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470 NON-ROUTINE MATHEMATICS OLYMPIAD PROBLEMS MATHEMATICS–X

NON-ROUTINE MATHEMATICS OLYMPIAD PROBLEMS1. 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 pattern

were 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

F2

6A4. 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 two

circles at points in the first quadrant. Find the y-intercept of the line.5. The mid-points of sides of a regular hexagon ABCDEF are joined to form a smaller hexagon. What

fraction 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 the

quadrilateral 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 byreversing 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

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MATHEMATICS–X NON-ROUTINE MATHEMATICS OLYMPIAD PROBLEMS 471

9. A rectangular sheet of sides a and 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. Anisha’s 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 to x, for eg. [2.67] = 2, [3] = 3, [–3.19] = – 4 etc., then

find the value of 1 1 1 1 2 1 99.....2 2 100 2 100 2 100

13. Let m be an integer such that 1 1000.m Find the probability of selecting at random an integer m suchthat 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.

BA

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19. Four positive integers a, b, c and d have a product of 8! (8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8) and satisfyab + a + b = 524bc + b + c = 146, andcd + c + d = 104Find 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 a2 = 2 and the nth term

1

2

1; 3.n

nn

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 vertexof 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 XC

bisects YZ. Find area of shaded region.X

Y Z

A B

C25. For how many positive integers n does 1 + 2 + .......+ n evenly 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 2005th term 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 0’s occur at the end of the decimal expansion of 100100 – 100!? (here, n! = 1 × 2 × 3 × ....× n)

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33. Let 4 4 4

1 1 1 ....1 2 3

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

and 4 4 4

1 1 1 ....1 3 5

B a similar sum for all odd positive integers. Express AB

as 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 n for 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 G39. 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 the

area 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 that 3 5

1nn

is also an integer..AMIT B

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43. Find the sum of all the digits in base 10 of the number (102007 + 1)2.44. In a ABC, 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 of x that satisfy the equation 22 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.

D I

EC

HF

GBA

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 yvalues that satisfy the given equations.

ANSWERS

1.6013 2. 20201 3. 108 3 36 4. 2 2 5.

34

6.115 7. 8 8. 2.37 9. 2 2( )

4b a ba

10. 3

11. 54 cm 12. 50 13. 0.667 14. 12

15. 10

16. 9 17. 32 18. 49 19. 10 20. 225

21. 1800 22. 36 23. 10 cm 24. 3 25. 526. 45 27. 190 28. 250 29. 17 30. 125

31. : 8 32. 24 33. 1615 34. 249 35. 5

36. 8 37. 12 8 2 38. 48 cm2 39. 106 40. 59

41. 30 42. n = – 9 or 3 43. 4 44. 16 45. 3 3 : 2

46. 3 47. 20 48. 10 49. 324 cm2 50. 52

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HINTS/SOLUTIONS1. 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 is 1 5 12 30.2 Therefore the altitude EF drawn to side AB is

6013 (using

1 AB×EF 30)2 .

So, 6013

is the radius of the inscribed circle centred at E.D C

A

E

B

12 5

13 F2. 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 n is

2 [1 + 3 + .... + (2n –1)] + 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 BCtan 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

BA E

F

30°30°

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 and y-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 = u and AB = y. Clearly, FGC ~ DEC, so,

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

B

CFDA u12

y GE

Also, BAC ~ FGC BA AC 8 81 FG GC 8y 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 ratio 1 3.2

MN 3AB 2

. Thus the desired quotient is

21 32

3 .4

MA

N

C

DE

F

B

O

6. In r ight BAE, 2 2BE 2 1 5. Since CFB ~ BAE, it follows that ar (CFB)22CB 2 1. ar ( BAE) . (2.1)

BE 25

=

4 .5 Then, area (CDEF) = ar (ABCD) – ar (BAE) – ar (CFB)

44 15

115

Ans.

7. Let N = 10 x + y. Then 10 x + y + 10 y + x = 11 (x + y) must be a perfect square. Since 1 x + y 18, itfollows that x + 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 the y-axis and x-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. Let x = BF and y =CF. Because x + y = BC = 2, the radius of the circle is

(1 ) ( 3 ) 3 32 2

x y 2.37 Ans.

D OC

A EB1 x

x

y

y F

60°

9. Let ABCD be the given rectangle with sides a and 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 b b a bb a a

B DO

A CE

b

Required Area = Area of OAC

2 22 21 1 .AC×OE

2 2 2b a ba b

a

= ( ) Ans.2 2b a + b4a

10. By heron’s 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 length 1314 6.528

and 1514 7.528

(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 CD E14

13 15

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

2n n

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

In the given problem, we want n such 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.AMIT B

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12. 1 1 1 1 2 1 491, 1, 1,...., 12 2 100 2 100 2 100

but after this 1 50 1 51 1 51 1 991, 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 3m mx x (using quadratic formula).

For x to be an integer m should 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 60° 1 .2

Otherwise, suppose that the co-ordinates of A are (x, y). Then 1(1 ) tan3xy x

and 2 2 1.x y

Substituting y from the first equation into the second, we have 2

2 1 1.3xx

Solving, we get

2x2 + x – 1 = 0.

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

So, x-coordinate of A is 1 .2

A( )x, y

C(1 0), B O D

y

1x

15. Suppose the number a has n digits each of which is d. Then a is divisible by 9 if and only if the sumd + d + ....+ d = nd of its digit is divisible by 9. Since 1 n 6, ( 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 is 1 6 32 9

Ans. 5

42

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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, x – y = 4, – x + y = 4 and – x –y = 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. Let x be 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.

ABC

D EG

F

14

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

and DE 4 x .

In ADF, we have AD = 1 + x and AF = 1 – x, soFD2 = (1 + x)2 – (1 – x)2 = 4x

and FD = 2 .x

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

23

x x 4 Ans.9

19. Note that : (a + 1) (b + 1) = ab + a + b + 1 = 524 + 1 = 525 = 3.52.7and, (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, a is 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, a – d = 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).2

n n n

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

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21. Here a1 = 1, a2 = 2.

Now, 323 4

1 2

11 2 1 3 13 ; 21 2

aaa a

a a

54

5 63 4

11 2 1 1 11 ; 13 2

aaa a

a a

6 77 8

5 6

1 11 1 2 12 ; 31 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 of 21AEB 6 8 24 cm ,and2

area of 21CFD 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)

2n n

1 + 2 + .... + n divides the positive integer 6n if and only if 6 12

( 1) / 2 1n

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 are

5 × 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, A–2, 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 > 16

130 > n > 4. So n can be any of the 125 integers strictly 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 thelarger 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

therefore 2

2 8(2 2 )r

r

π : 8.

2 2 r

2r

r

2r

32. As 100100 has many more 0’s at the end, we need here the number of 0’s at the end of 100!. This will be thenumber of factors of 5 in its decomposition into primes, since there will be plenty of 2’s to turn these in10’s. 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 6

B A

4 4 4 4

1 1 1 1 .......2 1 2 3

A

15

16 16AA A

AB

16 Ans.15

34. Let r denote the radius of the circle, and d1, and d2 the distances from the centre of the circle to the chordsof length 12 and 16 respectively. Each of the perpendiculars from the center to a chord makes a righttriangle, with other leg equal to half the length of the chord. Thus, 2 2 2

1 236 64 .d r d Assume thechords are an opposite sides of the centre. (If we put them on the same side, we will get a negative valuefor one of the d’s). Since d1+ d2 = 7, we obtain 7 (d1 – d2) = 64 – 36 = 28, hence d1 – d2 = 4. From this we

obtain, 1

112

d and 23 ,2

d and then 2 265 .4

r The distance from midway chord to the center is

1 2 2.2

d d If x denotes the length of the desired chord, then

22 22 ,

2x r

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) = xy or, (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) x –2 = 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 anderror method we find that the value of n is 8. [2n(n + 1) at n = 8 is 2 × 8 × 9 = (4 × 3)2 = 122].

37. Let r denotes 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 edgeof 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 CGNH 69Since EFGH is a square, that EF is parallel to HG, and so AEM = EBH, i.e. AME ~ EHB.

So, AM 12 AM 86 9

Thus, the area of 1AEF 12 82

248 cm Ans.AMIT B

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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 23 3 3 3 3 3 3 3

2 2 2 2

l b l b l b l bl b

2 54 .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 triangles 2 2 2 21 1 1 12 2 2 2

a 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 bb

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.

AMIT BAJA

J


42. Let 23 5 3 5. Then1 1

n nk kn n

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

2

2 22 2

5 8( 3) 5 13 3kk n k n

k k

.

If n is an integer then 3 – k2 must divide 8 and hence 3 – k2 belongs 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)2 contains 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. Since 1BD BC,2

we have, ar (ABD)

1 ( ABC).2

ar

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

B CD

G

A

Since 1GD AD3

we have,

1( BGD) ( ABD)3

ar ar

Hence, 1( BGD) ( ABC)6

ar ar

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

ABC and AD BC, then, 2

2 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 6a a

Area of ABC 21 3 3. . .2 2 4

a a a

Area of square is 2

.6a B CD

O

A

a

2a

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 for x.AMIT B

AJAJ


Case II. It is possible that the exponent be 0.therefore, x2 + 4x – 60 = 0 (x + 10) (x – 6) = 0 x = – 10 or x = 6.(Note that it is easy to verify that neither x = –10 nor x = 6 is a zero of x2 – 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 but x2 + 4x – 60 must also be even.

2 5 6 0x x

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

If x = 2; then 2 4 60x x is even, so x = 2 is a solution.

If x = 3, then 2 4 60x x is odd, so x = 3 is not a 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 = 90°and OCB is right angled at C.

D

A

B

O CExtend 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, b be 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 a is 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 C’s side at the bottom includes B’s as well as A’s, side of C = 9 + 1 = 10 cm.Now XY = 7 – 1 = 6 cm ZY = 10 – 6 = 4 cm

DI

EC

HF

GBAX

Z

YAMIT BAJA

J


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 cm2 Ans.

50. Consider the system of equations2 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 + xy2 – xy) (x + xy2 + xy) = 525Substituting from (2) gives.

(x + xy2 – xy) (35) = 525or, x + xy2 – xy = 15 ...(3)

Now, subtracting (3) from (2), 102 20xy xy

Substituting for x in (3) gives 10 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 real y values 122

5 Ans.2

AMIT BAJA

J

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