The Fourth Canadian Open Mathematics Challenge
The Canadian Mathematical Society
in collaboration with
The Center for Education
in Mathematics and Computing
The Fourth
Canadian Open
Mathematics Challenge
Wednesday, November 24, 1999
Solutions
c Canadian Mathematical Society 1999Part A
5 Note: All questions in part A were graded out of points.
1. Answer 495 The average on this question was 3.7.
Comments This question is most easily solved by adding the series from ’front to back’ or by using the formula. Students could also have just added the terms mechanically to get the same answer.
2. Answer
2 x x 3; x
2 R
or The average on this question was 3.9.
Comments
2 x
The intercepts of the function are -2, 0 or 3 which gives the answer
x
3
or . Generally speaking, students should have solved for the intercepts first and then used the diagram to read off the correct intervals. Students should take care in dealing with inequality signs. Many students had the signs going in the wrong direction.
2
3. Solution If we convert to base , this gives,
3 x 1 x
4
8
2 =
9
27
3
" # 1 x 2x
3
2
2
2 =
3
3
3 2x 3 3x
1
2
2
2 =
3
3
3 x+3
1
2
2 =
Therefore
3
3 x 3 = 1 x = 2:
Since bases are equal, ,
- The average on this question was 3.7.
Comments
3
2 This question was generally well done. It should be noted that converting to base
is by far the easiest way to approach the problem. The biggest mistake here was in the inappropriate use of the power rules for exponents.
4. Solution
- (x 2y z )(x 2y z ) = + +
15 Factoring equation 3, . x 2y z = 5 5(x 2y z ) = + + + 15 x 2y z = + +
3 Substituting , or, . 2x =
8 Subtracting this from (2): x =
4 Therefore, .
The average on this question was 3.4.
Comments A variety of good solutions were given by students who used the first two equations
y = 4 x z = 3 x
to arrive at and . This allowed substitutions into the third equation. The best way to proceed, howeve, was to factor the third equation as a
x 2y z =
5
difference of squares and then make the direct substitution on as
- shown above.
5. Solution
2
2 2 sin x(sin x 3) (sin + x + 3) = , (2 sin x 1)(sin x 3) =
- sin x =
Factoring gives, .
3 j sin xj
1 Either which is inadmissible since , or
2 2 sin x =
1
1 sin x = p
2
3
5
7 x = ; ; ;
Therefore, .
4
4
4
4 The average on this question was 1.8.
Comments This problem could have been solved by either factoring directly or using the factor
sin x =
3
theorem. Students should make the comment in their solutions that is
2
1 sin x =
inadmissible. From there, the recognition that has four solutions should
2 be an easy matter.
6. Solution
C
From draw D
1
2 E
1 a line perpen- dicular to both Z
2 D E E F and
C
and label the.
A k diagram as
2 shown. From
D
draw a line G
2 F perpendicular
J Y
GF
to to meet
J k the line at . D E = J F =
3 J Y =
3 2 =
1 GJ = k
1
4D J G
Since , . Thus, . Since is right
2
2
2
2
2
4 k 2k 1 = + 16 k 2k 4k + (k + + 1) + = 1) 1 =
- (k
16
angled, or, , ,
- (3 6) k =
4 GF =
6 D E F G (4) =
18 . This makes and the area of trapezoid is .
2 The average on this question was 2.1.
Comments This question could be done in a variety of ways. The easiest way is to use properties of tangents to a circle and the Pythagorean theorem. When solving problems of this type, it is almost always a matter of dropping perpendiculars and using simple prop- erties of circles and triangles to get the appropriate equations. There were a number of very unusual and insightful solutions to this problem
7. Solution
a 2r =
- a =
12 We are given that .
12 2r
Therefore . The
A
formula for the area ( ) of a
A = ar a
sector is, where
1 O
2 r
r is arc lenght and is radius. r
Using the formula for area, B
A
a1 A = (12 2r )r
2
2 A = r 6r + r =
3 To maximize the area we complete the square or use calculus to find . Thus the r = 3 radius that maximizes the area is .
The average for this question was 1.6.
Comments This question was nicely done by a large number of competitors. The formula for
1 A = ar
the area of a sector can be easily derived to be . From there if we use the a 2r =
12
relationship, it is not difficult to get the required expression for the area
- of the sector of the circle. It was graifying to see the number of students who solved this problem correctly.
8. Solution
- 14k
17 If we rewrite in the following way, k
9
- 14k 17 [14(k + + 9) 126℄
17 14(k 9) 143 143 = = = + + 14 ; k 6=
9 k 9 k 9 k 9 k 9 k
9 143 = 1:11:13 k 9 = q d d
Since , it is not difficult to see that where is a number
1:11:13 q d 1 d
contained in and neither nor equals . The smallest possible value for
q k =
31 and is 11 and 2 respectively. This makes .
The average on this question was 0.9.
Comments It was delighted to see how many competitors solved this problem and the variety of
djk 9 dj14k
- dj14k
17
solutions. Some students made the observation that and and so
17 14(k 9) dj143 or . This also leads quickly to the solution.
Part B
10 Note: All questions in part B were graded out of points.
1. (a) Solution
3
y
y = x b
- 4 y C
The line
meets the -axis at
y =
4
and the line at
B
. Since the slope of
this line is , we let
4 AC 3a AB
3 C
be and be x
4a
. Then the area of the triangle is given by B A
1 (3a)(4a) =
24
2
2 6a =
24
2 a = 4 a =
2
3a =
6 C (0;
b) AC =
6 4 b + + 4 =
6 Then . Now the coordinates of are , so . Then b =
2
10
or Comments
4 3a : 4a
3 The best way of solving this problem was to recognize that a slope of means that
the ratio of the height to the base for the triangle is . If the triangle has an area
b = 2 b =
10 of 24, then the actual lenghts of the sides are 6 and 8. This gives or .
In part (b), the triangle is right angled with a hypotenuse of 10. This gives the radius
5. Students should always draw diagrams for problems of this type. Many students lost marks here because they didn’t explain how they arrived at their answers and did not visualize a solution. (b) Solution Since either of the triangles have side lenghts 6 and 8 then the hypotenuse has a lenght of 10. The semi-circle must have a radius of 5. The average on this question was 3.4.
2. Solution Assume that we can expand and compare coefficients. Expanding,
2
3
2
- (x r )(x px + = x (p r )x + (pr + q + ) q )x + q r
Comparing coefficeints,
- p r = b
(1)
pr q =
- q r = d
(2)
(3)
- d bd d(b ) d b d + +
If is odd, so is . From this, and are both odd. From (3), if
q r
is odd then and are both odd.(4)
- b = p r pr + q = (q + + r ) p(1 r ) b + +
Adding (1) and (2), . Since is odd then
- (q r ) p(1 r ) q r q r
- p(1 r )
is also odd. From (4), if and are both odd then is even. This
r r
- p(1 r )
1
implies that must be odd but this is not possible because is odd and
is then even making both odd and even at the same time. This contradiction
3
2 x bx x d
implies that our original assumption was incorrect and thus cannot
2
- (x r )(x px q ) + + be expressed in the form .
The average on this question was 2.1.
Comments There were a variety of gorgeous solutions to this problem. We provide one solution in its entirety. The method of proof here is that of contradiction. In essence we as- sume that is possible to compare coefficients by expanding the left side. From this, we show that this leads to a contradiction. Since there is a contridictory conclusion, our original assumption must in fact have been false and thus it is not possible to compare coefficients, as is required. A large number of students developed a large varity of proofs, some of which were quite unique and very interesting and in fact correct.
3. Solution
4AB P A
Label as shown. From
2
4AB P 3 =
, b
2
2 p 2p os B
- \B AC
.
3
4
Because is a right angle, p p p B P Q C
os B = 3p
2
2 9 = p 2p
- 3p
so
1
2
2 9 = p
- 3
or, (1)
4AC Q
Following the same procedure in we have,
1
2
2 16 = p b
- 3
(2)
1
2
2
2
- 3
- 25 = 2p (b ) Adding (1) and (2) gives, .
1
2
2
2
2
2
2 b = 9p
3 p p p p = 5 (p > 0) B C =
- 25 = 2p (9p ) = 5p Since , .
3
5
45 Therefore, and or . p p p p =
5 AB =
12 AC =
33 Since , substituting in equation (1) and (2) gives and .
The average on this question was 1.0.
Comments This problem could be approached in a wide variety of ways. We provide one of the many possible proofs above. Many students attempted this proof by drawing lines
AB P Q
parallel to throught and and then using the side splitting theorem. This leads to a simple application of Pythagoras and a nice system of equations to solved. The number of students who presented unique solutions was quite gratifying.
4. Solution
P
4AB C \P AB = \P B A = \P C B = Let be a point inside such that . \AB C \B P C = 180 [( ) ℄ = 180 Let be .
.
- A N h P 180- a B C
P AB AB N
From draw a line perpendicular to meeting at . Applying the Sine Law
4P B C
in ,
P B a a = = sin sin(180 ) sin a sin
P B =
Therefore, (1)
sin
B N AB AB
4B P N os = = P B =
From , or (2)
P B
2P B 2 os
Equating our two expressions, (1) and (2), we have
a sin AB = sin
2 os
or
AB sin 2 sin os = a
AB sin sin 2 = a h
AB sin = h sin 2 =
Since then
a p p
3
3 h < a sin 2 <
Since , (by substitution)
2
2
2 2 < < 2 > > Then and or and .
3
6
3
3 >
But is impossible because the sum of the angles in the triangle is . Hence
3 a h
there is one value of for any given and . The average on this question was 0.1.
Comments
This was a very hard problem. It can be done in three or four ways. A number of
competitors got this question correct. We provide here just one solution. Some stu-
dents solved the problem but only one or two students made an attempt at dealing
p
3 h a
with restrictions, . This problem could have been attempted using coordi-
2 nates but was awkward and hard mechanically.