EWU IUMO 2009_Math Olympiad

8/9/2019 EWU IUMO 2009_Math Olympiad

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EWUIUMO2009 Question & Solve

Problem 1

Mrs. Mita is a lucky woman that she lives with all her daughters and granddaughters together in the

same house. The interesting thing is that each of her daughters has same number of sisters anddaughters. The age of Mrs. Mitas oldest daughter is 50 years and that of her youngest

granddaughter is only one year. Mrs. Mitas age is exactly the same as the total number of the

women members of the family but less than 75 years. How old is Mrs. Mita?Solution

Mrs.Mita

daughters1andsisters1havingeachdaughters x-x-x

tersgranddaugh)1( xx

Mrs. Mitas age = 751)1(12


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Problem 2

Given that baxxf +=)( and abxxg +=)( , where a and b are integers. If 8)1( =f and28))50(())50(( = fggf , what is the value of ba ?

Solution

81)1( =+=+= babaf

ba = 8 (1)abg +=50)50(

baabbabagf ++=++= 250)50())50(( (2)baf +=50)50(

ababababfg ++=++= 250)50())50(( (3)

From (2) and (3)

28)50(50))50(())50((2222 =+=++++= abbaababbaabfggf (4)

From (4) and (1)28561481664)8()8( 2222 =+=+++=+ bbbbbbbbbb (5)

From (5)

2

2814

562814

=

=

=

b

b

b

6288 === ba

426 ==ba (Ans)

Solve by: SANIUL ISLAM EEE EAST WEST UNIVERSITY


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Problem 3

Mitul, a 16 years old boy, was walking along a straight road through a dense forest. Suddenly, he

saw the following writing just below his feet:

The tallest tree of this forest is only 200 feet away from here along this road.

Mitul tried to see the tree from there and could just see the top of the tree after moving his eyes 50

up from the horizontal line of sight. Then he tried to see the other parts of the tree, but could not see

due to dense bushes. However, suddenly he could see the bottom of the tree through the bushes after

moving his eyes 1.7down from the horizontal line of sight. If Mitul was 6.2 feet tall, what was the

height of the tallest tree of the forest? [Calculate with four digit precision]

Solution

1h

50

020

H

2h7.1

3600.2381918.120050tan2001 ===h

9400.50297.02007.1tan2002 ===h

feetH 3000.2443600.2389400.5 =+= (Ans)



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Problem 4

In a math competition, eight participants are from Dhaka division, six from Khulna division, four

from Rajshahi division, and the remaining from Sylhet division. If the probability that the three toprankers are from Dhaka, Rajshahi, and Sylhet divisions each is 48/855, how many participants are

from Sylhet division?

Solution

Let x = number of participants from Sylhet division.Total number of participants = 8 + 6 + 4 + x = 18 + x

Number of choice of one top ranker from Dhaka division = 818 =C

Number of choice of one top ranker from Rajshahi division = 414 =C

Number of choice of one top ranker from Sylhet division = xCx =1

P[Three top rankers are from Dhaka, Rajshahi, and Sylhet division each]

855

483248

3

18

3

18==

=

++ C

x

C

xxx (1)

6

489686651

6

16355603064896

6)16)(35306(

6)16)(17)(18(

!3)!15()!18(

!3)!318()!18(

23322

2

318

+++=

+++++=

+++=+++=++=

++=+

xxxxxxxx

xxxxxxxx

xxCx

(2)

From (1) and (2)

855

48

489686651

63223

=+++

xxx

x

0)244853)(2(

0)2(2448)2(53)2(

04896244810653204896255451

0235008122592244848

23500841568244848164160

2

2

223

23

23

23

=+

=+

=++=++

=++

+++=

xxx

xxxxx

xxxxxxxx

xxx

xxxx

2

02

=

=

x

x

Or, 02448532 =+ xx

2

254.11253

2

24484)53(53 2 =

+=x

As x cannot be a fractional number or a negative number, 2=x is the answer.

Therefore, 2 participants are from Sylhet division.

Problem 5



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At East West University, students come from all six divisions of Bangladesh with equal chance.

What should be the minimum number of students enrolled at East West University to ensure that at

least 800 students are from a division?

Solution

Students come from all six divisions with equal chance.

Let us assume that exactly 800 1 = 799 students have come from each division making a totalstudent number 799 6 = 4794.

If another student is enrolled from any of the six divisions, total student from that division will be

799 + 1 = 800 satisfying the condition.

Therefore, at least 4794 + 1 = 4795 students must be enrolled at East West University to ensure that

at least 800 from a division.

[The same solution can be obtained using Generalized Pigeonhole Principle]



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Problem 6

Rita, a school going girl, is playing with her six feet long red ribbon and trying to make a right-

angle triangle so that the area enclosed within the triangle is maximum. What would be themaximum area of the right-angle triangle that Rita could make using her ribbon? [Calculate with

four digit precision]

Solution

x

y

Perimeter of the right-angle triangle = 6 feet

x

xy

xxy

xyyx

xyyx

xyyx

yxyxyxyx

yxyxyx

yxyx

yxyx

=

==

=+

=++++=+

+++=+

+=+

=+++

6

618

618)6(

6618

06618

02121236

2121236

)()(1236

)(6

6

2222

222

22

22

Area of the triangle,

x

xx

x

xxxyA

=

==

6

39

6

618

2

1

2

1 2

To make the area maximum

01812

054363

039693654

0)6(

)1)(39()69)(6(

06

39

2

2

22

2

2

2

=+

=+

=++

=

=

=

xx

xx

xxxxx

x

xxxx

x

xx

dx

d

dx

dA

2

4853.812

2

7214412 =

=x

2427.10=x or 7574.1=x

2427.10x , since the perimeter of the triangle is 6

Therefore, 7574.1=x



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7573.17574.16

7574.1618

6

618=

=

=x

xy

5441.17573.17574.15.02

1=== xyA sq. feet (ans)



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Problem 7

An ant starts walking from A and walks along a curved path up to B, then walks in a straight line up

to Cand then returns to A in another straight line. If we consider that the line AC lies on the x-axis

and A is on the origin, then the curved path AB can be expressed as a function23)( xxxf = . The

length of the line AC is 40 feet. If we draw a perpendicular line BD on the line AC, then the length

of the line DCis 30 feet. How much area in square feet does the ant enclose in its walk? [Calculateup to four digit precision]

A

B

DC

Solution

Coordinate ofA is (0, 0)

AD = ACDC= 40 30 = 10 Coordinate ofD is (10, 0)

Area of the region 6667.21663

10

4

10

34)(

3410

0

3410

0

23 ==

== xx

dxxxABD square feet

Length of the line 9001010)10(23 === fBD

Area of the right-angle triangle BCD = 0.5 DC BD = 0.5 30 900 = 13500 square feet

Total are of the region ABC= 2166.6667 + 13500 = 15666.6667 square feet (Ans)



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Problem 8

Prof. Khan is a microbiologist. He is experimenting with growth of a certain type of bacteria in a

given environment. He finds that the total number of bacteria after 100 t hours can beexpressed as

tt

eea

++= 2 ,

where initial number of bacteria is 4. The expression is an exponential expression and, as amicrobiologist, Prof. Khan dislikes complex mathematical expressions like such exponential

expressions. One of his friends, Prof. Rahman, a mathematician, suggests him to approximate the

exponential expression using an eighth-order polynomial expression. Do this to help Prof. Khan.

Solution

20160360124

!8!7!6!5!4!3!2!11

!8!7!6!5!4!3!2!112

2

8642

87654328765432

tttt

tttttttttttttttt

eea tt

++++=

++++++

++++++++++=

++=



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Problem 9

How many 6-digit numbers are there such that :

the digits of each number are all from the set { 1, 2, 3, 4, 5 }; andany digit that appears in the number appears at least twice?

( Example: 334343 is an admissible number, while 111355 is not.)

Solution:

Since each digit occurs at least twice, we have the following possibilities:

(i). Three digits occur twice each. We may choose three digits from { 1, 2, 3, 4, 5 }in3

5C =10

ways. If each occurs exactly twice, the number of such admissible 6-digit numbers is

.90010!2!2!2

!6=

(ii). Two digits occur exactly three times each. We can choose 2 digits in 1052=C ways.

Hence the number of admissible 6-digit numbers is .20010!3!3

!6=

(iii). One digit occurs four times and the other twice. We are choosing two digits again, which can

be done in 10 ways. The two digits are interchangeable. Hence the desired number of admissible 6-

digit numbers is .30010!2!4

!62 =

(iv). Finally all digits are the same. There are 5 such numbers. Thus the total number of admissible

6-digit numbers is 900 + 200 + 300 + 5 = 1405.



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Problem 10.

If a,b,c are three positive real numbers, prove that

.3111 222

++

+++

+++

ba

c

ac

b

cb

a

Solution:

We have the trivial inequalities ,212 aa + ,212 bb + and cc 212 + . Hence

.222111 222

ba

c

ac

b

cb

a

ba

c

ac

b

cb

a

++

++

+

++

+++

+++

Let b+c = x , c + a = y and a + b = z. Then

=+

++

++

=+

++

++ z

zyx

y

yxz

x

xzy

ba

c

ac

b

cb

a 222

.332223111 =++

++

++

+=

++

++

+

z

x

x

z

y

z

z

y

x

y

y

x

z

y

z

x

y

x

y

z

x

z

x

y

This is a consequence of MGMA .. . Hence the result follows.



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Problem 11.

Find the least possible value of a+b , where a, b are positive integers such that 11

divides a+13b and 13 divides a+11b.

Solution: Since 13 divides a+11b , we see that 13 divides a-2b and hence it also divides

6a 12b . This in turn implies that 13/ (6a+b) . Similarly( ) ( ) ( ).6/11126/112/11)13/(11 babababa ++++ Since gcd (11,13) =1we conclude that 143/(6a+b) . Thus we may write 6a+b=143k for some natural

number k . Hence6a+6b = 143k+5b =144k+6b-(k+b).

This shows that 6 divides k+b and hence .6+bk We therefore obtain

16865138)(51385143)(6 =+++=+=+ bkkbkba .It follows that 28+ba .

Taking a =23 and b =5 , we see that the conditions of the problem are satisfied .

Thus the minimum value of a +b is 28.



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Problem 12.

If x, y are integers and 17 divides both the expressionsyxyxyxandyxyxyx ++++ 2222 23752 , then prove that 17 divides

xy-12x +15y .

Solution: Observe that )12)((2322

+=++ yxyxyxyxyx . Thus17 divides either x-y or x-2y+1. Suppose that 17 divides x-y. In

this case yx (mod 17) and henceyyyyyyyxyxyx 2752752 22222 ++++ (mod 17).

Thus the given condition that 17 divides yxyxyx 752 22 ++ implies

that 17 also divides 2y and hence y itself . But then yx (mod 17)implies that 17 divides x also .Hence in this case 17 divides xy-12x+15y.

Suppose on the other hand that 17 divides x-2y+1 . Thus 12 yx (mod 17)and hence 65752 222 +++ yyyxyxyx (mod 17)

Thus 17 divides 12.652 + yxButyy (mod 17) also implies that

)65(21512 2 ++ yyyxxy (mod 17).

Since 17 divides 652 + yy , it follows that 17 divides xy-12x+15y.



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Problem 13

Let ABCD be a convex quadrilateral ; P, Q ,R, S be the midpoints of AB , BC ,CD,

DA respectively such that triangles AQR and CSP are equilateral . Prove that ABCDis a rhombus . Determine its angles .

Solution:We have QR = BD/2 =PS .Since AQR and CSP are both equilateral andQR=PS, they must be congruent triangles . This implies that AQ =QR = RA =CS =SP=PC Also

RQACEF == 060 .(See Fig.1)

Fig.1

Hence CS is parallel to QA . Now CS =QA implies that CSQA is a parallelogram . Inparticular SA is parallel to CQ and SA =CQ . This shows that AD is parallel to BC

and AD =BC .Hence ABCD is a parallelogram.

Let the diagonal AC and BD bisect each other at W. Then DW =DB/2=QR =CS =AR .

Thus in triangle ADC ,the medians AR,DW,CS are all equal. Thus ADC is equilateral.

This implies ABCD is a rhombus . Moreover the angles are 060 and 0120 .


D

R C

S Q

N

M

U

L

E

O

W

V F

A P B


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Problem 14

If is a real root of the equation 0235 =+ xxx ,prove that 36= (For any real number a,

we denote by [a] the greatest integer not exceeding a).

Solution: Suppose is a real root of the given equation .Then

0235

=+ (1)This gives 1135 =+ and hence ( ) 111 34 =++ .observe that

( )221 23234 +=+++ . If 01 ,then 02 + , giving ( ) 022 +

And hence ( ) ( ) 011 34 ++ . If 1 , then ( ) 01334 +=+ and hence 0134 ++ . Thisagain gives ( ) ( ) 011

34 ++ .

The above reasoning shows that for 0 ,we have 0135 + and hence cannot be equal to1. We conclude that a real root of 0235 =+ xxx is positive

(obviously 0 ).

Now using 0235 =+ , we get

2

246

+=

The statement 36= is equivalent to 43

6 .

Consider 4224 + .Since 0 , this is equivalent to 42 235 + .

Using the relation(1), we can write

025242222

++ o r.



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Treating this as a quadratic ,we get this is equivalent to 22

1 .Now observe that

if 2 then ( )( ) 25111 34 ++= which is impossible .If

2

10 ,

then ( ) ( ) 0111 34 ++= which again is impossible . We conclude

that 22

1 . Similarly 3224 + is equivalent to 032 235 +

which is equivalent to 0242 2 + .But this is ( ) 0122 which is valid .

Hence 436 and we get [ ] 36 = .



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Problem 15

Let A,B,C and D be four distinct points on a line, in that order. The circles with diameters AC and

BD intersect at the points X and Y. The line XY meets BC at the point Z. Let P be a point on the lineXY different from Z. The line CP intersects the circle with diameter AC at the points C and M, and

the line BP intersects the circle with diameter BD at the points B and N. Prove that the lines AM, DN

and XY are concurrent.

Solution:

Let the lines XY and DN meet at the point Q. Then triangles BPZ and QDZ are similar. Hence

ZP

ZB

ZD

ZQ= or .

.

ZP

ZBZDZQ = Now let the lines XY and AM meet at the point R. Then the triangles

CPZ and RAZ are similar. HenceZP

ZC

ZA

ZR= or

ZP

ZBZD

ZP

ZCZAZR

..== . Therefore, ZQ = ZR. That

is, Q and R are same points and so AM, DN and XY are concurrent.



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Problem 16

A father wishes to divide a square piece of land among his five sons. One son is his favorite, and he

wants to give him one quarter of the land, as shown. All others get equal share. He wants to do itusing no more than 8 straight cuts. How can he do it?

Answer



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Problem 18

An eight-digit number contains two 1's, two 2's, two 3's, and two 4's. The 1's are separated by one

digit, the 2's by two digits, the 3's by three digits, and the 4's by four digits. What is the number?

Answer: 41,312,432



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Problem 19

An egg salesman was asked how many eggs he had sold that day. He replied, "My first customer

said, 'I'll buy half your eggs and half an egg more.' My second and third said the same thing. When I

had filled all three orders I was sold out and I had not had to break a single egg all day."

Answer: Seven. He sold four eggs to the first customer, two to the second, and one to the third. Theproblem is solved most easily if you start with the last customer and work backwards.



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Problem 20

A chauffeur always arrives at the train station at exactly five o'clock to pick up his boss and drive

her home. One day his boss arrives an hour early, starts walking home, and is picked up by the

chauffeur on the way out to the train station. They arrive at home twenty minutes earlier than usual.How long did she walk before she met her chauffeur?

Answer: For fifty minutes. She saved the chauffeur ten minutes of traveling time each way and thus

was picked up at 4:50 pm rather than the usual time.



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Problem 21

A girl was eight years old on her first birthday. How could that be?

Answer: She was born on February 29, 1896. The year 1900 was not a leap year (only centuriesdivisible by 400 are leap years), so the next February 29 fell in 1904 when she was eight. She was

twelve on her second birthday.



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Problem 22

A staircase has 10 steps. One can climb the stair by 1-step jump, 2-step jumps or 3-step jumps. In

how many different ways can he climb up the staircase?

Answer: Sn=Sn-1+Sn-2+Sn-3

EWU IUMO 2009_Math Olympiad

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