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Contests
National and Regional Contests
Moldova Contests
EGMO TST - Moldova
2018 Moldova EGMO TST
8
8
Part of
2018 Moldova EGMO TST
Problems
(1)
Easy Geometry equality
Source: EGMO 2018 Moldova TST
2/11/2018
Let
A
B
C
ABC
A
BC
be a triangle with
A
B
=
c
AB=c
A
B
=
c
,
B
C
=
a
BC=a
BC
=
a
and
A
C
=
b
AC=b
A
C
=
b
. If
x
,
y
∈
R
x,y\in\mathbb{R}
x
,
y
∈
R
satisfy
1
x
+
1
y
+
z
=
1
a
\frac{1}{x} +\frac{1}{y+z} = \frac{1}{a}
x
1
+
y
+
z
1
=
a
1
,
1
y
+
1
x
+
z
=
1
b
\frac{1}{y} +\frac{1}{x+z} = \frac{1}{b}
y
1
+
x
+
z
1
=
b
1
,
1
z
+
1
y
+
x
=
1
c
\frac{1}{z} +\frac{1}{y+x} = \frac{1}{c}
z
1
+
y
+
x
1
=
c
1
. Prove that the following equality holds
x
(
p
−
a
)
+
y
(
p
−
b
)
+
z
(
p
−
c
)
=
3
r
2
+
12
R
∗
r
,
x(p-a) + y(p-b) + z(p-c) = 3r^2 + 12R*r ,
x
(
p
−
a
)
+
y
(
p
−
b
)
+
z
(
p
−
c
)
=
3
r
2
+
12
R
∗
r
,
Where
p
p
p
is semi-perimeter,
R
R
R
is the circumradius and
r
r
r
is the inradius.
geometry
circumcircle