BdMO Online Forum

help me solving this

$\text{Problem 8}$
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 $\angle OAB = 60$.

Got it bt what's about the 10??

Problem 10:
viewtopic.php?f=19&t=3844

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.