MATH SCHOOL

Olympiad Problems based on the Material for Second Semester of Second Grade Student

of Junior High School

By:

Group 4

Class 5A

Members:

Ni Luh Okassandiari (1313011026)

Ni Komang Ratna Srinadi (1313011034)

I Gusti Ayu Sekarini (1313011077)

JURUSAN PENDIDIKAN MATEMATIKA

FAKULTAS MATEMATIKA DAN ILMU PENGETAHUAN ALAM

UNIVERSITAS PENDIDIKAN GANESHA

SINGARAJA

2015

Ni Luh Okassandiari (1313011026)

1. Let ABCD be a unit square. Draw a quadrant of a circle with A as centre and B, D as

end points of the arc. Similarly, draw a quadrant of a circle with B as centre and A, C

as end points of the arc. Inscribe a circle touching the arcs AC and BD both

externally and also touching the side CD. Find the radius of the circle .

Solution:

Let O be the centre of . By symetry O is on the

perpendicular bisector of CD. Draw OL CD and

OK BC. Then OK = CL = CD/2 =1/2. If r is the

radius of , we see that BK = 1 – r, and OE = r.

Using Phytagoras’ theorem

2

22

2

111

rr

Simplification gives r = 1/16.

(The Canadian Regional Mathematics Olympiad (CRMO)-2012, Paper 3, Numb. 1)

Modification:

1. Let ABCD be a unit square. Draw a quadrant of a circle with A as centre and B, D as

end points of the arc. Similarly, draw a quadrant of a circle with B as centre and A, C

as end points of the arc. Inscribe a circle centred on the perpendicular of arc AC and

BD, and also touching the side CD. Find the area of the circle .

Solution:

To simplify the problem, we can use the

coordinate system.

Assume that A on point (0,0) then B will be

on (1,0). Thus, the equation of circle A is

122 yx and the equation of circle B is

11 22 yx .

Let O (x1,y1) be the centre of . By symetry

O is on the perpendicular bisector of CD, then

the coordinate of O must be (CD/2, y1) =

(1/2,y1). Since O is the perpendicular of circle

L

E

(0,0) (1,0)

(1,0)

A and B then O (1/2,y1) is fullfill both the equations of circle A and B.

2

3

43

14

1

1

1

2

1

2

1

2

1

2

yy

y

yx

Since we assume that the unit square is on the first quadrant then we use 2

31 y .

Then, the radius of is 2

31 OEr .

The area of

3

4

7

2

31

2

2 r .

2. The figure bellow is a rectangle which is formed by some squares made of matches.

As an example, figure 1 x 5 needs 16 matches, figure 2 x 5 needs 27 matches, as

follows.

Figure 1 x 5

Figure 2 x 5

The ammount of matches which are needed to form a rectangle of figure 51 x 5 are...

Solution:

Figure 1 x 5 needs 16 matches

Figure 2 x 5 needs 27 matches

By the same method we’ll get:

Figure 3 x 5 needs 38 matches

Figure 4 x 5 needs 49 matches

...

...

...

If we see accurately, the ammount of the matches forms an arithmatic sequence: 16,

27, 38, 49, ..., Un with the difference is 11.

To know the ammount of matches which are needed on figure 51 x 5, we have to find

the 51st term of the sequence, as follow:

bnaUn 1 111511651 U

56655016

Thus, the ammount of matches which are needed to form a rectangle of figure 51 x 5

are 566 matches.

(OSN Matematika SMP Tingkat Kota 2014 (bagian B), Numb. 3)

Modification:

2. To make a miniature of a 1 floor hotel with 1 floor consist of 5 rooms, an architect

needs 44 matches.

Thus, to make a miniature of 51 floors hotel with 1 floor consist of 5 rooms, the

architect needs... matches.

Solution:

................ 44 matches

................ 72 matches

................ 100 matches

...

...

...

If we see accurately, the ammount of the matches forms an arithmatic sequence: 44,

72, 100, ..., Un with the difference is 28.

To know the ammount of matches which are needed to make a miniature of 51 floors

hotel with 1 floor consist of 5 rooms, we have to find the 51st term of the sequence, as

follow:

bnaUn 1 281514451 U

1444140044

Thus, the architect needs 1444 matches.

3. DEB is a chord of a circle with DE = 3 and EB = 5. O is the centre of the circle.

Connect O and E so that the extension of OE intersect the circle on C. It’s given that

EC = 1. Determine the radius of the circle.

Solution:

Supposed the radius of the circle is r.

We can extend OC so that OC intersect the circle on F.

Then, CF is the diameter of the circle. CBFD is a

rectangular chord with E becomes the intersection of the

diagonals, then:

CE . EF = DE . EB

CE. (2r – CE) = DE . EB

1 . (2r – 1) = 3. 5

r = 8

(Canadian Mathematical Olympiad 1971)

Modification:

3. DEB is a chord of a circle with DE = 3 and EB = 5. O is the centre of the circle. Connect

O and E so that the extension of OE intersect the circle on C. It’s given that EC = 1.

Determine the area of OED .

Solution:

Supposed the radius of the circle is r.

We can extend OC so that OC intersect the circle on F.

Then, CF is the diameter of the circle. CBFD is a

rectangular chord with E becomes the intersection of the

diagonals, then:

CE . EF = DE . EB

CE. (2r – CE) = DE . EB

1 . (2r – 1) = 3. 5

r = 8

On OED , 92

837

2

ODEDOEs

Thus the area of OED by Herron’s formula:

361088939799 L

4. Positive integers are written on all the faces of a cube, one on each. At each corner

(vertex) of the cube, the product of the numbers on the faces that meet at the corner is

written. The sum of the numbers written at all the corners is 2004. If T denotes the sum

of the numbers on all the faces, find all the possible values of T.

Solution:

Let ABCDPQRS be a cube, and the numbers a, b, c, d, e, f be written on the faces

ABCD, BQRC, PQRS, APSD, ABQP, CRSD respectively. Then the products written at

the corners A, B, C, D, P, Q, R, S are respectively ade, abe, abf, adf, cde, bce, bcf, cdf.

The sum of these 8 numbers is:

adcdbcabfecdfbcfbcecdeadfabfabeade

dbcafe

This is given to be equal to 167.3.22004 2 . Observe that none of the factors ca ,

db , fe is equal to 1. Thus, dbcafe is equal to 4.3.167, 2.6.167,

2.3.334, or 2.2.501. Hence the possible values of T = a + b + c + d + e + f are

4+3+167 = 174, 2 + 6 + 167 = 175, 2 + 3 + 334 = 339, or 2 + 2 + 501 = 505.

Thus there are 4 possible values of T and they are 174, 175, 339, 505.

(Regional Mathematics Olympiad (RMO) 2004, Numb. 2)

S

D

R

Q P

C

A B

Modification:

4. Positive integers are written on all the faces of a cube, one on each. Each number on each

face has their pair on another face, so that the pair consist of two consecutive positive

integer. At each corner (vertex) of the cube, the product of the numbers on the faces that

meet at the corner is written. The sum of the numbers written at all the corners is 2015. If

T denotes the smallest numbers written on the faces, find the values of T.

Solution:

Let ABCDPQRS be a cube, and the numbers a, b, c, d, e, f be written on the faces

ABCD, BQRC, PQRS, APSD, ABQP, CRSD respectively. Then the products written at

the corners A, B, C, D, P, Q, R, S are respectively ade, abe, abf, adf, cde, bce, bcf, cdf.

The sum of these 8 numbers is:

adcdbcabfecdfbcfbcecdeadfabfabeade

dbcafe

This is given to be equal to 31.13.52015 . Observe that none of the factors ca ,

db , fe is equal to 1. Thus, dbcafe is equal to 5.13.31.

Since each number on each face has their pair on another face and each pair consist of

two consecutive positive integer, we can write

1615763231.13.5 dbcafe , thus the pairs are (e,f), (a,c),

and (b,d), or we can write as (2,3), (6,7), (15,16).

Thus, the the smallest numbers written on the faces is 2.

5. Two dices with the faces are coloured red or blue. The first dice consists of 5 red faces

and 1 blue face. When both of the dices throwed, the probability of the apperaring of two

faces with the same colour is ½. How many faces with red colour does the second dice

have?

S

D

R

Q P

C

A B

Solution:

Supposed the number of red face on the second dice is x then, the blue face must be 6 – x

The probability of the event that on both of the dices appear the same colour is:

2

1

6

6.

6

1

6.

6

5

xx

1865 xx

x = 3.

(Canadian Open Mathematics Challenge 1997)

Modification:

5. Three dices with the faces are coloured red, yellow, or green. The first dice consists of 1

red face, 2 green faces, and 3 yellow faces. The second dice consists of 2 red faces. The

third dice consist of 3 red face and the number of face with yellow colour is the

multiplicative of the number of the face with green colour. Each dice must consist all the

three colours. When both of the dices throwed, the probability of the apperaring of two

faces with the same colour is 3/36 How many faces with yellow colour does the second

dice have?

Solution:

1st dice 2

nd dice 3

rd dice

red 1 2 3

green 2 x Since each dice

must consist all

the three colours

then it must be 1

or 2.

It can’t be 2

because it’ll make

the yellow one

equal to 2 or 6.

Then, it must be 1

yellow 3 6 – (2+x) = 4-x Since the green

one is 1 then it

must be 2.

The probability of the event that on both of the dices appear the same colour is:

36

3

6

2.

6

4.

6

3

6

1.

6.

6

2

6

3.

6

2.

6

1

xx

18430 x

x = 3.

Ni Komang Ratna Srinadi (1313011034)

1. Original

Given three circles A, B, and C that intersect each other and passing through the

other centre point, with radius AB=CB=AB= 6 cm. Find the total shaded region!

Solution:

The length each side of the triangles are 6 cm.

So, the sum of the area of triangles is

327

364

13 2

L

The area of the segment is:

34530

364

16

360

605

ABC of area thesector a of area the5.

22

ABCTemL

The area of a circle is:

36622 rL

The total shaded region is

3186

3453032736

...

TemLLLAS

Source: Buku Kumpulan Soal Olimpiade Matematika SMP Tingkat

Kabupaten.2007. (Page:2. Number of problem:13)

Modification

Given three circles A, B, and C that intersect each other and any circle passing

through the others centre point, with radius AB=CB=AB= 8 cm. Find the total

center area (BAC)!

Solution:

The length each side of the triangles are 8 cm.

So, the sum of the area of triangles is

348

384

13 2

L

The area of the segment is:

3163

32

384

18

360

60

of area thesector a of area the.

22

ABCABCTemL

The total shaded region is

2 32

32348348

3163

323348

.3.

cm

TemLLAS

2. Original

Given a semi circle with radius AB as follows:

If the length of AD is 2cm shorter than BD, and the length of DC = 8 cm, then the

length of BD is…

Solution:

Assume that: BD = x, then AD = x-2

68482 222

22

xxxAC

DCADAC

A B

C

D

A B

C

D

8

x x-2

A B

C

D

651

2

2602

2

)64(1.4)2()2(

2

4

0642

0684462

684462

684222

)2(

684

684

2

2,1

2,1

2

2,1

2

2,1

2

22

22

2

2

2

x

x

x

a

acbbx

xx

xxxx

xxxx

xxxx

xx

xx

xx

x

AB

AC

AC

AD

The possible value of x is 651

Source: Buku Kumpulan Soal Olimpiade Matematika SMP Tingkat

Kabupaten.2007. (Page:77. Number of problem:9)

Modification

Given a semicircle of diameter AB as follows.

If the length of AD is 2cm shorter than BD, and the length of DC = 8 cm, then

determine the area of the smaller and the bigger segment areas!

A B

C

D

8

x x-2

Solution:

Assume that: BD = x, then AD = x-2

68482 222

22

xxxAC

DCADAC

651

2

2602

2

)64(1.4)2()2(

2

4

0642

0684462

684462

684222

)2(

684

684

2

2,1

2,1

2

2,1

2

2,1

2

22

22

2

2

2

x

x

x

a

acbbx

xx

xxxx

xxxx

xxxx

xx

xx

xx

x

AB

AC

AC

AD

The possible value of x is 651

The length of the radius is:

65

1651

1

2

22

2

2

2

x

xxxDBAD

OBAO

The area of the semi circle:

2

65

2

65.

2

SCL

4654

2

6518

2

8.

4654

4165

2

8)2(.

xCBDL

xACDL

The total area of the segments is:

26582

65

465446542

65

)...

cm

CBDLACDLSCLsegmentsL

3. Original

By using digits 2, 3, 4, 5, 6, 7, and 8 we will make some even hundreds numbers more

than 500. If no digits repeated in the numbers, then the number of numbers that can be

made is…

Solution :

The possible number as the hundreds (R) are 5, 6, 7, and 8, because it must > 500.

The possible number as the tens number (P) are any kind of them.

The possible number as the unit numbers (S) are 2, 4, 6, dan 8 because it must an even

number.

By filling slot method, we can find as follows:

a) For the hundreds 5 or 7, the possible unit numbers is 2, 4, 6 or 8.

2 5 4

We get the number of them are 2 x 5 x 4 = 40

b) For the hundreds 6 or 8, the possible unit numbers is 2 or 4.

2 5 3

We get the number of them are 2 x 5 x 3 = 30.

So, the number of the numbers that can be made are 40 + 30 = 70.

Source: Kurniawan.2007.OlimpiadeMatematika SMP. Jakarta : Erlangga. (Page:56. Number

of problem:22)

Modification

By usingdigits 0, 1, 2, 3, 4,..., 9 we will make odd hundreds numbers more than 500.

If no digits are repeated in the numbers, then the number of numbers we can produce

is…

Solution :

The possible number as the hundreds (R) are 5, 6, 7, 8 and 9 because it must > 500.

The possible number as the tens number (P) are any kind of natural numbers.

The possible number as the unit numbers (S) are 1,3,5,7, dan 9 because it must be odd

natural numbers.

By filling slot method we can find that:

a) For the hundreds 6 or 8, the possible unit numbers is 1,3,5,7 or 9

2 8 5

We get the number of them is 2 x 8 x 5 = 80

b) For the hundreds is the odd number (5, 7 or 9) the possible unit numbers is the

different odd number of the first digit.

3 8 4

We get the number of them is 3 x 8 x 4 = 96.

So, the number of the hundreds numbers that can be produce is 80 + 96 = 176

I Gusti Ayu Sekarini (1313011077)

Original

1. In the diagram, PQRS is a square with side length 2. Each of P, Q, R, and S is the

centre of a circle with radius 1. What is the area of the shaded region?

Source : Number 18 of Pascal Contest Grade 9- 2013, of University of Waterloo

(http://cemc.uwaterloo.ca/contests/past_contests.html)

Solution :

The area of the shaded region is equal to the area of square PQRS minus the

combined areas of the four unshaded regions inside the square.

Since square PQRS has side length 2, its area is 22 = 4.

Since PQRS is a square, then the angle at each of P, Q, R, and S is 90o .

Since each of P, Q, R, and S is the centre of a circle with radius 1, then each of the

four unshaded regions inside the square is a quarter circle of radius 1. (A central angle

of 90o gives a quarter of a circle.)

Thus, the combined areas of the four unshaded regions inside the square equals four

quarters of a circle of radius 1, or the area of a whole circle of radius 1. This area

equals . Therefore, the shaded region equals 4 – π.

Modifikasi

See the picture below. Calculate the area of ADEBC.

A B

E

= = 4 cm

D C

Solution :

A B

E

= = = = 4 cm

D F C

Lpersegi ABCD = S × S = 4 × 4 = 16 cm2

Lsegitiga DCE = ½ × DC × EF

EF = √

=√

= √

Lsegitiga DCE = ½ × DC × EF

= ½ × 4 × √

= 2 √ cm2

Ldaerah AEB = Lpersegi ABCD – ( Lsegitiga DCE + 2 Luas 1/12 lingkaran)

= 16 – ( 2 √ + 2 ( 1/12 × π × r2 ))

= 16 – ( 2 √ + 2 ( 1/12 × 3,14 × 42 ))

= 16 – ( 2 √ + 2 ( 1/12 × 3,14 × 16 ))

= 16 – ( 2 √ + 2 ( 4,2 ))

= 16 – ( 2 √ + 8,4 )

= 16 – 11,23

Ldaerah AEB = 4,77 cm2

Ltembereng = Ljuring - Lsegitiga DCE

= ( 1/6 × π × r2 ) - 2 √

= ( 1/6 × 3,14 × 16 ) - 2 √

= 8,38 - 2 √

= 1,45 cm2

the area of the shaded area = Lpersegi ABCD – (Lsegitiga DCE + Ltembereng + Ldaerah AEB )

= 16 – ( 2 √ + 1,45 + 4,77)

= 16 – ( 13,15 )

= 2,85 cm2

So, the area of ADEBC is 2,85 cm2 .

Original

2. Mike has two containers. One container is a rectangular prism with width 2 cm, length 4

cm, and height 10 cm. The other is a right cylinder with radius 1 cm and height 10 cm.

Both

containers sit on a at surface. Water has been poured into the two containers so that the

height of the water in both containers is the same. If the combined volume of the water in

the two

containers is 80 cm3, then the height of the water in each container is closest to

Source : Number 23 of Pascal Contest Grade 9- 2012, of University of Waterloo

(http://cemc.uwaterloo.ca/contests/past_contests.html)

Solution :

Suppose that the height of the water in each container is h cm. Since the first

container is a rectangular prism with a base that is 2 cm by 4 cm, then the volume of the

water that it contains, in cm3, is 2 x 4 x h = 8h.

Since the second container is a right cylinder with a radius of 1 cm, then the volume of the

water that it contains, in cm3, is π x 1

2 x h = πh.

Since the combined volume of the water is 80 cm3, then 8h + πh = 80.

Thus,

Of the given answers, this is closest to 7:2 (that is, the height of the water is closest to 7.2

cm).

Modifikasi

Mike has two containers. One container is a rectangular prism with width 3 cm, length 5

cm, and height 10 cm. The other is a right cylinder with radius 1 cm and height 10 cm.

Both

containers sit on a at surface. Water has been poured into the two containers so that the

height of the water in both containers is the same. If the combined volume of the water in

the two

containers is 50 cm3, then the height of the water in each container is closest to

Solution :

Suppose that the height of the water in each container is h cm. Since the first container is a

rectangular prism with a base that is 3 cm by 5 cm, then the volume of the water that it

contains, in cm3, is 3 x 5 x h = 15h.

Since the second container is a right cylinder with a radius of 1 cm, then the volume of the

water that it contains, in cm3, is π x 1

2 x h = πh.

Since the combined volume of the water is 80 cm3, then 15h + πh = 50.

Thus,

So, the height of the water is closest to 2.8 cm

Original

3. A drum-shaped cylinder radius 70 cm and filled with water as high 40 cm. A fixer

tile insert 110 pieces of ceramic tiles to the drum so high the water level increases

8cm. If the surface of each ceramic tile size 40 cm × 40 cm, what is the thick of the

ceramic tiles?

Source : Number 7 of Olimpiade Sains Nasional 2013, tingkat Provinsi

Solution :

Is known : r = 70 cm

tair mula-mula = 40 cm

π = 22

/7

110 the ceramic tiles → the size of the suface = 40 cm × 40 cm

tair tambahan = 8 cm

V110 ceramic = Vair tambahan

110 . 40 . 40 . tubin = π . r2. tair tambahan

110 . 40 . 40 . tubin = 22

/7 . 702 . 8

tubin = 22

/7 . 702 . 8

110 . 40 . 40

tubin = 0,7 cm

So thick ceramic tiles are 0,7 cm .

Modifikasi

A drum-shaped tube of radius 70 cm and filled with water as high as 20 cm. A fixer

tile insert 100 pieces of ceramic tiles to the drum, if the surface of each ceramic tiles

measuring 25 cm x 25 cm tiles with a thick of ceramics of 0.1 cm, what is the water

level increases.

Solution :

Is known : r = 70 cm

tair mula-mula = 20 cm

π = 22

/7

100 the ceramic tiles → the size of the suface = 25 cm × 25 cm

tubin = 0,1 cm

V110 ceramic = Vair tambahan

110 . 40 . 40 . 0,1 = π . r2. tair tambahan

110 . 40 . 40 . 0,1 = 22

/7 . 702 . tair tambahan

31.250 = 15.400 . tair tambahan

tair tambahan = 2,03 cm

so, the water level increases is 2,03 cm.

SURAT PERNYATAAN

Saya yang bertanda tangan di bawah ini,

Nama :Ni Luh Okassandiari

NIM :1313011026

Menyatakan dengan sebenarnya bahwa soal-soal yang saya kumpulkan untuk memenuhi

tugas Mata Kuliah Matematika Sekolah tahun akademik 2015/2016 benar-benar sesuai dengan

keteranganyang saya cantumkan di akhir setiap soal. Apabila saya melakukan penipuan atau

berbohong saya bersedia menanggung resiko yang diakibatkan oleh perbuatan tersebut.

Demikian pernyataan yang saya buat dengan sungguh-sungguh dan penuh tanggung jawab.

Singaraja, 21 September 2015

Yang membuat pernyataan,

Tanda tangan

(Ni Luh Okassandiari)

NIM :1313011026

SURAT PERNYATAAN

Saya yang bertanda tangan di bawah ini,

Nama : Ni Komang Ratna Srinadi

NIM : 1313011034

Menyatakan dengan sebenarnya bahwa soal-soal yang saya kumpulkan untuk memenuhi

tugas Mata Kuliah Matematika Sekolah tahun akademik 2015/2016 benar-benar sesuai dengan

keteranganyang saya cantumkan di akhir setiap soal. Apabila saya melakukan penipuan atau

berbohong saya bersedia menanggung resiko yang diakibatkan oleh perbuatan tersebut.

Demikian pernyataan yang saya buat dengan sungguh-sungguh dan penuh tanggung jawab.

Singaraja, 21 September 2015

Yang membuat pernyataan,

Tanda tangan

(Ni Komang Ratna Srinadi)

NIM :1313011034

SURAT PERNYATAAN

Saya yang bertanda tangan di bawah ini,

Nama : I Gusti Ayu Sekarini

NIM : 1313011077

Menyatakan dengan sebenarnya bahwa soal-soal yang saya kumpulkan untuk memenuhi

tugas Mata Kuliah Matematika Sekolah tahun akademik 2015/2016 benar-benar sesuai dengan

keterangan yang saya cantumkan di akhir setiap soal. Apabila saya melakukan penipuan atau

berbohong saya bersedia menanggung risiko yang diakibatkan oleh perbuatan tersebut.

Demikian pernyataan yang saya buat dengan sungguh-sungguh dan penuh tanggung jawab.

Singaraja, 21 September 2015

Yang membuat pernyataan,

Tanda tangan

(I Gusti Ayu Sekarini)

NIM :1313011077