lies inside triangle 
such that 
and 
.
is the midpoint of segment 
.
lies on segment 
with 
.
.
satisfying
for all real numbers 
and 
.
Y by Adventure10
Let 
be a non-obtuse triangle with 
and 
are the altitude and median, respectively. The angle bisector of 
intersects and at 
and 
, respectively. Assume that ![\[ \angle ABP=\angle PBQ=\angle QBC,\]](//latex.artofproblemsolving.com/3/a/4/3a4a3b7b42b0d3e083b5dc14c9c2ffcd28270b89.png)
(a) prove that is a right-angled triangle, and
(b) calculate
.
Soewono, Bandung
be a non-obtuse triangle with 
and 
are the altitude and median, respectively. The angle bisector of 
intersects and at 
and 
, respectively. Assume that ![\[ \angle ABP=\angle PBQ=\angle QBC,\]](http://latex.artofproblemsolving.com/3/a/4/3a4a3b7b42b0d3e083b5dc14c9c2ffcd28270b89.png)
(a) prove that
is a right-angled triangle, and(b) calculate
.Soewono, Bandung
![\[ \angle ABP=\angle PBQ=\angle QBC, \]](http://latex.artofproblemsolving.com/b/8/5/b85010d5c5bff218d39baa7d39c3502a4961f33a.png)
= 8.948922449° (yes, that ugly angle) and 
= 2.
and 
we have that 
and 
are isogonal conjugates. Therefore , 
is the reflection of 
over the bisector of 
and therefore we get that the circumcenter of 
.
.
,
.
and 
and therefore we obtain ![\[\frac{BP}{CH} = \frac{\sin \alpha}{\cos \alpha \cdot \sin 60 + \frac{\alpha}{3}} (*)\]](http://latex.artofproblemsolving.com/7/c/2/7c2fa935715478f0e02a250a14d09d090543149b.png)
we get 
so plugging this in 
gives us![\[ \frac{BP}{CH} = \frac{\sin 60 - \frac{2\alpha}{3}}{ \sin 30 - \frac{\alpha}{3} \cdot \sin 60 + \frac{\alpha}{3}} = \frac{\sin 60 - \frac{2\alpha}{3}}{ \sin 30 - \frac{\alpha}{3} \cdot \cos 30 - \frac{\alpha}{3}} = 2 \]](http://latex.artofproblemsolving.com/5/8/7/5878d43f1348f6180f46d34b8c96a657c560d621.png)
to 
.
,![\[\frac{CB}{CH} = \frac{AC}{AH} = \frac{CP}{PH} = 2\frac{CP}{PP'} = 2\frac{CB}{P'B} = 2\frac{CB}{PB},\]](http://latex.artofproblemsolving.com/8/c/2/8c29703eec6481d74596d916e2c1069c9dba67b2.png)
thus 
.
