Dynamic Geometry: P49

Geometry Level 4

The diagram shows the curves $y=\dfrac{1}{x}$ (blue) and $y=-\dfrac{1}{x}$ (pink) inscribing a variable circle which is tangent to each curve at a point. Construct a triangle with the center of the circle and the two tangent points as vertices, and the angle at the center of the circle be $\alpha$ . When the ratio of the area of the circle to the area of the triangle is $\dfrac{290\pi }{143}$ , the absolute value of $\cos \alpha$ can be expressed as $\dfrac{p}{q}$ , where $p$ and $q$ are coprime positive integers. Find $\sqrt{p+q}$ .

The answer is 13.

1 solution

Chew-Seong Cheong
Feb 27, 2021

Due to symmetry, the triangle in the circle is isosceles. Let the radius of the circle be $r$ . Then the ratio of the area of the circle to the area of the triangle is

$\begin{aligned} \frac {\pi r^2}{\frac 12 r^2 \sin \alpha} & = \frac {290 \pi}{143} \\ \frac 2{\sin \alpha} & = \frac {290}{143} \\ \sin \alpha & = \frac {143}{145} \\ \implies |\cos \alpha| & = \sqrt{1-\left(\frac {143}{145}\right)^2} = \frac {24}{145} \end{aligned}$

Therefore $\sqrt{p+q} = \sqrt{24+145} = \sqrt{169} = \boxed{13}$ .

Thank you for posting.

Valentin Duringer - 3 months, 2 weeks ago

Dynamic Geometry: P49

The answer is 13.

1 solution

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