Solving Triangles - Part 7

Geometry Level 3

In $\triangle ABC$ , $\angle B=\dfrac{2\pi}3$ , $D$ is a point on $AC$ so that $BD$ bisects $\angle B$ and $BD=2$ .

What is the minimum area of $\triangle ABC$ ?

4\sqrt{3}

6\sqrt{3}

3\sqrt{3}

5\sqrt{3}

1 solution

Chew-Seong Cheong
Jul 30, 2019

Let $BC = b$ and the altitude $AE=h$ . Let $DF$ be the altitude from $D$ to $BC$ . Then $BF = 1$ and $DF=\sqrt 3$ . We note that $\triangle ACE$ and $\triangle DCF$ are similar. Therefore,

$\begin{aligned} \frac {AE}{CE} & = \frac {DF}{CF} \\ \frac h{\frac h{\sqrt 3}+b} & = \frac {\sqrt 3}{b-1} \\ h & = \frac {\sqrt 3 b}{b-2} \end{aligned}$

Therefore, the area of $\triangle ABC$ :

$\begin{aligned} A_\triangle & = \frac {hb}2 = \frac {\sqrt 3b^2}{2(b-2)} \\ & = \frac {\sqrt 3} 2 \left(\frac {(b - 2)^2 + 4b - 4}{b-2}\right) \\ & = \frac {\sqrt 3} 2 \left({\color{#3D99F6}b-2} + 4 + \frac 4{\color{#3D99F6}b-2} \right) & \small \color{#3D99F6} \text{By AM-GM inequality: } \\ & \ge \frac {\sqrt 3} 2 \left({\color{#3D99F6}2\sqrt{\frac {4(b-2)}{b-2}}} + 4 \right) & \small \color{#3D99F6} \text{Equality occurs when }b=4 \\ & = \boxed{4\sqrt 3} \end{aligned}$

Reference : AM-GM inequality

@Alice Smith , backslash "\" is extensively used in LaTex. You can enter \triangle for $\triangle$ , \angle $\angle$ , \pi $\pi$ , \alpha $\alpha$ , \beta $\beta$ , \in $\in$ , \cup $\cup$ , \frac \pi 2 $\frac \pi 2$ (the braces { } are unnecessary), \frac {2\pi}3 $\frac {2\pi}3$ . Better to use "area of $\triangle ABC$ " instead of $S_{\triangle ABC}$ because not all use $S$ to indicate area.

Chew-Seong Cheong - 1 year, 10 months ago

Solving Triangles - Part 7

1 solution

0 pending reports