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Question

Lemoine point $L$ of $\triangle ABC$ satisfies $|BL|+|CL|\le\frac2{\sqrt3}|BC|.$

Attempt:

By sine rule, we have $|BL|/|BC|=\sin\angle BCL/\sin\angle BLC$ and $|CL|/|BC|=\sin\angle CBL/\sin\angle BLC$. Thus, the inequality is equivalent to$$\sin\angle BCL+\sin\angle CBL\le\frac2{\sqrt3}\sin\angle BLC$$ Since Lemoine point is the isogonal conjugate of the centroid $G$, we have$$\angle BCL=\angle GCA$$and$$\angle CBL=\angle GBA$$and$$\angle BLC=\pi-(\angle GBA+\angle GCA)$$ Thus, the inequality becomes$$\sin\angle GBA+\sin\angle GCA\le\frac2{\sqrt3}\sin(\angle GBA+\angle GCA)$$

So we eliminated the point $L$ and the inequality is now in terms of the centroid $G$.

By Sum-to-product identity rewrite left-hand side $$ 2 \sin \left(\frac{\angle GBA+\angle GCA}{2}\right) \cos \left(\frac{\angle GBA-\angle GCA}{2}\right) \leq \frac{2 \sin (\angle GBA+\angle GCA)}{\sqrt{3}} $$ By double-angle formula rewrite right-hand side, then factor out $\sin\left(\frac{\angle GBA+\angle GCA}{2}\right)$ $$\sin\left(\frac{\angle GBA+\angle GCA}{2}\right)\left(\cos \left(\frac{\angle GBA+\angle GCA}{2}\right)-\frac{1}{2} \sqrt{3} \cos \left(\frac{\angle GBA-\angle GCA}{2}\right)\right) \geq 0$$ Dividing by $\sin\left(\frac{\angle GBA+\angle GCA}{2}\right)>0$ $$\cos \left(\frac{\angle GBA+\angle GCA}{2}\right)-\frac{1}{2} \sqrt{3} \cos \left(\frac{\angle GBA-\angle GCA}{2}\right)\geq 0$$

But I don't know how to proceed from here. Can someone help me?

hbghlyj
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1 Answers1

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Put $A=(x,y),B=(-1,0),C=(1,0)$. Then $L$ has rational Cartesian coordinates since its barycentrics are $a^2:b^2:c^2$: $$L=\frac{4(x,y)+((x-1)^2+y^2)(-1,0)+((x+1)^2+y^2)(1,0)}{4+(x-1)^2+y^2+(x+1)^2+y^2}=\frac{(4x,2y)}{3+x^2+y^2}$$ The points $P=(p,q)$ for which $|BP|+|CP|\le\frac2{\sqrt3}|BC|$ in this coordinatisation satisfy $p^2+4q^2\le\frac43$. Substituting $L$'s coordinates into this inequality leaves $$\frac{16(x^2+y^2)}{(3+x^2+y^2)^2}\le\frac43$$ $$\frac s{(3+s)^2}\le\frac1{12},s=x^2+y^2\ge0$$ which is easily verified; equality holds iff $s=3$.

Parcly Taxel
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