Orthocenter inequality

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Source: TWN APMO pretest

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#1 h Jul 26, 2020, 2:20 AM

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Let $H$ be the orthocenter of an acute triangle $ABC$ .
The lines $AH$ , $BH$ , $CH$ intersects the circumcircle of $ABC$ at point $A'$ , $B'$ , $C'$ ,respectively.
Find the range of $S=\frac{AH}{AA'}+\frac{BH}{BB'}+\frac{CH}{CC'}$ .

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#2 h Jul 26, 2020, 5:58 AM

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The value of $S$ is determined by two angles $B$ and $C$ .
The range of $B$ and $C$ is
$\{(B,C):0<B,C<\pi/2,\; B+C > \pi/2\} \subset \mathbb{R}^2$ which is a connected region. Therefore, the range of $S$ is connected, hence it's an interval.

Let's start the main arguments.
$R$ : the radius of the circumcircle. Then,
$AH = 2R \cos A,\quad AA' = 2R \sin(\angle ABA') = 2R \sin \left (B + \frac{\pi}{2} - C \right) = 2R \cos (B-C)$ Let $x=\cot B \cot C$ , $y=\cot C \cot A$ , $z=\cot A \cot B$ . Note that $x+y+z=1,\; x,y,z>0$ .
Let $f(x) = \frac{1}{1+x},\; x\geq 0$ . Note that $f(x) = \frac{1}{1+x},\; x\geq 0$ is a strictly convex function.
Lower Bound
$\begin{align*} S =& \sum_{\textrm{cyc}} \frac{AH}{AA'} = \sum_{\textrm{cyc}} \frac{\cos A}{\cos(B-C)} = -\sum_{\textrm{cyc}} \frac{\cos(B+C)}{\cos(B-C)} \\ =& \sum_{\textrm{cyc}} \frac{1-\cot B \cot C}{1+\cot B \cot C} =-3 + 2\sum_{\textrm{cyc}} f(x) \underset{\textrm{Jensen}}{\geq} -3 + 2 \cdot 3 \cdot f(1/3)= \frac{3}{2} \end{align*}$ Equality holds iff $A=B=C$ , i.e. $\;\Delta ABC$ is equilateral.

Upper Bound
Since $f(x) = \frac{1}{1+x},\; x\geq 0$ is a strictly convex function,
$f(x)+f(y)< f(x+y)+f(0),\quad x,y>0$ So,
$S = -3 + 2(f(x)+f(y)+f(z))< -3 + 2(f(0)+f(0)+f(x+y+z)) = -3 + 2(2f(0)+f(1))=2$ Every value arbitrary close to $2$ is obtained as $A\rightarrow \pi/2,\; B,C\rightarrow \pi/4$ , so $x\rightarrow 1,\;y,z\rightarrow 0$ .

Answer : $\displaystyle \left[\frac{3}{2}, 2 \right)$

This post has been edited 3 times. Last edited by EagleEye, Jul 26, 2020, 6:02 AM

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#3 h Jul 27, 2020, 4:42 AM • 2 Y

Y by Mango247, Mango247

Thanks for solution.

I'm wondering that are there any approaches without using trig functions?

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sqing

#4 h Jul 27, 2020, 5:14 AM

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Good.
In acute triangle $ABC$ . Prove that $\frac{3}{2}\leq \frac{\cos A}{\cos(B-C)}+\frac{\cos B}{\cos(C-A)} +\frac{\cos C}{\cos(A-B)} <2$ here
Crux Mathematicorum, Vol. 50(7), September 2024 Q4962:
In acute triangle $ABC$ . Prove that $\frac{\cos A}{\cos(B-C)}+\frac{\cos B}{\cos(C-A)} +\frac{\cos C}{\cos(A-B)} \leq 2(\cos^2 B+\cos^2 A+\cos^2 C)$

This post has been edited 1 time. Last edited by sqing, Sep 21, 2024, 9:09 AM

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#5 h Jul 27, 2020, 5:18 AM • 1 Y

Y by Inconsistent

@2above A solution for the lower bound without trig functions:

Define $h_A$ , $h_B$ , and $h_C$ to be the lengths of the altitudes from A, B and C respectively.

From reflecting the orthocenter we have $\frac{AH}{AA'} = \frac{AH}{2h_A-AH} = \frac{2h_A}{2h_A-AH}-1 = \frac{4[ABC]}{4[ABC]-a \cdot AH}-1$ .
So we want to find the minimum of $\frac{4[ABC]}{4[ABC]-a \cdot AH}+ \frac{4[ABC]}{4[ABC]-b \cdot BH}+ \frac{4[ABC]}{4[ABC]-c \cdot CH}-3$ By AM-HM and noticing $a\cdot AH + b\cdot BH + c\cdot CH = 4[ABC]$ , we get $\frac{4[ABC]}{4[ABC]-a \cdot AH}+ \frac{4[ABC]}{4[ABC]-b \cdot BH}+ \frac{4[ABC]}{4[ABC]-c \cdot CH}-3 \geq \frac{9}{2} -3 = \frac{3}{2}$ With equality when $\triangle ABC$ is equilateral.

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