So I've recently come across this pretty neat geometrical way for solving depressed cubic equations where we use $x^3$ term as a cube $bx$ term as another cube and solve for x using some neat geometry, same goes for cubic and quadratic for quadratic we use $x-\frac b{2a}$ substitution to solve for $x$, in cubic we use $x-\frac b{3a}$ substitution then turn into depressed cubic and solve the equation, now similarly with a quartic can be solved by using $x-\frac b{4a}$ substitution and then some algebra can give the value of $x$. But I was wondering if there is a geometrical way to solve the quartic equation?
Using geometry to solve quartic equations
abstract-algebrageometry
Related Solutions
Here is a solution not using trigonometry:
Assume $H$ is the midpoint of $AB$. Let's extend $BK$ such that it intersects $JH$ at $L$. Then, it is obvious that $\angle LAB=40^{\circ}$, so $\angle KAL=20^{\circ}$, which means $KA$ is the angle bisector of $\angle BAL$. Hence, by the angle bisector theorem, we have: $$\frac{LK}{KB}=\frac{AL}{AB}.$$
Note that this theorem can be proved without using trigonometry; for example take a look at this link.
On the other hand, consider the figure below, and observe that $\angle LAS=30^{\circ}$. Thus $LS=\frac{AL}{2}$.
Therefore, since $\triangle JLS$ and $\triangle JHA$ are similar, we have:
$$\frac{JL}{LS}=\frac{JA}{AH} \\ \implies \frac{JL}{JA}=\frac{AL}{AB} \implies \frac{JL}{JB}=\frac{AL}{AB}=\frac{LK}{KB},$$
which implies that $JK$ is the angle bisector of $\angle BJL$. So, $\angle BJK=10^{\circ}$.
We are done.
Best Answer
We can indeed lead to the solution of the quartic equation from common concepts of geometry. We achieve this by appropriately extending an idea of Viete that allows the trigonometric solution of the depressed cubic equation if all its roots are real. I have a long time to write here and I do not remember how the formulas are inserted in the text in Latex format, so I will limit myself to an enlightening image.
Each of the three constructions of the image is fully defined by three parameters:
The radius of the circle, which is common to all three structures and is calculated directly through the coefficients of the depressed quartic equation we want to solve.
The length of the orange line, which starts from the center of the circle.
An angle (that is equal to at most half a right angle).
Each of these three constructions defines the exact position of the roots of the quartic equation on the x-axis (you can observe that the number of parameters of each construction is equal to the number of coefficients of the depressed quartic). However, in order to be able to solve the equation in a decent and elegant way, it is necessary to use in our algebraic relations elements from all three constructions. To do this, we first find the values of the roots of the resolved cubic shown in the image. Then, with the help of these roots we find the trigonometric values of the angles of the three constructions described above, in order to extract through them the values of the roots of the quartic equation.
You may have noticed that this method is only applicable to depressed equations that have two positive and two negative real roots. This is because in all other cases, some of the parameters of our constructions receive complex values, so it is not possible to design them, in whole or in part. Nevertheless, the complex values are still manageable algebraically, so that, in the final analysis, with the same substantially trigonometric relations we can derive both the real and complex roots of any depressed quartic.
My presentation is inadequate, but it allows you to try this method.
IMAGE https://ibb.co/LSHdq3B