Given three points $F_A, F_B, F_C$ that are the feet of perpendiculars from unknown vertices $A,B,C$ on the three sides of $\triangle ABC$, find the three vertices.
My effort:
I can think of two ways to tackle this problem. The first is to have variable normal vectors of the three sides, and find the intersections, and check for perpendicularity. This gives a nonlinear system of $3$ equations in $3$ angles of the the normal vectors.
The second method is using the orthocenter. Let the unknown orthocenter be $H(x,y)$, then
The vertices can be expressed as
$ C = F_A + t_1 (F_A – H)^* = F_B + t_2 (F_B – H)^* = F_C + t_3 (F_C – H) $
$ A = F_B + t_4 (F_B – H)^* = F_C + t_5 (F_C – H)^* = F_A + t_6 (F_A – H) $
where $V^*$ is the vector obtained from $V$ by rotating it by $90^\circ$.
The above generates a nonlinear system of $8$ equations in $8$ unknowns.
I look forward to additional methods or shortcuts to this problem.
Best Answer
Triangle $F_AF_BF_C$ is called the orthic triangle of $\triangle ABC$. We can use properties of orthic triangle to find its mother triangles.
Theorem: For an acute $\triangle ABC$, vertices $A$, $B$ and $C$ are the excentres of its orthic triangle.
Proof:
$ACF_AF_C$ is cyclic $\implies\angle BF_AF_C=\angle BAC$
$ABF_AF_B$ is cyclic $\implies\angle CF_AF_B=\angle BAC$
(exterior angles of cyclic quadrilateral)
Therefore $\angle BF_AF_C=\angle CF_AF_C$. Thus $BC$ bisects the exterior angle of $\triangle F_AF_BF_C$ at $F_A$.
Similarly, sides of triangle $ABC$ bisect the exterior angles of triangle $F_AF_BF_C$. Hence $A$, $B$, $C$ are the excentres of $\triangle F_AF_BF_C$.
Given that, we can find the points $A$, $B$ and $C$.[1]
Also note that, if you swap points $(A,B,C)$ and $H$ you get the same orthic triangle $F_AF_BF_C$. Thus you can find a family of $4$ triangles (one acute and three obstuse) which fulfills the given condition.
[1]: Coordinate of the excentre of a triangle