The vertices of the equilateral triangle $ABC$ are at distances of $2, 3, 5$ units from a point $D$ lying in the plane of the triangle. Calculate the length of the side of the triangle.
First, let's examine what we can say about the position of point $D$. Suppose that point $D$ is inside the triangle, and let's say it is 2 units from $A$, 3 units from $B$, and 5 units from $C$. Since the triangle is equilateral, no vertex has a special role. Point $D$ on the first diagram is the intersection of the circles with radii 5 units centered at $C$, 3 units centered at $B$, and 2 units centered at $A$. Since $CD = 5$, $D$ can only be inside the triangle if $AC$, the side of the triangle, is longer than 5 units, but then the circles centered at $A$ and $B$ with radii 2 and 3 units, respectively, would not intersect. Therefore, $D$ cannot be an interior point. Similarly, it can be shown that $D$ cannot be on the sides $AC$ or $BC$ either.

If $D$ were on the segment $AB$, it would be the intersection of the circles centered at $A$ and $B$, which would mean $AB = 5$ units, but then $D$ could not be 5 units from $C$. Therefore, $D$ must be outside the triangle.

We will present three different solutions for the rest of the problem.

**Solution I.** Place the equilateral triangle in a Cartesian coordinate system such that vertex $A$ is at the origin, and the side $AB$ lies on the $x$-axis. Let the side length of the triangle be $a$, then the coordinates of the vertices are $A(0; 0)$, $B(a; 0)$, and $C\left(\frac{a}{2}; \frac{\sqrt{3}}{2} a\right)$. Let the coordinates of point $D$ be $(x; y)$, and $|\overrightarrow{DA}|=2$, $|\overrightarrow{DB}|=3$, $|\overrightarrow{DC}|=5$. Writing these in coordinates, we get the system of equations:

$$
x^{2}+y^{2}=4, \quad (1) \quad (x-a)^{2}+y^{2}=9, \quad (2) \quad \left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{\sqrt{3}}{2} a\right)^{2}=25. \quad (3)
$$

Perform the squaring and simplifications:

$$
x^{2}-2ax+a^{2}+y^{2}=9, \quad (2') \quad x^{2}-ax+y^{2}-\sqrt{3}ay+a^{2}=25. \quad (3')
$$

Substituting $x^{2}+y^{2}=4$ from (1) into (2') and (3'):

$$
x=\frac{a^{2}-5}{2a} \quad \text{and} \quad y=\sqrt{4-x^{2}}=\sqrt{4-\left(\frac{a^{2}-5}{2a}\right)^{2}}
$$

Substituting the expressions for $x$ and $y$ into (3'), we get the quartic equation for $a$:

$$
4a^{4}-152a^{2}+1444=0
$$

From which the side length of the triangle is $a=\sqrt{19} \approx 4.359$ units.

Zoltán Bakonyi (Bp., Veres Péter Gimn., IV. o.t.)

**Solution II.** Draw the triangle $ABC$ and the point $D$ outside it, such that $AD=2$, $BD=3$, and $CD=5$.

Rotate the triangle $CBD$ around point $C$ by $60^\circ$ so that $B$'s image, $B'$, falls on $A$, and $D$'s image is $D'$. Thus, $CD=CD'=5$, and triangle $DCD'$ is isosceles, and since $\angle DCD' = 60^\circ$, $\angle CD'D = \angle D'DC = 60^\circ$. Therefore, triangle $DCD'$ is equilateral, and $DD'=5$. Since $DA + AD' = 2 + 3 = 5$, points $D$, $A$, and $D'$ are collinear, and $\angle AD'C = 60^\circ$. Apply the cosine rule to triangle $AD'C$ on side $AC = a$:

$$
a^{2}=3^{2}+5^{2}-2 \cdot 3 \cdot 5 \cdot \cos 60^\circ
$$

From which $a=\sqrt{19}$, as we already saw.

Péter Fazekas (Bp., Apáczai Cs. J. Gimn., III. o.t.)

**Solution III.** Ptolemy's theorem states that in a cyclic quadrilateral, the product of the diagonals is equal to the sum of the products of the opposite sides. The converse of the theorem is also true, i.e., if the product of the diagonals of a quadrilateral is equal to the sum of the products of the opposite sides, then the quadrilateral is cyclic. (The theorem and its proof can be found, for example, in H. S. M. Coxeter-S. L. Greitzer: The New Mathematical Discoveries, p. 74, Gondolat Publishing, Budapest, 1977.)

In our case, $5a = 2a + 3a$, so $ADBC$ is a cyclic quadrilateral, and since $\angle ACB = 60^\circ$, $\angle BDA = 120^\circ$. Apply the cosine rule to triangle $BDA$:

$$
a^{2}=2^{2}+3^{2}-2 \cdot 2 \cdot 3 \cdot \cos 120^\circ = 2^{2}+3^{2}+6 = 19
$$

Thus, $a=\sqrt{19}$.

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