Question

The equation of an altitude of an equilateral triangle is $\sqrt{3}x+y=2\sqrt{3}$, and one of the vertices is $(3,\sqrt{3})$.
The possible number of triangles is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

Let the triangle be $ABC$ with $C\equiv (3,\sqrt{3})$ and the altitude drawn through the vertex (meeting $BC$ at $D$ ) be $\sqrt{3}x+y-2\sqrt{3}=0$.
If $B$ is $\left(x_b,y_b\right)$, then we have
$\frac{x_b-3}{\sqrt{3}}=\frac{y_b-\sqrt{3}}{1}=\frac{2(3\sqrt{3}+\sqrt{3}-2\sqrt{3})}{4}=-\sqrt{3}$
or $x_b=0,y_b=0$
and the coordinates of $D$ are $(3/2,\sqrt{3}/2)$.
Let the coordinates of vertex $A$ be $\left(x_a,y_a\right)$. Then,
$\frac{x_a-(3/2)}{-1/2}=\frac{y_a-(\sqrt{3}/2)}{\sqrt{3}/2}=\pm 3$
or $\left(x_a,y_a\right)\equiv (0,2\sqrt{3})$ or $(3,-\sqrt{3})$
Hence, the remaining vertices are $(0,0)$ and $(0,2\sqrt{3})$
or $(0,0)$ and $(3,-\sqrt{3})$.
Also, the orthocenter is $(1,\sqrt{3})$
or $(2,0)$.