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470 NON-ROUTINE MATHEMATICS OLYMPIAD PROBLEMS MATHEMATICSX

NON-ROUTINE MATHEMATICS OLYMPIAD PROBLEMS

1. What is the radius of a circle inscribed in a rhombus with diagonals of length 10 and 24?2. Figures 0, 1, 2, and 3 consist of 1, 5, 13 and 25 non-overlapping unit squares respectively. If the patternwere continued, how many non-overlapping unit squares would there be in figure 100?

Figure 0 Figure 1 Figure 2 Figure 33. In rectangle ABCD, C is trisected by CF and CE, where E is on AB, F is on AD, BE = 6, and AF = 2. Find

area of rectangle ABCD.

D C

BE

F

26

A

4. A circle of radius 2 has centre at (2, 0). A circle of radius 1 has centre at (5, 0). A line is tangent to the twocircles at points in the first quadrant. Find they-intercept of the line.

5. The mid-points of sides of a regular hexagon ABCDEF are joined to form a smaller hexagon. Whatfraction of the area of ABCDEF is enclosed by the smaller hexagon?

6. ABCD is a 2 2 square, E is the mid-point of AD, and F is on BE. If CF BE, find the area of thequadrilateral CDEF.

A E D

F

B C

7. How many two digit positive integers N have the property that the sum of N and the number obtained by

reversing the order of the digits of N is a perfect square?8. A circle with centre O is tangent to the co-ordinate axes and to the hypotenuse of the 30 60 90

triangle ABC as shown, where AB = 1. Find the radius of the circle.

C

A60

B1

O


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MATHEMATICSX NON-ROUTINE MATHEMATICS OLYMPIAD PROBLEMS 471

9. A rectangular sheet of sides aand b(a.> b) is folded along its diagonal (as shown). Find the area of theoverlapped region.

B C

DA a

b

D

CB

A

10. In ABC, AB = 13, BC = 14, and AC = 15. Let D denote the mid-point of BC and let E denote theintersection of BC with the bisector of BAC. Find area of ADE.

11. On a large piece of a paper, Anisha creates a rectangular spiral by drawing line segments of lengths (incm), of 1, 1, 2, 2, 3, 3, .... as shown. Anishas pen runs out of ink after the total of all the lengths she hasdrawn is 3000 cm. What is the length of longest line segment that she draws?

3

3

2

21

1

4

4

12. If [x] denotes the largest integer less than or equal tox, for eg. [2.67] = 2, [3] = 3, [3.19] = 4 etc., then

find the value of1 1 1 1 2 1 99

.....2 2 100 2 100 2 100

13. Let mbe an integer such that 1 1000.m Find the probability of selecting at random an integermsuchthat the quadratic equation 6x2 5mx+ m2= 0 has atleast one integer solution.

14. The circle shown is a unit circle centred at the origin. The segment BC is a diameter and C is the point(1, 0). The angle has measure 30. What is the x-coordinate of the point A?

A

C(1, 0)B O

15. How many positive integers less than one million have all digits equal and are divisible by 9?

16. Find area of a triangle whose sides are 5, 6 and 13 .

17. The graph of |x| + |y| = 4 encloses a region in the plane. What is the area of the region?18. A circle centred at A with a radius of 1 and a circle centred at B with a radius of 4 are externally tangent.

A third circle is tangent to the first two circles and to one of their common external tangents as shown.Find the radius of the third circle.

B

A


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19. Four positive integers a, b, cand dhave a product of 8! (8! = 1 2 3 4 5 6 7 8) and satisfy

ab+ a+ b= 524

bc+ b+ c= 146, andcd+ c+ d= 104

Find the value of a d.

20. The sum of 18 consecutive positive integers is a perfect square. Find the smallest possible value of thissum.

21. Build a sequence starting with the first term a1 = 1, the second term a

2 = 2 and the nth term

1

2

1; 3.n

n

n

aa n

a

Find the sum of the sequence to first 1000 terms.

22. Square EFGH is inside square ABCD so that each side of EFGH can be extended to pass through a vertex

of ABCD. Square ABCD has side length 50, and BE = 1. What is the area of the inner square EFGH?

A B

C

E

G

H

D

F

23. The area of trapezium ABCD is 164 cm2. The altitude is 8 cm, AB = 10 cm, CD = 17 cm. Find BC.

24. The area of XYZ is 8. Points A and B are mid-points of congruent segments XY and XZ. Altitude XCbisects YZ. Find area of shaded region.

X

Y Z

A B

C25. For how many positive integers ndoes 1 + 2 + .......+ nevenly divide 6n?

26. How many three-digit numbers satisfy the property that the middle digit is the average of the first andthe last digit?

27. The digits 1, 2, 3, 4, 5, 6, 7 and 9 are used to form four two-digit prime numbers, with each digit usedexactly once. What is the sum of these four primes?

28. The first term of a number sequence is 2005. Each succeeding term is the sum of the cubes of the digitsof the previous term. What is the 2005 thterm of the sequence?

29. The positive integers A, B, A B, A + B are all prime numbers. Find the sum of these four primes.

30. How many positive integers n satisfy the following condition :

(130 n)50> n100> 2200?

31. A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribedin this square. What is the ratio of the area of the smaller circle to the area of the larger square?

32. How many 0s occur at the end of the decimal expansion of 100100 100!? (here, n! = 1 2 3 .... n)


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33. Let4 4 4

1 1 1....

1 2 3A denote the sum of the reciprocals of the fourth powers of all positive integers,

and4 4 4

1 1 1....

1 3 5B a similar sum for all odd positive integers. Express

A

Bas a fraction.

34. Parallel chords in a circle have length 12 and 16, and the distance between them is 7. Another chord ismidway between them. What is its length?

35. How many ordered pairs (x, y) of integers satisfy 1 1 1 ?2x y

(Note that both positive and negative

integers are allowed).36. What is the smallest value of nfor which the product (22 1) (32 1) (42 1) ..... (n2 1) is a perfect square?37. In the given diagram, a circular arc centred at one vertex of a square of side length 4 passes through two

other vertices. A small circle is tangent to the large circle and to two sides of the square. What is theradius of the small circle?

38. ABC is an isosceles with AB = AC and BC = 30 cm. Square EFGH, which has a side length of 12 cm, isinscribed in ABC, as shown. Find area of the AEF.

A

B C

E F

H G

39. How many times does the numeral 2 appear in a book having page number 1 to 250?

40. Points P, Q, R and S divide the sides of a rectangle ABCD in the ratio 1 : 2 as shown. What fraction of thearea of the rectangle is the area of the parallelogram PQRS?

D C

BA

R

S

P

Q

41. Identical isosceles right triangles are removed from opposite corners of a square resulting in a rectangle.If the sum of the cut-off pieces is 450, find the length of the diagonal of the rectangle.

42. Find all the integers n, such that3 5

1

n

n

is also an integer..


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43. Find the sum of all the digits in base 10 of the number (102007+ 1)2.

44. In aABC, the mid-point of the side BC is D and the centroid is G. Find the ratio of the area of BGD

to the area of ABC.45. A circle is inscribed in an equilateral triangle and a square is inscribed in the circle. Find the ratio of the

area of the triangle to the area of the square.

46. Find the sum of all values ofxthat satisfy the equation22 4 60( 5 5) 1.x xx x

47. A wheel of radius 8 rolls along the diameter of a semicircle of radius 25 until it bumps into this semicircle.What is the length of the portion of the diameter that cannot be touched by the wheel?

48. The measures of length of the sides of a triangle are integers and that of its area is also an integer. Oneside is 21 and the perimeter is 48. Find the measure of the shortest side.

49. In the given figure, A, B, C, D, E, F, G, H, I are all squares of which A, B have respectively area 1 and 81cm2. Find the area of the square I.

DI

EC

HF

GB

A

50. For the system of equations 2 2 2 2 4 525x x y x y and 2 35,x xy xy find the sum of the real y

values that satisfy the given equations.

ANSWERS

1.60

132.20201 3. 108 3 36 4. 2 2 5.

3

4

6.11

57.8 8.2.37 9.

2 2( )4

ba b

a 10.3

11. 54 cm 12.50 13.0.667 14.1

215.10

16. 9 17.32 18.4

919.10 20.225

21. 1800 22.36 23.10 cm 24.3 25.5

26. 45 27. 190 28.250 29.17 30.125

31. : 8 32.24 33.16

1534. 249 35.5

36. 8 37.12 8 2 38.48 cm2 39.106 40.5

9

41. 30 42.n= 9 or 3 43.4 44.1

645. 3 3 : 2

46. 3 47.20 48.10 49.324 cm2 50.5

2


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HINTS/SOLUTIONS

1. Let E be the intersection of the diagonals of rhombus ABCD. Because diagonals of rhombus are perpen-dicular and bisect each other, AEB is a right triangle with sides 5, 12 and 13 (using Pythagoras theorem)

and its area is1

5 12 30.2

Therefore the altitude EF drawn to side AB is60

13(using

1ABEF 30)

2 .

So,60

13is the radius of the inscribed circle centred at E.

D C

A

E

B

12 5

13F

2. Calculating the number of squares in the first few figures uncovers a pattern. Figure 0 has 2(0) + 1 = 2(02)+ 1 squares, figure 1 has 2 (1) + 3 = 2(12) + 3 squares, figure 2 has 2 (1 + 3) + 5 = 2(22) + 5 squares, and figure3 has 2(1 + 3 + 5) + 7 = 2(32) + 7 squares. In general, the number of unit squares in figure nis

2 [1 + 3 + .... + (2n1)] + 2n+ 1 = 2 (n2) + 2n+ 1

Therefore, the figure 100 has 2(1002) + 2(100) + 1 =20201 squares.

3. In the 30 60 90 triangle CBE,BC BC

tan 60 3 BC 6 3BE 6

. So, FD = AD AF = 6 3

2. In the 30 60 90 triangle CFD, CD = FD 3 = 18 2 3 . So, area of rectangle ABCD = (BC) (CD)

= 6 3 (18 2 3 ) = 108 3 36 151.

D C

BAE

F

3030

30

60

60

62

4. Let D and F denote the centres of the circles. Let C and B be the points where the x-axis andy-axisintersect the tangent line, respectively. Let E and G denote the points of tangency as shown. We knowthat AD = DE = 2, DF = 3, and FG = 1. Let FC = uand AB =y. Clearly, FGC ~ DEC, so,

3 3.1 2u u u Hence, GC 8 . (using Pythagoras theorem in FGC).

B

CFDA u1

2y G

E

Also, BAC ~ FGC BA AC 8

81 FG GC 8

y 2 2


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5. Let M and N denote the mid-points of AB and AF, respectively. Then MN AM 3. Since AMO is a

30 60 90 triangle and MN = MO. It follows that the hexagons are similar, with similarity ratio1

3.2

MN 3

AB 2

. Thus the desired quotient is

21

32

3.

4

MA

N

C

DE

F

B

O

6. In right BAE, 2 2BE 2 1 5. Since CFB ~ BAE, it follows that ar ( CFB)

22CB 2 1

. ar ( BAE) . (2.1)BE 25

=

4.

5Then, area (CDEF) = ar (ABCD) ar (BAE) ar (CFB)

44 1

5

11

5Ans.

7. Let N = 10x+y. Then 10x+y+ 10y+x= 11 (x+y) must be a perfect square. Since 1 x+y 18, itfollows thatx+y= 11. There are eight such numbers : 29, 38, 47, 56, 65, 74, 83 and 92.

8. Let D and E denote the points of tangency on they-axis andx-axis, respectively, and let BC be tangentto the circle at F. Tangents to a circle from a point are equal, so BE = BF and CD = CF. Letx= BF andy=CF. Becausex+y= BC = 2, the radius of the circle is

(1 ) ( 3 ) 3 3

2 2

x y 2.37Ans.

D O

C

AEB1 x

x

y

y F

60

9. Let ABCD be the given rectangle with sides aand b(a>b), which is folded along the diagonal AC. here,

2 2AC a b . Clearly OBA ODC

OA = OC OAC is an isosceles triangle.


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Let OE AC. Clearly OE will also the median.Again OEA ~ ABC

2 2

2 21

OE EA OE .2 OEAB BC 2

a bb a b

b a a

B D

O

A CE

b

Required Area = Area of OAC

2 22 21 1 .ACOE

2 2 2

b a ba b

a

= ( ) Ans.2 2b

a + b4a

10. By herons formula, the area of ABC is (21)(8)(7)(6) 84 , so the altitude from vertex A is

2(84)12.

14

The mid-point D divides BC into two segments of length 7, and the bisectors of BAC divides BC into

segments of length13

14 6.528

and

1514 7.5

28

(since the angle bisector divides the opposite

sides into lengths proportional to the remaining two sides). Thus the ADE has base DE = 7 6.5 = 0.5

and altitude 12, so its area is 1 12 0.52

3. Ans.

A

B CDE

14

13 15

11. As, 1 + 2 + 3 + .... + n=( 1)

2

n n

2(1 + 2 + 3 + .... + n) = n(n+ 1)

In the given problem, we want nsuch that

2 (1 + 2 + 3 + .... + n) 3000

2( 1) 3000 3000n n n [ n+ 1 n].

Now, 3000 54.7

Let n= 54, we find (54) (55) = 2970 < 3000, which is correct estimation.

(If we try n= 55, then (55) (56) = 3080 > 3000)

Longest length completed was 54 cm.


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12.1 1 1 1 2 1 49

1, 1, 1,...., 12 2 100 2 100 2 100

but after this1 50 1 51 1 51 1 99

1, 1 1,...., 1.2 100 2 100 2 100 2 100

i.e. 1 occurs 50 times.

So, required sum

50 times

0 0 .... 0 1 1 .... 1 = 50 Ans.

13. The given quadratic equation has two solutions and2 3

m mx x (using quadratic formula).

Forx to be an integer mshould be divisible by 2 and 3.There are 500 multiples of 2, 333 multiples of 3, and 166 multiples of 6 between 1 and 1000. Therefore, the

required probability is

500 333 166

0.667.1000p

14. Using degree measure theorem, we observe that = 2.= 2 30 = 60. Then, x= cos 601

.2

Otherwise, suppose that the co-ordinates of A are (x, y). Then1

(1 ) tan3

xy x

and 2 2 1.x y

Substituting y from the first equation into the second, we have

2

2 1 1.3

xx

Solving, we get

2x2+x 1 = 0.

(2x 1) (x+ 1) = 0. The only positive root is1

.2

So, x-coordinate of A is1

.2

A( )x, y

C(1 0),BO D

y

1

x

15. Suppose the number a has ndigits each of which is d. Then ais divisible by 9 if and only if the sumd+ d+ ....+ d= ndof its digit is divisible by 9. Since 1 n6, ( as number is less than one million) oneconcludes that d = 3, 6, 9 are the only possibilities. With d= 3, we get the two numbers 333 and 333 333divisible by 9. with d= 6, we get 2 numbers 666 and 666 666 divisible by 9. With d= 9, we get six numbers9, 99, 999, ...., 999999 divisible by 9, so altogether there are 2 + 2 + 6 = 10 solutions.

16. Build such a triangle from triangles we know more about. Start with 3, 4, 5 triangle. Note that2 213 2 3 so we might try to build a triangle in the problem from 2; 3; 13 and 3; 4; 5. We can

append them along the edge of length 3. Notice that since both triangles are right, we can append them

so that the union is a triangular region with sides of length 4+ 2, 5 and 13 , whose area is1

6 32

9

Ans.5

42

313


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17. We know; 0

| |; 0

x xx

x x

So, the equation |x| + |y| = 4 give rise to four linear equations viz.x+y= 4,xy= 4, x+y= 4 and xy= 4. These four represents the straight lines in a plane. Thus, the enclosed region is a square with

vertices (4, 0), (0, 4), (4, 0) and (0, 4), which has area 2(4 2) = 32.

(0,4)

(4,0)

(0, 4)

(4, 0)

18. Let C be the intersection of the horizontal line through A and the vertical line through B. In right ABC,BC = 3 and AB = 5, so AC = 4. Letxbe radius of third circle, and D be the centre. Let E and F be the pointsof intersection of the horizontal line through D with the vertical lines through B and A, respectively, asshown.

A

B

C

D EG

F

1 4

In BED, we have BD = 4 +xand BE = 4 x, so DE2= (4 +x)2 (4 x)2= 16x

and DE 4 x .In ADF, we have AD = 1 +xand AF = 1 x, so

FD2= (1 +x)2 (1 x)2= 4x

and FD = 2 .x

Hence 4 = AC = FD + DE = 2 x + 4 x = 6 x

2

3x x

4Ans.

9

19. Note that : (a+ 1) (b+ 1) = ab+a+ b+ 1 = 524 + 1 = 525 = 3.52.7

and, (b+ 1) (c+ 1) = bc+ b+c+ 1 = 146 + 1 = 147 = 3.72

Since, (a+ 1) (b+ 1) is a multiple of 25 and (b+ 1) (c+ 1) is not a multiple of 5, it follows that (a+ 1) mustbe a multiple of 25. Since (a+ 1) divides 525, ais one of the 24, 74, 174, or 524. Among these only 24 is adivisor of 8!, so, a= 24. This implies that b+ 1 = 21 and b= 20. From this it follows that c+ 1 = 7 andc= 6. Finally, (c+ 1) (d+ 1) = 105 = 3.5.7. So d+ 1 = 15 and d= 14. Therefore, ad= 24 14 = 10 Ans.

20. Let n, n+ 1, ...., n+ 17 be the 18 consecutive integers. Then the sum is n+ (n+ 1) + ...... + (n+ 17)

17 1818 (1 2 ... 17) 18 9(2 17).

2n n n

Since, 9 is a perfect square, 2n+ 17 must also be a perfect square. The smallest value of nfor which thisoccurs is n= 4, so 9 (2n+ 17) = 9 25 =225 Ans.


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21. Here a1= 1, a

2= 2.

Now,32

3 41 2

11 2 1 3 1

3 ; 21 2

aa

a aa a

54

5 6

3 4

11 2 1 1 11 ; 1

3 2

aaa a

a a

6 77 8

5 6

1 11 1 2 12 ; 3

1 1

a aa a

a a

and so on.

Note that the sequence has groups of repeated terms.

The sequence is 1, 2,3, 2,1 , 1,2,3,2,1, 1, 2,3, 2,1 , ...

Sum of first 1000 terms 1000 (1 2 3 2 1)5

= 200 9 = 1800 Ans.

22. The symmetry of the figure implies that ABH, BCE, CDF and DAG are congruent right triangles.

So, 2 2BH CE BC BE 50 1 49 7

and EH = BH BE = 7 1 = 6.

Hence, the square EFGH has area = 62= 36 Ans.

23. In right AEB and DFC, 2 2AE 10 8 36 6 cm, and 2 2FD 17 8 225 15 cm.

So, the area of21

AEB 6 8 24 cm , and 2

area of21CFD 15 8 60 cm .

2

B C

10 178 8

FEA D

Now, area of rectangle BCFE = 164 cm2 (24 cm2+ 60 cm2)

= 164 cm2 84 cm2= 80 cm2

= BC BE = BC 8

80BC8

10 cm Ans.

24. Segments AD and BE are drawn perpendicular to YZ. Segments AB, AC and BC divide XYZ into fourcongruent triangles. Vertical line segments AD, XC and BE divide each of these in half. Three of the eight

small triangles are shaded, or 38

of XYZ. The shaded area is 3 (8) 3.8

Ans.

X

BA

Y ZC ED


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25. Because 1 + 2 + .... + n=( 1)

2

n n

1 + 2 + .... + ndivides the positive integer 6nif and only if6 12

( 1) / 2 1

n

n n n

is an integer..

There are five such positive values of n, namely, 1, 2, 3, 5 and 11.

26. The first and last digits must be both odd or both even for their average to be an integer. There are5 5 = 25 odd-odd combinations for the first and last digits. There are 4 5 = 20 even-even combinationsthat do not use zero as the first digit. Hence the total is 25 + 20 = 45.

27. The digits 2, 4, 5 and 6 cannot be the units digit of any two-digit prime, so these four digits must be thetens digit, and 1, 3, 7 and 9 are the unit digits. The sum is thus

10 (2 + 4 + 5 + 6) + (1 + 3 + 7 + 9) = 190

here, one set that satisfies the conditions is {23, 47, 59, 61}.28. The sequence begins 2005, 133, 55, 250, 133, ..... .

Thus, after the initial term 2005, the sequence repeats the cycle 133, 55, 250. Because 2005 = 1 + 3 668,thus 2005th term is the same as the last term of the repeating cycle, 250.

29. The numbers A B and A+ B are both odd or both even. However, they are also both prime, so they mustboth be odd. Therefore one of A and B is odd and the other even. Because A is a prime between A B andA + B, A must be the odd prime. Therefore, B = 2, the only even prime. So, A2, A, and A + 2 areconsecutive odd primes and thus must be 3, 5 and 7. The sum of the four primes 2, 3, 5 and 7 is the primenumber 17.

30. The condition is equivalent to 130 n> n2> 24= 16,

So, 130 n> n2

and n2

> 16130 > n> 4. So ncan be any of the 125 integersstrictly between 130 and 4.

31. Let the radius of the smaller circle be r. Then the side length of the smaller square is 2 r. The radius of the

larger circle is half the length of the diagonal of the smaller square, so it is 2r. Hence the larger square

has side of length 2 2r. The ratio of the area of the smaller circle to the area of the larger square is

therefore2

2 8(2 2 )

r

r

: 8.

2 2 r

2r

r

2r

32. As 100100has many more 0s at the end, we need here the number of 0s at the end of 100!. This will be thenumber of factors of 5 in its decomposition into primes, since there will be plenty of 2s to turn these in10s. There are 20 multiples of 5 in 100! and four of them have a second factor of 5 (i.e. the multiples of 25).So, total number of zeros will be 20 + 4 =24.


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33. Observe that,4 4 4

1 1 1....

2 4 6B A

4 4 4 41 1 1 1

.......2 1 2 3

A

15

16 16

AA A

A

B

16Ans.

15

34. Let rdenote the radius of the circle, and d1, and d

2the distances from the centre of the circle to the chords

of length 12 and 16 respectively. Each of the perpendiculars from the center to a chord makes a right

triangle, with other leg equal to half the length of the chord. Thus, 2 2 21 236 64 .d r d Assume the

chords are an opposite sides of the centre. (If we put them on the same side, we will get a negative valuefor one of the ds). Since d

1+ d

2= 7, we obtain 7 (d

1 d

2) = 64 36 = 28, hence d

1 d

2= 4. From this we

obtain,1

11

2d and 2

3,

2d and then 2

265.

4r The distance from midway chord to the center is

1 2 2.2

d d Ifxdenotes the length of the desired chord, then

2

2 22 ,2

xr

and so x 249 .

6

rr

8

Od1

d2

2

1.53.5

5.5

midway chord

35. For , 0x y , the equation reduces to 2 (x+y) =xyor, (x 2) (y 2) = 4.

Now, factors of 4 are 1, 2, 4. So, possible solutions are as follows :

(i) x 2 = 1 ; y 2 = 4 x= 3,y= 6

(ii) x 2 = 1;y 2 = 4 x= 1, y = 2(iii)x 2 = 2 ;y 2 = 2 x= 4,y= 4

(iv)x 2 = 2 ;y 2 = 2 x= 0,y= 0, which is not possible.

(v)x2 = 4;y 2 = 1 x= 6,y= 3

(vi)x 2 = 4;y 2 = 1 x = 2,y= 1

Thus the total number of solutions is 5.


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36. The given product is

(22 1) (32 1) (42 1) (52 1) ...... (n2 1)

= (2 1) (2 + 1) (3 1) (3 + 1) (4 1) (4 + 1) (5 1) (5 + 1) ...... (n 1) (n+ 1)

= (1 3) (2 4) (3 5) (4 6)...(( 1) ( + 1))n n

multiplying using the rule above, the product reduces to

2 32 42 52 ..... (n 1)2 n(n+ 1)

= 2 n(n+ 1) [3 4 5 .... (n 1)]2

= 2n(n+ 1) a perfect square number.

For this whole product to be a square number, we must have 2n(n+ 1) as a square number. Using trial and

error method we find that the value of nis 8.[2n(n+ 1) at n= 8 is 2 8 9 = (4 3)2= 122].

37. Let rdenotes the desired radius. The little 45 45 90 triangle in the top right corner with hypotenusegoing from the centre of the small circle to the top right vertex of the square and one leg along the top edge

of the square implies 4 2 4 2.r r (This is two ways of viewing the length of that hypotenuse).

Thus,4( 2 1) 2 1

( 2 1) 4( 2 1)2 1 2 1

r r

24( 2 1)4(2 1 2 2)

2 1

r

4= 4 (3 2 2 ) = 12 8 2 .

38. Let M be the mid-point of EF and N be the mid-point of HG. By symmetry, N is also the mid-point of BC.Also, the line through A and M will also pass through N, and will be perpendicular to both EF and BC.Since the side length of the square is 12, then EM = HN = 6 and EH = 12. Since, it is given that BC = 30,then BN = 15 and so BH = 9.

A

E M F6

12

B CGNH69

Since EFGH is a square, that EF is parallel to HG, and soAEM = EBH, i.e. AME ~ EHB.

So,AM 12

AM 86 9

Thus, the area of1

AEF 12 82

248cm Ans.


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39. The numeral 2 appear :

Between Pages 1 to 9 1 time

10 to 19 1 time

20 to 29 11 times

30 to 99 7 times

100 to 119 2 times

120 to 129 11 times

130 to 199 7 times

200 to 209 11 times

210 to 219 11 times

220 to 229 21 times

230 to 249 22 times

and on page 250 1 time

So, Total number of times= 106.

(Take care of 22, 122, 202, 212, 222, 232, 242)

40. Area of ||gm PQRS = Area of ABCD (area of s DRS, RCQ, PBQ, APS)

2 1 1 2 2 1 1 2

3 3 3 3 3 3 3 32 2 2 2

l b l b l b l b

l b

2 5

4 .9 2 9

lblb lb

Area of PQRS 59

(area of ABCD)

required fraction 5=9

.

41. From the figure, we find 2, 2a x b y (45 45 90)

Sum of the areas of the four triangles2 2 2 21 1 1 1

2 2 2 2a a b b

2 2 2 21 (2 2 )2

a b a b

2 2 2 2( 2) ( 2) 2( ) 450x y x y (given)

a b

b

a

b

a

ab

45

45

2x

2y

Now, length of diagonal 2 2 2 2(2 ) (2 ) 4( )x y x y

2 22 2 225x y

2 15 30 Ans.


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42. Let 23 5 3 5

. Then1 1

n nk k

n n

2 2 2( 1) 3 5 3 5k n n k n k n

2

2 2

2 2

5 8( 3) 5 1

3 3

kk n k n

k k

.

If nis an integer then 3 k2must divide 8 and hence 3 k2belongs to {1, 2, 4, 8}. It follows that3 k2= 1 or 3 k2= 2 and consequently n= 9 or 3.

43. (102007+ 1)2= (102007)2+ 2 102007+ 1

= 104014+ 2 102007+ 1

Thus, the decimal expansion of (102007+ 1)2contains 2 ones, 1 two and all other digits are zero. Sum of thedigits = 1 + 1 + 2 = 4 Ans.

44. ABD and ABC have the same altitude from A. Since1

BD BC,

2

we have, ar (ABD)

1( ABC).

2ar

Now, BDG and BDA have the same altitude from B.

B CD

G

A

Since1

GD AD3

we have,

1( BGD) ( ABD)

3ar ar

Hence,1

( BGD) ( ABC)6

ar ar

45. The incentre, circumcentre, orthocentre and centroid of an equilateral triangle coincide. If ais the side of

ABC and AD BC, then,22 3AD = . .

4 2aa a Now,, 1 1 3OD AD .

3 3 2 2 3aa . The diago-

nal of the square inscribed in the circle is the diameter of the incircle. So, the diagonal of the square is

2;

2 3 3

a a side of the square is .

3. 2 6

a a

Area of ABC 21 3 3

. . .2 2 4

a a a

Area of square is2

.6

a B CD

O

A

a

2

a

Thus, the desired ratio is

2

23 :4 6

aa 3 3 : 2 Ans.

46. We consider the solution in three cases :

Case I. It is possible for the base to be 1.

therefore, x2 5x+ 5 = 1

2 5 4 0 ( 1)( 4) 0x x x x

1or 4.x x

Both these values are acceptable forx.


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Case II. It is possible that the exponent be 0.

therefore,x2+ 4x 60 = 0

(x+ 10) (x 6) = 0 x= 10 orx= 6.(Note that it is easy to verify that neitherx= 10 norx= 6 is a zero ofx2 5x+ 5 = 0 , so thatthe form 0 does not occur).

Case III. It is possible that the base is 1 and the exponent is even.

therefore, 2 5 5 1x x butx2+ 4x 60 must also be even.

2 5 6 0x x

( 2)( 3) 0 2 or 3.x x x x

Ifx= 2; then 2 4 60x x is even, sox= 2 is a solution.

Ifx= 3, then 2 4 60x x is odd, sox= 3 is nota solution.

Therefore, the sum of all the solutions = 1 + 4 10 + 6 + 2 = 3 Ans.

47. Draw OBC, where O is the centre of the large circle, B is the centre of the wheel, and C is the point oftangency of the wheel and the diameter of the semi-circle. Since BC is a radius of the wheel, OCB = 90and OCB is right angled at C.

D

A

B

O C

Extend OB to meet the semi-circle at D. Then BD = BC = 8, since they are both radii of the wheel, and OB= 25 8 = 17. Now, In OBC, OC2= 172 82= 225 OC = 15.Then AC = 25 15 = 10. The length of the portion of the diameter that cannot be touched by the wheelis a length equivalent to 2AC or 20.

48. Let a, bbe the other two sides. a+ b= 27 and s, the semi-perimeter of the triangle is 24.

Area 24(24 )( 3)(24 21)a a and hence 72 (a 3) (24 a) must be a perfect square. If ais the

shortest side, then a 13. By trial and error method we find a= 10. If a= 10, Area 24 14 3 7 84 ,is an integer.

49. The square A has side 1 cm and B has side 9 cm (since their areas are 1 and 81 respectively). Since theside, on the right, of B, has its top most 1 cm as a side of A the square G has a side of 9 1 = 8 cm. Similarly,F has a side of 7 cm. Since Cs side at the bottom includes Bs as well as As, side of C = 9 + 1 = 10 cm.

Now XY = 7 1 = 6 cm ZY = 10 6 = 4 cm

DI

EC

H

F

GB

AX

Z

Y


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Hence the square D has a side of 10 + 4 (= side of C + side of E) = 14 cm. The side of square I (the righthand side of I) is the sum of the same sides of D as well as E which is 14 + 4 = 18 cm.

Area of square I = 18 18 cm2= 324 cm2Ans.50. Consider the system of equations

2 2 2 2 4 525x x y x y ...(1)

and 2 35x xy xy ...(2)

The expression on the LHS of eqn. (1) can be rewritten as 2 2 2 2 4 2 2 2 2( )x x y x y x xy x y

2 2( )( )x xy xy x xy xy

Thus, (x+xy2xy) (x+xy2+xy) = 525

Substituting from (2) gives.

(x+xy2

xy) (35) = 525or, x+xy2xy= 15 ...(3)

Now, subtracting (3) from (2),10

2 20xy xy

Substituting forxin (3) gives10

10 10 15yy

10y2 25y+ 10 = 0 2y2 5y+ 2 = 0 (2y 1) (y 2) = 0

1 or 22

y y

The sum of the realyvalues1

22

5

Ans.2

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