ABC is a triangle. P is a point inside ABC such that its distances from the sides of Triangle ABC are x, y, z. If a, b, c and k are given set of constants, prove that the locus of P such that ax+by+cz=k is either an empty set or a line segment or coincides with the sets of all points in triangle ABC
Now, for the locus of point P to set of all points in triangle, the constants would have have to sides of the triangle and k would have to be twice the area of triangle ABC.
I can't figure anything out for the locus to be straight line however and that lead me to thinking ' What if I shouldn't be relating the distances to the area of the triangle'
One other approach I tried was to assume that locus coninciding with all points would only apply for equilateral triangles, for locus to be a straight line the triangle would have to be a isoceles and for locus to be an empty set, the triangle would be scalene one.
It would be incredibly helpful if someone could point whether I should relate the various cases of the locus to types of triangles or to given constants or to something else.
Best Answer
First, we simplify the problem by allowing for signed distances, and consider points outside the triangle.
We will show that the set of points that satisfy $ax+by+cz=k$ is either the null set, a line or the entire plane. Then, when restricted to the (interior) of the triangle, this becomes a null set, a line segment, or the entire interior, so the result follows.
Hint: For a point $P$, if the side lengths of the triangle are $l_a, l_b, l_c$, then $l_a x + l_b y + l_c z = 2 \Delta $.
Hint: Hence, it follows that $ax+by+cz=k$ is the entire plane iff $a:b:c:k = l_a : l_b: l_c : 2 \Delta$.
If $a:b:c = l_a : l_b: l_c $, then we either get the entire plane or the null set.
Henceforth, assume $a:b:c \neq l_a : l_b: l_c $.
Hint: For fixed values of $a, b, c, k$, if $P_1$ and $P_2$ are distinct points that satisfy the equation, then so does the entire line segment $P_1 P_2$.
Corollary: The solution set is a (affine) sub-space. It remains to show that the solution set cannot be a single point.
Hint: For fixed values of $a, b, c, k$, if there is (at least) 1 point that satisfies it, then there are at least 2 distinct points that satisfy it.
Suppose we have the solution $P_1 = (x_1, y_1, z_1)$, then the point $P_2 = (x_1 + bl_c - cl_b, y_1 +cl_a - al_c, z_1 + al_b-bl_a)$ will also satisfy both equations (expand and cancel terms). These are distinct points if $a:b:c \neq l_a : l_b: l_c $.