Area Of A Triangle

Using the Law of sines for $\triangle{ABD}$ we have:

$\dfrac{2}{\sqrt{3}}AD = \dfrac{2y}{\sin(180^{\circ} - m)} = \dfrac{2y}{\sin(m)} \implies$ $AD = \dfrac{\sqrt{3}y}{\sin(m)}$

and using the Law of sines for $\triangle{DBC}$ we have:

$\dfrac{2}{\sqrt{3}}DC = \dfrac{y}{\sin(m)} \implies DC = \dfrac{\sqrt{3}y}{2\sin(m)}$

$\implies \dfrac{AD}{DC} = 2$ .

Let $DC = x \implies AD = 2x$ .

Using the law of cosines for $\triangle{ABD}$ we have:

$4x^2 = 4y^2 + 100^2 - 400y\cos(60^{\circ} = 4y^2 - 200y + 100^2$

$\implies x^2 = y^2 - 50y + 25^2 * 4$

and using the law of cosines for $\triangle{DBC}$ we have:

$x^2 = y^2 + 100^2 - 200y\cos(60^{\circ}) = y^2 - 100y + 25^2 * 4^2$

$\implies y^2 - 50y + 25^2 * 4 = y^2 - 100y + 25^2 * 4^2 \implies 50y - 4 * 25^2 * 3 = 0 \implies$

$y = \dfrac{(2)(50)(25)(3)}{50} = 150$

$\implies x^2 = 150^2 - 50(150) + 25^2 * 4 = 25^2 * 6^2 - 25^2 * 12 + 25^2 * 4 = 25^2(28) =$

$25^2 * 2^2 * 7 \implies x = 50\sqrt{7} \implies AC = 3x = 150\sqrt{7}$

and $AD = 2x = 100\sqrt{7} = \dfrac{\sqrt{3}y}{\sin(m)} = \dfrac{150\sqrt{3}}{\sin(m)} \implies$

$\sin(m) = \dfrac{3}{2}\sqrt{\dfrac{3}{7}}$ and $AD = 100 \implies$ height $h$ of $\triangle{ABC}$ is $h = 100(\dfrac{3}{2})\sqrt{\dfrac{3}{7}}$

$= 150\sqrt{\dfrac{3}{7}}$

$\implies A_{\triangle{ABC}} = \dfrac{1}{2}(150\sqrt{7})(150\sqrt{\dfrac{3}{7}}) =$ $2 * 25^2 * 3^2\sqrt{3} =$

$2 (75)^2\sqrt{3} = a * b^a * \sqrt{c} \implies a + b + c = \boxed{80}$ .

Let $BC = x$ ; then $AB = 2x$ . Let $\angle BAC = \theta$ . By sine rule ,

$\begin{cases} \dfrac {AB}{BD} = \dfrac {\sin \angle ADB}{\sin \angle BAD} & \implies \dfrac {2x}{100} = \dfrac {\sin (120^\circ - \theta)}{\sin \theta} & ...(1) \\ \dfrac {BC}{BD} = \dfrac {\sin \angle BDC}{\sin \angle BCD} & \implies \dfrac x{100} = \dfrac {\sin (60^\circ + \theta)}{\sin (60^\circ -\theta)} & ...(2) \end{cases}$

$\begin{aligned} \frac {(1)}{(2)}: \quad 2 & = \frac {\blue{\sin(120^\circ - \theta)}}{\sin \theta} \times \frac {\sin(60^\circ - \theta)}{\sin (60^\circ+\theta)} & \small \blue{\text{Note that }\sin (180^\circ - \phi) = \sin \phi} \\ & = \frac {\blue{\sin(60^\circ + \theta)}}{\sin \theta} \times \frac {\sin(60^\circ - \theta)}{\sin (60^\circ+\theta)} \\ & = \frac {\sin(60^\circ - \theta)}{\sin \theta} \\ & = \frac {\sqrt 3}2 \cot \theta - \frac 12 \end{aligned}$

$\implies \tan \theta = \dfrac {\sqrt 3}5 \implies \sin \theta = \dfrac {\sqrt{21}}{14}, \cos \theta = \dfrac {5\sqrt 7}{14}$ and from $(1)$ :

$\begin{aligned} \frac {2x}{100} & = \frac {\sin (60^\circ + \theta)}{\sin \theta} = \frac {\frac {\sqrt 3}2 \times \frac {5\sqrt 7}{14} + \frac 12 \times \frac {\sqrt{21}}{14}}{\frac {\sqrt {21}}{14}} = 3 \\ \implies x & = 150 \\ A_{\triangle ABC} & = \frac 12 AB \times BD \sin 60^\circ + \frac 12 BC \times BD \sin 60^\circ \\ & = \frac 12 (300+150) (100) \left(\frac {\sqrt 3}2\right) \\ & = 2(75^2)\sqrt 3 \end{aligned}$

Therefore $a+b+c = 2+75+3 = \boxed {80}$ .