Viviani's Theorem, Three Distances and One Total, math that surprises you lesson

Pick a point inside an equilateral triangle and measure how far it is from each of the three sides. Move the point straight toward one side and that distance shrinks while the other two grow, and wherever the point goes, the three distances add up to the same number, the height of the triangle. This page explains why by cutting the triangle into three smaller triangles and counting area, and it shows what goes wrong in other triangles and for points outside.

Key terms

Viviani’s theorem

Definition Let $T$ be an equilateral triangle with side length $s$ and height $h$ . Take any point $P$ inside $T$ or on its boundary, and let $d_1$ , $d_2$ and $d_3$ be the perpendicular distances from $P$ to the three sides. Viviani’s theorem says that

d_1 + d_2 + d_3 = h

for every such point $P$ . The reason is an area count. The segments from $P$ to the three corners cut $T$ into three triangles. Each of them has one side of $T$ as its base, of length $s$ , and one of the distances as its height, so together they have area

\tfrac12 s\,(d_1 + d_2 + d_3).

These three triangles fill $T$ exactly, so this is also the area $\tfrac12 s h$ of $T$ , and dividing both by the positive number $\tfrac12 s$ gives the theorem. The distances must be measured to the sides along perpendiculars, not to the corners. For a point outside the triangle the three distances add up to more than $h$ . In a triangle whose sides are not all equal the sum is not the same at every point, and among all triangles only the equilateral ones have a constant sum. The theorem is named after the seventeenth-century Italian mathematician Vincenzo Viviani.

In plain words Whenever the point moves, whatever distance it gains from some sides it loses from the others, and the gains and losses cancel exactly. A point at a corner shows the total plainly, because it touches two sides and is a full height away from the third.

Example Take the triangle with corners $A = (-1, 0)$ , $B = (1, 0)$ and $C = (0, \sqrt3)$ , so that $s = 2$ and $h = \sqrt3 \approx 1.732$ . At the center $(0, \sqrt3/3)$ the three distances are equal by symmetry, each $\sqrt3/3$ , and they add up to $\sqrt3$ . The point $P = (1/2, \sqrt3/4)$ is off center. Its distance to the bottom side $AB$ is its $y$ -coordinate, $\sqrt3/4$ , and the formula in the perpendicular distance card gives $\sqrt3/8$ to the side $BC$ and $5\sqrt3/8$ to the side $CA$ . As fractions of the height these are

\frac{h}{4} + \frac{h}{8} + \frac{5h}{8} = h .

At the corner $C$ the distances are $0$ , $0$ and $\sqrt3$ , again a total of $h$ .

Equilateral triangle

Definition A triangle is equilateral when its three sides have the same length $s$ . Its three angles are then all $60^\circ$ . Its height $h$ is the perpendicular distance from a corner to the line through the opposite side, and it is the same from every corner. By symmetry the perpendicular from a corner meets the opposite side at its midpoint, so the Pythagorean theorem gives $h^2 + (s/2)^2 = s^2$ , and therefore

\begin{gathered} h = \frac{\sqrt3}{2}\, s, \\[0.4em] \text{Area} = \frac12 s h = \frac{\sqrt3}{4}\, s^2 . \end{gathered}

In plain words All sides are equal, all angles are equal, and the triangle looks the same from each corner. Viviani’s theorem uses exactly this sameness, since every side can serve as a base of the same length.

Example For side $s = 2$ the height is $\sqrt3 \approx 1.732$ and the area is $\sqrt3$ . For side $s = 6$ the height is $3\sqrt3 \approx 5.196$ and the area is $9\sqrt3 \approx 15.59$ .

Perpendicular distance

Definition The perpendicular distance from a point $P$ to a line $\ell$ is the length of the shortest segment from $P$ to a point of $\ell$ . That shortest segment meets $\ell$ at a right angle, at a point called the foot of the perpendicular. When the line has the equation $ax + by + c = 0$ , with $a$ and $b$ not both $0$ , and $P = (x_0, y_0)$ , the distance is

\frac{\lvert a x_0 + b y_0 + c \rvert}{\sqrt{a^2 + b^2}} .

In Viviani’s theorem the distance to a side means the distance to the line through that side. For a point inside an equilateral triangle or on its boundary, the foot always lies on the side itself, because every angle of the triangle is smaller than a right angle.

In plain words It is how far you would walk to reach the line if you headed straight for it. The distance to an end of the side, or to the line along any slanted direction, is a different quantity and is never shorter.

Example In the triangle with corners $A = (-1, 0)$ , $B = (1, 0)$ and $C = (0, \sqrt3)$ , the side $BC$ lies on the line $\sqrt3\, x + y - \sqrt3 = 0$ . For $P = (1/2, \sqrt3/4)$ the formula gives

\begin{aligned} d &= \frac{\lvert \sqrt3/2 + \sqrt3/4 - \sqrt3 \rvert}{\sqrt{3 + 1}} \\[0.4em] &= \frac{\sqrt3/4}{2} = \frac{\sqrt3}{8} . \end{aligned}

Distances to the corners do not share Viviani’s property. From the center $(0, \sqrt3/3)$ each corner is $2\sqrt3/3 \approx 1.155$ away, for a total of $2\sqrt3 \approx 3.464$ . From the corner $C$ the three corners are $0$ , $2$ and $2$ away, for a total of $4$ .

Area decomposition

Definition If a region is cut into finitely many pieces that cover it and overlap only along their edges, then the area of the region is the sum of the areas of the pieces. Edges have no area, so it does not matter which piece they are counted in. The area of a triangle is half a base times the height, where the height is the perpendicular distance from the opposite corner to the line through the base. In Viviani’s argument the segments from $P$ to the corners cut the triangle $ABC$ into the triangles $PBC$ , $PCA$ and $PAB$ . When $P$ lies on the boundary, one or two of them are flat with area $0$ , and the count still works. In any triangle with side lengths $a$ , $b$ and $c$ and area $K$ , the same decomposition gives

a\, d_a + b\, d_b + c\, d_c = 2K

for every point inside the triangle or on its boundary, where $d_a$ is the distance to the side of length $a$ , and likewise for $d_b$ and $d_c$ . Only when $a = b = c$ can the common length be divided out to leave a plain sum of distances.

In plain words One region counted in two ways must give the same area. Here one way is the whole triangle and the other is three slices meeting at the point. Moving the point shifts area from one slice to another but never changes the total.

Example Take the right triangle with corners $(0, 0)$ , $(4, 0)$ and $(0, 3)$ . Its sides have lengths $4$ along the bottom, $3$ along the left and $5$ along the slanted side $3x + 4y = 12$ , and its area is $6$ . At the point $(1, 1)$ the distances to these three sides are $1$ , $1$ and $1$ , with sum $3$ . At the point $(2, 1/2)$ they are $1/2$ , $2$ and $4/5$ , with sum $3.3$ . The plain sum has changed, but the weighted sums agree,

\begin{aligned} 4 \cdot 1 + 3 \cdot 1 + 5 \cdot 1 &= 12, \\[0.4em] 4 \cdot \tfrac12 + 3 \cdot 2 + 5 \cdot \tfrac45 &= 12, \end{aligned}

and both equal $2K$ . At the corners the plain sum is the height from that corner, which is $12/5 = 2.4$ at $(0, 0)$ , $4$ at $(4, 0)$ and $3$ at $(0, 3)$ .

Check yourself

A point inside an equilateral triangle with side $6$ is $1$ unit from one side and $2$ units from another. How far is it from the third side?

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The height of the triangle is

h = \frac{\sqrt3}{2} \cdot 6 = 3\sqrt3,

about $5.196$ . By Viviani’s theorem the three distances add up to $h$ , so the third distance is $3\sqrt3 - 3 \approx 2.196$ . The answer does not depend on where exactly the point sits, only on the two distances given.

Change a rule and move the point outside. In the triangle with corners $A = (-1, 0)$ , $B = (1, 0)$ and $C = (0, \sqrt3)$ , take $P = (0, -1)$ , below the bottom side. Add up its perpendicular distances to the three lines through the sides. Why does the area argument give a different total?

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The distance to the bottom line $y = 0$ is $1$ . The line through $B$ and $C$ is $\sqrt3\, x + y - \sqrt3 = 0$ , so the distance to it is

\frac{\lvert 0 - 1 - \sqrt3 \rvert}{2} = \frac{1 + \sqrt3}{2},

and by symmetry the distance to the line through $C$ and $A$ is the same. The total is $1 + (1 + \sqrt3) = 2 + \sqrt3 \approx 3.732$ , more than the height $\sqrt3 \approx 1.732$ .

The three triangles $PAB$ , $PBC$ and $PCA$ no longer fill $ABC$ without overlap. The triangles $PBC$ and $PCA$ together cover $ABC$ and also the triangle $PAB$ that sticks out below it, so the area of $ABC$ is the area of $PBC$ plus the area of $PCA$ minus the area of $PAB$ . Here $s = 2$ , so each of these triangles has area $\tfrac12 \cdot 2 \cdot d = d$ , the distance that serves as its height, and the area count reads

2 \cdot \frac{1 + \sqrt3}{2} - 1 = \sqrt3 .

So the distance to the bottom side has to be subtracted, not added. If a distance gets a minus sign whenever its side’s line separates the point from the triangle, the total is $h$ for every point of the plane. The plain total is therefore larger than $h$ at every point outside the triangle, here by $2 \cdot 1 = 2$ .

Extend the pattern to a regular hexagon with side $2$ . Show that the perpendicular distances from any point inside it to its six sides always add up to the same total, and find that total.

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Join the point to the six corners. This cuts the hexagon into six triangles that fill it without overlap, each with one side of the hexagon, of length $2$ , as its base and one of the distances as its height. Their areas add up to

\begin{aligned} &\tfrac12 \cdot 2 \cdot (d_1 + \dots + d_6) \\[0.4em] &= d_1 + \dots + d_6 . \end{aligned}

The segments from the center to the corners cut the hexagon into six equilateral triangles of side $2$ , each of area $\sqrt3$ , so the hexagon has area $6\sqrt3$ . Hence the six distances always add up to $6\sqrt3 \approx 10.39$ . Two checks agree. At the center each distance is the height $\sqrt3$ of one of those small triangles, and six of them make $6\sqrt3$ . Also, opposite sides are parallel and $2\sqrt3$ apart, and the distances from a point between two parallel lines add up to the gap between them, so the three pairs give $3 \cdot 2\sqrt3$ . The argument needs equal side lengths, not equal angles, so it works in every convex polygon whose sides all have the same length.