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Geometry Level pending

In $\triangle{ABC}$ above, $\overline{AB} \cong \overline{DC} = 1$ and $m\angle{DAC} = 30^{\circ}$ and $\angle{BAD}$ is a right angle.

If $A_{\triangle{ABD}} = \dfrac{a}{b}\sqrt{b^{\frac{b}{c}} - a}$ , where $a, b$ and $c$ are coprime positive integers, find $a + b + c$ .

1 solution

Rocco Dalto
Feb 5, 2021

In $\triangle{ADC}, \:\ \theta - \alpha = 30^{\circ} \implies \alpha = \theta - 30$ and $m = \sqrt{x^2 - 1}$ .

Using the law of sines on $\triangle{ADC} \implies \dfrac{1}{\sin(30^{\circ})} = \dfrac{m}{\sin(\theta - 30^{\circ})} \implies$

$m = 2\sin(\theta - 30^{\circ}) = \sqrt{3}\sin(\theta) - \cos(\theta) \implies m = \sqrt{3}(\dfrac{1}{x}) - \dfrac{m}{x}$

$= \dfrac{1}{x}(\sqrt{3} - m) \implies mx = \sqrt{3} - m \implies (x + 1)m = \sqrt{3} \implies m = \dfrac{\sqrt{3}}{x + 1} \implies$

$\dfrac{\sqrt{3}}{x + 1} = \sqrt{x^2 - 1} \implies 3 = (x^2 - 1)(x + 1)^2 = (x - 1))x^3 + 3x^2 + 3x + 1)$

$\implies x^4 + 2x^3 - 2x - 4 = 0 \implies x^3(x + 2) - 2(x + 2) = 0 \implies$

$(x^3 - 2)(x + 2) = 0$ and $x \neq -2 \implies x = 2^{\frac{1}{3}} \implies m = \sqrt{x^2 - 1} = \sqrt{2^{\frac{2}{3}} - 1} \implies$

$A_{\triangle{ABD}} = \dfrac{1}{2}\sqrt{2^{\frac{2}{3}} - 1} = \dfrac{a}{b}\sqrt{b^{\frac{b}{c}} - a} \implies a + b + c = \boxed{6}$ .