Answers: ( HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 15 December 2017
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1 Answers: (0- HKMO Het Events) reted y: Mr. Frncis Hung Lst updted: 5 Decemer 07 - Individul - Group Individul Events I Simplify I I 6 A prllelogrm is cut into 78 pieces of equilterl tringles with sides unit. If the perimeter of the prllelogrm is P units, find the mximum vlue of P. equilterl tringles joint to form smll prllelogrm The given prllelogrm is cut into 89 smll prllelogrms, nd 89 is prime numer. The dimension of the given prllelogrm is unit 89 units. P ( + 89) 80 units Figure shows right-ngled tringle AD where is point on A nd A. Given tht A nd AD 0, find the vlue of. In AD, AD tn AD In AD, A tn0 tn However, A A + + tn tn 0 I4 Given tht x + 99 y, where x, y re positive integers. Find the vlue of x. 9 5, 5 99 x 0 04, , x 5 I5 Given tht y (x + )(x + )(x + )(x + 4) + 0, find the minimum vlue of y. Reference 99HG5, 99 HG6, 995 FI4.4, 996 FG0., 000 FG., 004 FG., 0 FI. y (x + )(x + 4)(x + )(x + ) + 0 (x + 5x + 4)(x + 5x + 6) + 0 (x + 5x) + 0(x + 5x) (x + 5x) + 0(x + 5x) (x + 5x + 5) The minimum vlue of y is 0. Pge
2 Answers: (0- HKMO Het Events) reted y: Mr. Frncis Hung Lst updted: 5 Decemer 07 I6 In convex polygon with n sides, one interior ngle is selected. If the sum of the remining n interior ngle is 0, find the vlue of n. Reference: 989 HG, 990 FG0.-4, 99 HG, 00 FI ( ) < 0 < 80(4 ) 60 n 4 I7 Figure shows circle psses through two points nd, nd point A is lying outside the circle. Given tht is dimeter of the circle, A nd A intersect the circle t D nd E respectively nd re of ADE A 45, find. re of ED In AD, A 45 (given) AD 90 (dj. on st. line, in semi-circle) AD sin 45 () A ADE A (ext. cyclic qud.) AED A (ext. cyclic qud.) DAE A (common ) ADE ~ A (equingulr) AD AE (corr. of sides, ~'s) () A A A 45 Figure D E re of re of ADE A AD AE sin 45 A Asin 45 AD A y () y () re of re of ADE ED Pge
3 Answers: (0- HKMO Het Events) reted y: Mr. Frncis Hung Lst updted: 5 Decemer 07 I8 Solve x x. 0 < x < < x < 6 Then x x ( x ) + x x 4 6x x + 0 (x 4 x) 6x (x x)(x + x + ) 6x We wnt to fctorise the ove eqution s (x x + )(x + x + + ) 0 (x + x) + (x x) 6x () nd ( + ) 90 () From (), + 6 (), 0 (4) (5), su. (5) into (): L.H.S. ( 0) 90 R.H.S. (x x )(x + x + ) 0 x x 0 or x + x x,, 5, < 0 nd 5 0 < x < 6 x 5 only. Method 5 5 > 6 Let y y (y ) y, then clerly y > x nd y y > y 4 6y + y () () (): y 4 x 4 6(y x ) + y + x 0 (y + x)(y x)(y + x ) 6(y + x)(y x) + (y + x) 0 (y + x)[(y x)(y + x ) 6(y x) + ] 0 y + x 0 (y x)(y + x 6) + 0 () Assume x nd y re positive integers. Then () ecomes (y x)(y + x 6) y x (4) nd y + x 6 (5) From (4), y x + (6) Su. (6) into (5): (x + ) + x 6 x + x + + x 6 0 x + x 60 0 x + x 0 0 x 5 or 6 (rejected) Pge
4 Answers: (0- HKMO Het Events) reted y: Mr. Frncis Hung Lst updted: 5 Decemer 07 Pge 4 I9 Figure shows pentgon ADE. A DE AE + D nd A 90, find the re of the pentgon. Drw the ltitude N DE. Let AE y, D y ut AE out nd then stick the tringle to s shown in the figure. AE F (y construction) F AE y (corr. sides, 's) DE DF D D (common side) E F (corr. sides, 's) DE DF (S.S.S.) The re of the pentgon re of DE + re of DF re of DF 9 sq. units y - y y F N D E A I0 If nd re rel numers, nd + +. Find the mximum vlue of () 0 0 y (), 0 () ( + ) (y () nd ()) ( + ) The mximum vlue of +.
5 Answers: (0- HKMO Het Events) reted y: Mr. Frncis Hung Lst updted: 5 Decemer 07 Group Events G Given tht the length of the sides of right-ngled tringle re integers, nd two of them re the roots of the eqution x (m + )x + 4m 0. Find the mximum length of the third side of the tringle. Reference: 000 FI5., 00 FI., 00 FI., 0 FI. Let the sides of the right-ngled tringle e, nd c. If, re the roots of the qudrtic eqution, then + m + nd 4m m (4 ) (4 ) 4(4 ) 8 ( 4)(4 ) 8 ( 4)( 4) c The mximum vlue of the third side is. G Figure shows trpezium AD, where A, D 5 nd the digonls A nd D meet t O. If the re of AO is 7, find the re of the trpezium AD. Reference 99 HI, 997 HG, 000 FI., 00 FI., 004 HG7, 00 HG4 A // D AO ~ OD (equingulr) Are of OD 5 Are of OD 5 Are of AO 7 9 Are of OD 75 Are of AOD DO Are of AOD 5 Are of AO O 7 Are of AOD 45 Are of O O Are of O 5 Are of AO OA 7 Are of O 45 The re of the trpezium AD G Let x nd y e rel numers such tht x + xy + y 0. Find the mximum vlue of x xy + y. 0 x + xy + y (x + y) xy xy (x + y) (*) Let T x xy + y 0 x + xy + y (x xy + y ) + xy T + xy xy 0 T 0 y (*) 0 T The mximum vlue of x xy + y is 609. G4 If, re roots of x + 0x + 5 0, find the vlue of ( )( ). Reference: 99 HG, 00 HI ( )( ) ( )( + ) 4( + )( + ) 4( ) 4(5 0 + ) Pge 5
6 Answers: (0- HKMO Het Events) reted y: Mr. Frncis Hung Lst updted: 5 Decemer 07 G5 As shown in Figure, AD is squre of side 0 units, E nd F re the mid-points of D nd AD respectively, E nd F intersect t G. Find the length of AG. Join F nd AG. E DE DF FA 5 lerly E DF AF (S.A.S.) Let G x AF (corr. s 's) E 90 G 90 x G 90 (s sum of ) onsider AG nd F. A cos x F AG x + FG F G cos x AG ~ F (rtio of sides, included ngle) A AG (corr. sides, ~ s) F F F F (corr. sides, DF AF) F D A E G x x AG A 0 Method Define rectngle coordinte system with A origin, A positive x-xis, AD positive y-xis. (0, 0), (0, 0), E (5, 0), D (0, 0), F (0, 5) 0 5 Eqution of F: y 5 x 0 y x + 5 () Eqution of E: y 0 x 0 y x + 0 () 5 0 () (): x + 5 x + 0 x 6 G6 Su. x 6 into (): y 8 AG Let nd re positive rel numers, nd the equtions x + x + 0 nd x + x + 0 hve rel roots. Find the minimum vlue of +. Discriminnts of the two equtions () () () ( 4)( ) 0 ( 4)[( + ) + ] 0 ( 4) 0 0 or 4 > 0 4 only When 4, su. into (): 4 0 ( + )( ) 0 or > 0 only The minimum vlue of Pge 6
7 Answers: (0- HKMO Het Events) reted y: Mr. Frncis Hung Lst updted: 5 Decemer 07 G7 Given tht the length of the three sides of A form n rithmetic sequence, nd re the roots of the eqution x x + 47x 60 0, find the re of A. Let the roots e d, nd + d. d d 4 () ( d) + ( + d) + ( d)( + d) 47 d 47 d () ( d)( + d) 60 d 60 () Su. () nd () into (): L.H.S R.H.S. The sides of the tringle re, 4 nd 5. The re of A 4 6 sq. units. G8 In Figure, A is n isosceles tringle with A A, 40. The rdius of the inscried circle of A is 4. Find the length of A. Reference 007 FG4.4 Let I e the centre. The inscried circle touches A nd A t F nd E respectively. Let A A x. Let D e the mid-point of. AD AD (S.S.S.) F A E x AD AD x corr. s, 's, dj. s on st. line I 4 touches the circle t D 0 0 D (converse, tngent rdius) ID IE IF rdii 4 IEA, IF A (tngent rdius) AD x 0 (Pythgors' theorem) S A S I + S IA + S IA (where S stnds for res) 40 x x 4 x 4 0 x 0 40 x x x 0x x 5 x 0 0 x 5(x 0) 0 + x 4x A x Pge 7
8 Answers: (0- HKMO Het Events) reted y: Mr. Frncis Hung Lst updted: 5 Decemer 07 G9 At most how mny numers cn e tken from the set of integers:,,,, 0, 0 such tht the sum of ny two numers tken out from the set is not multiple of the difference etween the two numers? In order to understnd the prolem, let us tke out few numers nd investigte the property. Tke, 5, , 5 4, 6 4k, for ny integer k + 9 0, 9 8, 0 8k, for ny integer k , 9 5 4, 4 4k, for ny integer k Tke, 6, , 6, 9 + 8, 8 5, 5k for ny integer k , 8 6, 4 7 Tke, 8, , 8 6, 40 6k for ny integer k , 40 8, 68 k for ny integer k , 40 8, 5 8k for ny integer k We cn tke three numers, 5, 9 or, 8, 40 (ut not, 6, 8). Tke the rithmetic sequence,, 5,, 0. (007 numers) The generl term T(n) n for n 007 T(n) + T(m) n + m (n + m ) T(n) T(m) n m (n m) T(n) + T(m) [T(n) T(m)]k for some integer k. For exmple, (5 )4. The sequence,, 5,, 0 does not stisfy the condition. Tke the rithmetic sequence, 4, 7,, 0. (67 numers) The generl term T(n) n for n 67 T(n) + T(m) n + m 4 (n + m ) T(n) T(m) n m (n m) T(n) + T(m) [T(n) T(m)]k for ny non-zero integer k We cn tke t most 67 numers to stisfy the condition. Pge 8
9 Answers: (0- HKMO Het Events) reted y: Mr. Frncis Hung Lst updted: 5 Decemer 07 G0 For ll positive integers n, define function f s (i) f () 0, (ii) f () + f () + + f (n ) + f (n) n f (n),n >. Find the vlue of f (0). Reference: 04 FG.4 f f () + f () + + f (n ) (n )f (n) f (n) f f n n f f () 0 f () 0 0 f f f (4) 0 0 f f f It is oserved tht the nswer is 0 divided y the n th tringle numer. lim: f (n) 0 for n n n n,,, 4, proved ove. Suppose f (k) 0 for k,,, m for some positive integer m. k k f f (m+ ) f f m mm m mm 4 m m 0 mm m mm 0 0 m m It is lso true for m. y the principle of mthemticl induction, the formul is true. f (0) Pge 9
10 Answers: (0- HKMO Het Events) reted y: Mr. Frncis Hung Lst updted: 5 Decemer 07 Geometricl onstruction. Line segment PQ nd n ngle of size re given elow. onstruct the isosceles tringle PQR with PQ PR nd QPR. R F 4 E D 6 7 J 5 P H Q Steps. Let the vertex of the given ngle e E.. Use P s centre, PQ s rdius to drw circulr rc QR.. Use E s centre, certin rdius to drw n rc, cutting the given ngle t D nd F respectively.. Use P s centre, the sme rdius in step to drw n rc, cutting PQ H. 4. Use D s centre, DF s rdius to drw n rc. 5. Use H s centre, DF s rdius to drw n rc, cutting the rc in step t J. 6. Join PJ, nd extend PJ to cut the rc in step t R. 7. Join QR. PQR is the required tringle. Pge 0
11 Answers: (0- HKMO Het Events) reted y: Mr. Frncis Hung Lst updted: 5 Decemer 07. onstruct rectngle with A s one of its sides nd with re equl to tht of A elow. Theory Let the height of the rectngle e h. Let the height of the tringle e k. Are of rectngle re of tringle Ah A k h k The height of rectngle is hlf of the height of the tringle. Q D P 4 A E Steps.. Drw line segment E A. (E lies on A, E is the ltitude of A). Drw the perpendiculr isector PQ of E, D is the mid-point of E.. Drw line segment AQ A, cutting PQ nd Q. 4. Drw line segment P A, cutting PQ nd P. APQ is the required rectngle. Pge
12 Answers: (0- HKMO Het Events) reted y: Mr. Frncis Hung Lst updted: 5 Decemer 07. The figure elow shows two stright lines A nd A intersecting t the point A. onstruct circle with rdius equl to the line segment MN so tht A nd A re tngents to the circle. Lemm: 如圖, 已給一綫段 A, 過 作一綫段垂直於 A 作圖方法如下 : () 取任意點 ( 在 A 之間的上方 ) 為圓心, 為半徑作 A 一圓, 交 A 於 P () 連接 P, 其延長綫交圓於 Q; 連接 Q Q 為所求的垂直綫 作圖完畢 證明如下 : PQ 為圓之直徑 ( 由作圖所得 ) PQ 90 ( 半圓上的圓周角 ) A 證明完畢 Steps.. Drw the ngle isector AQ. Use A s centre, MN s P S rdius to drw n rc.. Use the lemm to drw AP A, AP cuts the rc in 4 step t P. 4. Drw PQ AP, PQ cuts the ngle isector t Q. A 5. Drw QR PQ, QR cuts A t R. 6. Use Q s centre, QR s 5 rdius to drw circle. This is the required circle. M R N Proof: ARQ 90 (s sum of polygon) APQR is rectngle. A is tngent touching the circle t R (converse, tngent rdius) Let S e the foot of perpendiculr drwn from Q onto A, QS A. AQR AQS (A.A.S.) SQ SR (corr. sides, 's) S lies on the circle nd QS A A is tngent touching the circle t S (converse, tngent rdius) P Q 6 Q Pge
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