math_person_101
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- Mar 12, 2014
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My son and I were working on the problems for a math contest that was written a few weeks ago. There are a few questions that are bugging us and here's one of them. This was a grade 11 math contest in Ontario, Canada.
Points P(r,s) and Q(t,u) are on the parabola with equation y = x^2 - 1/5 mx + 1/5 n so that PQ=13 and the slope of PQ is 12/5. For how many pairs (m,n) of positive integers with n<=1000 is r + s + t + u = 27?
The multiple choice answers are A) 28 B) 26 C) 27 D) 29 E) 25
We realized that PQ is a the hypotenuse of a right triangle with vertical leg 12 and horizontal 5. That means that t = r + 5 and u = s + 12. Using a bit of algebra then,
r + s = 5.
I suppose an equation for the line with P and Q would be y-s = 13(x-r) or y-s = 13(x-(5-s)) . Then y = 13(x-(5-s)) + s
I could then combine the two equations so 13(x-(5-s)) + s = x^2 - 1/5 mx + 1/5 n
Am I going in the right direction? Where do I go from here?
Help ;-)
Points P(r,s) and Q(t,u) are on the parabola with equation y = x^2 - 1/5 mx + 1/5 n so that PQ=13 and the slope of PQ is 12/5. For how many pairs (m,n) of positive integers with n<=1000 is r + s + t + u = 27?
The multiple choice answers are A) 28 B) 26 C) 27 D) 29 E) 25
We realized that PQ is a the hypotenuse of a right triangle with vertical leg 12 and horizontal 5. That means that t = r + 5 and u = s + 12. Using a bit of algebra then,
r + s = 5.
I suppose an equation for the line with P and Q would be y-s = 13(x-r) or y-s = 13(x-(5-s)) . Then y = 13(x-(5-s)) + s
I could then combine the two equations so 13(x-(5-s)) + s = x^2 - 1/5 mx + 1/5 n
Am I going in the right direction? Where do I go from here?
Help ;-)