Almost pi!

Geometry Level 3

Construct unit circle $O$ with diameter $\overline{AB}$ , and unit circle $A$ . These circles intersect at two points, so let one be $C$ and construct unit circle $C$ . Let point $D$ be the intersection between circles $A$ and $C$ that is outside circle $O$ .

Construct line $l$ tangent to circle $O$ which passes through $A$ . Then let $\overline{DO}$ and $l$ intersect at $E$ , and construct point $H$ on line $l$ such that $EH = 3$ and $AH < 3$ .

$\overline{BH}$ has a length very close to $\pi$ which can be written as $\sqrt{\frac{a-b\sqrt{c}}{d}}$ , where $a$ , $b$ , and $d$ are relatively prime integers and $c$ is a square-free integer. Find $a+b+c+d$ .

The answer is 52.

1 solution

Zain Majumder
Oct 6, 2018

The left figure is a diagram of the construction, with most points labeled. Note that since line $l$ is tangent to circle $O$ , $\overline{AH}$ and $\overline{AO}$ are perpendicular. In addition, two blue equilateral triangles $\triangle DAC$ and $\triangle OAC$ with side length $1$ have been added.

The right figure is a closer look at the two triangles. $AO$ and $CD$ are parallel because of the $60^{\circ}$ alternate interior angles. Since $l$ is perpendicular to $AO$ , it is also perpendicular to $CD$ , so $E$ is the incenter of $\triangle DAC$ . Therefore, $EA = \frac{\sqrt{3}}{3}$ , and $AH = EH-EA = 3-\frac{\sqrt{3}}{3}$ .

Applying Pythagorean Theorem to $\triangle HAB$ yields $BH = \sqrt{(3-\frac{\sqrt{3}}{3})^2 + 2^2} = \boxed{\sqrt{\frac{40-6\sqrt{3}}{3}}} \approx 3.14153333871 \approx \pi$ .

Almost pi!

The answer is 52.

1 solution

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