geometric inequality with altitudes! CMO 2015 P2

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aditya21

717 posts

aditya21

#1 h Apr 24, 2015, 8:21 PM • 1 Y

Y by Adventure10

Let $ABC$ be an acute-angled triangle with altitudes $AD,BE,$ and $CF$ . Let $H$ be the orthocentre, that is, the point where the altitudes meet. Prove that $\frac{AB\cdot AC+BC\cdot CA+CA\cdot CB}{AH\cdot AD+BH\cdot BE+CH\cdot CF}\leq 2.$

This post has been edited 7 times. Last edited by djmathman, Jun 16, 2015, 9:26 PM
Reason: rearranged numerator

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ABCDE

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ABCDE

#2 h Apr 24, 2015, 8:39 PM • 3 Y

Y by A64298347, Adventure10, Mango247

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aditya21

717 posts

aditya21

#3 h Apr 24, 2015, 8:43 PM • 4 Y

Y by IMO2019, Jason99, Adventure10, Mango247

here is my solution =
let us denote $a,b,c$ as sides of triangle and $R$ as its circumradius
than by easy trigonometry we get $AH=2RcosA , AD=2RsinBsinC$ etc.
so $AH.AD=4R^2cosAsinBsinC=cbcosA$ etc.

than our inequality converts to proving that
$ab+bc+ca\leq 2(cbcosA+abcosC+cacosB)$

now by cosine law ,we have $2cbcosA=b^2+c^2-a^2$ etc.
substituting,our inequality becomes

$ab+bc+ca\leq a^2+b^2+c^2$
which is obvious as it is equivalent to $\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\ge 0$

so we are done

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JuanOrtiz

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JuanOrtiz

#4 h May 27, 2015, 3:24 AM • 3 Y

Y by MillenniumFalcon, Adventure10, Mango247

Notice $2AH*AD=2AE*AC=2bc*cosA=c^2+b^2-a^2$ and so the problem translates to $a^2+b^2+c^2 \ge ab+bc+ca$ , done.

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MexicOMM

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MexicOMM

#5 h Jun 1, 2015, 5:47 AM • 2 Y

Y by Adventure10, Mango247

You three had the exact same solution, and still posted it. A new approach, instead of using $cos$ , we see that $2AH \cdot AD= AF \cdot AB + AE \cdot AC$ , analogously $2BH \cdot BE= BF \cdot AB + BD \cdot BC$ , and $2CH \cdot CF= CE \cdot AC + CD \cdot BC$ , but $AF+BF=AB$ , $AE+CE=AC$ , and $CD+BD=BC$ , so we need to prove $ab+bc+ca \le a^2+b^2+c^2$ , which is well known.

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AMN300

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AMN300

#6 h May 12, 2016, 7:55 PM • 2 Y

Y by Adventure10, Mango247

not really contributing anything here, but might as well post my solution
also, this seems too easy for an oly problem?
solution

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Virgil Nicula

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Virgil Nicula

#7 h May 12, 2016, 10:36 PM • 2 Y

Y by Adventure10, Mango247

See here the proposed problem P2.

This post has been edited 3 times. Last edited by Virgil Nicula, May 12, 2016, 10:58 PM

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programjames1

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programjames1

#8 h Jun 6, 2018, 5:46 PM • 2 Y

Y by Adventure10, Mango247

ABCDE wrote:

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How does Power of a Point give you $AH\cdot AD=AF\cdot AB?$

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Kaskade

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Kaskade

#9 h Jun 6, 2018, 5:53 PM • 2 Y

Y by Adventure10, Mango247

programjames1 wrote:

ABCDE wrote:

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How does Power of a Point give you $AH\cdot AD=AF\cdot AB?$

Quadrilateral $HDBF$ is cyclic (from $\angle BFH = \angle HDB = 90$ ), and the sides $HD$ and $BF$ extended meet at $A$ .

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Jzhang21

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Jzhang21

#10 h Dec 27, 2018, 5:00 PM • 2 Y

Y by Adventure10, Mango247

Note that it suffices to prove $2(AH\cdot AD+BH\cdot BE+CH\cdot CF)\geq ab+bc+ca.$

Claim 1: $AD\cdot AH=\frac{-a^2+b^2+c^2}{2}, BH\cdot BE=\frac{a^2-b^2+c^2}{2},$ and $CH\cdot CE=\frac{a^2+b^2-c^2}{2}$ .
Proof: Note that $AD=b\sin C$ and $CDHE$ is a cyclic quadrilateral. Then, $\angle AHE=\angle C$ so $AH=\frac{AE}{\sin C}=\frac{c\cos A}{\sin C}.$ Hence, $AD\cdot AH=b\sin C\cdot \frac{c\cos A}{\sin C}=bc\cos A=bc(\frac{-a^2+b^2+c^2}{2bc})=\frac{-a^2+b^2+c^2}{2},$ as desired. The others follow similarly. $\Box$

By Claim 1, we simplify the equation to $2(\frac{-a^2+b^2+c^2}{2}+\frac{a^2-b^2+c^2}{2}+\frac{a^2+b^2-c^2}{2})=a^2+b^2+c^2\geq ab+bc+ca.$ This follows by the Trivial Inequality $\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\geq 0.$ $\blacksquare$

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Pluto1708

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Pluto1708

#11 h Dec 27, 2018, 5:54 PM • 2 Y

Y by Adventure10, Mango247

Easy problem for Canado MO

Just note that the $DHS$ Joke is equivalent to $\sum{AH \cdot AD}$ = $\sum{bc \cdot \cos{A}}$ = $\frac{a^2 + b^2 + c^2}{2}$
Therefore the expression becomes $a^2 + b^2 + c^2 \ge ab + bc + ca$ which is trivial .

This post has been edited 1 time. Last edited by Pluto1708, Dec 27, 2018, 5:57 PM
Reason: \delta

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Maffeseater

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Maffeseater

#12 h Apr 11, 2019, 2:26 PM • 2 Y

Y by Adventure10, Mango247

ABCDE wrote:

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What is PoP in this context? Might be obvious but I simply can't make any connexions to any known results :maybe:

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Pluto1708

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Pluto1708

#13 h Apr 11, 2019, 3:16 PM • 2 Y

Y by Adventure10, Mango247

That's pop applied to cylic quad $BFEC$

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khanhnx

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khanhnx

#14 h Apr 15, 2019, 9:11 AM • 1 Y

Y by Adventure10

We have: $AB$ . $CA$ + $BC$ . $CA$ + $CA$ . $BC$ $\le$ $AB^2$ + $BC^2$ + $CA^2$ = $AF$ . $AB$ + $BF$ . $AB$ + $BD$ . $BC$ + $CD$ . $BC$ + $CE$ . $CA$ + $AE$ . $CA$ = 2 ( $AH$ . $AD$ + $BH$ . $BE$ + $CH$ . $CF$ )
So: $\dfrac{AB . CA + BC . CA + CA . BC}{AH . AD + BH . BE + CH . CF}$ $\le$ 2
Equality holds when: $AB$ = $BC$ = $CA$ or $\triangle$ $ABC$ is equilateral

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heheXD1

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heheXD1

#15 h Mar 9, 2021, 9:38 PM

Y by

aditya21 wrote:

here is my solution =
let us denote $a,b,c$ as sides of triangle and $R$ as its circumradius
than by easy trigonometry we get $AH=2RcosA , AD=2RsinBsinC$ etc.
so $AH.AD=4R^2cosAsinBsinC=cbcosA$ etc.

than our inequality converts to proving that
$ab+bc+ca\leq 2(cbcosA+abcosC+cacosB)$

now by cosine law ,we have $2cbcosA=b^2+c^2-a^2$ etc.
substituting,our inequality becomes

$ab+bc+ca\leq a^2+b^2+c^2$
which is obvious as it is equivalent to $\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\ge 0$

so we are done

"by easy trigonometry we get $AH=2RcosA , AD=2RsinBsinC$ ." I've been looking at this for like 2 days and I don't see how this is true... Could someone please provide some insight or hints.

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NoctNight

108 posts

NoctNight

#16 h Oct 14, 2021, 10:43 AM

Y by

By power of a point on cyclic quads $BDHF$ and $CDHE$ , $AH\cdot AD=AF\cdot AB=AE\cdot AC$ . Multiplying the denominator on the LHS to the RHS gives
$\sum_{cyc} AB\cdot BC\leq \sum_{cyc} 2(AH\cdot AD)=\sum_{cyc} AF\cdot AB+AE\cdot AC=\sum_{cyc} AB^2$ AM-GM completes the proof.

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Mogmog8

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Mogmog8

#17 h Jan 31, 2022, 5:24 PM • 2 Y

Y by centslordm, megarnie

By PoP and AM-GM, $\begin{align*}2(AH\cdot AC+BH\cdot BE+CH\cdot CF)&=AF\cdot AB+AE\cdot AC+BF\cdot BA+BD\cdot BC+CE\cdot CA+CD\cdot CB\\&=BC(CD+DB)+AC(DE+AE)+AB(AF+BF)\\&=BC^2+AC^2+AB^2\\&\ge AB\cdot AC+BC\cdot CA+CA\cdot CB\end{align*}$ $\square$

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MagicalToaster53

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MagicalToaster53

#18 h Jun 9, 2024, 10:16 PM

Y by

Let $a = \overline{BC}, b = \overline{CA}, c = \overline{AB}$ . Without loss of generality, assume $a \leq b \leq c$ . Observe that $AF = AC \cos \angle A$ , and also $\triangle AFH \sim \triangle ADB$ so that $\frac{AF}{AH} = \frac{AD}{AB} \implies \frac{AB}{AH} = \frac{AD}{AF} = \frac{AD}{AC \cos \angle A} \implies AH \cdot AD = \frac{b^2 + c^2 - a^2}{2}.$ Similarly, $BH \cdot BE = \frac{c^2 + a^2 - b^2}{2}, \text{ and } \frac{a^2 + b^2 - c^2}{2}.$ Therefore our initial inequality is equivalent to $\sum_{cyc}^{3} ab \leq \sum_{cyc}^{3} a^2,$ which is trivial by Chebyshev's rearrangement inequality. $\blacksquare$
Remark

This post has been edited 1 time. Last edited by MagicalToaster53, Jun 9, 2024, 10:16 PM

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