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Đề thi Toán quốc tế CALGARY năm 2018

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4

2

nd

JUNIOR

HIGH

SCHOOL

MATHEMATICS

CONTEST



MAY

2

,

2018.



-SOLUTIONS-NAME: GENDER:


PLEASE PRINT (First name Last name) (optional)


SCHOOL: GRADE:


(9,8,7,. . . )


• You have 90 minutes for the examination. The test has
two parts: PART A — short answer; and PART B —
long answer. The exam has 9 pages including this one.


• Each correct answer to PART A will score 5 points.
You must put the answer in the space provided. No
part marks are given. PART A has a total possible
score of 45 points.


• Each problem in PART B carries 9 points. You should
show all your work. Some credit for each problem is
based on the clarity and completeness of your answer.
You should make it clear why the answer is correct.
PART B has a total possible score of 54 points.


• You are permitted the use of rough paper.
Geome-try instruments are not necessary. References
includ-ing mathematical tables and formula sheets are not



permitted. Simple calculators without programming
or graphic capabilities are allowed. Diagrams are not
drawn to scale: they are intended as visual hints only.


• When the teacher tells you to start work you should
read all the problems and select those you have the
best chance to do first. You should answer as many
problems as possible, but you may not have time to
answer all the problems.


MARKERS’ USE ONLY


PART A
×5
B1
B2
B3
B4
B5
B6
TOTAL
(max: 99)


BE SURE TO MARK YOUR NAME AND SCHOOL
AT THE TOP OF THIS PAGE.


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PART A:

SHORT ANSWER QUESTIONS

(Place answers in
the boxes provided)


A1



18


A1

How many 4-digit numbers can be made by arranging the digits 2, 0, 1, and 8? (A
4-digit number cannot start with 0.)


A2


11m2


A2

A rectangle whose length and width are positive integers has perimeter 24 metres
and its area (in square metres) is a prime number. What is its area in square metres?


A3


8


A3

Two brothers and two sisters are waiting in line for ice cream. How many ways can
they line up so that the two brothers are not next to each other and the two sisters
are not next to each other?


A4


11


A4

Notice that1<sub>2</sub>−1<sub>3</sub> = 1<sub>6</sub> and 1<sub>3</sub>−1<sub>4</sub> = <sub>12</sub>1. If <sub>x</sub>1−<sub>x</sub><sub>+1</sub>1 = <sub>132</sub>1 wherexis a positive integer,
what isx?


A5



5.844L


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A6


18◦


A6

A circle with centreOhas diameter AB. A pointP is placed on the circle such that


∠AOP = 36◦ (as in the diagram below). What is ∠OP B in degrees?


O


A


B


P


36◦


A7


35m


A7

There is a path 4 metres wide with streetlights on either side spaced 6 metres apart.
If a dog runs from one streetlight to the next as shown, what is the total distance it
runs in metres?


6m 6m 6m



6m 6m 6m


3m


4m


A8


55


A8

We knowx andyare positive integers such that 8x+ 9y= 127. Find the value ofx.


A9


8cm2


A9

The following diagram is a 5cm×4cm box containing several quarter circles of radius
2cm. Find the area of the shaded region in square cm.


1cm 2cm 2cm


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PART B:

LONG ANSWER QUESTIONS



B1

In the following diagram any two circles which are adjacent along the same side are
1cm apart, measured from their centres. List all different distances (in cm) from the
centre of one circle to the centre of another.


1cm


1cm



Solution:


1cm


1cm


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B2

Penny’s age is the sum of the ages of her two brothers. Six years ago, her age was
theproduct of the ages of her two brothers. How old is Penny and her brothers?


Solution:


Letaandbbe the ages of Penny’s two brothers six years ago, where we may assume
thata≥b. Then, Penny’s age was ab, so her age now is ab+ 6. But her brothers’
ages now must bea+6 andb+6, so Penny’s age now must bea+6+b+6 =a+b+12.
Thus we know that,


ab+ 6 =a+b+ 12,


which simplifies to


ab−a−b= 6.


By adding 1 to both sides and factoring, this equation becomes
(a−1)(b−1) = 7.


The only integers solution to this equation that satisfya≥b >0 is when


a−1 = 7 and b−1 = 1.



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B3

The edge-length of the base of a square pyramid is 8 cm. The length of each of the
four slanting edges is 9 cm. What is the pyramid’s height in cm?


8cm


8cm
9cm
?


Solution:


The length of the diagonal of the square base is 8√2 by Pythagoras’ Theorem.
Dividing this by 2, we can find the height by considering the triangle,


4√2cm
9cm
?


So by Pythagoras’ Theorem,




81−16·2 =√49 = 7.


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B4

A rectangular pool table has dimensions 1.4m by 2.4m and six pockets, as shown (the
center pockets on the horizontal sides are placed at the midpoint of their respective
side). A ball is projected from the lower left corner at an angle of 45◦ with the
sides of the table and bounces at the marked points shown in the figure, always at
45◦. The marked points are placed 0.4m around the table clockwise. Will the ball
continue indefinitely or will it fall into a pocket? If the latter is the case, which

pocket?


45◦
45◦
45◦


45◦


45◦


Solution:


2


5


17


9


10


14


12
7


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B5

Can 2018 be written as the difference of two square integers? If yes provide an
example, if no explain why.



Solution:


Suppose 2018 =x2−y2 for some positive integersx andy. Then as 2018 factors as
2·1009,


(x−y)(x+y) = 2·1009.


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B6

Triangle ABC has edge-lengths BC = 13, AB = 14, CA = 15. The straight line


DE intersectsAC in the ratio 10 : 5 and BC in the ratio 9 : 4.


Is DE parallel to AB, or do the lines extending segments AB and DE intersect


AB in some point F, to the left or right? If DE does intersect AB, calculate the
appropriate length,AF orBF.


F?
F?
F?
F?
A B
C
E D
5 4
10
14
9
Solution:


DG, CH, EI are perpendiculars from D, C, E, onto AB. If CH = h, HA = x,



HB=y, then, By Pythagoras’ Theorem,


h2+x2= 152, h2+y2= 132, x2−y2 = 152−132 = 56
andx+y= 14, so thatx=HA= 9 and y=HB = 5.


DG= 9


13h >
10


15h=EI


so that F is on the far left. Triangles DGF, EIF are similar. AI = 10<sub>15</sub>9 = 6.


HG= <sub>13</sub>45 = 20<sub>13</sub>, so thatAG= 9 +20<sub>13</sub> = 137<sub>13</sub>.
LetAF =z, so that GF =z+ 137/13 and


z+ 137/13


z+ 6 =


F G
F I =


DG


EI =


9h/13


10h/15 =


27
26.
27z+ 162 = 26z+ 274, so thatAF =z= 112.


Check by Menelaus:


10
5 ·


4
9 ·


14 +z


−z =−1


so 45z= 40(14 +z). It follows thatz= 112. By either method AF=112.


F


C


E D


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