BDMO regional 2015

For discussing Olympiad level Geometry Problems
Math Mad Muggle
Posts: 29
Joined: Mon Jan 23, 2017 10:32 am
Location: Rajshahi,Bangladesh

BDMO regional 2015

Unread post by Math Mad Muggle » Fri Feb 03, 2017 12:51 am

help me solving this
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dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm
Location: Dhaka,Bangladesh

Re: BDMO regional 2015

Unread post by 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

Math Mad Muggle
Posts: 29
Joined: Mon Jan 23, 2017 10:32 am
Location: Rajshahi,Bangladesh

Re: BDMO regional 2015

Unread post by Math Mad Muggle » Mon Feb 06, 2017 10:24 am

Got it bt what's about the 10??

Absur Khan Siam
Posts: 61
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

Re: BDMO regional 2015

Unread post by Absur Khan Siam » Mon Feb 06, 2017 10:54 pm

"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

aritra barua
Posts: 50
Joined: Sun Dec 11, 2016 2:01 pm

Re: BDMO regional 2015

Unread post by 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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