What is the Lemoine point useful for? What is the Lemoine point useful for?
Can someone give concrete examples /common example what math problems can be solved with usage of Lemoine point?
 A: I'll call it the symmedian point, following Honsberger 1995 Chapter 7. Here's a summary of what the chapter proves:


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*In $\triangle ABC$, the symmedian point $K$ is the isogonal conjugate of the centroid $G$, and the point of concurrence of the symmedians, which are isogonal conjugates of the medians. In particular, $AP$ is a symmedian iff the distances of $AB,\,AC$ to $P$ are in the same ratios as these sides' lengths, iff $BP:PC=c^2:b^2$. Similarly, $K$'s distance from each side is proportional to its length.

*If $\angle BAC=\frac{\pi}{2}$, $K$ is the midpoint of the altitude to the hypotenuse $BC$.

*The symmedians of $\triangle ABC$ bisect the sides of its orthic triangle.

*The tangents to $\triangle ABC$'s circumcircle at $B,\,C$ meet on the symmedian through $A$.

*Mark $\triangle ABC$'s points of contact with its incircle. These are the vertices of its Gergonne triangle. The latter's symmedian point is the original triangle's Gergonne point. It's the point of concurrence of the lines joining the Gergonne triangle's vertices, construed as points on $\triangle ABC$'s sides, to the opposite vertices of $\triangle ABC$.

*Through the Gergonne point, draw three lines parallel to the sides of the Gergonne triangle. These lines meet $\triangle ABC$'s sides in six points. These are cyclic, on a circle concentric with $\triangle ABC$'s incircle.

*A line from the midpoint of $AB$ to the midpoint of the altitude to $AB$ goes thrugh $K$.

*The pedal triangle of $K$ has $K$ as its centroid.

A: If a triangle and its symmedian point are given, you can construct points of the circumcircle using only a straightedge.
Of course, you cannot construct the symmedian point using only a straightedge. But let us assume that the symmedian point has somehow been provided already.
This is illustrated in the figure below. As long as the green line $p$ passes through the symmedian point $K$, the associated green point $P$ ends up on the circumcircle of $ABC$.

To explain, I need Poncelet's concept of trilinear polarity. Nowadays that would probably better be termed trilateral duality, but herein I will use the terms as introduced on the above-mentioned Wikipedia page.
From the Definition, you can see that trilinear polars and poles can be constructed using a straightedge only.
Also noted is that, given a circumconic $k$, the trilinear polars of all its points concur in a point $K$.
Concretely, in a homogenous triangle coordinate system, which represents the vertices $A,B,C$ as $(1:0:0), (0:1:0), (0:0:1)$ respectively, a circumconic's equation takes the form $k(u,v,w) = Rvw + Swu + Tuv = 0$ with constant coefficients $R,S,T$.
Then the point of concurrence of all its trilinear polars is $K=(R:S:T)$.
Note that there are several such triangle coordinate systems, depending on which triangle center shall be represented as $(1:1:1)$, but the above correspondence holds for any such coordinate system.
Now draw a line $p$ through $K$ but not through $A$, $B$, or $C$, and construct the trilinear pole $P$ of $p$, using straightedge only. That point $P$ must end up on the circumconic $k$ associated with $K$. Varying the slope of the line $p$ also varies the location of $P$ on the circumconic $k$.
For an instance of the circumconic $k$, let us consider the circumcircle.
In barycentric coordinates, its equation is $k(u,v,w) = a^2vw + b^2 wu + c^2 uv = 0$, where $a,b,c$ are the triangle's side lengths. Therefore, the associated point $K$ has barycentric coordinates $(a^2:b^2:c^2)$, which are those of the symmedian point.
