For any point $X$ inside an acute-angled triangle $ABC$ we define $$f(X)=\frac{AX}{A_1X}\cdot \frac{BX}{B_1X}\cdot \frac{CX}{C_1X}$$where $A_1, B_1$, and $C_1$ are the intersection points of the lines $AX, BX,$ and $CX$ with the sides $BC, AC$, and $AB$, respectively. Let $H, I$, and $G$ be the orthocenter, the incenter, and the centroid of the triangle $ABC$, respectively. Prove that $f(H) \ge f(I) \ge f(G)$ . (D. Bazylev)
Problem
Source: 2012 Belarus TST 3.1
Tags: geometry, ratio, orthocenter, incenter, Centroid
tk1
31.12.2020 03:15
For $X\equiv G$ it is obvious that $f(G)=8.\ (1)$ For $X\equiv H$ we find $\frac{AH}{A_1H}=\frac{AC_1/\sin B}{CA_1/\tan B}=\frac{AC_1}{CA_1}\cdot\frac{1}{\cos B}.$ Similarly for the other fractions, and using Ceva's theorem we get $f(H)=\frac{1}{\cos A\cdot \cos B\cdot \cos C}.\ (2)$ For $X\equiv I$ we have $\frac{AI}{A_1I}=\frac{c}{BA_1}=\frac{c}{ac/(b+c)}=\frac{b+c}{a}.$ Similarly for the other fractions, and we get $f(I)=\frac{1}{abc}\cdot (a+b)(b+c)(c+a).\ (3)$ From AM-GM we easily get $(a+b)(b+c)(c+a)\ge 8abc,$ and therefore from $(1)$ and $(3)$ we see that $f(I)\ge f(G).$ Equality occurs for an equilateral triangle. We are left to show that $f(H)\ge f(I).$ This is equivalent to $$\frac{abc}{\cos A\cdot\cos B\cdot\cos C}\ge (a+b)(b+c)(c+a)\Longleftrightarrow \tan A\cdot \tan B\cdot\tan C\ge (\sin A+\sin B)(\sin B+\sin C)(\sin C+\sin A).$$ But for an acute-angled trangle we know that $\tan A\cdot\tan B\cdot\tan C\ge 3\sqrt{3},$ so it is sufficient to show that $S:=(\sin A+\sin B)(\sin B+\sin C)(\sin C+\sin A)\le 3\sqrt{3}.$ But $\sin A+\sin B=2\cos \tfrac{C}{2}\cos\tfrac{A-B}{2}\le 2\cos \tfrac{C}{2},$ and therefore $S\le 8\cos\tfrac{A}{2}\cdot\cos\tfrac{B}{2}\cdot\cos\tfrac{C}{2}=4\cos\tfrac{A}{2}\left(\cos\tfrac{B-C}{2}+\cos\tfrac{B+C}{2}\right)\le 4\cos\tfrac{A}{2}\left(1+\sin\tfrac{A}{2}\right)\le 3\sqrt{3}.$ The last step is easy to prove by setting, for example,$x=\sin\tfrac{A}{2}$ and squaring. Equality occurs for an equilateral triangle.