-circle at T. () Join T. N () Join T. Figure (7) Use as centre, T as radius to draw an arc, cutting the given circle at M and N. (8) Join M, N. T = 90 ( in semi-circle) T ~ T (equiangular) T = (ratio of sides, ~ s) T = T y (7), = T = M = N M and N are the tangents from external point. (converse, intersecting chords theorem) Figure http://www.hkedcity.net/ihouse/fh7878/ age

nswers: (008-09 HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 8 ugust 08 Geometrical construction. In Figure,, and are three fixed non-collinear points. onstruct a circle passing through the given three points. The perpendicular bisectors of, and are concurrent at the circumcentre O. It is sufficient to locate the circumcentre by any two perpendicular bisectors. () Join, and. () raw the perpendicular bisector of. M is the mid-point of. () raw the perpendicular bisector of. N is the mid-point of. The two perpendicular bisectors intersect at O. () Use O as centre, O as radius, draw a circle. OM OM (S..S.) ON ON (S..S.) O = O = O (corr. sides, 's) The circle pass through, and. Let L be the mid-point of. Join OL. OL OL (S.S.S.) LO = LO (corr. s, 's) LO + LO = 80 (adj. s on st. line) LO = LO = 90 OL is the perpendicular bisector of. The perpendicular bisectors of, and are concurrent at a point O. M Figure S L O N. In Figure, is the base of a triangle, and the length of is the sum of the lengths of and. Given that = 0, construct the triangle. () Use as centre, as radius to draw an arc, use as centre, as radius to draw another arc. The two arcs intersect at. is an equilateral triangle. = 0 () With as centre, as radius, draw an arc, cutting produced at. () Join. () raw the perpendicular bisector of which cuts at. N is the mid point of. () Join. N N (S..S.) = (corr. sides, 's) + = + is the required triangle. N 0 Figure http://www.hkedcity.net/ihouse/fh7878/ age

nswers: (008-09 HKMO Heat Events) reated by: Mr. Francis Hung Last updated: 8 ugust 08. In Figure, is an acute angle triangle. is a point on. onstruct a triangle XY such that X is a point on, Y is a point on and the perimeter of XY is the least. () Use as centre, as radius to draw an arc. () Use as centre, as radius to draw another arc. These two arcs intersect at. (S.S.S.) = (corr. s, s) () Use as centre, as radius to draw an arc () Use as centre, as radius to draw another arc. These two arcs intersect at. (S.S.S.) = (corr. s, s) () Use as centre, as radius to draw an arc, which cuts at. ' X = X (common side) = (radii) X X (S..S.) X = X (corr. sides, s) () Use as centre, as radius to draw an arc, which cuts at. Y = Y (common side) = (radii) Y Y (S..S.) X Y 8 7 8 Y = Y (corr. sides, s) ' (7) Join, which cuts at X and at Y. (8) Join X, Y. The position of and are fixed irrespective the size and the shape of XY. erimeter of XY = X + XY + Y X + XY + Y is the minimum when, X, Y, are collinear. XY is the required triangle. http://www.hkedcity.net/ihouse/fh7878/ age