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#1 h Dec 5, 2010, 11:23 AM • 3 Y

Y by justJen, Adventure10, Mango247

Let $ABCDEF$ be a convex hexagon in which diagonals $AD, BE, CF$ are concurrent at $O$ . Suppose $[OAF]$ is geometric mean of $[OAB]$ and $[OEF]$ and $[OBC]$ is geometric mean of $[OAB]$ and $[OCD]$ . Prove that $[OED]$ is the geometric mean of $[OCD]$ and $[OEF]$ .
(Here $[XYZ]$ denotes are of $\triangle XYZ$ )

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#2 h Dec 5, 2010, 12:33 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

I use the fact that if $A$ , $A'$ are on other sides of $BC$ and $AA'$ meets $BC$ at $K$ , then $\frac{[ABC]}{[A'BC]}=\frac{AK}{KA'}$ which can be proved easily by projecting $A, A'$ on $BC$ and using similarity for two right triangles. Now, in the hexagon, let $AC$ meet $OB$ at $M$ , $CE$ meet $OD$ at $N$ and $EA$ meet $OF$ at $P$ . By Ceva's theorem for triangle $ACE$ , we have
$\frac{AM}{MC}\cdot \frac{CN}{NE}\cdot \frac{EP}{PA}=1$
And from the fact, we have
$\frac{a}{b}\cdot \frac{c}{d}\cdot \frac{e}{f}=1$
where
$a=[OAB]; b=[OBC]; c=[OCD]; d=[ODE]; e=[OEF]; f=[OFA]$
So,
$\frac{a^2c^2e^2}{b^2d^2f^2}=1$
We also have $ae=f^2; ac=b^2$
$\Longrightarrow ce=d^2$

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#3 h Dec 5, 2010, 12:50 PM • 2 Y

Y by Adventure10, Mango247

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#4 h Dec 5, 2010, 3:21 PM • 1 Y

Y by Adventure10

Goutham wrote:

I use the fact that if $A$ , $A'$ are on other sides of $BC$ and $AA'$ meets $BC$ at $K$ , then $\frac{[ABC]}{[A'BC]}=\frac{AK}{KA'}$ which can be proved easily by projecting $A, A'$ on $BC$ and using similarity for two right triangles. Now, in the hexagon, let $AC$ meet $OB$ at $M$ , $CE$ meet $OD$ at $N$ and $EA$ meet $OF$ at $P$ . By Ceva's theorem for triangle $ACE$ , we have
$\frac{AM}{MC}\cdot \frac{CN}{NE}\cdot \frac{EP}{PA}=1$
And from the fact, we have
$\frac{a}{b}\cdot \frac{c}{d}\cdot \frac{e}{f}=1$
where
$a=[OAB]; b=[OBC]; c=[OCD]; d=[ODE]; e=[OEF]; f=[OFA]$
So,
$\frac{a^2c^2e^2}{b^2d^2f^2}=1$
We also have $ae=f^2; ac=b^2$
$\Longrightarrow ce=d^2$

Beautiful Proof!
Mine was same as cmtappu96

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3130 posts

#5 h Dec 5, 2010, 3:38 PM • 2 Y

Y by Adventure10, Mango247

firstuser wrote:

Goutham wrote:

I use the fact that if $A$ , $A'$ are on other sides of $BC$ and $AA'$ meets $BC$ at $K$ , then $\frac{[ABC]}{[A'BC]}=\frac{AK}{KA'}$ which can be proved easily by projecting $A, A'$ on $BC$ and using similarity for two right triangles. Now, in the hexagon, let $AC$ meet $OB$ at $M$ , $CE$ meet $OD$ at $N$ and $EA$ meet $OF$ at $P$ . By Ceva's theorem for triangle $ACE$ , we have
$\frac{AM}{MC}\cdot \frac{CN}{NE}\cdot \frac{EP}{PA}=1$
And from the fact, we have
$\frac{a}{b}\cdot \frac{c}{d}\cdot \frac{e}{f}=1$
where
$a=[OAB]; b=[OBC]; c=[OCD]; d=[ODE]; e=[OEF]; f=[OFA]$
So,
$\frac{a^2c^2e^2}{b^2d^2f^2}=1$
We also have $ae=f^2; ac=b^2$
$\Longrightarrow ce=d^2$

Beautiful Proof!
Mine was same as cmtappu96

I got it in the last 10 minutes or so. Couldn't complete it

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274 posts

#6 h Dec 6, 2010, 6:55 AM • 1 Y

Y by Adventure10

Goutham wrote:

firstuser wrote:

Goutham wrote:

I use the fact that if $A$ , $A'$ are on other sides of $BC$ and $AA'$ meets $BC$ at $K$ , then $\frac{[ABC]}{[A'BC]}=\frac{AK}{KA'}$ which can be proved easily by projecting $A, A'$ on $BC$ and using similarity for two right triangles. Now, in the hexagon, let $AC$ meet $OB$ at $M$ , $CE$ meet $OD$ at $N$ and $EA$ meet $OF$ at $P$ . By Ceva's theorem for triangle $ACE$ , we have
$\frac{AM}{MC}\cdot \frac{CN}{NE}\cdot \frac{EP}{PA}=1$
And from the fact, we have
$\frac{a}{b}\cdot \frac{c}{d}\cdot \frac{e}{f}=1$
where
$a=[OAB]; b=[OBC]; c=[OCD]; d=[ODE]; e=[OEF]; f=[OFA]$
So,
$\frac{a^2c^2e^2}{b^2d^2f^2}=1$
We also have $ae=f^2; ac=b^2$
$\Longrightarrow ce=d^2$

Beautiful Proof!
Mine was same as cmtappu96

I got it in the last 10 minutes or so. Couldn't complete it

sem here, only that happened for the 2nd geometry problem

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409 posts

#7 h Dec 6, 2010, 11:05 AM • 2 Y

Y by Adventure10, Mango247

cmtappu96 wrote:

[geogebra]718b96c4a887549bde132c20accd5fdc17325fc7[/geogebra]

for any $\triangle XYZ, [XYZ] = \frac{1}{2} \sin X \cdot yz$

Now $[OAF]^2 = [OAB][OEF]$
$\frac{1}{4} \sin^2 \gamma \cdot a^2f^2 = \frac{1}{4} \sin \beta \cdot \sin \alpha \cdot afbe$
$\sin^2 \gamma \cdot af = \sin \alpha \cdot \sin \beta \cdot be$
Similarly we get using $[OBC]^2 = [OAB][OCD]$ ,
$\sin^2 \beta \cdot bc = \sin \alpha \cdot \sin \gamma \cdot ad$
Multiplying these equations, we get,
$\sin^2 \beta \cdot \sin^2 \gamma \cdot abcf = \sin^2 \alpha \sin \beta \sin \gamma \cdot abde$
$\sin \beta \cdot \sin \gamma \cdot cf = \sin^2 \alpha \cdot de$

Multiplying by $\frac{1}{4} de$ , we get

$\left(\frac{1}{2} \sin \beta \cdot ef \right) \left(\frac{1}{2} \sin \gamma \cdot cd \right) = \left(\frac{1}{2} \sin \alpha \cdot de\right)^2$
$[OCD][OEF] = [OED]^2$

I did solve the problem in the same manner in just 5 mins in the paper.

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#8 h Apr 25, 2011, 2:37 PM • 2 Y

Y by Adventure10, Mango247

Even I got it in the last five minutes. I was about to give in my paper, thinking that I would not be able to get it, especially with just 5 minutes left, when I remembered Area = absinC/2

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#9 h Jun 11, 2011, 4:49 AM • 2 Y

Y by Adventure10, Mango247

Let $OA =a, OB=b, OC =c, OD =d, OE=e, OF =f$ and $\angle AOB = \angle EOD =\alpha$ , $\angle BOC = \angle FOE = \beta$ and $\angle COD = \angle AOF = 180 -\alpha -\beta$

Note that $[OAB][OCD][OEF] = [OBC][ODE][OAF]$

iff $ab \sin \alpha \cdot cd \sin (\alpha + \beta) \cdot ef \sin B = bc \sin \beta \cdot de \sin \alpha \cdot af \sin (\alpha + \beta)$ which is obviously true.

Hence

$[ODE] = \frac {[OAB][OCD][OEF]}{[OBC][OAF]}$

$= \frac {[OAB][OCD][OEF]}{\sqrt{[OAB][OCD][OAB][OEF]}} = \sqrt{[OCD][OEF]}$

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aayush-srivastava

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#10 h Nov 24, 2015, 12:14 PM • 2 Y

Y by Adventure10, Mango247

http://olympiads.hbcse.tifr.res.in/uploads/rmo-2010

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#11 h Oct 16, 2021, 2:48 AM

Y by

Hi all, this may be late, but here's a solution I found. Thanks!

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