Geometric Inequality

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#1 h Sep 29, 2006, 7:45 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an acute triangle with circumcenter $O$ and circumradius $R$ . $AO$ meets the circumcircle of $BOC$ at $A'$ , $BO$ meets the circumcircle of $COA$ at $B'$ and $CO$ meets the circumcircle of $AOB$ at $C'$ . Prove that $OA'\cdot OB'\cdot OC'\geq 8R^{3}.$ Sorry if this has been posted before since this is a very classical problem, but I failed to find it with the search-function.

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#2 h Sep 29, 2006, 11:35 PM • 2 Y

Y by Adventure10, Mango247

well,
let $A''=AO\cap BC$ $B''=BO\cap AC$ $C''=CO\cap AB$
then $OA'\cdot OA''=R^{2}$
so we need to prove $R^{3}\geq 8OA''OB''OC''$
which is equivalent to
$(x+y)(y+z)(z+x)\geq 8xyz$ where $x=S_{AOB},y=S_{BOC},z=S_{COA}$

or we can get it from the inequality:
$xyz=(a-x)(b-y)(c-z)\rightarrow 8xyz\le abc$

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zhaoli

#3 h Sep 30, 2006, 12:44 PM • 1 Y

Y by Adventure10

I have generalized the problem 1 year ago:

http://www.mathlinks.ro/Forum/viewtopic.php?t=55206

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#4 h Sep 30, 2006, 1:57 PM • 3 Y

Y by minlat7, Adventure10, Mango247

It is well known ( at least to me ) the generalization which mentioned by zhaoli in his post. So, if P is an arbitrary point inwardly to a given triangle ABC and DEF it’s cevian triangle, then we have that:

(AP / PD) + (BP / PE) + (CP / PF) >= 6, (1) and (AP / PD)·(BP / PE)·(CP / PF) >= 8, (2)

I have found the elementary proofs of (1), (2), in a Hellenic textbook by J. Panakis – Triangle’s Geometry – volume A – page 107 - ''Gutenberg'' publications – Athens 1969.

Some other interest (I think) results, not well known (I don’t know any reference in Hellenic publication), are the follows:

1). - [(AP / PD)·(BP / PE)·(CP / PF)] - [(AP / PD) + (BP / PE) + (CP / PF)] = 2, (3). I named this equality as Van Aubel’s equality.

2). - (AP / PD) + (BP / PE) + (CP / PF) = (AP' / P'D') + (BP' / P'E') + (CP' / P'F'), (4).

3). - (AP / PD)·(BP / PE)·(CP / PF) = (AP' / P'D')·(BP' / P'E')·(CP' / P'F'), (5).

Where P', is the isotomic of P, with respect to the triangle ABC and D'E'F', it’s cevian triangle.

Kostas Vittas.

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#5 h Sep 30, 2006, 6:48 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Yimin Ge wrote:

Let $ABC$ be an acute triangle with circumcenter $O$ and circumradius $R$ . $AO$ meets the circumcircle of $BOC$ at $A'$ ,
$BO$ meets the circumcircle of $COA$ at $B'$ and $CO$ meets the circumcircle of $AOB$ at $C'$ . Prove that $OA'\cdot OB'\cdot OC'\geq 8R^{3}\ .$

Lemma. In an acute triangle $ABC$ exists the inequality $\boxed{\ \cos (A-B)\cdot\cos (B-C)\cdot\cos (C-A)\ge 8\cos A\cdot\cos B\cdot\cos C\ }\ .$

Indeed, using the well-known inequality $\{x,y,z\}\subset (0,\infty )\Longrightarrow (x+y)(y+z)(z+x)\ge 8xyz$ for $x: =\sin 2A>0$ a.s.o.

obtain $\prod (\sin 2A+\sin 2B)\ge 8\prod \sin 2A$ $\Longleftrightarrow$ $\prod \sin (A+B)\cdot\prod \cos (A-B)\ge \prod \sin 2A$ $\Longleftrightarrow$

$\prod \sin A\cdot\prod\cos (A-B)\ge 8\prod\sin A\cdot\prod\cos A$ $\Longleftrightarrow$ $\prod \cos (A-B)\ge 8\prod \cos A\ .$

Proof of the proposed problem. Prove easily that $m(\widehat{OBA'})=90^{\circ}+C-B$ and $m(\widehat{OA'B})=90^{\circ}-A\ .$

From the triangle $BOA'\ ,$ using the Sinus' theorem obtain $\frac{OA'}{\cos (B-C)}=\frac{OB}{\cos A}\ ,$ i.e. $OA'=R\cdot\frac{\cos (B-C)}{\cos A}\ .$

Therefore, using the above lemma obtain the required inequality : $OA'\cdot OB'\cdot OC'=R^{3}\cdot\frac{\prod \cos (B-C)}{\prod \cos A}\ge 8R^{3\ .}$

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#6 h Sep 30, 2006, 10:20 PM • 2 Y

Y by Adventure10, Mango247

cool inequality zhaoli!
the inequalities mentioned by vittasko are well known also to me but the trick with inversion to change it to the generalisation of this problem is very NICE !

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#7 h Oct 1, 2006, 9:03 PM • 1 Y

Y by Adventure10

http://www.mathlinks.ro/Forum/viewtopic.php?t=55206:

zhaoli wrote:

Let $ABC$ be an acute-angled triangle, $P$ be an arbitrary point interior to $ABC$ . $AP$ meet the circle $BPC$ again in $A_{0}$ , $BP$ meet the circle $CPA$ again in $B_{0}$ , and $CP$ meet the circle $APB$ again in $C_{0}$ . Prove that

(1) $PA_{0}\cdot PB_{0}\cdot PC_{0}\geq 8PA \cdot PB \cdot PC$ .

(2) $\frac{PA_{0}}{PA}+\frac{PB_{0}}{PB}+\frac{PC_{0}}{PC}\geq 6$ .

Thanks, zhaoli, for this nice generalisation.

I didn't bother to read your solution since I'm not very experienced with inversion (I'll try to find it myself later). This ist my solution for the original problem which also works for your generalisation.

By ptolemy, we have $PA_{0}=\frac{PB\cdot CA_{0}+PC\cdot BA_{0}}{BC}.$ Let $x=\sin\angle BPC_{0}=\sin\angle CPB_{0},$ $y=\sin\angle BPA_{0}=\sin\angle APB_{0},$ $z=\sin\angle CPA_{0}=\sin\angle APC_{0}$ . Then we have, by law of sines: $\frac{CA_{0}}{BC}=\frac{z}{x}, \frac{BA_{0}}{BC}=\frac{y}{x}.$ so we get $PA_{0}=\frac{z}{x}PB+\frac{y}{x}PC.\quad (1)$ .

Multiplying $(1)$ cyclically yields ${PA_{0}\cdot PB_{0}\cdot PC_{0}\geq \prod_{cyc}(\frac{z}{x}PB+\frac{y}{x}PC}) \geq PA\cdot PB\cdot PC$ by AMGM.

Summung $(1)/PA$ cyclically yields ${\sum_{cyc}\frac{PA_{0}}{PA}\geq \sum_{cyc}(\frac{z}{x}\frac{PB}{PA}+\frac{y}{x}\frac{PC}{PA}}) \geq 6$ by AMGM.
qed

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#8 h Jun 9, 2010, 2:42 PM • 3 Y

Y by minlat7, Adventure10, Mango247

Yimin Ge wrote:

Let $ABC$ be an acute triangle with circumcenter $O$ and circumradius $R$ . $AO$ meets the circumcircle of $BOC$ at $A'$ , $BO$ meets the circumcircle of $COA$ at $B'$ and $CO$ meets the circumcircle of $AOB$ at $C'$ . Prove that $OA'\cdot OB'\cdot OC'\geq 8R^{3}.$

We can prove easily (as Mr. Virgil did) that the inequality is equivalent to ${\prod \cos (B-C)} \geq 8 \prod{\cos A}$
Let $AA' \cap BC=X$ . Sine law in $\triangle ABX$ yields $\frac{c}{\cos (B-C)}=\frac{BX}{\cos C} \iff \frac{\cos (B-C)}{\cos C}=\frac{c}{BX}$
Again from sine law we get $\frac{BX}{XC}=\frac{c\cos C}{b \cos B} \iff \frac{BX}{a}=\frac{c \cos C}{b \cos B+c \cos C} \\ \iff \frac{\cos (B-C)}{\cos C}=\frac{c}{BX}=\frac{b \cos B+c\cos C}{a \cos C }$
So the required inequality transforms into $\prod ( b \cos B+c\cos C) \geq 8abc \prod \cos A$
Now we set $a \cos A=x, b \cos B=y, c \cos C=z$ . So we need to prove that $(x+y)(y+z)(z+x) \geq 8xyz$
Which is apparently true by AM-GM, and we are done!

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Sardor

#9 h Jun 27, 2014, 11:21 AM • 1 Y

Y by Adventure10

My solution (i think it's very simple):
Applyin Ptolemy's theorem for the cyclic quadrilateral $BOCA'$ we have $OA' \cdot BC= CO \cdot BA' + BO \cdot CA' =R(BA'+CA')$ or
$\frac{OA'}{R}= \frac{CA'}{BC} + \frac{BA'}{BC}=\frac{sin2B}{sin2A}+\frac{sin2C}{sin2A}=\frac{sin2B+sin2C}{sin2A}$ and similary we have
$\frac{OB'}{R}=\frac{sin2C+sin2A}{sin2B}$ and $\frac{OC'}{R}=\frac{sin2B+sin2A}{sin2C}$ . Finally we have $OA' \cdot OB' \cdot OC'=(\frac{sin2C+sin2A}{sin2B})(\frac{sin2B+sin2A}{sin2C})(\frac{sin2C+sin2B}{sin2A}) R^3 \ge 8R^3$ , because we have by AM-GM $(x+y)(y+z)(z+x) \ge 8xyz$ for positive real numbers $x,y,z$ . DONE !

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#10 h Sep 11, 2020, 10:13 AM • 1 Y

Y by Leqso

My Solution using Inversion:

Consider an Inversion centered at $O$ with radius $R$ .Easy to see that since $A^{'}=(BOM)\cap AO\Rightarrow X=BM\cap AO$ ( $X$ is image of $A^{'}$ under this transformation,similarly $Y$ and $Z$ ).Now using Inversion Distance formula,we have: $AA^{'}=\frac{R^{2}}{OA\times OX}\times AX\Rightarrow OA^{'}=AA^{'}-AO=\frac{R^{2}\times AX}{OA\times OX}-R$ .Analogously,we get:

$OB^{'}=\frac{R^{2}\times BY}{OB\times OY}-R$ , $OC^{'}=\frac{R^{2}\times CZ}{OC\times OZ}-R$ .

So we have to show that $\prod R(\frac{R\times AX}{OA\times OX}-1)\geq 8R^{3}\Leftrightarrow \prod (\frac{AX}{OX}-1)\geq 8\Leftrightarrow \prod \frac{AO}{OX}\geq 8$

But the last one is obvious and the equality holds when $AX$ , $BY$ , $CZ$ are all medians in $\triangle ABC$ .

$Q.E.D$

This post has been edited 1 time. Last edited by A.L.E.X, Sep 11, 2020, 10:13 AM
Reason: blue

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#11 h Jan 26, 2022, 2:33 PM

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Let $H$ be the orthocenter of $\triangle ABC$ , let $\triangle DEF$ be the orthic triangle of $\triangle ABC$ and let $K,L,M$ be reflection of $H$ over $D,E,F$ , respectively.
Inversion with radius $\sqrt{bc}$ at $A$ swaps $A'$ and $H$ , $D$ and antipode of $A$ wrt $(ABC)$ , thus we need to show that $\begin{align*} \left(\frac{AD\cdot 2R}{AH}-R\right)\left(\frac{BE\cdot 2R}{BH}-R\right)\left(\frac{CF\cdot 2R}{CH}-R\right)\geq 8R^3, \end{align*}$ which is equivalent to $\begin{align*} \frac{AK}{AH}\cdot \frac{BL}{BH}\cdot\frac{CM}{CH}\geq 8. \end{align*}$ Now we consider $\triangle KLM$ as our "base triangle". Indeed $H$ is the incenter of $\triangle KLM$ and $A,B,C$ midpoints of arcs $LM,MK,KL$ , respectively. Let $X=\overline{KH}\cap\overline{LM}$ , $X=\overline{LH}\cap\overline{MK}$ and $Z=\overline{MH}\cap\overline{KL}$ . Thus, we have $\frac{AK}{AH}=\frac{AH}{AX}=\frac{AM}{AX}=\frac{KL}{LX}.$ Similarly we get that $\tfrac{BL}{BH}=\tfrac{LM}{MY}$ and $\tfrac{CM}{CH}=\tfrac{MK}{KZ}$ . By Ceva, $LX\cdot MY\cdot KZ=XM\cdot YK\cdot ZL$ , therefore by AM-GM, $\begin{align*} KL\cdot LM\cdot MK&=(LX+XM)\cdot(MY+YK)\cdot(YZ+ZL)\\&\geq 8\sqrt{LX\cdot XM\cdot MY\cdot YK\cdot YZ\cdot ZL}\\&=8\cdot LX\cdot MY\cdot KZ, \end{align*}$ as desired.

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