‫توزيع عالمات مسابقة الرياضيات‬
‫ العادية‬2013
Questions
Answers Keys
Grades
5 18  2 98
15 2 14 2 1
=

2 3  24  4 2 12 2  4 2 8
1.a
A
1.b
A=0.125=1.25  10-1
0.75
0.25
I
II
2
tan 2  
3
x
1
X
2
3
1.a
1.b
2
III
3
4.a
4.b
1
IV
2
1.a
V
sin 2 
, 1  sin 2    cos 2 . Hence tan2α (1-sin2α) = sin2α.
cos 2 
0.5
3  5 3  5   20
7
5
0.5
7
0  2  2  7  4 10  6  9  7 12
 4.8
40
Number of read books
0
2
Frequency
2
7
% Relative frequency
5
17.5

Central angle
18
63
0.75
4
10
25
6
9
22.5
7
12
30
Total
40
100
90 
81
108
360
2
The number of students is 21
0.25
3x 2  27
6 2  18
0.5
0.5
(2x-3-x+6)(2x-3+x-6) = 3(x  3)(x  3)
0.5
B(x)  2(x  3)  (x  9)  (x  3)[2(x  3)  (x  3) = (x-3)(x-9)
F(x) is defined for x  3 and x  9
3(x  3) 3(x  3)
F(x)=
;
= -1; then 4x = 0; x =0
x 9
x 9
0.5
2
2
 x  y  6
, …….., x=10,
y=16

8x  5y  0
x 5
Refer toThales:  and PK=6 :y-x=6 So,we get the given system.OP = y =6
y 8
.or……
The coordinates of A and B verify the equation of (d).
0.25
0.75
1
1
0.5
1.b
Figure
A, B, C and (d)
0.75
9y
A'
B
J
8
7
6
5
4
3
2
1
-8 -7 -6 -5 -4 -3 -2 -1 0
-1
-2
-3
A
-4
-5
-6
2.a
2.b
3
4
5.a
5.b
6.a
6.b
C
1 2 3 4 5 6 7 8 9 10 x
D
BC= 104 = 2 26
BC2=AB2+AC2 ;
ABC is right.
7  3
3 1
xJ 
 2, y J 
 2 ; J  2; 2  .
2
2
J midpoint of [BC]
The perpendicular bisector of [AB] is perpendicular to [AB] at its midpoint (-4 ; 0)
.Its slope =1 and its equation is y = x + 4.

BC 10; 2 
 
AD  BC, D  9; 5
ABA'C parallelogram + right angle: rectangle.
A',, C and D are collinear . CA'=CD or……
2
3
4.a
4.b
5
6
0.5
0.75
0.5
0.5
0.5
0.5
0.25
1
VI
0.5
0.5
(MB) and (NA) are perpendicular to (MN).
EN EB 1




BME
 ENA
 900 AEN
 MEB
,


EM EA 2
EPBM is a parallelogram+ right angle: rectangle..
0.5
BPE is right, [EB] diameter then P is a point on (C1)
In the triangle IMP, [MK] is a median
0.5
0.75
  90o and L moves
(BP) is perpendicular to (MB) then perpendicular to (AN) so ALB
on the circle with diameter [AB].
1
0.5
1