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BdMO Online Forum • View topic - BDMO regional 2015

BDMO regional 2015

For discussing Olympiad level Geometry Problems
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BDMO regional 2015

Post Number:#1  Unread postby Math Mad Muggle » Fri Feb 03, 2017 12:51 am

help me solving this
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Re: BDMO regional 2015

Post Number:#2  Unread postby dshasan » Fri Feb 03, 2017 11:57 am

$\text{Problem 8}$
Join $B,O$. In $\bigtriangleup ABO$ and $\bigtriangleup BOC$, $\angle COB = \angle ABO$ and
$\angle CBO = \angle AOB$ and $BO$ is the common side. So, $\bigtriangleup ABO$ is congruent to $\bigtriangleup BOC$. So, $\angle OAB = \angle OCB \Rightarrow \angle OAB = \angle CBO
\Rightarrow \angle OAB = \angle AOB$. Therefore, $\bigtriangleup ABO$ is equilateral and $\angle OAB = 60$.
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton
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Re: BDMO regional 2015

Post Number:#3  Unread postby Math Mad Muggle » Mon Feb 06, 2017 10:24 am

Got it bt what's about the 10??
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Re: BDMO regional 2015

Post Number:#4  Unread postby Absur Khan Siam » Mon Feb 06, 2017 10:54 pm

"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
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Re: BDMO regional 2015

Post Number:#5  Unread postby aritra barua » Thu Feb 23, 2017 4:49 pm

For problem 10,note that SP=7 units because ST is the perpendicular bisector of triangle SPR.Apply angle bisector theorem taking PT=RT=a and PQ=b,find b=18a/7.Then apply Stewart's theorem;which mentions 324a^2/7+36a^2=16(49+63),derive a point, (24a)^2=112^2;a=14/3.Therefore,PR=2a=28/3 which takes the form x/y,x+y=28+3=31.
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