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### BdMO-2014 circled pentagon

Posted: **Fri Mar 29, 2019 11:07 am**

by **marek12**

$ABCDE$ 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?

### Re: BdMO-2014 circled pentagon

Posted: **Wed Apr 10, 2019 10:12 am**

by **samiul_samin**

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?

### Re: BdMO-2014 circled pentagon

Posted: **Fri Apr 12, 2019 11:50 pm**

by **marek12**

Yes, but i think is something wrong with it