Regional Mathematical Olympiad 2010

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Regional Mathematical Olympiad (First Round) Chandigarh, 2010

1. How many even numbers between 4000 and 7000 are there, having distinct digits?

GENERAL SOL:

Since the numbers are between 4000

and 7000

So, the first digit can either 4,5 or 6

The number should also be even.

So last digit can be 0, 2, 4, 6 or 8.

CASE 1:

If the last digit is 0, 2 or 8

The last digit can be filled in one way.

First digit can be filled in 3 ways (4, 5 or 6).

The remaining two digits can be filled in

8 or 7 ways.

Total numbers ending with 0, 2 or 8

= (3×8×7)×3 = 504

CASE 2:

If the last digit is 4 or 6.

PIONEER’S SOL:

First digit can be either 4, 5 or 6.

♦If Ist digit is 4.Then last digit can be 0, 2, 6, 8.

Ways = 7×8×4

♦If Ist digit is 5. Then last digit can be 0, 2, 4, 6, 8.

Ways = 8×7×5

♦ If first digit is 6. Then last digit can be 0, 2, 4, 8.

Ways = 8×7×4

Total ways = 8×7(5+4+4) = 8×7×13 = 728.

Total numbers = (2×8×7)×2

= 112 × 2 = 224

Total even numbers with distinct digits

= 224+ 504 =728.



2. If Tn = 1+2+3+4 …….+ n, for n = 1, 2, 3 ……. Determine the value of

3 20102

2 3 2010

T TT.......................

T 1 T 1 T 1

− − −

Sol:

( )

( )

( )

( )

( ) ( )

( )( )n

2n

n n 1n n 1 n n 1 n n 1T 2

n n 1T 1 n n 1 2 n 1 n 2n 2n n 21

2

+

+ + += = = =

+− + − − −− − −−

… (i)

2 4

2 4

3 5

3 5

T T2.3 4.5

T 1 1.4 T 1 3.6

T T3.4 5.6.................

T 1 2.5 T 1 4.7

∴ = =− −

= =− −

n

n

T 2.3 3.4 4.5 5.6 2010 2011.........T 1 1.4 2.4 3.6 4.7 2009 2012

× = × × × − ×

( )( )

( )( )

( )( ) ( )

( ) ( )

2.3.4.5.6.......2010 3.4.5.6.......2011

1.2.3.4.5.6.......2009 4.5.6.7.......2011

2010 1.2.3.......2009 .3 4.5.6.7.......2011

1.2.3........2009 .2012 4.5.6.7.......2011

3 2010 3015

2012 1006

=

=

×= =

3. Let a, b, c be positive integers such that b a 3

c b 3

+

+is a rational number. Show that

2 2 2a b c

a b c

+ +

+ +is an integer.

Sol:

( )( )

( )

( )

22

22

2 2 2 2

b a 3 c b 3b a 3 c b 3

x c b 3 c b 3 c b 3

bc 3ab 3 ac bbc 3b 3ac 3ab

c 3b c 3b

+ −+ −= × =

+ − −

− + −− + −= =

− −

For x to be rational



2

2

ac b 0

b ac

a,b,c are in G.P.

− =

=

∴

Since a, b, c are positive integers.

Let b = ar, c = ar2

Where, r must also be integer.

( )( )

( )( )( )

2 2 2 2 2 2 2 4

2

2 3

2 2 4 2

32

2 3 3 2

2 3 3

23 32

3

a b c a a r a r

a b c a ar ar

(r ) 1a.1.

a 1 r r r 1

r 1a 1 r r1.

r 1

(r ) 1 r 1 (r ) 1 r 1a. a.

(r 1)(r 1)r 1 r 1 r 1

a r 1 r 1 rr 1 r 1a. a a r r

(r 1) r 1r 1

+ + + +∴ =

+ + + +

−

+ + − = =

−+ +

−

− − − −= × = ×

+ −− − −

+ + −− += × = = + −

+ +−

Since a is integer and r is integer ∴ it is an integer.

4. The right angled triangle having sides 3, 7 , 4 has the strange property that two sides are of

integer lengths and the third has length square root of the sum of these integers.

(a) Find another such triangle.

(b) Prove that there are infinitely many such triangles, and show how to construct these sides.

Sol:

Let two sides be x and y such that y > x

By Pythagoras theorem:



( )( )

( )( ) ( )

( )( )

22 2

2 2

2 2

x x y y

x x y y

x y x y 0

x y x y x y 0

x y x y 1 0

x y 0 or x y 1 0

x y or x y 1

x y(Not possible) or y x 1

+ + =

+ + =

− + + =

+ − + + =

+ − + =

+ = − + =

= − = −

= − = +

⇒ y = x+1

Thus, there are infinite such triangles.

(a) Another such triangle is 5, 6, 11

5. Let ABCDE be a regular pentagon. Assume that there is a circle which is tangent to CD at D and

is tangent to AB at A. Let F be the midpoint of the minor arc AD. Prove that AFDE is a rhombus.

Sol:

Since F is the midpoint of minor arc AD

∴FE will passes through center of the circle. i.e. O will lie on FE.

Since ABCDE is a regular polygon

∴ AB = BC = CD = DE = EA and ∠EAB = ∠ABC = ∠BCD = ∠CDE = ∠DEA = 108°

Now, Since tangent is perpendicular to the radius

∴ OA is the radius and ∠1+∠2 = 90° ….. (i)

∠EAB = 108°

∠EAB + ∠1 = 180° [Linear pair]

⇒ ∠1 = 72°

From (i), ∠2 = 18°

Similarly ∠13 = 18°

Now in ∆EAO and ∆EDO

AE = ED, ∠2 = ∠13

and AO = OD [Radius]

∴ ∆EAO ≅ ∆EDO [SAS]

∴ ∠5 = ∠6 = 54°



Since AE = ED

∴ ∠EAM = ∠EDM [angles opposite to equal sides are equal]

∴ ∠2 + ∠3 = ∠13 +∠14

∴ ∠3 = ∠14 = x

In ∆EAD, 108°+18°+x+18°+x = 180°

⇒ x = 18°

∴ ∠3 = ∠14 = 18°

Now in ∆EAM, 54°+36°+∠11 = 180°

⇒ ∠11 = 90°

⇒ EM ⊥ AD ……. (ii)

In ∆OAM, 18° + 90°+∠7 = 180°

⇒ ∠7 = 72°


In ∆EAM and ∆EDM, EA = ED

∠EMA = ∠EMD = 90°

EM = EM [Common]

∴ ∆EAM ≅ ∆EDM

∴ AM = MD [CPCT] ….. (iii)

∠4 =1

2∠7 [Angle at center is twice the angle in the remaining part of the circle]

⇒ ∠4 = 36°


Now in ∆AMF, 36°+90°+∠10 = 180°

∴ ∠10 = 54°

Similarly ∠9=54°

Now in ∆AFM and ∆AEM

∠4 = ∠EAM = 36°

∠11 = ∠AMF = 90°

AM = AM [common]

∴ ∆AFM ≅ ∆AEM



⇒ FM = ME [CPCT] ….. (iv)

From (ii), (iii) and (iv), AFDE is a rhombus (Diagonals of rhombus bisect each other at right angles)

6. A square is inscribed in an equilateral triangle. Find the ratio of the area of the square to the

area of the triangle.

Sol:

Let the side of the square be a

In ∆CEF:

0 EFtan60

FC

a3

FC

aFC ....(i)

3

=

=

=

Since ∆DBG ∼ ∆EFG

aBG .....(2)

3∴ =

From (1) and (2)

2aBC a .....(3)

3= +

Area of Square = a2

Area of Equilateral ∆ 23(side)

4=



( )

22 2

2 22 33 2a 3 2

a a 1 a4 43 3 4 3

+ = + = + =

Now,

( )

2

2

2

Area of square a 4 3

Area of equialteral Δ 7 4 3

2 3 a4 3

= =++

7. Prove that there are only two real numbers such that

(x – 1) (x – 2) (x – 3) (x – 4) (x – 5) (x –6) = 720

Sol:

(x – 1) (x – 2) (x – 3) (x – 4) (x – 5) (x –6) = 720 = 6!

Since L.H.S. is the product of six consecutive numbers and RHS is the product of first six natural

numbers.

Thus, LHS has to be product of first 6 natural numbers. For this to be possible x can have only 2

values i.e. x = 0, x = 7

Explanation: (three cases arises)

x > 0

x – 1 = 6

x = 7

x < 0

x – 1 = 1

x = 2 (Not possible)

(Make the expression zero)

x = 0

8. Five points on a circle are numbered 1, 2, 3, 4, 5 in clockwise order. A frog jumps in the

clockwise order from one number to another as follows:

(a) If it is at odd number it moves one place.

(b) If it is at even number it moves two places.

If the frog is initially at 5, where it will be after 2010 jumps?



Sol:

The path is 5 → 1 → 2 → 4 → 1→ 2 →5 →………

As 2010 is multiple of 3, so the frog must be on 4th position.

Regional Mathematical Olympiad 2010

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