Problems in Geometry
(2)
1.
Given a rectangle
ABCD
prove that

PA

2
+

PC

2
=

PB

2
+

PD

2
for any point
P.
2.
Given an acute angled positively oriented triangle
ABC,
let
O
and
H
be the circum
center and the orthocenter of
ABC,
respectively. Let
AH
intersect
BC
in
˜
A.
(A) Prove that

AH

= 2
R
cos
A.
1
(B) Prove that

˜
AH

= 2
R
cos
B
cos
C .
2
(C) Prove that
OH
is perpendicular to
AH
iff tan
B
tan
C
= 3
.
3
3.
Let
P
be a point on the circumcircle of a triangle
ABC
such that
AP
is the internal
bisector of the angle
A .
Prove that the following conditions are equivalent
4 5
(i)

PA

= 2

PB

,
(ii) 2 sin
A
2
= cos
B

C
2
,
(iii) 2
a
=
b
+
c .
4.
In a triangle
ABC
prove that the following assertions are equivalent:
(i)
A
= 90
◦
,
(ii)
r
+
r
b
+
r
c
=
r
a
,
(iii)
r
b
r
c
=
rr
a
.
6
5.
Generalise the previous result by proving that in a triangle
ABC
the following condi
tions are equivalent:
(i) cos
A
=
α

1
α
+ 1
,
(ii)
r
b
+
r
c
=
α
(
r
a

r
)
,
(iii)
r
b
r
c
=
αrr
a
.
where
α
6
=

1 is a constant. Compute the angle
A
in a triangle
ABC
in which
3
r
+
r
b
+
r
c
= 3
r
a
.
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 Spring '11
 cemtezer
 Geometry, Pythagorean Theorem, Law Of Cosines, triangle

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