STEP I 1994 question 9 solution

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STEP I question 9

Let (x,y) be a point on the cannon-ball's trajectory, then:

Now if the distance from the origin is , then we have the point (x,h), so writing the x and y coordinates with the trig function the subject:

But , so:

, as required.

Since this equation arose from considering a point we assumed to be on the trajectory, then there has to exist a real value of t at which the cannon-ball is at this point. For a real value of t, we need a real value of t^2, so the solutions to the quadratic must be real, so using the discriminant:

, as required.

Okay, now we need to show that there is an angle of firing such that we have when y=h, so going back to the equations for x and y we get:

and

substitute in the time from the x into the y:

For there to exist an angle of firing as described, we again need real roots to this quadratic, , which is of course true (and it really is an angle of firing).

Solution by ad absurdum