Loci of Points Inspired by Viviani's Theorem

TL;DR

This paper explores geometric loci related to a triangle where the sum of distances or squared distances to sides remains constant, extending Viviani's theorem to new shapes like line segments and ellipses.

Contribution

It introduces new geometric loci based on Viviani's theorem, including the characterization of these loci as line segments, the entire triangle, or ellipses.

Findings

Loci of points with constant sum of distances form line segments or the entire triangle.

Loci of points with constant sum of squared distances form ellipses.

The work extends Viviani's theorem to broader geometric contexts.

Abstract

We consider loci of points such that their sum of distances or sum of squared distances to each of the sides of a given triangle is constant. These loci are inspired by Viviani's theorem and its extension. The former locus is a line segment or the whole triangle and the latter locus is an ellipse.

Tables

Table 1. Table 1: Examples

$a$	$b$	${\alpha }^2=a^2+3b^2$	${\beta }^2=2a^2$	Ellipse	$k$	Figure
$1$	$1$	$4$	$2$	$\frac{x^2}{4}+\frac{y^2}{2}=1$	$\frac{9}{2}$	5
$\sqrt{3}$	$1$	$6$	$6$	$x^2+y^2=6$	$10$	6

Equations

$$\mathcal{V}=x+y-\frac{3x+4y-12}{5}=\frac{2}{5}x+\frac{1}{5}y+\frac{12}{5}.$$

$$x^2+y^2+{\left(\frac{3x+4y-12}{5}\right)}^2=5.$$

$$34x^2+41y^2+24xy-72x-96y+19=0.$$

$$\underset{i=1}{\sum ^3}{d}_i^2=\frac{(ax+cy-ac{)}^2}{a^2+c^2}+\frac{(ax-by+ab{)}^2}{a^2+b^2}+y^2.$$

$$\frac{(ax+cy-ac{)}^2}{a^2+c^2}+\frac{(ax-by+ab{)}^2}{a^2+b^2}+y^2=k.$$

$$\mathcal{A}{x}^2+\mathcal{B}xy+\mathcal{C}{y}^2+\mathcal{D}x+\mathcal{E}y+\mathcal{F}=0,$$

$$\mathcal{A}=\frac{a^2}{p}+\frac{a^2}{q},\mathcal{B}=\frac{2ac}{p}-\frac{2ab}{q},\mathcal{C}=\frac{c^2}{p}+\frac{b^2}{q}+1$$

$$\delta ={\mathcal{B}}^2-4\mathcal{A}\mathcal{C}=-4\frac{a^2}{pq}\left(b^2+2bc+c^2+p+q\right)\text{,}$$

$$\frac{(ax+by-ab{)}^2}{a^2+b^2}+\frac{(ax-by+ab{)}^2}{a^2+b^2}+y^2=k.$$

$$\frac{x^2}{a^2+3b^2}+\frac{(y-\frac{2a{b}^2}{a^2+3b^2}{)}^2}{2a^2}=\frac{(a^2+b^2)k-2a^2{b}^2}{2a^2(a^2+3b^2)}+\frac{2b^4}{(a^2+3b^2{)}^2}.$$

$$\frac{(a^2+b^2)k-2a^2{b}^2}{2a^2(a^2+3b^2)}+\frac{2b^4}{(a^2+3b^2{)}^2}=1,$$

$$k=\frac{2a^2(a^4+7a^2{b}^2+10b^4)}{(a^2+b^2)(a^2+3b^2)}.$$

$$\mathcal{A}=\mathcal{C}\text{ and }\mathcal{B}=0.$$

$$\frac{a^2}{p}+\frac{a^2}{q}=\frac{c^2}{p}+\frac{b^2}{q}+1\text{ and }\frac{2ac}{p}-\frac{2ab}{q}=0.$$

$$\frac{2ac}{p}-\frac{2ab}{q}=0\iff (a^2-bc)(c-b)=0.$$

$$\frac{(a^2+b^2)k-2a^2{b}^2}{2a^2(a^2+3b^2)}+\frac{2b^4}{(a^2+3b^2{)}^2}\ge 0.$$

$$k\ge \frac{2a^2{b}^2}{a^2+3b^2}.$$

Full text

Loci of Points Inspired by Viviani’s Theorem

Elias Abboud

Faculty of education, Beit Berl College, Doar Beit Berl 44905, Israel

eabboud@beitberl.ac.il

Viviani (1622-1703), who was a student and assistant of Galileo, discovered that equilateral triangles satisfy the following property: the sum of the distances from the sides of any point inside an equilateral triangle is constant.

Viviani’s theorem can be easily proved by using areas. Joining a point $P$ inside the triangle to its vertices divides it into three parts. The sum of their areas will be equal to the area of the original one. Therefore, the sum of the distances from the sides will be equal to the altitude, and the theorem follows.

Given a triangle (which includes both boundary and inner points), $\Delta,$ and a positive constant, $k,$ we shall consider the following questions:

Question 1.** ***What is the locus $T_{k}(\Delta)$ of points $P$ in $\Delta$ such that the sum of the distances of $P$ to each of the sides of $\Delta$ equals $k$? *

Question 2. What is the locus $S_{k}(\Delta)$ of points $P$ in the plane such that the sum of the squares of the distances of $P$ to each of the sides of $\Delta$ equals $k$?

By Viviani’s theorem, the answer to the first question is straightforward for equilateral triangles: If $k$ equals the altitude of an equilateral triangle, then the locus of points is the entire triangle. On the other hand, if $k$ is not equal to the altitude of the equilateral triangle then the locus of points is empty.

Since the distance function is linear with two variable and because of squaring the distances, the locus of points in the second question, if not empty, is a quadratic curve. In our case, we shall see that this quadratic curve is always an ellipse. This fact will be the basis of a new characterization of ellipses.

We start first with the locus of points which have constant sum of distances to each of the sides of a given triangle and consider separately the cases of isosceles and scalene triangles. Then, we turn to the locus of points which have constant sum of squared distances to each of the sides of a given triangle. We prove the main theorem which gives a characterization of ellipses and conclude by discussing minimal sum of squared distances.

Constant sum of distances

Samelson [6, p. 225] gave a proof of Viviani’s theorem that uses vectors and Chen & Liang [3, p. 390-391] used this vector method to prove a converse: if inside a triangle there is a circular region in which the sum of the distances from the sides is constant, then the triangle is equilateral. In [2], this converse is generalized in the form of two theorems that we will use in addressing Question 1. To state these theorems, we need the following terminology: Let $\mathcal{P}$ be a polygon consisting of both boundary and interior points. Define the distance sum function $\mathcal{V}:\mathcal{P}\rightarrow\mathbb{R},$ where for each point $P\in\mathcal{P}$   the value $\mathcal{V}(P)$ is defined as the sum of the distances from $P$ to the sides of $\mathcal{P}.$

Theorem 1

Any triangle can be divided into parallel line segments on which $\mathcal{V}$ is constant. Furthermore, the following conditions are equivalent:

• •

$\mathcal{V}$* is constant on the triangle $\Delta.$*

• •

There are three non-collinear points, inside the triangle, at which $\mathcal{V}$ takes the same value.

• •

$\Delta$* is equilateral.*

Theorem 2

(a) Any convex polygon $\mathcal{P}$ can be divided into parallel line segments, on which $\mathcal{V}$ is constant.

(b) If $\ \mathcal{V}$ takes equal values at three non-collinear points, inside a convex polygon, then $\mathcal{V}$ is constant on $\mathcal{P}$ .

The discussion in [2, p. 207-210] lays out a connection between linear programming and Theorems 1 and 2. Theorem 1 is proved explicitly by defining a suitable linear programming problem and Theorem 2 is proved by means of analytic geometry, where the distance sum function $\mathcal{V}$ is computed directly and shown to be linear in two variables. We shall apply these theorems to find the loci $T_{k}(\Delta)$ for isosceles and scalene triangles $\Delta.$

Isosceles triangle

For isosceles triangles we exploit their reflection symmetry to find directly the line segments on which $\mathcal{V}$ is constant, and for scalene triangles we apply any of the above theorems to show that $\mathcal{V}$ is constant on certain parallel line segments. To find the direction of these line segments we compute explicitly the equation of $\mathcal{V}$. The idea is illustrated through an example.

Since an isosceles triangle has a reflection symmetry across the altitude, we conclude that; if the sum of distances of point $P$ from the sides of the triangle is $k$, then the reflection point $P^{\prime}$ across the altitude satisfies the same property. For such triangles we have the following proposition.

Proposition 1

If $\Delta$ is a non-equilateral isosceles triangle and $k$ ranges between the lengths of the smallest and the largest altitudes of the triangle, then $T_{k}(\Delta)$ is a line segment, whose end points are on the boundary of the triangle, parallel to the base.

Proof.

In Fig. 1, if the line segment $DE$ is parallel to the base $BC$ then, when $P$ moves along $DE$, the length $a$ remains constant and $b+c=DP\sin\alpha+PE\sin\alpha=DE\sin\alpha,$ which is also constant. Hence, $a+b+c=k$ is constant on the segment $DE.$ By Theorem 1, there are no other points in $\Delta$ with distance sum $k$ unless the triangle is equilateral.

If $a$ approaches zero then the length of $DE$ approaches the length of the base $BC,$ and $k=a+DE\sin\alpha$ approaches the length of the altitudes from $B$ (or $C).$ On the other hand, if the length of $DE$ approaches zero then $k$ approaches the length of the altitude from $A$. Hence, $T_{k}(\Delta)$ is defined whenever $k$ ranges between the lengths of the smallest and the largest altitudes of the triangle.

Scalene triangle

Similarly, by Theorem 1, the locus of points $T_{k}(\Delta)$ for a scalene triangle $\Delta$ is a line segment, provided $k$ ranges between the lengths of the smallest and the largest altitudes of the triangle. Otherwise, $T_{k}(\Delta)$ will be empty. Indeed, the connection with linear programming allows us to deduce the result. The convex polygon is taken to be the feasible region, and the distance sum function $\mathcal{V}$ corresponds to the objective function. The parallel line segments, on which $\mathcal{V}$ is constant, correspond to the isoprofit lines. The mathematical theory behind linear programming states that an optimal solution to any problem will lie at a corner point of the feasible region. If the feasible region is bounded, then both the maximum and the minimum are attained at corner points. Now, the distance sum function $\mathcal{V}$ is a linear continuous function in two variables. The values of $\mathcal{V}$ at the vertices of the feasible region, which is the triangle $\Delta,$ are exactly the lengths of the altitudes. Moreover, this function attains its minimum and its maximum at the vertices of the triangle, and ranges continuously between its extremal values. Hence, it takes on every value between its minimum and its maximum. Therefore, we have the following proposition.

Proposition 2

If $\Delta$ is a scalene triangle and $k$ ranges between the lengths of the smallest and the largest altitudes of the triangle, then $T_{k}(\Delta)$ is a line segment, whose end points are on the boundary of the triangle.

The question is: How can we determine this segment for a general triangle? We shall illustrate the method by the following example. Given the vertices of a triangle, we first compute the equations of the sides then we find the distances from a general point $(x,y)$ inside the triangle to each of the sides. In this way we obtain the corresponding distance sum function $\mathcal{V}$. Taking $\mathcal{V}=c,$ we get a family of parallel lines which, for certain values of the constant $c,$ intersect the given triangle in the desired line segments.

Example 1

Let $\Delta$ be the right angled triangle with vertices $(0,0),(0,3)$ and $(4,0),$ respectively (see Fig. 2). If the constant sum of distances from the sides is $k,$ $2.4\leq k\leq$ $4,$ then the locus is a line segment inside the triangle parallel to the line $2x+y=0$.

In Fig. 2, $T_{2.4}(\Delta)=\{A\},T_{2.8}(\Delta)=DG,T_{3.2}(\Delta)=EH,T_{3.6}(\Delta)=FI$ and $T_{4}(\Delta)=\{C\}.$ Note that at the extremal values of $k,$ the segments shrink to a corner point of the triangle $\Delta.$

Indeed, the smallest altitude of the triangle is $2.4$ and the largest altitude is $4.$ Hence, by the previous proposition, the locus of points is a line segment. To find this line segment, we compute the distance sum function $\mathcal{V}.$ The equation of the hypotenuse is $3x+4y=12.$ Hence,

[TABLE]

Therefore, the lines $\mathcal{V}=c$ are parallel to the line $2x+y=0$ and the result follows.

Note that the proofs of Theorems 1 and 2 follow the same line.

Constant sum of squares of distances

Motivated by the previous results, we deal with the second question of finding the locus of points which have a constant sum of squares of distances from the sides of a given triangle. Referring to Example 1, $DE^{2}+DF^{2}+DG^{2}=5$ (the number 5 is chosen arbitrary) implies the following equation of a quadratic curve;

[TABLE]

Simplifying, one gets the equivalent equation

[TABLE]

This is exactly the equation of the ellipse shown in Fig. 3.

A new characterization of the ellipse

In view of the previous example we may prove the following theorem, which states that among all quadratic curves the ellipse can be characterized as the locus of points that have a constant sum of squares of distances from the sides of an appropriate triangle.

Theorem 3

The locus $S_{k}(\Delta ABC)$ of points that have a constant sum of squares of distances from the sides of an appropriate triangle $ABC$ is an ellipse (and vice versa).

Proof.

We shall prove the following two claims:

(a) Given a triangle, the locus of points which have a constant sum of squares of distances from the sides is an ellipse. These loci, for different values of the constant, are homothetic ellipses with respect to their common center (their corresponding axes are proportional with the same factor).

(b) Given an ellipse, there is a triangle for which the sum of the squares of the distances from the sides, for all points on the ellipse, is constant.

Using analytic geometry, we choose for part (a) the coordinate system such that the vertices of the triangle lie on the axes. Suppose the coordinates of the vertices of the triangle are $A(0,a),B(-b,0),C(c,0),$ where $a,b,c>0$ (see Fig. 4). Let $(x,y)$ be any point in the plane and let $d_{1},d_{2},d_{3}$ be the distances of $(x,y)$ from the sides of the triangle $ABC$. It follows that

[TABLE]

Hence, $\underset{i=1}{\overset{3}{\sum}}d_{i}^{2}=k$ (constant) if and only if the point $(x,y)$ lies on the quadratic curve:

[TABLE]

In general, a quadratic equation in two variables,

[TABLE]

represents an ellipse provided the discriminant $\delta=\mathcal{B}^{2}-4\mathcal{AC}<0$ is negative.

To prove our claim, note that from equation (1) we have:

[TABLE]

where, $p=$ $a^{2}+c^{2}$ and $q=$ $a^{2}+b^{2}.$ Therefore,

[TABLE]

and the result follows.

Now, we prove part (b). Applying rotation or translation of the axes we may assume that the ellipse has a canonical form $\frac{x^{2}}{\alpha^{2}}+\frac{y^{2}}{\beta^{2}}=1,$ $\alpha\geq\beta>0.$ If we find positive real numbers $a,b$ with $\alpha^{2}=a^{2}+3b^{2},\beta^{2}=2a^{2},$ then the isosceles triangle with vertices $A^{\prime}(0,a-l),B^{\prime}(-b,-l),C^{\prime}(b,-l),$ gives the required property in part (b), where $l=\frac{2ab^{2}}{a^{2}+3b^{2}}.$

Indeed, because of the symmetry of the ellipse, we choose first an isosceles triangle with vertices $A(0,a),B(-b,0),C(b,0)$ and compute the sum of the squared distances from its sides. Substituting $c=b$ in equation (1), gives the equation of the locus $S_{k}(\Delta ABC)$ as

[TABLE]

Equivalently, we get a translation of a canonical ellipse:

[TABLE]

Thus, it is enough to take

[TABLE]

and therefore,

[TABLE]

Translating downward by $l=\frac{2ab^{2}}{a^{2}+3b^{2}},$ we get that the ellipse $\frac{x^{2}}{a^{2}+3b^{2}}+\frac{y^{2}}{2a^{2}}=1$ is the locus of points, $S_{k}(\Delta A^{\prime}B^{\prime}C^{\prime}),$ which have a constant sum of squares of distances from the sides of the triangle with vertices $A^{\prime}(0,a-l),B^{\prime}(-b,-l),C^{\prime}(b,-l).$ Moreover, this constant is given by equation (5).

Table 1 includes two examples that demonstrate Theorem 3.

Notice that, though a given triangle and a given constant $k$ yield at most one ellipse, the same ellipse can be obtained using different triangles. This can be easily seen in Fig. 5. By the above computations, the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ is the locus of points $S_{4.5}(\Delta),$ where $\Delta$ is the triangle with vertices $A^{\prime}=$ $(0,0.5),B^{\prime}=(-1,-0.5)$ and $C^{\prime}=(1,-0.5).$ If we reflect the whole figure across the $x-$axis, then the same ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$ will be the locus of points $S_{4.5}(\widetilde{\Delta}),$ where $\widetilde{\Delta}$ is the reflection of $\Delta$ across the $x-$axis with vertices: $(0,-0.5),(-1,0.5)$ and $(1,0.5).$

Notice also that, in the second example, the locus of points is a circle. In general, the locus of points is a circle exactly when, in equation (2),

[TABLE]

Equivalently, from (3) we have;

[TABLE]

Substituting the values of $p$ and $q$ and simplifying we get;

[TABLE]

Consequently, two cases have to be considered:

First, $a^{2}=bc,$ in which case the triangle $\Delta ABC$ is right angled. This case does not occur, since the conditions in (6) lead to a contradiction as follows: $\frac{2ac}{p}-\frac{2ab}{q}=0$ implies $\frac{c}{p}=\frac{b}{q}$ or equivalently $\frac{bc}{p}=\frac{b^{2}}{q}.$ Since, $a^{2}=bc$ we get $\frac{a^{2}}{p}=\frac{b^{2}}{q}.$ Hence, the first condition in (6) simplifies into $\frac{a^{2}}{q}=\frac{c^{2}}{p}+1.$ This equation together with the relations $a^{2}=bc$ and $\frac{c}{p}=\frac{b}{q},$ yield a contradiction.

Second, $c=b,$ in which case the triangle $\Delta ABC$ is isosceles. In this case, we have $p=q$. Substituting in the first condition of (6) we get $a^{2}+3b^{2}=2a^{2},$ which is equivalent to $a=\sqrt{3}b.$ Hence, the vertices of the triangle are $A(0,\sqrt{3}b),B(-b,0)$ and $C(b,0)$. This implies that the triangle is equilateral.

Therefore, we have the following result.

Conclusion 1

The locus $S_{k}(\Delta)$ of points that have a constant sum of squares of distances from the sides of a given triangle $\Delta$ is a circle if and only if the triangle $\Delta$ is equilateral.

Minimal sum of squared distances

Another interesting question is to find the minimal sum of squared distances, from the sides of a given triangle. For the isosceles triangle with vertices $A(0,a),B(-b,0)$ and $C(b,0),$ equation (4) indicates that, as the right hand side approaches $0,$ the ellipse degenerates to one point: $(0,\frac{2ab^{2}}{a^{2}+3b^{2}}).$ Thus, the locus of points is defined exactly when

[TABLE]

Equivalently,

[TABLE]

Therefore the following consequence holds.

Conclusion 2

For the isosceles triangle with vertices $A(0,a),B(-b,0)$ and $C(b,0),$ the minimal sum of squared distances from the sides is $\frac{2a^{2}b^{2}}{a^{2}+3b^{2}}$ attained at the point $(0,\frac{2ab^{2}}{a^{2}+3b^{2}}),$ inside the triangle.

When $a^{2}=3b^{2}$ then the triangle is equilateral with vertices $A(0,\sqrt{3}b),B(-b,0)$ and $C(b,0).$ In this case, the minimal sum of squared distances is $b^{2}$ attained at the point $(0,\frac{\sqrt{3}}{3}b)$ which is exactly the incenter of the equilateral triangle. In Fig. 7, $b=1$ and the loci of points are circles which degenerate to the incenter of the equilateral triangle as $k$ approaches $1.$

Concluding remarks

Other related results are the following: Kawasaki [4, p. 213], with a proof without words, used only rotations to establish Viviani’s theorem. Polster [5], considered a natural twist to a beautiful one-glance proof of Viviani’s theorem and its implications for general triangles.

The restriction of the locus, in the first question, to subsets of the closed triangle is intended to avoid the use of the signed distances. These signed distances, when considered, allows us to search loci of points outside the triangle for ”large” values of the constant $k.$ The reader is encouraged to do some examples.

Theorem 2, allows us to address Question 1 to any convex polygon in the plane.

The question about loci of points, with constant sum of squares of distances, can be generalized to any polygon or any set of lines in the plane. In this case, one should distinguish whether all the lines are parallel or not. The reader is encouraged to work out some examples using GeoGebra.

Conclusion 2 can be restated for general triangles, by a change of the coordinate system. This demands familiarity with the following theorem (which is beyond our discussion): Every real quadratic form $q=X^{T}AX$ with symmetric matrix $A$ can be reduced by an orthogonal transformation to a canonical form.

Finally, this characterization of the ellipse was exploited to build an algorithm for drawing ellipses using GeoGebra (see [1]).

Acknowledgement: The author is indebted to the referees, who read the manuscript carefully, and whose valuable comments concerning the style, the examples, the theorems and the proofs improved substantially the exposition of the paper. Special thanks are also due to the editor for his valuable remarks.

This work is part of research which was supported by Beit Berl College Research Fund.