BDMO 2017 National round Secondary 5

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Absur Khan Siam
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Re: BDMO 2017 National round Secondary 5

Unread post by Absur Khan Siam » Mon Mar 05, 2018 10:52 pm

Tasnood wrote:
Mon Mar 05, 2018 10:34 pm
Sorry for Interrupt. My answer was same of #Nahin but this solution is new to me.
How can we find the radius of two circles same?
$AC = AB$ , radius of the $\widehat{BC}$
$BC = AB$ , radius of the $\widehat{AB}$

Thus, $AC = BC$
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

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samiul_samin
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Re: BDMO 2017 National round Secondary 5

Unread post by samiul_samin » Mon Mar 05, 2018 11:18 pm

Durjoy Sarkar wrote:
Mon Mar 05, 2018 10:06 pm
radius of two big circle are same. let X be the point where little circle is tangent.
it is well known the center of little circle, tangent point are lies on $OB$.
$MO=OX$
$BX= BO+OX=BO+MO $
The placement of point $X$ is not clear to me.I didn't understand the $2$nd line of it.

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Tasnood
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Re: BDMO 2017 National round Secondary 5

Unread post by Tasnood » Mon Mar 05, 2018 11:43 pm

Again a question. How can we say that $A$-centered circle will go through $B$?

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ahmedittihad
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Re: BDMO 2017 National round Secondary 5

Unread post by ahmedittihad » Tue Mar 06, 2018 2:13 pm

Both have $AB$ as radius.....
Frankly, my dear, I don't give a damn.

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Tasnood
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Re: BDMO 2017 National round Secondary 5

Unread post by Tasnood » Wed Mar 07, 2018 10:07 am

Durjoy Sarkar wrote:
Mon Mar 05, 2018 10:06 pm
radius of two big circle are same. let X be the point where little circle is tangent.
it is well known the center of little circle, tangent point are lies on $OB$.
$MO=OX$
$BX= BO+OX=BO+MO $
You proved $BX=OB+OM$, not $AB=OB+OM$

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Rangon Roy Utsab
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Re: BDMO 2017 National round Secondary 5

Unread post by Rangon Roy Utsab » Thu Mar 08, 2018 10:50 pm

Tasnood wrote:
Wed Mar 07, 2018 10:07 am
Durjoy Sarkar wrote:
Mon Mar 05, 2018 10:06 pm
radius of two big circle are same. let X be the point where little circle is tangent.
it is well known the center of little circle, tangent point are lies on $OB$.
$MO=OX$
$BX= BO+OX=BO+MO $
You proved $BX=OB+OM$, not $AB=OB+OM$
Look carefully and you will notice that $BX$ and $AB$ are radii of the same circle centered $B$

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