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!

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    $\begingroup$ I think you should edit the title, because it is just weird. $\endgroup$ – user230734 May 24 '15 at 0:26
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    $\begingroup$ I quite like the evocative title. $\endgroup$ – pjs36 May 24 '15 at 1:05
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    $\begingroup$ How to prove their existence ? - Build the triangles, and the proof will come. $\endgroup$ – Lucian May 24 '15 at 4:43
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    $\begingroup$ I think you should not edit the title, because it is just weird. $\endgroup$ – JiK May 24 '15 at 11:26
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    $\begingroup$ this question comes close to another problem of bundeswettbewerb mathematik in germany, round 2, deadline for entries Sept. 1st, 2015. please stick to the rules and don't ask for solutions in the community. thank you to all users for not posting comments or solutions to the above mentioned problem. Karl Fegert Grading commission bundeswettbewerb mathematik germany $\endgroup$ – user246719 Jun 8 '15 at 16:55

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!


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.

  • $\begingroup$ and +1 for at least mentioning equilateral triangles. $\endgroup$ – A. Rex May 26 '15 at 4:39
  • $\begingroup$ Could you please explain why by multiplying our $\alpha, \beta,\gamma$ with the corresponding height we obtain the distance of our point $p$ from one side? $\endgroup$ – user161516 May 28 '15 at 7:54
  • $\begingroup$ @Silenttiffy I updated the answer, I hope that help. $\endgroup$ – achille hui May 28 '15 at 10:10
  • $\begingroup$ @achillehui While your argument wonderfully explains why the formula is correct for $\alpha, \beta$ I can't make a connection to $\gamma$ because unlike the others $\gamma$ can not be understood as a component parallel to one side. Generally speaking, what is the use of $\gamma$ when it is completely dependent on $\alpha$ and $\beta$? $\endgroup$ – user161516 May 28 '15 at 21:56
  • $\begingroup$ @Silenttiffy If you relabel you vertices from $A,B,C$ to $A' = B, B' = C, C' = A$, then the corresponding $\alpha' = \beta, \beta' = \gamma, \gamma' = \alpha$. So the line for constant $\gamma$ = line of constant $\beta'$ is parallel to the line $C'A' = AB$. The point of baricentric coordinates is the formulas remains valid under relabeling. To attack a problem, you have the freedom to choose either $\alpha,\beta$; $\beta, \gamma$ or $\gamma,\alpha$ as independent variables for analysis. One typically pick the one to make the problem easiest. $\endgroup$ – achille hui May 28 '15 at 22:10

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.)

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    $\begingroup$ :))) $3$ identical answers within 5 minutes :) $\endgroup$ – Alexey Burdin May 24 '15 at 1:10
  • $\begingroup$ @Alexey: I would've beaten you if I didn't take so long editing. ;) I should probably just delete this answer. (Well ... hmmm ... someone just up-voted me. I guess there's perceived value in this near-duplicate, so I'll leave it. :) $\endgroup$ – Blue May 24 '15 at 1:14
  • $\begingroup$ All 3 answers mention different details. Ours don't mention palallelity, and yours do mention lines in the rest 6 regions on the plane. My answer is most poor one :) For editing: I put the \$ signs after all editing, otherwise it takes too slow indeed. $\endgroup$ – Alexey Burdin May 24 '15 at 1:19

Let the triangle vertices' Cartesian coordinates be $A(x_1,y_1),\ B(x_2,y_2),\ C(x_3,y_3)$.
Let the line equations of $AB, BC, CA$ be $A_3x+B_3y+C_3=0,\ A_2x+B_2y+C_2=0,\ A_1x+B_1y+C_1=0$ respectively.
We know that the distance from the line $Ax+By+C=0$ to a point $(x,y)$ is $$\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}\hbox{.} $$ Inside the triangle, the signs of $A_ix+B_iy+C$ do not change (wlog we assume these all three $\ge 0$), so the sum of distances is $$\sum\limits_{k=1}^3 \frac{A_kx+B_ky+C}{\sqrt{A_k^2+B_k^2}}$$ and it must equal to a $d$: $$\sum\limits_{k=1}^3 \frac{A_kx+B_ky+C}{\sqrt{A_k^2+B_k^2}}=d\iff$$ $$x\sum\limits_{k=1}^3 \frac{A_k}{\sqrt{A_k^2+B_k^2}}+ y\sum\limits_{k=1}^3 \frac{B_k}{\sqrt{A_k^2+B_k^2}}+ \sum\limits_{k=1}^3 \frac{C}{\sqrt{A_k^2+B_k^2}}-d=0\hbox{,}$$ which is surely a line.


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