BdMO-2014 circled pentagon

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marek12
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Joined: Thu Jun 23, 2016 1:55 pm

BdMO-2014 circled pentagon

Unread post by marek12 » Fri Mar 29, 2019 11:07 am

$ABCDE$ is a circled pentagon. $AC= 2$, $AD= 3$, $BD=5$, $BE =1$ and $CD/DE = 10/3$, $BC/CE = a/b$ (where $a$ and $b$ are coprime) How to find the value $a-b$?


|AB| < |AC| ... so |AD| + |AB| < |AD| + |AC| = 2 + 3 = 5 = |BD| so is there a triangle ΔABD?

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samiul_samin
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Joined: Sat Dec 09, 2017 1:32 pm

Re: BdMO-2014 circled pentagon

Unread post by samiul_samin » Wed Apr 10, 2019 10:12 am

marek12 wrote:
Fri Mar 29, 2019 11:07 am
$ABCDE$ is a circled pentagon. $AC= 2$, $AD= 3$, $BD=5$, $BE =1$ and $\dfrac{CD}{DE} = \dfrac{10}{3}$, $\dfrac{BC}{CE} = \dfrac{a}{b}$ (where $a$ and $b$ are coprime) How to find the value $a-b$?


$|AB| < |AC| \cdots$ so $|AD| + |AB| < |AD| + |AC| = 2 + 3 = 5 = |BD|$ so is there a triangle $ΔABD?$
Regional problem?

marek12
Posts: 2
Joined: Thu Jun 23, 2016 1:55 pm

Re: BdMO-2014 circled pentagon

Unread post by marek12 » Fri Apr 12, 2019 11:50 pm

Yes, but i think is something wrong with it :?

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