历年 Canadian Open Mathematics Challenge加拿大数学公开赛
Part A Solutions:
A4) Solution：Factoring equation 3, (x + 2y - z)(x + 2y + z) = 15.
Substituting x + 2y - z = 5, 5(x + 2y + z) = 15 or, x + 2y + z = 3.
Subtracting this from (2): 2x = 8
Therefore, x = 4.
The average on this question was 3.4.
- Comments： A variety of good solutions were given by students who used the first two equations to arrive at y = 4 - x and z = 3 - x. This allowed substitutions into the third equation. The best way to proceed, however, was to factor the third equation as a difference of squares and then make the direct substitution on x + 2y - z = 5 as shown above.
Part B Solutions:
B2)Assume that we can expand and compare coefficients. Expanding, (x + r)(x2 + px + q) = x3 + (p + r)x2 + (pr + q)x + qr
p + r = b (1)
pr + q = c (2)
qr = d (3)
If bd + cd is odd, so is d(b + c). From this, d and b + c are both odd. From (3), if d is odd then q and r are both odd.(4)
Adding (1) and (2), b+c = p+r+pr+q = (q+r)+p(1+r). Since b+c is odd then (q+r)+p(1+r) is also odd. From (4), if q and r are both odd then q+r is even. This implies that p(1 + r) must be odd but this is not possible because r is odd and r + 1 is then even making p(1 + r) both odd and even at the same time. This contradiction implies that our original assumption was incorrect and thus x3 +bx2 +cx+d cannot be expressed in the form (x + r)(x2 + px + q).
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 assume 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.