Given a triangle,a similar triangle $ABC$ is drawn where $A$ is a fixed point and $B$ is any point on a given line.Find the locus of $C$.

Question

Given a triangle,a similar triangle $ABC$ is drawn where $A$ is a fixed point and $B$ is any point on a given line.Find the locus of $C.$

I thought over it but there is no concrete method visible to find out the locus of C.

Answer 1

Let us consider in the complex plane.

We may suppose that $A(\alpha)$ is at $\alpha=i$ and that $B(\beta)$ is on the real axis ($\beta=p$ where $p\in\mathbb R$).

Also, if we set $q=\frac{A'B'}{C'A'},\theta=\angle{C'A'B'}$, then for $C(\gamma)$, we have $$\gamma-\alpha=q(\cos\theta+i\sin\theta)(\beta-\alpha)$$ where we suppose that $\triangle{A'B'C'}$ is given and that $\triangle{ABC}$ is similar to $\triangle{A'B'C'}$.

Then, $$\begin{align}\gamma &=i+q(\cos\theta+i\sin\theta)(p-i)\\&=i+q(p\cos\theta-i\cos\theta+p\sin\theta i+\sin\theta)\\&=(pq\cos\theta+q\sin\theta)+i(1-q\cos\theta+pq\sin\theta)\end{align}$$ So, if we set $\gamma=x+yi$ where $x,y\in\mathbb R$, then we have $$x=pq\cos\theta+q\sin\theta,\quad y=1-q\cos\theta+pq\sin\theta$$

Eliminating $p$ from these gives $$(x-q\sin\theta)q\sin\theta=q\cos\theta(y-1+q\cos\theta),$$ i.e. $$y-(1-q\cos\theta)=\tan\theta\ (x-q\sin\theta).$$

By the way, we can replace $\theta$ with $-\theta$.

Hence, the answer is that the locus of $C$ is two lines :

$$y-(1-q\cos\theta)=\tan\theta\ (x-q\sin\theta),$$ $$y-(1-q\cos(-\theta))=\tan(-\theta)\ (x-q\sin(-\theta)).$$

Answer 2

We may assume $A=0\in{\mathbb C}$, $B=b+t\, i$ with $b>0$ and $t\in{\mathbb R}$. On the other hand, let $\triangle(A_*B_*C_*)\subset {\mathbb C}$ be the given sample triangle, and define $c\in{\mathbb C}\setminus{\mathbb R}$ by $$C_*-A_*= c(B_*-A_*)\ .$$ The locus in question is then parametrized by $$t\mapsto C(t):=c(b+t\,i)\qquad(-\infty<t<\infty)\ .$$ It is a line through the point $C_0:=c \,b$, and orthogonal to $c$.

Answer 3

Assume that $ABC$ is a triangle and $B$ belongs to some line $l$.

Let $D$ be the intersection $\neq B$ between $l$ and the circumcircle of $ABC$.

Let $C'$ be some point on the line $DC$, and $B'\in l$ a point such that $\widehat{B'AC'}=\widehat{BAC}$.

By angle chasing we have that $B'A C'$ is similar to $BAC$, since both $BACD$ and $B'AC'D$ are cyclic quadrilaterals. It follows that the wanted locus is a line.