I6 As shown in the figure, ABCD is a convex quadrilateral and
1 730639 2 201499
3
22
4 232 5 1016064 14-15
36 Individual
6
32
7 126
8
1
9
13
10
55
1
3
3
1
2
3
4
4
18
5 –15 14-15 120900
2 Group 6 2014
6 140
7 8 454
9
10
2
14
23 2015
Individual Events
I1 How many pairs of distinct integers between 1 and 2015 inclusively have their products as
multiple of 5? Multiples of 5 are 5, 10, 15, 20, 25, 30, , 2015. Number = 403 Numbers which are not multiples of 5 = 2015 – 403 = 1612 Let the first number be x, the second number be y.
Number of pairs = No. of ways of choosing any two numbers from 1 to 2015 – no. of ways of choosing such that both x, y are not multiples of 5. 2015 1612 2015 2014 1612 1611 5 2014 4 1611 = = = 403
C C 2 2
2
2
2
2
= = 403
403 10 2 5 1007 2 1611 (5035 – 3222) = 4031813 = 730639 2015
I2 Given that
10 = . 000 1 . Find the value of n.
n times
201500
10 = . 000
1
n times n = 201500 – 1 = 201499
I3 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, 360 – x, is also an integer. Using the fact that the sum of exterior angle of a convex polygon is 360 .
360 Each exterior angle = , which is an integer.
n n must be an positive integral factor of 360. n = 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360 However, n = 1 and n =2 must be rejected because the least number of sides is 3.
There are 22 possible value of n.
I4 As shown in the figure, EGB = 64,
A + B + C + D + E + F = ? reflex BGF = reflex CGE = 180 + 64 = 244 Consider quadrilateral ABGF, A +B +reflex BGF +F = 360 ( sum of polygon) Consider quadrilateral CDEG, C +D +E+reflex CGE = 360 ( sum of polygon) Add these two equations, A + B + C + D +E +F = 720 – 2(244) =232
I5 It is given that a 1 , a 2 , , a n , is a sequence of positive real numbers such that a 1 = 1 and
1 a + n +1 = a n a . Find the value of a 2015 . n
4
1
9 = 2 + =
a
2
4
4
9
3
1
16
a + = + =
3
4
2
4
4 2
n
1 Claim: a = for n
n 1
4 Pf: By M.I. n = 1, 2, 3, proved already. 2
k
1
Suppose a = for some positive integer k. k
4 2 2 2 1 k 1 k 1 k 1 2 k 1 1 k 1
1
a = a a 1 + +1 = + = = k k k
4
4
2
4
4
4 By M.I., the statement is true for n 1 2 2016
2 a 2015 = = 1008 = 1016064
4 I6 As shown in the figure, ABCD is a convex quadrilateral and AB + BD + CD = 16. Find the maximum area of ABCD.
Let AB = a, BD = b, CD = c, ABD = , BDC = Area of ABCD = area of ABD + area of BCD
1
1 = ab sin bc sin
2
2
1
1 ab bc , equality holds when = 90, = 90
2
2
1
1 = b a c = b 16 b
2
2 2 1 b 16 b
(A.M. G.M., equality holds when b = 8, a + c = 8)
2
2
=
32 The maximum area of ABCD = 32
I7 Let x , y, z > 1, p > 0, log x p = 18, log y p = 21 and log xyz p = 9. Find the value of log z p.
Reference: 1999 FG1.4, 2001 FG1.4
log p log p log p = 18, = 21, = 9 log x log y log xyz log x 1 log y 1 log x log y log z
1 , ,
log p 18 log p 21 log p
9 log x log y log z 1 1 log z
1 log p log p log p
18 21 log p
9 log z
1 = log p 126 log p log p = = 126
z
log z
3 7
1 2 2014 4 2014 8 2014
I8 Find the value of .
2 2 4 4 8 8 4029 2014 2015 2014 2015 2014 2015 3 7
1 2 2014 4 2014 8 2014 2 2 4 4 8 8 2014 2015 2014 2015 2014 2015 2014 2015 3 7
1
1 2 2014 4 2014 8 2014 =
1 2 2 4 4 8 8 2015 2014 2014 2015 2014 2015 2014 2015 2014 2015 3 7 2 2014
2 2014 4 2014 8 2014 =
1 2 2 2 2 4 4 8 8 2015 2014 2014 2015 2014 2015 2014 2015 3 3 7 4 2014
4 2014 8 2014 =
1 4 4 4 4 8 8 2015 2014 2014 2015 2014 2015 7 7 8 2014
8 2014 = 1 = 1
8 8 8 8 2015 2014 2015 2014 2 2 I9 Let x be a real number. Find the minimum value of x
4 x 13 x 14 x 130 .
Reference 2010 FG4.2
Consider the following problem: 8 Q(7, 9) Let P(2, 3) and Q(7, 9) be two points. R(x, 0) is a variable point on x-axis.
To find the minimum sum of distances PR + RQ. 2 2 6 Let y = sum of distances = x
2 9 x
7
81 4 If we reflect P(2, 3) along x-axis to P’(2, –3), M(2, 0) is the foot
P(2, 3)
of perpendicular, 2 then PMR P’MR (S.A.S.)
y = PR + RQ = P’R + RQ P’Q (triangle inequality) 2 2
7
2
9 3 = 13 y 5 M(2, 0) 2 2 R(x, 0)
The minimum value of x
4 x 13 x 14 x 130 is 13. -2
P'(2, -3)
I10 B, H and I are points on the circle. C is a point outside the circle. BC is tangent to the circle at B . HC and IC cut the circle at D and G respectively. It is given that HDC is the angle bisector
area of BDH of . BCI, BC = 12, DC = 6 and GC = 9. Find the value of area of DHIG
By intersecting chords theorem,
2 I H CH CD = BC
2
6 CH = 12
CH = 24 G
DH = 24 – 6 = 18
9
6 Let
BCD = = GCD ( HDC is the angle bisector) D 1
12 B BC CD sin
C S BCD 2
12
4
(1) 1 S GC CD sin CDG
9
3 2 Consider
BCD and BDH They have the same height but different bases. S DH
18
BDH
3 (2) S CD BCD
6
Consider CDG and CIH
DCG = ICH (common s) CDG = CIH (ext. , cyclic quad.) CGD = CHI (ext. , cyclic quad.) CDG ~ CIH (equiangular) 2 2 S CIH CH
24
64
S CG CDG
9
9
S DHIG 64
9
55
S
9
9 CDG
S
9
CDG
(3) S DHIG
55 S area of BDH S S BCD BDH CDG
9
36
(1) (2)(3): = = 4 3 = area of DHIG S S S CDG BCD DHIG
3
55
55
Group Events
1
1
1
1 G1 Find the value of . 1860 1865 1865 1870 1870 1875 2010 2015
Reference: 2010 HI3
1
1
1
1
1860 1865 1865 1870 1870 1875 2010 2015
1
1
1
1
1
=
25 372 373 373 374 374 375 402 403
1
1
1
1
1
1
1
1
1
=
25 372 373 373 374 374 375 402 403
1
1
1
= 25 372 403
31 =
3747900
1 =
120900
G2 Given an equilateral triangle ABC with each side of length 3 and P is an interior point of the
triangle. Let PX, PY and PZ be the feet of perpendiculars from P to AB, BC and CA respectively, find the value of PX + PY + PZ. (Reference 1992 HG8, 2005 HG9)
A Let the distance from P to AB, BC, CA be h , h , h respectively.
1
2
3
1
1
1
1 2
9
3
area of ABC
3 h 1 3 h 2 3 h = = 3 3 sin 60
2
2
2
2
4
3
3 P
PX + PY + PZ = h 1 + h 2 + h 3 =
2 B
C G3 The coordinates of P are ( 3 + 1, 3 + 1). P is rotated 60
anticlockwise about the origin to Q. Q is then reflected along the
2 y -axis to R. Find the value of PR . Reference: 2007 HI10 Let the inclination of OP be .
3
1 tan = = 1 = 45 3
1 Inclination of OQ = 45 + 60 = 105 Angle between OQ and positive y-axis = 105 – 90 = 15 Inclination of OR = 90 – 15 = 75 POR = 75 – 45 = 30 2 2 OP = OR = 3
1 1 1 = 6
2
Apply cosine rule on POR 2 2
2
PR =
6 2 6 2
2 6
2 6 2 cos
30
3 = 6 2
2
12 2 2 = 8
4
3 2
3
2
=
4
2 b
2
- G4 Given that a + 9 ab – 3b, where a and b are real numbers. Find the value of ab.
2 Reference: 2005 FI4.1, 2006 FI4.2, 2009 FG1.4, 2011 FI4.3, 2013 FI1.4, 2015 FI1.1
2
2
4a + 2b + 36 4ab – 12b
2
2
2
(4a – 4ab + b ) + (b + 12b + 36) 0
2
2
(2a – b) + (b + 6) 0 2a – b = 0 and b + 6 = 0 b = –6 and a = –3
ab = 18 2 2 G5 Given that the equation has two real roots. Find the sum of x 15 x 58 2 x 15 x
66 the roots.
2 Let y = x + 15x
2
(y + 58) = 4(y + 66)
2 y + 116y + 3364 = 4y + 264
2 y + 112y + 3100 = 0
(y + 62)(y + 50) = 0
2
2 x + 15x + 62 = 0 or x + 15x + 50 = 0
= 225 – 248 < 0 or = 225 – 200 > 0 The first equation has no real roots and the second equation has two real roots Sum of the two real roots = –15
G6 Given that the sum of two interior angles of a triangle is n , and the largest interior angle is 30 greater than the smallest one. Find the largest possible value of n.
Let the 3 angles of the triangle be x , y and x – 30, where x y x – 30 (1)
x + y + x – 30 = 180 (
s sum of ) y = 210 – 2x (2) Sub. (2) into (1): x 210 – 2x x – 30
x
70 and 80 x 80 x 70 (3)
n = x + y = x + 210 – 2x by (2)
x = 210 – n (4) Sub. (4) into (3): 130
n 140 The largest possible value of n = 140
G7 Four circles with radii 1 unit, 2 units, 3 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 O and the radius be r. Let the centres of the circles with radii 2, 3, 1 be A, B and C respectively. AB = 3 + 2 = 5, AC = 2 + 1 = 3, BC = 3 + 1 = 4
2
2
2
2
2 AC + BC = 3 + 4 = 25 = AC
ABC is a with C = 90 (converse, Pythagoras’ theorem)
OA = r + 2, OB = r + 3, OC = r + 1 Let the feet of drawn from O to BC and AC respectively.
Let CQ = x, CP = y; then AQ = 3 – x, BP = 4 – y.
2
2
2 In OCQ, x + y = (r + 1) (1) (Pythagoras’ theorem)
2
2
2 In OAQ, (3 – x) + y = (r + 2) (2) (Pythagoras’ theorem)
2
2
2 In OBP, x + (4 – y) = (r + 3) (3) (Pythagoras’ theorem)
1 (2) – (1): 9 – 6x = 2r + 3 x = 1 r (4)
3
1 (3) – (1): 16 – 8y = 4r + 8 y = 1 r (5)
2
2 2 B
1
1 2
Sub. (4), (5) into (1): 1
1
1
r r r
3
2
2
1 2
1 2 2
1
1
1
2 r r r r r r
3
9
4
23 2
11
4 - y r r
1
36
3
r + 3
2
23r + 132r – 36 = 0
P O
(23r – 6)(r + 6) = 0
r + 2 y r + 1
6
r = or –6 (rejected) A x 3 - x
C Q
23 Method 2 We shall use the method of circle inversion to solve this problem.
Lemma 1 In the figure, a circle centre at N, with D
5 radius touches another circle centre at F, with
6
6 radius externally. ME is the common tangent of 5 the two circles. A third circle with centre at P touches the given two circles externally and also the line ME.
EF is produced to D so that DE = 6. Join DP. O lies on NM, Y, I lies on FE so that NM OP, PIFE, NY FE. Prove that
30 6 (a) the radius of the smallest circle is ;
121 (b) ME = 2;
501840 (c) DP =
121 5 5 6 Proof: Let the radius of the smallest circle be a. 6 F
5
6
5
6 N
Y
Then PN = + a, PF = + a, NM = , FE =
6
5
6
5 O
I P
5
6
6
5
11 NO = – a, FI = – a, FY = – =
M E
6
5
5
6
30
5 5 10a
2
2
2 In PNO, OP = ( + a) – ( – a) = (Pythagoras’ theorem)
6
6
3
6 6 24a
2
2
2 In PIF, PI = ( + a) – ( – a) = (Pythagoras’ theorem)
5
5
5
2
2
2
2
2
2 In NYF, NY + FY = NF (Pythagoras’ theorem) (OP + PI) + FY = NF 2 2 2
10 a 24 a
11
5
6
3
5
30
6
5
2 2 2 2
10
24 50
72
5 2
6
2
11
2
a
4 a 4 a 4 a
4
3
5
15
15
15
242
30
a
4 a = 15 121
10 24 a
11
2
11
2
30 ME = OP + PI = a = a = = 2
3
5
15
11
15
30 696 = DE – IE = 6 =
DI
121 121
24a
30
12 PI = = 24 =
5 5 121
11
2
2
2 In + PI = DP (Pythagoras’ theorem)
DPI, DI 2 2 696
12 2
DP
121
11
2 501840 DP = 501840 DP = 2
121 121 Lemma 2 Given a circle C with centre at O and radius r.
A A O C P P O O O P
1 P P
1 B B
Figure 1 Figure 3
Figure 2 Figure 4
P and P 1 are points such that O, P, P 1 are collinear.
2 If OP = r , then P is the point of inversion of P respect to the circle C. (Figure 1)
OP
1
1 P is also the point of inversion of P 1 . O is called the centre of inversion.
If P lies on the circumference of the circle, then OP = r, OP
1 = r, P and P 1 coincides.
If OP < r, then OP
1 > r; if OP > r, then OP 1 < r; if OP = 0, then OP 1 = ; OP = , OP 1 = 0.
If OP < r and APB is a chord, then the inversion of APB is the arc AP ; the inversion of the
1 B
straight line AB is the circle AP B which has a common chord AB. (Figure 2)
1 If OP > r, the inversion of a line (outside the given circle) is another smaller circle inside the
given circle passing through the centre O. (Figure 3) If OP = 0, the inversion of a line through the centre is itself, the line AB. (Figure 4) c 10
c c 1 A 6 A c 3 c 7 c O O O c 2 O B c 9 B
Figure 7 Figure 6
Figure 5 Figure 8
Given another circle C
1 which intersects the original circle at A and B, but does not pass
through O. Then the inversion of C with respect to the given circle is another circle C
1
2
passing through A and B but does not pass through O. (Figure 5) Given another circle C 3 which intersects the original circle at A and B, and passes through O. Then the inversion of C 3 with respect to the given circle is the straight line through A and B. (Figure 6) Given a circle C outside but does not intersect the original circle. The inversion of C respect
6
6
to the given circle is another circle C
7 inside but does not pass through O. Conversely, the
inversion of C
7 is C 6 . (Figure 7)
Given a concentric circle C
9 with the common centre O inside the given circle C. Then the
inversion of C
9 is another concentric circle C 10 outside C. Conversely, the inversion of C 10 is C . (Figure 8)
9
c C c
14 O
c c P
16 P c O c O
21 C
22 P P O c
12 Figure 12 c
19 Figure 10
Figure 9 Figure 11
Given a circle C
12 inside the given circle C but does not intersect the original circle, and
passes through O. Then the inversion of C with respect to the given circle is the straight line
12
outside C. (Figure 9) Given a circle C 14 inside the given circle C passes through O and touches C internally at P. Then the inversion of C
14 with respect to the given circle is the tangent at P. Conversely, the
inversion of the tangent at P is C
14 . (Figure 10) Given a circle C inside the given circle C which encloses O but touches C internally at P.
16 Then the inversion of C 16 with respect to the given circle is a circle C 19 encloses C and
touches C at P. Conversely, the inversion of C
19 is C 16 . (Figure 11)
Given a circle C
21 inside the given circle C which does not enclose O but touches C internally
at P. Then the inversion of C
21 with respect to the given circle is a circle C 22 which touches C
externally at P. Conversely, the inversion of C is C . (Figure 12)
22
21 C
3 V C D
1 U B G C
5 F H K C 8 C Q C
4
2 S E N C
7 R C W J T 9 L M C A C
6 X
I Z Suppose the circle C with centre at A and the circle C with centre at B touch each other at E.
1
2 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 C 3 . Use D as centre, DE as radius to draw a circle C
4 . C 4 and C 3 intersect at WV.
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 C .
5 BE = 3, AE = 2 (given), DE = 6, EZ = 4, DZ = 6 + 4 = 10.
= HZ = HW = 5
HD Let the diameter of C be x, i.e. GE = x, DG = 6 – x.
5 HG = HD – DG = 5 – (6 – x) = x – 1
2
2
2
2 In DGW, WG = 6 – (6 – x) = 12x + x (1) (Pythagoras’ theorem)
2
2
2
2 In HGW, WG = 5 – (x – 1) = 24 + 2x – x (2) (Pythagoras’ theorem)
6
2
2
(1) = (2): 24 + 2x – x = 12x + x x = 2.4 The radius of C is 1.2 =
5
5
2 DG DZ = (6 – 2.4)10 = 36 = DE G is the point of inversion of Z w.r.t. C .
4 Clearly E is the point of inversion of E w.r.t. C 4 .
The inversion of C
2 w.r.t. C 4 is C 5 .
The inversion of C w.r.t. C is the tangent UEX (see Figure 10).
1
4 Let the circle, with centre at C and radius 1 be C .
6 Join DC. DC cuts C 6 at J. DC is produced to cut C 6 again I. Then IJ = diameter of C 6 = 2.
4 In ABC, let ABC = , cos = , CBD = 180 – (adj. s on st. line)
5 4 221
2
2
2
221 CD = In BCD, CD = 3 + 4 – 2 34 cos(180 – ) = 25 24 =
5
5
5 221 221 DJ = DC – CJ = – 1; DI = DC + CI = + 1
5
5 Invert C 6 w.r.t. C 4 to C 7 centre at N. Suppose DI intersects C 7 at K and L in the figure.
221 221
2
2 DI DK = 6 and DJ DL = 6 DK
1 36 and DL 1
36
5
5
36
5 221
36
5 221 DK = = 1 and DL = = 1
6
5
6
5
221 221
1
1
5
5
5 221 5 221
5 5 The radius of C
LK = DL – DK = 1 1 – = 7 is .
6
5
6
5
3
6
Now construct a smaller circle C
8 centre P, touches C 5 and C 7 externally and also touches XU.
is not shown in the figure.
P
30 501840 By the result of Lemma 1, the radius of C is ; ME = 2 and DP =
8
121 121 DP cuts C at Q, DP is produced further to cut C again at R.
8
8
501840 30 501840
30
- – = DP – PQ =
- DQ ; DR = DP + PR = 121 121 121 121
Now invert C
8 w.r.t C 4 to give C 9 . This circle will touch C 1 , C 2 and C
6 externally.
DR intersects C 9 at S, produce DR further to meet C 9 again at T. Then
501840 30 501840
30
2
2 DS DR = 6 and DT DQ = 6 DS
36 and DT
36 121 121 36 121 501840
30 36 121 501840
30 DS = = ; DT = = 115 115
501840 30 501840
30 501840 30 501840
30
12
- – ST = DT – DS = = = diameter of C
9
115 115
23
6 The radius of C 9 is .
23
2
2
3
3 G8 Given that a, b, x and y are non-zero integers, where ax + by = 4, ax +by = 22, ax +by = 46
4
4
5
5 and ax + by = 178. Find the value of ax + by .
2
2
3
3
4
4 ax + by = 4 (1), ax +by = 22 (2), ax +by = 46 (3), ax + by = 178 (4).
5
5 Let ax + by = m (5)
2
2
(x + y)(2): (x + y)(ax + by ) = 22(x + y)
3
3 ax + by + xy(ax + by) = 22(x + y)
Sub. (1) and (3): 46 + 4xy = 22(x + y) 23 + 2xy = 11(x + y) (6)
3
3
(x + y)(3): (x + y)(ax + by ) = 46(x + y)
4
4
2
2 ax + by + xy(ax + by ) = 46(x + y)
Sub. (2) and (4): 178 + 22xy = 46(x + y) 89 + 11xy = 23(x + y) (7) 11(7) – 23(6): 450 + 75xy = 0 xy = –6 (8) 11(6) – 2(7): 75(x + y) = 75 x + y = 1 (9)
4
4
(x + y)(4): (x + y)(ax + by ) = 178(x + y)
5
5
3
3 ax + by + xy(ax + by ) = 178(x + y)
Sub. (3) and (5): m + 46xy = 178(x + y) Sub. (8) and (9): m + 46 (–6) = 1781 m = 454
Given that, in the figure, ABC is an equilateral triangle
G9
with AF = 2, FG = 10, GC = 1 and DE = 5. Find the value of HI. (Reference: 2017 HG9)
AF + FG + GC = 2 + 10 + 1 = 13
AB = BC = CA = 13 (property of equilateral triangle) Let AD = x, then BE = 13 – 5 – x = 8 – x Let HI = y, BH = z, then IC = 13 – y – z By intersecting chords theorem,
AD AE = AFAG x (x + 5) = 2
12
2 x + 5x – 24 = 0
(x – 3)(x + 8) = 0
A
2 x = 3 or –8 (rejected)
F x BE = 8 – x = 5
D BH
BI = BEBD
z (z + y) = 5 10 = 50 (1)
5 CI CH = CGCF
10
(13 – y – z)(13 – z) = 11
E
169 – 13(y + 2z) + z(y + z) = 11 (2)
8 - x
Sub. (1) into (2): 169 – 13(y + 2z) + 50 = 11
G
1 y + 2z = 16 13-y-z C z y y = 16 – 2z (3) B H
I Sub. (3) into (1): z(z + 16 – 2z) = 50
2 z – 16z + 50 = 0 z =
8 14 or 8
14 From (3), 2z 16 z 8 8 14 is rejected z = 8 14 only HI = y = 16 – 2z = 16 – 2( 8 14 ) =
2
14
2 G10 Let a n and b n be the x-intercepts of the quadratic function y = n(n – 1)x – (2n – 1)x + 1,
where n is an integer greater than 1. Find the value of a +
2 b 2 + a 3 b
3 + a 2015 b 2015 .
Reference: 2005 HI5
The quadratic function can be written as y = (nx – 1)[(n – 1)x – 1]
1
1 The x-intercepts are and .
n n
1
1
1
1
a n b n = = for n > 1 n n
1 n 1 n
1
1
1
1
1
- a = 1
- a
2 b
2 3 b 3 + a 2015 b 2015
2
2 3 2014 2015 1 2014
= 1 = 2015 2015
Geometrical Construction 1. Construct an isosceles triangle which has the same base and height to the following triangle.
B E 3
4 1 4
2 A C D
Steps (1) Construct the perpendicular bisector of AC, D is the mid point of AC. (2) Copy ACB. (3) Draw CBE so that it is equal to ACB, then BE // AE (alt. s eq.)
BE and the bisector in step 1 intersect at E.
(4) Join AE , CE. Then AEC is the required isosceles triangle with AE = CE.
2. Given the following line segment MN represent a unit length, construct a line segment of
1 length .
5 A
4 G 4
6 7 5
1 H 2 F E 3 C 8 M N
Steps (1) Use N as centre, NM as radius to draw a circular arc. (2) Use M as centre, MN as radius to draw a circular arc, cutting the arc in step 1 at E.
MNE is an equilateral triangle. MNE = 60 (3) Use E as centre, MN as radius to draw a circular arc, cutting the arc in step 1 at F. FNE is an equilateral triangle. FNE = 60
(4) 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. (5) Join NG and produce it longer. NG intersects the arc in step 1 at H.
NG is the bisector of ENF. MNG = 60 + 30 = 90
(6) Use H as centre, HN as radius to draw a semi-circle, cutting NG produced at A. AN = 2 2 2 = 5
(7) Join AM , cutting the semicircle in step (6) at C. AM = 1
2 (8) Join NC . Then MC is the required length. Proof: ACN = 90 ( in semi-circle) It is easy to show that CMN ~ NMA (equiangular)
MN MC
(cor. sides, ~ ’s)
MN AM
1
=
MC
5
3. Construct a square whose area is equal to the difference between the areas of the following two squares ABCD and PQRS.
5 F 7
6 G 6 CE = PQ 2 4 3 E 7
1 4 C B R Q M P S A D
Steps (1) Draw the perpendicular bisector of BC , M is the mid-point of BC. (2) Use M as centre, MB as radius to draw a semi-circle outside the square ABCD. (3) Use C as centre, PQ as radius to draw an arc, cutting the semicircle in (2) at E. (4) Join and produce it longer. Join BE.