BDMO regional 2015
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help me solving this
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Re: BDMO regional 2015
$\text{Problem 8}$
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 $\angle OAB = 60$.
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
- Charles Caleb Colton
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Re: BDMO regional 2015
Got it bt what's about the 10??
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Re: BDMO regional 2015
Problem 10:
viewtopic.php?f=19&t=3844
viewtopic.php?f=19&t=3844
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
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Re: BDMO regional 2015
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.