2006 SMT/Team Problems/Problem 13

Solution

A line that passes through the origin has an equation of $y=mx$ . If the line $y=mx$ is tangent to the hyperbola than the equation $(mx)^2=x^2-x+1$ will have only one solution. This means that the discriminant of the equation $x^2(m^2-1)+x-1=0$ will be equal to zero. Solving:

$\begin{align*} b^2-4ac &= 0\\ 1+4(m^2-1) &=0\\ (m^2-1) &=\frac{-1}{4}\\ m^2 &=\frac{3}{4}\\ m&=\pm\frac{\sqrt{3}}{2} \end{align*}$

We can ignore the negative root of the equation because the line $y=mx$ is tangent to the parabola in the first quadrant. Therefore $m =\frac{\sqrt{3}}{2}$ .

We now need to find the cosine of the angle formed between the line $y=\frac{\sqrt{3}}{2}x$ and the x-axis. We can do this by forming a right triangle using an arbitrary point on the line, and the x-axis. We can then solve for the cosine of the angle.

Picking the point $(2,\sqrt{3})$ , we find that the hypotenuse of the right triangle formed using the x-axis as a side is $\sqrt{7}$ . Therefore, the cosine of the angle formed between the line and the x-axis is $\frac{2}{\sqrt{7}} = \boxed{\frac{2\sqrt{7}}{7}}$

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2006 SMT/Team Problems/Problem 13

Solution