= Find the value of n.

nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0 09 099 00 - Individual 9 0 0900 - Group 0 0 9 0 0 Individual Events I How many pairs of distinct integers between and 0 inclusively have their products as multiple of? Multiples of are, 0,, 0,, 0,, 0. Number = 0 Numbers which are not multiples of = 0 0 = Let the first number be x, the second number be y. Number of pairs = No. of ways of choosing any two numbers from to 0 no. of ways of choosing such that both x, y are not multiples of. 0 0 0 0 = = = 0 0 00 = 0(0 ) = 0 = 09 0 I Given that 0 I = 0 000 0 = 0.000 0 n times = 0.000 0. Find the value of n. n times n = 000 = 099 Let x be the measure of an interior angle of an n-sided regular polygon, where x is an integer, how many possible values of n are there? If x is an integer, then each exterior angle, 0 x, is also an integer. Using the fact that the sum of exterior angle of a convex polygon is 0. 0 Each exterior angle =, which is an integer. n n must be an positive integral factor of 0. n =,,,,,,, 9, 0,,,, 0,, 0,, 0,, 0,, 90, 0, 0, 0 However, n = and n = must be rejected because the least number of sides is. There are possible value of n. I s shown in the figure, EG =, + + + D + E + F =? reflex GF = reflex GE = 0 + = onsider quadrilateral GF, + +reflex GF +F = 0 ( sum of polygon) onsider quadrilateral DEG, +D +E+reflex GE = 0 ( sum of polygon) dd these two equations, + + + D +E +F = 0 () = http://www.hkedcity.net/ihouse/fh age

nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0 I It is given that a, a,, a n, is a sequence of positive real numbers such that a = and a n+ = a n + a n. Find the value of a 0. 9 a = + = 9 a = + + = laim: a n = n for n f: y M.I. n =,,, proved already. Suppose a k = k for some positive integer k. a k+ = a k + a k = k k + = k k = k y M.I., the statement is true for n 0 a 0 = = 00 = 00 I s shown in the figure, D is a convex quadrilateral and + D + D =. Find the maximum area of D. Let = a, D = b, D = c, D =, D = rea of D = area of D + area of D = ab sin bcsin ab bc, equality holds when = 90, = 90 = ba c= b b b b (.M. G.M., equality holds when b =, a + c = ) = The maximum area of D = I Let x, y, z >, p > 0, log x p =, log y p = and log xyz p = 9. Find the value of log z p. Reference: 999 FG., 00 FG. log p log p log p =, =, = 9 log x log y log xyz log x log y log x log y log z,, log p log p log p 9 log x log y log z log z log p log p log p log p 9 log z = log p log z p = log p = log z http://www.hkedcity.net/ihouse/fh age

nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0 I 0 0 0 Find the value of 09 0 0 0 0 0 0. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 = 0 0 0 0 0 0 0 0 0 0 0 = 0 0 0 0 0 0 0 0 = 0 0 0 0 = I9 Let x be a real number. Find the minimum value of x x x x 0. Reference 00 FG. onsider the following problem: Q(, 9) Let (, ) and Q(, 9) be two points. R(x, 0) is a variable point on x-axis. To find the minimum sum of distances R + RQ. Let y = sum of distances = x 9 x If we reflect (, ) along x-axis to (, ), M(, 0) is the foot of perpendicular, then MR MR (S..S.) y = R + RQ = R + RQ Q (triangle inequality) y 9 = The minimum value of x x x x 0 is. - (, ) M(, 0) '(, -) R(x, 0) I0, H and I are points on the circle. is a point outside the circle. is tangent to the circle at. H and I cut the circle at D and G respectively. It is given that HD is the angle bisector of I, =, D = and G = 9. Find the value of y intersecting chords theorem, HD = H = H = DH = = Let D = = GD ( HD is the angle bisector) S D Dsin () S G Dsin 9 DG onsider D and DH They have the same height but different bases. S DH DH () S D D onsider DG and IH area of area of DH. DHIG H I G 9 D http://www.hkedcity.net/ihouse/fh age

nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0 DG = IH (common s) DG = IH (ext., cyclic quad.) GD = HI (ext., cyclic quad.) DG ~ IH (equiangular) S IH H S DG G 9 9 SDHIG 9 SDG 9 9 S DG 9 () SDHIG area of DH S D S ()()(): = area of DHIG S S DG DH D S DG 9 = = S DHIG http://www.hkedcity.net/ihouse/fh age

nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0 Group Events G Find the value of. 0 0 0 00 0 Reference: 00 HI 0 0 0 00 0 = 0 0 = 0 0 = 0 = 900 = 0900 G Given an equilateral triangle with each side of length and is an interior point of the triangle. Let X, Y and Z be the feet of perpendiculars from to, and respectively, find the value of X + Y + Z. (Reference 99 HG, 00 HG9) Let the distance from to,, be h, h, h respectively. h h h =area of = X + Y + Z = h + h + h = sin 0 9 G The coordinates of are ( +, + ). is rotated 0 anticlockwise about the origin to Q. Q is then reflected along the y-axis to R. Find the value of R. Reference: 00 HI0 Let the inclination of be. tan = = = Inclination of Q = + 0 = 0 ngle between Q and positive y-axis = 0 90 = Inclination of R = 90 = R = = 0 = R = = pply cosine rule on R cos0 R = = = = http://www.hkedcity.net/ihouse/fh age

nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0 G Given that a b + + 9 ab b, where a and b are real numbers. Find the value of ab. Reference: 00 FI., 00 FI., 009 FG., 0 FI., 0 FI., 0 FI. a + b + ab b (a ab + b ) + (b + b + ) 0 (a b) + (b + ) 0 a b = 0 and b + = 0 b = and a = ab = G Given that the equation x x x x has two real roots. Find the sum of the roots. Let y = x + x (y + ) = (y + ) y + y + = y + y + y + 00 = 0 (y + )(y + 0) = 0 x + x + = 0 or x + x + 0 = 0 = < 0 or = 00 > 0 The first equation has no real roots and the second equation has two real roots Sum of the two real roots = G Given that the sum of two interior angles of a triangle is n, and the largest interior angle is 0 greater than the smallest one. Find the largest possible value of n. Let the angles of the triangle be x, y and x 0, where x y x 0 () x + y + x 0 = 0 (s sum of ) y = 0 x () Sub. () into (): x 0 x x 0 x 0 and 0 x 0 x 0 () n = x + y = x + 0 x by () x = 0 n () Sub. () into (): 0 n 0 The largest possible value of n = 0 G Four circles with radii unit, units, units and r units are touching one another as shown in the figure. Find the value of r. Let the centre of the smallest circle be and the radius be r. Let the centres of the circles with radii,, be, and respectively. = + =, = + =, = + = + = + = = is a with = 90 (converse, ythagoras theorem) = r +, = r +, = r + Let the feet of drawn from to and respectively. Let Q = x, = y; then Q = x, = y. In Q, x + y = (r + ) () (ythagoras theorem) In Q, ( x) + y = (r + ) () (ythagoras theorem) In, x + ( y) = (r + ) () (ythagoras theorem) () (): 9 x = r + x = r () () (): y = r + y = r () http://www.hkedcity.net/ihouse/fh age

nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0 r r r r r r 9 r r 0 r + r = 0 (r )(r + ) = 0 r = or (rejected) Sub. (), () into (): r r r - y Method We shall use the method of circle inversion to solve this problem. Lemma In the figure, a circle centre at N, with radius touches another circle centre at F, with radius externally. ME is the common tangent of r + y r + r + x Q - x D the two circles. third circle with centre at touches the given two circles externally and also the line ME. EF is produced to D so that DE =. Join D. lies on NM, Y, I lies on FE so that NM, IFE, NYFE. rove that 0 ; (a) the radius of the smallest circle is (b) ME = ; 00 (c) D = roof: Let the radius of the smallest circle be a. Then N = + a, F = + a, NM =, FE = N = a, FI = a, FY = = 0 In N, = ( + a) ( a) 0a = N M (ythagoras theorem) F Y I E In IF, I = ( + a) ( a) a = (ythagoras theorem) In NYF, NY + FY = NF (ythagoras theorem) ( + I) + FY = NF 0a a 0 0 0 a a a a 0 a a = 0a a 0 ME = + I = = a = = http://www.hkedcity.net/ihouse/fh age

nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0 0 9 DI = DE IE = = a 0 I = = = In DI, DI + I = D (ythagoras theorem) 9 D D 00 00 = D = Lemma Given a circle with centre at and radius r. Figure Figure Figure Figure and are points such that,, are collinear. If = r, then is the point of inversion of respect to the circle. (Figure ) is also the point of inversion of. is called the centre of inversion. If lies on the circumference of the circle, then = r, = r, and coincide. If < r, then > r; if > r, then < r; if = 0, then = ; =, = 0. If < r and is a chord, then the inversion of is the arc ; the inversion of the straight line is the circle which has a common chord. (Figure ) If > r, the inversion of a line (outside the given circle) is another smaller circle inside the given circle passing through the centre. (Figure ) If = 0, the inversion of a line through the centre is itself, the line. (Figure ) c c c c c Figure Figure Figure Figure Given another circle which intersects the original circle at and, but does not pass through. Then the inversion of with respect to the given circle is another circle passing through and but does not pass through. (Figure ) Given another circle which intersects the original circle at and, and passes through. Then the inversion of with respect to the given circle is the straight line through and. (Figure ) Given a circle outside but does not intersect the original circle. The inversion of respect to the given circle is another circle inside but does not pass through. onversely, the inversion of is. (Figure ) Given a concentric circle 9 with the common centre inside the given circle. Then the inversion of 9 is another concentric circle 0 outside. onversely, the inversion of 0 is 9. (Figure ) c 0 c c 9 http://www.hkedcity.net/ihouse/fh age

nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0 c c c c c 9 Figure Figure 0 Figure 9 Figure Given a circle inside the given circle but does not intersect the original circle, and passes through. Then the inversion of with respect to the given circle is the straight line outside. (Figure 9) Given a circle inside the given circle passes through and touches internally at. Then the inversion of with respect to the given circle is the tangent at. onversely, the inversion of the tangent at is. (Figure 0) Given a circle inside the given circle which encloses but touches internally at. Then the inversion of with respect to the given circle is a circle 9 encloses and touches at. onversely, the inversion of 9 is. (Figure ) Given a circle inside the given circle which does not enclose but touches internally at. Then the inversion of with respect to the given circle is a circle which touches externally at. onversely, the inversion of is. (Figure ) c c c D V G U K F H Q N S E J R W 9 T L M X I Z Suppose the circle with centre at and the circle with centre at touch each other at E. Draw a common tangent XEU. Let EZ and ED be the diameters of these two circles. Let H be the mid-point of DZ. Use H as centre HD as radius to draw a circle. Use D as centre, DE as radius to draw a circle. and intersect at W and V. Join VW. VW intersects DZ at G. Let F be the mid-point of EG. Use F as centre, FE as radius to draw a circle. E =, E = (given), DE =, EZ =, DZ = + = 0. http://www.hkedcity.net/ihouse/fh age 9

nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0 HD = HZ = HW = Let the diameter of be x, i.e. GE = x, DG = x. HG = HD DG = ( x) = x In DGW, WG = ( x) = x + x () (ythagoras theorem) In HGW, WG = (x ) = + x x () (ythagoras theorem) () = (): + x x = x + x x =. The radius of is. = DGDZ = (.)0 = = DE G is the point of inversion of Z w.r.t.. learly E is the point of inversion of E w.r.t.. The inversion of w.r.t. is. The inversion of w.r.t. is the tangent UEX (see Figure 0). Let the circle, with centre at and radius be. Join D. D cuts at J. D is produced to cut again I. Then IJ = diameter of =. In, let =, cos =, D = 0 (adj. s on st. line) In D, D = + cos(0 ) = = D = DJ = D J = ; DI = D + I = + Invert w.r.t. to centre at N. Suppose DI intersects at K and L in the figure. DIDK = and DJDL = DK and DL DK = = and DL = = LK = DL DK = = The radius of is. Now construct a smaller circle centre, touches and externally and also touches XU. is not shown in the figure. 0 00 y the result of Lemma, the radius of is ; ME = and D = D cuts at Q, D is produced further to cut again at R. 00 0 00 0 DQ = D Q = ; DR = D + R = + Now invert w.r.t to give 9. This circle will touch, and externally. DR intersects 9 at S, produce DR further to meet 9 again at T. Then DSDR = and DTDQ = 00 0 00 0 DS and DT DS = ST = DT DS = = 00 0 00 0 The radius of 9 is. 00 0 ; DT = = 00 0 00 0 = = diameter of 9 00 0 http://www.hkedcity.net/ihouse/fh age 0

nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0 G Given that a, b, x and y are non-zero integers, where ax + by =, ax +by =, ax +by = and ax + by =. Find the value of ax + by. ax + by = (), ax +by = (), ax +by = (), ax + by = (). Let ax + by = m () (x + y)(): (x + y)(ax + by ) = (x + y) G9 ax + by + xy(ax + by) = (x + y) Sub. () and (): + xy = (x + y) + xy = (x + y) () (x + y)(): (x + y)(ax + by ) = (x + y) ax + by + xy(ax + by ) = (x + y) Sub. () and (): + xy = (x + y) 9 + xy = (x + y) () () (): 0 + xy = 0 xy = () () (): (x + y) = x + y = (9) (x + y)(): (x + y)(ax + by ) = (x + y) ax + by + xy(ax + by ) = (x + y) Sub. () and (): m + xy = (x + y) Sub. () and (9): m + ( ) = m = Given that, in the figure, is an equilateral triangle with F =, FG = 0, G = and DE =. Find the value of HI. (Reference: 0 HG9) F + FG + G = + 0 + = = = = (property of equilateral triangle) Let D = x, then E = x = x Let HI = y, H = z, then I = y z y intersecting chords theorem, DE = FG x(x + ) = x + x = 0 (x )(x + ) = 0 x = or (rejected) E = x = HI = ED D z(z + y) = 0 = 0 () IH = GF ( y z)( z) = E 9 (y + z) + z(y + z) = () Sub. () into (): 9 (y + z) + 0 = - x y + z = y = z () z H Sub. () into (): z(z + z) = 0 z z + 0 = 0 z = or From (), z z is rejected z = only HI = y = z = ( ) = x F y 0 G -y-z I http://www.hkedcity.net/ihouse/fh age

nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0 G0 Let a n and b n be the x-intercepts of the quadratic function y = n(n )x (n )x +, where n is an integer greater than. Find the value of a b + a b + + a 0 b 0. Reference: 00 HI The quadratic function can be written as y = (nx )[(n )x ] The x-intercepts are n and. n a n b n = = n n n n for n > a b + a b + + a 0 b 0 = 0 0 0 = = 0 0 http://www.hkedcity.net/ihouse/fh age

nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0 Geometrical onstruction. onstruct an isosceles triangle which has the same base and height to the following triangle. E D Steps () onstruct the perpendicular bisector of, D is the mid point of. () opy. () Draw E so that it is equal to, then E // E (alt. s eq.) E and the bisector in step intersect at E. () Join E, E. Then E is the required isosceles triangle with E = E. http://www.hkedcity.net/ihouse/fh age

nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0. Given the following line segment MN represent a unit length, construct a line segment of length. G E H F M N Steps () Use N as centre, NM as radius to draw a circular arc. () Use M as centre, MN as radius to draw a circular arc, cutting the arc in step at E. MNE is an equilateral triangle. MNE = 0 () Use E as centre, MN as radius to draw a circular arc, cutting the arc in step at F. FNE is an equilateral triangle. FNE = 0 () Use E as centre, NM as radius to draw a circular arc. Use F as centre, NM as radius to draw a circular arc. The two arcs intersect at G. EFG is an equilateral triangle. () Join NG and produce it longer. NG intersects the arc in step at H. NG is the bisector of ENF. MNG = 0 + 0 = 90 () Use H as centre, HN as radius to draw a semi-circle, cutting NG produced at. N = () Join M, cutting the semicircle in step () at. M = = () Join N. Then M is the required length. roof: N = 90 ( in semi-circle) It is easy to show that MN ~ NM (equiangular) M MN (cor. sides, ~ s) MN M M = http://www.hkedcity.net/ihouse/fh age

nswers: (0- HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0. onstruct a square whose area is equal to the difference between the areas of the following two squares D and QRS. F E = Q E G M R Q S D Steps () Draw the perpendicular bisector of, M is the mid-point of. () Use M as centre, M as radius to draw a semi-circle outside the square D. () Use as centre, Q as radius to draw an arc, cutting the semicircle in () at E. () Join E and produce it longer. Join E. E = 90 ( in semi-circle) () Use E as centre, E as radius to draw an arc, cutting E produced at F. () Use as centre, E as radius to draw an arc. Use F as centre, FE as radius to draw an arc. The two arcs intersect at G. () Join FG and G. Then by ythagoras theorem, E = E ; EFG is the required square. http://www.hkedcity.net/ihouse/fh age