Locus of points satisfying a given condition.

Question

Find the locus of the points P in the plane of an equilateral triangle ABC for which the triangle formed with PA, PB, and PC has constant area.

I have used the Heron's formula but no avail ( I already suspected that this problem could not be easily solved by this formula but I had no other idea since the only thing I know about the triangle is its sides PA PB PC ). I have no idea how to proceed. So any help will be appreciated.

Update : I have seen that futurologist says that the locus will be any circle centred at the circumcentre of ΔABC. But can anyone say why this is so?

Answer 1

Construction. Assume without loss of generality that $P$ is an arbitrary point in the plane such that $PA = \max\{PA, PB, PC\}$ and $PC = \min\{PA, PB, PC\}$. Then perform a $60^{\circ}$ counter-clockwise rotation around the point $A$ and let $P^*$ be the image of $P$ under the rotation. Then $PA=P^*A$ and $\angle \, PAP^* = 60^{\circ}$. Hence triangle $APP^*$ is equilateral and $PA=P^*A = PP^*$. Moreover, since triangle $ABC$ is equilateral itself, point $C$ is the image of point $B$ under the counter-clockwise $60^{\circ}$ rotation. Consequently, triangle $ACP^*$ is the rotation image of $ABP$ and so these two triangles are congruent. Thus, $PB=P^*C$.

Observe that triangle $CPP^*$ is a triangle with edges $PP^* = PA, \,\, P^*C = PB$ and $PC$.

Statement 1. Point $P$ satisfies the property $PA = PB + PC$ if and only if $P$ lies on the circumcircle of the equilateral triangle $ABC$, on the arc between points $B$ and $C$ not containing $A$.

Proof: Assume first that $PA = PB + PC$. Then $PP^* = P^*C + PC$ which is possible if and only if $C$ lies on the edge $PP^*$. But then $\angle \, CPA = \angle \, P^*PA = 60^{\circ}$ because $P^*PA$ is equilateral. Consequently $\angle\, CPA = \angle \, CBA = 60^{\circ}$ which means that point $P$ lies on the circumcircle of triangle $ABC$.

Conversely, let $P$ be on the circumcircle of the equilateral triangle $ABC$. Then $\angle\, CPA = \angle \, CBA = 60^{\circ}$. On the other hand by construction triangle $P^*PA$ is equilateral so $60^{\circ} = \angle \, P^*PA = \angle CPA$ which is possible if and only if point $C$ lies on the segment $P^*P$ (here is where we use the condition that $P$ is located on the arc $BC$ disjoint form $A$ which guarantees that $C$ and $P^*$ are on the same side of line $AP$). But then, $PP^* = P^*C + PC$ and since it has already been established that $PP^* = PA$ and $P^*C = PB$, we conclude that $PA = PB + PC$. $\square$

Statement 2. Let $P_1$ and $P_2$ be two points from the plane. Then two triangles, the first formed by edges of length $P_1A, P_1B, P_1C$ and the second formed by edges of length $P_2A, P_2B$ and $P_2C$, have the same signed area if and only if $P_1$ and $P_2$ lie on the same circle concentric with the circle circumscried around the equilateral triangle $ABC$. In the case of unsigned area (absolute value of area) the circles are two.

Proof: As before, pick an arbitrary point $P$ in the plane such that $PA = \max\{PA, PB, PC\}$ and $PC = \min\{PA, PB, PC\}$. Carry out the construction in the beginning. Then as proved in construction, triangle $CPP^*$ has edge lengths $PC, \, P^*C = PB$ and $PP^* = PA$.

From now on, given a polygon in the plane, by $S_{\text{polygon}}$ I denote the area of that polygon.

So we are interested in the area $S_{CPP^*}$ of triangle $CPP^*$.

Let $O$ be the center of the equilateral triangle $ABC$. Denote $AO = BO = CO = h_0$, the length $PO = r$ and let angle $\angle \, 180^{\circ} - \alpha$, i,e, $\alpha$ is the measure of the exterior angle of triangle $AOP$ at vertex $O$. Assume that the point $P$ is chosen so that $CPP^*$ is inside $APP^*$, meaning the area of $CPP^*$ is positive. Then

$$S_{CPP^*} = S_{APP^*} - (S_{ACP} + S_{ACP^*})$$

As proved in construction triangle $ACP^*$ is congruent to triangle $ABP$, so $S_{ACP^*}=A_{ABP}$ and thus \begin{align} S_{ACP} + S_{ACP^*} &= S_{ACP} + S_{ABP} = S_{ABPC} = S_{BPO} + S_{CPO} + S_{ABO} + S_{CAO}\\ &= S_{BPO} + S_{CPO} + 2S_0 \end{align} where $S_0 = S_{ABO} = S_{BCO} = S_{CAO}$ because the three triangles $ABO, BCO, CAO$ are congruent since $O$ is the center of equilateral triangle $ABC$. Observe that according to our notations $\angle \, BOP = 60^{\circ} + \alpha$ and $\angle \, COP = 60^{\circ} - \alpha$. Calculate the areas $$S_{BPO} = \frac{1}{2}\, BO \cdot PO \, \,\sin(60^{\circ}+\alpha) = \frac{1}{2}\, h_0 \, r \, \sin(60^{\circ}+\alpha)$$ $$S_{CPO} = \frac{1}{2}\, CO \cdot PO \, \,\sin(60^{\circ}-\alpha) = \frac{1}{2}\, h_0 \, r \, \sin(60^{\circ}-\alpha)$$ Since by construction triangle $APP^*$ is equilateral $$S_{APP^*} = \frac{1}{2}\, PA\cdot P^*A \, \,\sin(60^{\circ}) = \frac{\sqrt{3}}{4}\, PA^2$$ By the cosine law applied to triangle $AOP$ \begin{align}AP^2 &= AO^2 + PO^2 - 2 \, AO \cdot PO \, \cos(180^{\circ} - \alpha)\\ &= h_0^2 + r^2 + 2 \, h_0 \, r \, \cos(\alpha)\end{align} Thus $$S_{APP^*} = \frac{\sqrt{3}}{4}\,\big(h_0^2 + r^2 + 2 \, h_0 \, r \, \cos(\alpha)\big)$$ Putting all of this together \begin{align} S_{CPP^*} &= S_{APP^*} - (S_{ACP} + S_{ACP^*}) = S_{APP^*} - (S_{BPO} + S_{CPO} + 2S_0) = S_{APP^*} - (S_{BPO} + S_{CPO}) - 2S_0\\ &= \frac{\sqrt{3}}{4}\,\big(h_0^2 + r^2 + 2 \, h_0 \, r \, \cos(\alpha)\big) - \left( \frac{1}{2}\, h_0 \, r \, \sin(60^{\circ}+\alpha) + \frac{1}{2}\, h_0 \, r \, \sin(60^{\circ}-\alpha)\right) - 2 S_0\\ &= \frac{\sqrt{3}}{4}\,\big(h_0^2 + r^2 + 2 \, h_0 \, r \, \cos(\alpha)\big) - \frac{1}{2}\, h_0 \, r \, \Big( \sin(60^{\circ}+\alpha) + \sin(60^{\circ}-\alpha)\Big) - 2 S_0 \end{align} By trigonometry, $$\sin(60^{\circ}+\alpha) + \sin(60^{\circ}-\alpha) = \frac{\sqrt{3}}{2} \cos(\alpha)$$ and consequently, \begin{align} S_{CPP^*} &= S_{APP^*} - (S_{BPO} + S_{CPO}) - 2S_0\\ &= \frac{\sqrt{3}}{4}\,\big(h_0^2 + r^2 + 2 \, h_0 \, r \, \cos(\alpha)\big) - \frac{1}{2}\, h_0 \, r \, \Big( \sin(60^{\circ}+\alpha) + \sin(60^{\circ}-\alpha)\Big) - 2 S_0\\ &= \frac{\sqrt{3}}{4}\,\big(h_0^2 + r^2 \big) + \frac{\sqrt{3}}{2} \, h_0 \, r \, \cos(\alpha) - \frac{\sqrt{3}}{2}\, h_0 \, r \, \cos(\alpha) - 2 S_0\\ &= \frac{\sqrt{3}}{4}\,\big(h_0^2 + r^2 \big) - 2 S_0\\ &= \frac{\sqrt{3}}{4}\,\big(h_0^2 + OP^2 \big) - 2 S_0 \end{align} And there you have it, if two points $P_1$ and $P_2$ in the plane determine two triangles of the same signed area (see construction), that area equals $$S_{AP_1P_1*} = S_{AP_2P_2^*} = \frac{\sqrt{3}}{4}\,\big(h_0^2 + OP_1^2 \big) - 2 S_0 = \frac{\sqrt{3}}{4}\,\big(h_0^2 + OP_2^2 \big) - 2 S_0$$ and therefore $OP_1 = OP_2 = r$ so they lie on the circle centered at $O$ and of radius $r$. Conversely, if $P_1$ and $P_2$ lie on the circle centered at $O$ and of radius $r$, then $OP_1 = OP_2 = r$ so the corresponding triangle areas (see construction) are both equal to $\frac{\sqrt{3}}{4}\,\big(h_0^2 + r^2 \big) - 2 S_0$, so $S_{AP_1P_1^*} = S_{AP_2A_P^*} = \frac{\sqrt{3}}{4}\,\big(h_0^2 + r^2 \big) - 2 S_0$. The case of $CPP^*$ not containing $C$ is analogous.

Answer 2

Let $PA=a$, $PB=b$ and $PC=c$ be the distances of point $P$ from the vertices of equilateral triangle $ABC$. By Heron's formula, if $S$ is the area of the triangle of sides $a$, $b$ and $c$ we have $$ \tag{1} 16S^2=(a+b+c)(a+b-c)(a-b+c)(-a+b+c)=4b^2c^2-(b^2+c^2-a^2)^2. $$ Choose now the coordinates of $A$, $B$ and $C$ so that the center of $ABC$ is at the origin: $$ A=(r,0),\quad B=\left(-{1\over2}r,{\sqrt3\over2}r\right), \quad C=\left(-{1\over2}r,-{\sqrt3\over2}r\right), $$ where $r$ is the radius of $ABC$. If $P=(x,y)$ we then have $$ a^2=(x-r)^2+y^2,\quad b^2=\left(x+{1\over2}r\right)^2+\left(y-{\sqrt3\over2}r\right)^2,\quad c^2=\left(x+{1\over2}r\right)^2+\left(y+{\sqrt3\over2}r\right)^2. $$ Substituting that into $(1)$ one gets: $$ 16S^2=3(x^2+y^2-r^2)^2, \quad\hbox{that is:}\quad x^2+y^2=r^2\pm{4\over\sqrt3}S. $$ The latter is the equation of two circles centered at the origin, which are then the required locus. They are both real if $r^2>{4\over\sqrt3}S$, that is if $S<{1\over3}S_{ABC}$, and have radii $\sqrt{r^2\pm{4\over\sqrt3}S}$. If instead $S>{1\over3}S_{ABC}$, then only one of them is real.

EDIT.

The diagram below shows that for $S<{1\over3}S_{ABC}$ the locus is actually formed by two circles. The upper triangle on the right has sides congruent to $PA$, $PB$ and $PC$, the lower triangle has sides congruent to $QA$, $QB$ and $QC$. Both triangles have the same area.

Answer 3

Area of a triangle $\Delta $ with sides $(a,b,c)$ by simplified Heron/Brahmagupta formula written in the required form:

$$ 16 \Delta^2 = 2(a^2b^2 + b^2c^2+c^2a^2) -(a^4+b^4+c^4) \tag{1}$$

We take an equilateral triangle of circum-radius $2R$ with vertices (taken double for convenience)

$$ (2R,0), (-R,\pm \sqrt {3} R) \tag{2}$$

and compute sides of triangle of sides as three distances between the above vertices and variable point $ P (x,y) $ for required locus of same or constant area $\Delta$ enclosed.

$$ a^2 = (x-R)^2 + y^2;\, b^2 =(x+R)^2 +( y-\sqrt 3 R )^2;\,c^2 =(x+R)^2 +( y+ \sqrt 3 R )^2 \, ; \tag{3}$$

Plug into (1) and simplify algebraically, (CAS aided) ..

$$ |\Delta| = \dfrac{\sqrt {3}}{4} [(x^2 +y^2) - (2 R)^2] \tag{4} $$

$$ |\Delta| = \dfrac{\sqrt {3}}{4} [(x^2 +y^2) - (r_c)^2] \tag{5} $$

which are all Circle loci centered at origin and circum-radius $2R=r_c$

It is noticed that area of the triangle vanishes(becomes zero) when $P$ is taken as one vertex because sides $ ( \sqrt {3} R, \sqrt {3}R,0) $ cannot enclose any area. We have positive area for point $P$ outside the circum-circle as shown and negative when inside and zero on the circum-circle.

This can be verified with a short trig check for the three rays inside the circle, as $ (b+c-a) $ is one factor of area.

$$ b+c-a = PD\, [\cos(\pi/3 +u) + \cos(\pi/3 -u)- \cos u ] = PD \, ( 2 \cos \pi/3 \cos u - \cos u ) =0 \tag {6} $$

Area cannot be enclosed if on re-arrangement $a,b,c$ lie along a straight line, as it happens here. (or if $ a < b+ c $ which does not arise here). $60^0$ angles marked in red.

$$ 4R \Delta = abc$$

In fact by virtue of the above there is no need for any further calculation, we have constant areas $A$ for magnified circium radii !! due to geometric similarity...

Finally, if side length of each side of equilateral triangle $ ABC= L$ is given and the area $ |\Delta| =A $ of a triangle connecting sides to vertex $P$ is also given as $ A,$ so that circum-radius is $ r_c= \dfrac{L}{\sqrt3},$ then by virtue of (4) we have for positive Area Regime (Faint Blue and brown shaded annular rings):

$$ \boxed{ r_{locus} = \sqrt{ \frac{L^2}{3} +\frac{4A}{\sqrt3} } } \tag{6} $$

EDIT1:

Negative area Regime ( Dark and light Blue annular rings) has meaning/sense here as at a certain $ r_{locus} $ area is zero. The length and area cannot be arbitrarily/independently given beyond a certain limit in this regime.

The graph shows positive regime above x-axis and negative regime below x-axis as a parabola of quadratic equation solution for both signed areas $A$, continuously with respect to changing locus radius.

$$ \boxed{ r_{locus} = \sqrt{ \frac{L^2}{3} -\frac{4A}{\sqrt3} } } \tag{7} $$

A plot of Equation (7) shows that computed area of triangle with sides $ PA,PB,PC, $

$$ |A| \gt \sqrt3 L^2/12 $$

is not permissible for real $\, r_{locus}. $