A triangle has sides of 79, 65, and 47 units in length. How far is the point from the vertices from which we see the sides of the triangle at equal angles?
I. Solution. In triangle $ABC$, let $BC=a=79$, $CA=b=65$, $AB=c=47$ units, and the point in question be $P$, with $PA=x$, $PB=y$, $PC=z$. The point cannot be outside the triangle in any case, for otherwise the sum of the viewing angles of two sides would equal the viewing angle of the third side, whether we try to place $P$ in the vertex domain of any angle or in the infinite "umbrella" domain adjacent to any side; thus, the three viewing angles cannot be equal. Similarly, $P$ cannot be on the perimeter of the triangle either.

Therefore, the sum of the viewing angles is $360^{\circ}$, and each has a value of $120^{\circ}$. We obtain $P$ by construction, taking the $120^{\circ}$ viewing arcs of two side segments on the side where the triangle is, and their common point is $P$. (Figure 1.)
![](https://cdn.mathpix.com/cropped/2024_05_02_a4c18032b9285a720edfg-1.jpg?height=650&width=828&top_left_y=500&top_left_x=626)

Figure 1

Apply the cosine theorem successively to the sides opposite $P$ in triangles $PBC$, $PCA$, and $PAB$. The cosine of the angle at $P$ is $(-0.5)$ in each, so

$$
\begin{aligned}
& x^{2}+y^{2}+xy=c^{2} \\
& y^{2}+z^{2}+yz=a^{2} \\
& z^{2}+x^{2}+zx=b^{2}
\end{aligned}
$$

Subtract (3) from (2) and (1) from (3):

$$
\begin{aligned}
& y^{2}-x^{2}+z(y-x)=(y-x)(x+y+z)=a^{2}-b^{2} \\
& z^{2}-y^{2}+x(z-y)=(z-y)(x+y+z)=b^{2}-c^{2}
\end{aligned}
$$

Here it is useful to abandon the search for a general solution, as the given side lengths provide significant simplification, with the right-hand side of the last two equations being equal:

$$
a^{2}-b^{2}=79^{2}-65^{2}=2016=65^{2}-47^{2}=b^{2}-c^{2}
$$

Thus, from the equality of the left-hand sides of (4) and (5), given $x+y+z>0$,

$$
y-x=z-y>0
$$

which means $x, y$, and $z$ form an increasing arithmetic sequence. Therefore, the sum of the three unknowns in (4) and (5) is $3y$, and substituting this and the numerical values,

$$
y-x=z-y=\frac{2016}{3y}=\frac{672}{y}
$$

From this, the extreme terms of the sequence can be expressed in terms of the middle term:

$$
x=y-\frac{672}{y}=\frac{y^{2}-672}{y}, \quad z=y+\frac{672}{y}=\frac{y^{2}+672}{y}
$$

and substituting these into (3) gives a quadratic equation in $y^2$:

$$
\begin{gathered}
\left(y^{2}+672\right)^{2}+\left(y^{2}-672\right)^{2}+\left(y^{4}-672^{2}\right)=(65y)^{2} \\
3y^{4}-65^{2}y^{2}+672^{2}=0 \\
y^{2}=\frac{1}{6}(4225 \pm \sqrt{12431617})=\left\{\begin{array}{c}
1291.8 \\
611.5
\end{array}\right.
\end{gathered}
$$

It is sufficient to take the positive values of $y$:

$$
y_{1}=35.94, \quad y_{2}=10.79
$$

For the substitution into (6), $672/y_{1}=18.70$, $672/y_{2}=62.26$, and the sought distances are:

$$
x=PA=17.24, \quad y=PB=35.94, \quad z=PC=54.61 \text{ units. }
$$

The other solution: $x_{2}=-51.46, y_{2}=10.79, z_{2}=73.04$, if we take the absolute value of $x