Small shaded region

Geometry Level 3

In right $\triangle{ABC}, \:\ \overline{AC} = \sqrt{3}j, \overline{AB} = j$ and $\overline{AC}$ and $\overline{AB}$ are tangent to circle $O$ at points $D$ and $A$ respectively.

Let $A_{d}$ be the area of the pink shaded region above.

If the value of $j$ for which $A_{d} = \dfrac{\sqrt{3}}{2} - \dfrac{\pi}{4}$ can be expressed as $j =\alpha + \sqrt{\beta}$ , where $\alpha$ and $\beta$ are coprime positive integers, find $\alpha + \beta$ .

The answer is 5.

1 solution

Rocco Dalto
Mar 22, 2021

$\tan(\theta) = \dfrac{1}{\sqrt{3}} \implies \theta = \dfrac{\pi}{6}$ and $\dfrac{j - r}{r} = \sec(\theta) = \dfrac{2}{\sqrt{3}} \implies$

$(2 + \sqrt{3})r = \sqrt{3}j \implies r = \dfrac{\sqrt{3}j}{2 + \sqrt{3}} = \sqrt{3}(2 - \sqrt{3})j \implies$

$\overline{OB} = j - r = (2 - \sqrt{3})j \implies \overline{BD}^2 = \overline{OB}^2 - r^2 = (2 - \sqrt{3})^2j^2 \implies$

$\overline{BD} = (2 - \sqrt{3})j \implies A_{\triangle{ABC}} = \dfrac{\sqrt{3}}{2}(2 - \sqrt{3})^2j^2$

and

$A_{sector} = \dfrac{3}{2}(2 - \sqrt{3})^2\dfrac{\pi}{6}j^2$

$\implies A_{d} = A_{\triangle{ABC}} - A_{sector} = (2 - \sqrt{3})^2(\dfrac{\sqrt{3}}{2} - \dfrac{\pi}{4})j^2 = \dfrac{\sqrt{3}}{2} - \dfrac{\pi}{4}$

$\implies j = \dfrac{1}{2 - \sqrt{3}} = 2 + \sqrt{3} = \alpha + \sqrt{\beta} \implies \alpha +\beta = \boxed{5}$ .

Small shaded region

The answer is 5.

1 solution

0 pending reports