BDMO 2018 junior 10

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
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Safwan
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Location: BAU,Mymensingh,Bangladesh

Re: BDMO 2018 junior 10

Unread post by Safwan » Tue Jan 29, 2019 8:04 pm

Plz can u post the elaborate explanation?

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samiul_samin
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Re: BDMO 2018 junior 10

Unread post by samiul_samin » Wed Jan 30, 2019 12:15 pm

Safwan wrote:
Tue Jan 29, 2019 8:04 pm
Plz can u post the elaborate explanation?
Here is an elaborated solution.
Solution
STEP $1$:
Join $E$,$D$
.$AO$ & $ED$ intersects at $F$
According to the question
$\angle{BAC}=\angle{ABC}=\angle{ACB}=60$

By easy angle chasing we get,

$\angle{ECB}=20^{\circ}$

$\angle{ABD}=20^{\circ}$

$\angle{DBC}=40^{\circ}$

$\angle{ECA}=40^{\circ}$

$\angle{BOC}=120^{\circ}$

$\angle{DOC}=60^{\circ}$

$\angle{BDC}=80^{\circ}$

$\angle{BEC}=80^{\circ}$

$\angle{ADO}=100^{\circ}$

$\angle{EOD}=120^{\circ}$

$\angle{EOD}+\angle{EAD}=180^{\circ}$

So $OEAD$ is a inscribed in a circle.
So,$\angle{EDO}=\angle{EAO}$

STEP $2$:

Using cosine law,$\angle {ADE}=90^{\circ}$
So,
:arrow: $\angle{EDO}=10^{\circ}$

:arrow: $\angle{EAO}=10^{\circ}$

:arrow: $\angle{OAD}=50^{\circ}$

:arrow: $\angle{OAD}+\angle{ACO}=90^{\circ}$

:arrow: $\angle{AOC}=90^{\circ}$
Note:Draw a perfect figure to understand the solution.
If there is any mistake please reply.

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Safwan
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Location: BAU,Mymensingh,Bangladesh

Re: BDMO 2018 junior 10

Unread post by Safwan » Wed Jan 30, 2019 7:05 pm

THNX a LOT!

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