I think I see mysterious lines inside triangles—how to prove their existence?

Lately I've been fooling around with points inside a triangle and the sum of their distances from all sides.

This was when I noticed a weird behaviour: For each point I chose there always seemed to be a straight line going through my chosen point and the entire triangle where every point had the same sum of distances from all sides! And as if that's not enough, if I select a different point the line through this point looks parallel to all the other lines created in the same manner but through other points.

So is there a way to prove my observation?

Question: Do all points inside a triangle that have the same sum of distances from all sides lie on a line and is there a way to give a mathematical equation for said line? (Disregarding equilateral triangles)

Because I used numerical means to find this pattern I am not sure whether it even exists. Any kind of help will be appreciated!

This is easy to prove if you know a bit about vectors and dot products.

For any vector $\vec{a} \neq \vec{0}$ and any real number $b$, the set of points $\vec{x}$ which satisfy $\vec{a} \cdot \vec{x} = b$ forms a line that is perpendicular to $\vec{a}$. Every line in the plane can be written in this manner. Also, the distance between a point $\vec{y}$ and the line $\vec{a} \cdot \vec{x} = b$ is given by $\frac{|\vec{a} \cdot \vec{y} - b|}{\|\vec{a}\|}$.

Call the sides of the triangle side $1$, side $2$, and side $3$. For $i = 1,2,3$, pick a unit vector $\vec{a}_i \neq \vec{0}$ that is normal to side $i$ and points inward, and a number $b_i$ such that the points $\vec{x}$ on side $i$ satisfy $\vec{a}_i \cdot \vec{x} = b_i$

Then, the distance between a point $\vec{x}$ and side $i$ is given by $|\vec{a}_i \cdot \vec{x}-b_i|$. If $\vec{x}$ is inside the triangle, then $\vec{a}_i \cdot \vec{x}-b_i > 0$ (since we picked $\vec{a}_i$ to point inwards), and so, the distance from $\vec{x}$ to side $i$ is simply $\vec{a}_i \cdot \vec{x}-b_i$.

Thus, the total distance from a point $\vec{x}$ (inside the triangle) to the three sides of the triangle is $(\vec{a}_1 \cdot \vec{x}-b_1) + (\vec{a}_2 \cdot \vec{x}-b_2) + (\vec{a}_3 \cdot \vec{x}-b_3)$ $= (\vec{a}_1+\vec{a}_2+\vec{a}_3) \cdot \vec{x} - (b_1+b_2+b_3)$.

If we let $C$ be any constant, then the sum of the distances from a point $\vec{x}$ inside the triangle to the sides will equal $C$ provided $(\vec{a}_1+\vec{a}_2+\vec{a}_3) \cdot \vec{x} - (b_1+b_2+b_3) = C$, i.e. $(\vec{a}_1+\vec{a}_2+\vec{a}_3) \cdot \vec{x} = b_1+b_2+b_3+C$.

As long as $\vec{a}_1+\vec{a}_2+\vec{a}_3 \neq \vec{0}$, then this is a line which is normal to $\vec{a}_1+\vec{a}_2+\vec{a}_3$. Furthermore, no matter what we pick the total distance $C$, this line will be normal to $\vec{a}_1+\vec{a}_2+\vec{a}_3$. Hence, all the lines formed in this manner are perpendicular to a common vector, and thus, are parallel.

If you aren't familiar with vectors and dot products, you can write out $\vec{a}_1 = (a_{11},a_{12})$, $\vec{a}_2 = (a_{21},a_{22})$, $\vec{a}_3 = (a_{31},a_{32})$, $\vec{x} = (x,y)$, and carry out the same computations. You'll still end up with the equation of a line.

If the edges of the triangle lies on lines $A_i x + B_i y + C = 0$, then the sum of the distances from point $P=(p,q)$ to these lines is given by

$$d = \pm_1\;\frac{A_1 p + B_1 q + C_1}{\sqrt{A_1^2+B_1^2}} \;\pm_2\;\frac{A_2p+B_2 q+C_2}{\sqrt{A_2^2+B_2^2}} \;\pm_3\;\frac{A_3p + B_3 q + C_3}{\sqrt{A_3^2+B_3^2}} \qquad (\star)$$ where each "$\pm_i$" is "$+$" for all points on one side of the $i$-th line, and "$-$" for all points on the other side.

Note that all $P$ in the interior of the triangle lie on a particular side of each edge-line; each $\pm_i$ remains fixed. Therefore, $(\star)$ represents a linear equation for the interior points with a particular total distance $d$ from the edges of the triangle; that is to say: The solution points do, indeed, lie on a line. Congratulations on your perceptiveness!

(For each of the seven regions of the plane determined by the extended sides of the triangle, you get such a constant-sum-of-distances line.)

Let us look at the problem in a slightly different angle. If one know baricentric coordinate system, it will be "obvious" why this is true.

Let $\vec{A}, \vec{B}, \vec{C}$ be any three non-collinear points, they form the vertices of a non-degenerate triangle $\triangle ABC$. For any point $\vec{p} \in \mathbb{R}^2$, there exists a unique pair of real numbers $\alpha, \beta$ such that

$$\vec{p}-\vec{C} = \alpha( \vec{A}-\vec{C}) + \beta( \vec{B} - \vec{C} ) \quad\iff\quad \vec{p} = \alpha \vec{A} + \beta \vec{B} + (1 - \alpha - \beta)\vec{C} $$ Let $\gamma = 1 - \alpha - \beta$, the triplet $(\alpha,\beta,\gamma)$ is called the baricentric coordinates for $\vec{p}$. Furthermore, the points $\vec{p}$ lies inside or on $\triangle ABC$ if and only if $\alpha, \beta, \gamma \ge 0$

Let $h_A, h_B, h_C$ be the height of $\triangle ABC$ for corresponding vertices. The distance between $\vec{p}$ and the sides $BC$, $CA$, $AB$ are $h_A |\alpha|$, $h_B |\beta|$ and $h_C|\gamma|$ respectively. The loucs for a point whose sum of distances to the 3 sides equal to $d$ is then given by:

$$h_A |\alpha| + h_B|\beta| + h_C|\gamma| = d$$

For points inside $\triangle ABC$, the problem of finding the locus is equivalent to solving following pair of linear equations:

$$\begin{array}{rrrl} \alpha +& \beta +& \gamma &= 1\\ h_A \alpha +& h_B \beta +& h_C \gamma &= d \end{array} $$ When $\triangle ABC$ is not equilateral, this pair of equations has rank 2 which has either zero or infinite many solutions. Furthermore if $(\alpha, \beta, \gamma)$ is a solution, other solution will have the form:

$$(\alpha',\beta',\gamma') = (\alpha,\beta,\gamma) + \lambda (h_B-h_C,h_C-h_A,h_A-h_B)\quad\text{ for some } \lambda \in \mathbb{R}$$

Translate this back to points on $\mathbb{R}^2$. This mean is $\vec{p}$ is a point inside $\triangle ABC$, the locus of point $\vec{p}'$ have same sum of distances has the form:

$$\vec{p}' = \vec{p} + \lambda\vec{u}, \quad\text{ for some }\;\lambda \in \mathbb{R}$$ i.e. the locus is a line along the direction $\displaystyle\;\vec{u} = (h_B-h_C)\vec{A} + (h_C-h_A)\vec{B} + (h_A-h_B)\vec{C}$.

Please note that this $\vec{u}$ is independent of choice of $d$ and hence $\vec{p}$. What this means is for all points inside $\triangle ABC$, not only the locus of same distances are all lines, all those lines are parallel to each other!

Clarifications

About the question why multiplying $\alpha$ with height $h_A$ gives us the distance to line $BC$. For any point $p$, let $d_p$ be the distance of $p$ to the line $BC$. By definition, we have $$\vec{p} - \vec{C} = \alpha(\vec{A}-\vec{C}) + \beta(\vec{B}-\vec{C}).$$ For any fixed $\alpha$, let $\vec{p}_0 = \vec{C} + \alpha(\vec{A}-\vec{C})$. For any point $p$ with same $\alpha$, we have $$\vec{p} - \vec{p_0} = \beta (\vec{B} - \vec{C})$$ When viewed from $p_0$, $p$ is along the direction $\vec{B}-\vec{C}$. This means the locus of $p$ for fixed $\alpha$ is a line parallel to the side $BC$. As a result, $d_p$ is constant over such a line and $d_p$ depends only on $\alpha$. As long as $p$ doesn't crosses the line $BC$, it is clear this dependence on $\alpha$ is linear. Notice

• When $\alpha = 0$, the line of constant $\alpha$ coincides with line $BC$, so $d_p = 0$ there.
• When $\alpha = 1$, the line of constant $\alpha$ crosses $A$, so $d_p = h_A$ there.

Combine these, we find the proportional constant is $h_A$ when $\alpha \ge 0$. This means as long as $p$ is on the same side as $A$ with respect to line $BC$, $d_p = h_A \alpha = h_A |\alpha|$. By symmetry, $d_p = h_A |\alpha|$ on the other side too.