## BdMO-2014 circled pentagon

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
marek12
Posts: 2
Joined: Thu Jun 23, 2016 1:55 pm

### BdMO-2014 circled pentagon

\$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?

samiul_samin
Posts: 1004
Joined: Sat Dec 09, 2017 1:32 pm

### Re: BdMO-2014 circled pentagon

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

Yes, but i think is something wrong with it 