Примена једначине 3-степена у геометрији // Applying a 3rd degree Equation in Geometry

           From a LinkedIn post of Reygan Diionisio , see here. You can also watch it on YouTube.

ANSWER CiP

$$x=3$$

                    Solution CiP

          With the notations in the figure below

     We will apply the Law of cosine to triangles. More precisely, a consequence of it.

$$\cos \measuredangle ACB=\frac{a^2+b^2-c^2}{2ab} \tag{1}$$

where, for triangle $ABC$ we note $a=BC,\;b=AC,\;c=AB.$

Applying (1) to triangle $ABE$ we have

$$\cos \measuredangle{1}=\cos \widehat{AEB}=\frac{AE^2+BE^2-AB^2}{2\cdot AE \cdot BE}=$$

$=\frac{(x+5)^2+x^2-49}{2x(x+5)}=\frac{2x^2+10x-24}{2x(x+5)}=\frac{x^2+5x-12}{x(x+5)}. \tag{2}$

Applying (1) to triangle $CDE$ we have

$$\cos \measuredangle{2}=\cos \widehat{CED}=\frac{CE^2+ED^2-CD^2}{2\cdot CE \cdot ED}=$$

$=\frac{25+x^2-49}{2\cdot 5 \cdot x}=\frac{x^2-24}{10x}. \tag{3}$

Given that angles $\measuredangle{1}$ and $\measuredangle{2}$ are supplementary, we have
$$\cos \measuredangle{1}=-\cos \measuredangle{2}$$

so, from (2) and (3), we have the equation

$$\frac{x^2+5x-12}{x(x+5)}=-\frac{x^2-24}{10x}$$

$\Leftrightarrow\;10\cdot(x^2+5x-12)=-(x+5)(x^2-24)\Leftrightarrow x^3+15x^2+26x-240=0 \tag{4}$

We easily find that the equation (4) has the rational root $x_1=3$ through the Horner's Method:

$$\begin{array}{c|c} \;&1&15&26&-240\\ \hline 3&1&18&80&0 \end{array}$$

Also from the table, second line, we see that the other two roots are given by the equation

$x^2+18x+80=0$, but they are $x_2=-10,\;x_3=-8$ but they have no geometric significance, being negative.

$\blacksquare$

          Remark CiP  For $x=3$, from the relation (2) we obtain $\cos \measuredangle{1}=\frac{1}{2}$ so $\measuredangle{1}=60^{\circ}$