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-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.
A1
18
A2
11m2
A3
8
A4
11
A5
5.844L
A6
18◦
∠AOP = 36◦ (as in the diagram below). What is ∠OP B in degrees?
O
A
B
P
36◦
A7
35m
6m 6m 6m
6m 6m 6m
3m
4m
A8
55
A9
8cm2
1cm 2cm 2cm
1cm
1cm
Solution:
1cm
1cm
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.
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.
45◦
45◦
45◦
45◦
45◦
Solution:
2
5
17
9
10
14
12
7
Solution:
Suppose 2018 =x2−y2 for some positive integersx andy. Then as 2018 factors as
2·1009,
(x−y)(x+y) = 2·1009.
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
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