## Task 6 - 321236

Let $k_{1}, k_{2}$, and $k_{3}$ be three concentric circles with radii $r_{1}=5, r_{2}=3 \sqrt{2}$, and $r_{3}=1$, respectively.

Determine the maximum possible value for the area of a triangle $A B C$ with the property that $A$ lies on $k_{1}$, $B$ lies on $k_{2}$, and $C$ lies on $k_{3}$.

Hint: It can be assumed as a known fact that among all triangles with the mentioned property, there exists a triangle with the maximum possible area.
Let point $D$ be the center of the concentric circles. It is also the orthocenter of triangle $ABC$. This means that if we draw lines through the center $D$ and the vertices, these lines intersect the sides of the triangle at right angles. Why this must be the case can be easily understood:

If we consider two points as fixed, for example, $A$ and $B$, then the area of the triangle is equal to the base $AB$ times half the distance of point $C$ from the side $AB$. If point $C$ can only rotate around $D$ at a distance $r_3$, the distance of point $C$ from $AB$ is maximized (and thus the area as well) when the line $CD$ is perpendicular to $AB$. The same applies analogously to the other points.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-2322.jpg?height=808&width=826&top_left_y=1749&top_left_x=615)

Let $x_3$ be the extension of the segment $CD$ to the segment $AB$. Then we have:

$$
x_3^2 + p_3^2 = r_1^2
$$

$$
x_3^2 + q_3^2 = r_2^2
$$

The area of the triangle to be maximized is

$$
A = \frac{1}{2} (x_3 + r_3) (p_3 + q_3)
$$

Solving (1) and (2) for $p_3$ and $q_3$ and substituting into (3), we get:

$$
A(x) = \frac{1}{2} (x_3 + r_3) \left( \sqrt{r_1^2 - x_3^2} + \sqrt{r_2^2 - x_3^2} \right)
$$

Differentiating with respect to $x_3$ and setting to zero to find the maximum (we can eliminate the factor $\frac{1}{2}$ by setting to zero):

$$
\begin{aligned}
& 0 = \left( \sqrt{r_1^2 - x_3^2} + \sqrt{r_2^2 - x_3^2} \right) + (x_3 + r_3) \left( \frac{-x_3}{\sqrt{r_1^2 - x_3^2}} + \frac{-x_3}{\sqrt{r_2^2 - x_3^2}} \right) \\
& 0 = \left( \sqrt{r_1^2 - x_3^2} + \sqrt{r_2^2 - x_3^2} \right) - x_3 (x_3 + r_3) \frac{\sqrt{r_1^2 - x_3^2} + \sqrt{r_2^2 - x_3^2}}{\sqrt{r_1^2 - x_3^2} \cdot \sqrt{r_2^2 - x_3^2}} \\
& 0 = 1 - \frac{x_3 (x_3 + r_3)}{\sqrt{r_1^2 - x_3^2} \cdot \sqrt{r_2^2 - x_3^2}} \\
& x_3 (x_3 + r_3) = \sqrt{r_1^2 - x_3^2} \cdot \sqrt{r_2^2 - x_3^2} \\
& x_3^4 + 2 r_3 x_3^3 + r_3^2 x_3^2 = (r_1^2 - x_3^2)(r_2^2 - x_3^2) \\
& x_3^4 + 2 r_3 x_3^3 + r_3^2 x_3^2 = x_3^4 - (r_1^2 + r_2^2) x_3^2 + r_1^2 r_2^2 \\
& 2 r_3 x_3^3 + (r_1^2 + r_2^2 + r_3^2) x_3^2 - r_1^2 r_2^2 = 0
\end{aligned}
$$

This results in a cubic equation for $x_3$. Substituting the numerical values, we get:

$$
\begin{gathered}
2 x_3^3 + 44 x_3^2 - 450 = 0 \\
x_3^3 + 22 x_3^2 - 225 = 0
\end{gathered}
$$

We can and must guess the solution $x_3 = 3$ and factor the equation:

$$
(x_3 - 3)(x_3^2 + 25 x_3 + 75) = 0
$$

The other solutions of the quadratic term are not valid because they are both negative. With $x_3 = 3$, we also get

$$
p_3 = \sqrt{25 - 9} = 4
$$

and

$$
q_3 = \sqrt{18 - 9} = 3
$$

Thus, the maximum triangle area is

$$
A_{\max} = \frac{1}{2} (3 + 1)(3 + 4) = 14
$$

Additional note on the general solution: Multiplying the above general cubic equation by $r_3^2$, we get:

$$
2 r_3^3 x_3^3 + (r_1^2 + r_2^2 + r_3^2) r_3^2 x_3^2 - r_1^2 r_2^2 r_3^2 = 0
$$

Setting $x_3 r_3 = c$, we obtain:

$$
2 c^3 + (r_1^2 + r_2^2 + r_3^2) c^2 - r_1^2 r_2^2 r_3^2 = 0
$$

As can be seen, this equation is symmetric with respect to $r_1, r_2$, and $r_3$. Therefore, $c$ is an invariant, and it holds that

$$
x_1 r_1 = x_2 r_2 = x_3 r_3 = c
$$

The solution of the cubic equation leads to the casus irreducibilis, so that $c$ cannot be expressed as a sum of square roots and cube roots, but only with the cosine of angle trisectors. Given the known fact that the trisection of an angle is not constructible with elementary construction tools, the triangle with the maximum area is also not constructible with elementary tools. Setting

$$
\begin{gathered}
r_q = \sqrt{\frac{r_1^2 + r_2^2 + r_3^2}{3}} \\
r_m = (r_1 r_2 r_3)^{\frac{1}{3}}
\end{gathered}
$$

as the quadratic and geometric mean of the radii, the cubic equation for the invariant $c$ is:

$$
2 c^3 + 3 r_q^2 c^2 - r_m^6 = 0
$$

The solution of the cubic equation is then (the complete calculation is not shown here):

$$
c = r_q^2 \left[ \cos \left( \frac{1}{3} \arccos \left( 2 \left( \frac{r_m}{r_q} \right)^6 - 1 \right) \right) - \frac{1}{2} \right]
$$

To find a symmetric formula for the triangle area, we use the following relationship:

$$
\begin{gathered}
A_{ABC} = \frac{1}{2} (A_{BDCA} + A_{CDAB} + A_{ADBC}) \\
A_{ABC} = \frac{1}{4} (r_1 (p_1 + q_1) + r_2 (p_2 + q_2) + r_3 (p_3 + q_3)) \\
A_{ABC} = \frac{1}{4} (r_1 \sqrt{r_2^2 - x_1^2} + r_1 \sqrt{r_3^2 - x_1^2} + r_2 \sqrt{r_1^2 - x_2^2} + r_2 \sqrt{r_3^2 - x_2^2} + r_3 \sqrt{r_1^2 - x_3^2} + r_3 \sqrt{r_2^2 - x_3^2}) \\
A_{ABC} = \frac{1}{4} (\sqrt{r_1^2 r_2^2 - c^2} + \sqrt{r_1^2 r_3^2 - c^2} + \sqrt{r_1^2 r_2^2 - c^2} + \sqrt{r_2^2 r_3^2 - c^2} + \sqrt{r_1^2 r_3^2 - c^2} + \sqrt{r_2^2 r_3^2 - c^2}) \\
A_{ABC} = \frac{1}{2} (\sqrt{r_1^2 r_2^2 - c^2} + \sqrt{r_1^2 r_3^2 - c^2} + \sqrt{r_2^2 r_3^2 - c^2}) \\
A_{ABC} = \frac{1}{2} \sum_{i=1}^{3} \sqrt{\left( \frac{r_m^3}{r_i} \right)^2 - c^2}
\end{gathered}
$$