I need help to solve this problem. I have no idea how to solve. Is there previously solved copy of this problem?
Thank you very much.
Let $H$ be the orthocenter of a triangle $ABC$ with circumradius $R$. Show that the distance $HK$ between $H$ and the Lemoine point $K$ of the triangle $ABC$ satisfies $$HK^2=4R^2-\frac{a^4+b^4+c^4}{a^2+b^2+c^2}-\frac{3a^2b^2c^2}{\left(a^2+b^2+c^2\right)^2}\,,$$ where $a:=BC$, $b:=CA$, and $c:=AB$.
You can solve this question by analytic geometry, using two lemmas:
Lemma 1: In any triangle ABC these relations hold true:
$$a^2+{AH}^2=b^2+{BH}^2=c^2+{CH}^2=4R^2$$
Lemma 2: If $A=(x_1,y_1)$, $B=(x_2,y_2)$, $C=(x_3,y_3)$ are the coordinates of vertices of triangle ABC, then the coordinates of its Lemoine point are $$K=({a^2x_1+b^2x_2+c^2x_3\over a^2+b^2+c^2},{a^2y_1+b^2y_2+c^2y_3\over a^2+b^2+c^2})$$
Now, if you put the origin of cartesian orthogonal axes at orthocenter H,
$$HK^2=({a^2x_1+b^2x_2+c^2x_3\over a^2+b^2+c^2})^2 +({a^2y_1+b^2y_2+c^2y_3\over a^2+b^2+c^2})^2,$$ $$(a^2+b^2+c^2)^2HK^2=(a^2x_1+b^2x_2+c^2x_3)^2 +(a^2y_1+b^2y_2+c^2y_3)^2,$$ $$(a^2+b^2+c^2)^2HK^2=a^4(x_1^2+y_1^2)+b^4(x_2^2+y_2^2)+c^4(x_3^2+y_3^2)+(2x_1x_2+2y_1y_2)a^2b^2+(2x_1x_3+2y_1y_3)a^2c^2+(2x_2x_3+2y_2y_3)b^2c^2,$$ $$(a^2+b^2+c^2)^2HK^2=a^4{AH}^2+b^4{BH}^2+c^4{CH}^2+({AH}^2+{BH}^2-c^2)a^2b^2+({AH}^2+{CH}^2-b^2)a^2c^2+({BH}^2+{CH}^2-a^2)b^2c^2,$$ $$(a^2+b^2+c^2)^2HK^2=(a^4+a^2b^2+a^2c^2){AH}^2+(b^4+a^2b^2+b^2c^2){BH}^2+(c^4+a^2c^2+b^2c^2){CH}^2-3a^2b^2c^2$$
Having done that, we use lemma 1 stated above and after further algebraic simplification we finally get
$${HK}^2=4R^2-{a^4+b^4+c^4\over a^2+b^2+c^2}-{3a^2b^2c^2\over (a^2+b^2+c^2)^2}$$
QED.