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Problem
Problem
If H is the orthocentre of triangle ABC,
R
=
circumradius
R = \text{circumradius}
R
=
circumradius
and
P
=
A
H
+
B
H
+
C
H
P = AH + BH + CH
P
=
A
H
+
B
H
+
C
H
, then a.
P
=
2
(
R
+
r
)
P = 2(R + r)
P
=
2
(
R
+
r
)
max. of P is
3
R
3R
3
R
d.
P
=
2
(
R
−
r
)
P = 2(R - r)
P
=
2
(
R
−
r
)
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