On the area of a pedal triangleIvan Borsenco
Geometry has been always the area of mathematics that attracted problemsolvers with its exactness and intriguing results. The article presents one of such beautiful results  the Euler’s Theorem for the pedal triangle and itsapplications. We start with the proof of this theorem and then we discussOlympiad problems.
Theorem 1.
Let
C
(
O,R
) be the circumcircle of the triangle
ABC
.Consider a point
M
in the plane of the triangle. Denote by
A
1
,B
1
,C
1
theprojections of
M
on triangle’s sides. The following relation holds
S
A
1
B
1
C
1
S
ABC
=

R
2
−
OM
2

4
R
2
.
Proof:
First of all note that quadrilaterals
AB
1
MC
1
,
BC
1
MA
1
,
CA
1
MB
1
are cyclic. Applying the extended Law of Cosines in triangle
AB
1
C
1
we get
B
1
C
1
=
AM
sin
α
. Analogously
A
1
C
1
=
BM
sin
β
and
B
1
C
1
=
CM
sin
γ
. Itfollows that
B
1
C
1
BC
=
AM
2
R , A
1
C
1
AC
=
BM
2
R , A
1
B
1
BC
=
CM
2
R .
Suppose
AM,BM,CM
intersect the circumcircle
C
(
O,R
) at points
X,Y,Z
.
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Angle chasing yields
∠
A
1
B
1
C
1
=
∠
A
1
B
1
M
+
∠
MB
1
C
1
=
∠
A
1
CM
+
∠
MAC
1
=
∠
ZYB
+
∠
BYX
=
∠
ZYX.
Similarly,
∠
B
1
C
1
A
1
=
∠
YZX
and
∠
B
1
A
1
C
1
=
∠
YXZ
. Thus triangles
A
1
B
1
C
1
and
XYZ
are similar and
A
1
B
1
XY
=
R
A
1
B
1
C
1
R .
Because triangles
MAB
and
MYX
are also similar we have
XY AB
=
MX MB.
Combining the above results we obtain
S
A
1
B
1
C
1
S
ABC
=
RR
A
1
B
1
C
1
·
A
1
B
1
·
B
1
C
1
·
A
1
C
1
AB
·
BC
·
AC
=
MX MB
·
MA
2
R
·
MB
2
R
=
MA
·
MX
4
R
2
=

R
2
−
OM
2

4
R
2
.
As we can see, the proof does not depend on the position of
M
(inside oroutside the circle).
Corollary 1.
If
M
lies on the circle, the projections of
M
onto triangle’ssides are collinear. This fact is known as Simson’s Theorem.One more theorem we want to present without proof is the famous Lagrange Theorem.
Theorem 2.
Let
M
be a point in the plane of triangle
ABC
with barycentric coordinates (
u,v,w
). For any point
P
in the plane of
ABC
the followingrelation holds
u
·
PA
2
+
v
·
PB
2
+
w
·
PC
2
= (
u
+
v
+
w
)
PM
2
+
vwa
2
+
uwb
2
+
uvc
2
u
+
v
+
w .
The proof can be found after applying Stewart’s Theorem a few times. Thecorollary of this theorem in which we are interested is the case when
P
coincides with the circumcenter
O
. We get
R
2
−
OM
2
=
vwa
2
+
uwb
2
+
uvc
2
(
u
+
v
+
w
)
2
.
From this fact and the Euler’s Theorem for pedal triangle we obtain the nextresult:
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Theorem 3.
Let
M
be a point in the plane of triangle
ABC
withbarycentric coordinates (
u,v,w
). Denote by
A
1
,B
1
,C
1
the projections of
M
onto triangle’s sides. Then
S
A
1
B
1
C
1
S
ABC
=
vwa
2
+
uwb
2
+
uvc
2
4
R
2
(
u
+
v
+
w
)
2
.
From the above results we can see that Theorem 1 and Theorem 3 give usinsight on the area of pedal triangles. The Euler’s Theorem for the area of a pedal triangle is a useful tool in solving geometry problems. The ﬁrst oneapplication we present is about Brocard points.Let us introduce the deﬁnition of a Brocard point. In triangle
ABC
thecircle that passes through
A
and is tangent to
BC
at
B
, circle that passesthrough
B
and is tangent to
AC
at
C
and the circle that passes through
C
and is tangent to
AB
at
A
are concurrent. The point of their concurrencyis called Brocard point. In general, we have two Brocard points, the secondone being obtained by reversing clockwise or counterclockwise the tangency of circles.
Problem 1.
Let Ω
1
and Ω
2
be the two Brocard points of triangle
ABC
.Prove that
O
Ω
1
=
O
Ω
2
,
where
O
is the circumcenter of
ABC
.
Solution:
From the deﬁnition of the Brocard points we see that
∠
Ω
1
AB
= Ω
1
BC
= Ω
1
CA
=
w
1
,
and similarly
∠
Ω
2
BA
= Ω
2
AC
=Ω
2
CB
=
w
2
. Let us prove that
w
1
=
w
2
.Observe that
S
B
Ω
1
C
S
ABC
=
B
Ω
1
·
BC
sin
w
1
AB
·
BC
sin
β
= sin
2
w
1
sin
2
β
and, analogously,
S
C
Ω
1
A
S
ABC
= sin
2
w
1
sin
2
γ , S
A
Ω
1
B
S
ABC
= sin
2
w
1
sin
2
α .
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(2007)
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Summing up the areas we obtain1 =
S
B
Ω
1
C
+
S
C
Ω
1
A
+
S
A
Ω
1
B
S
ABC
= sin
w
21
sin
2
α
+ sin
w
21
sin
2
β
+ sin
w
21
sin
2
γ
or1sin
w
21
= 1sin
2
α
+ 1sin
2
β
+ 1sin
2
γ .
The same expression we get summing the areas for Ω
2
:1sin
w
22
= 1sin
2
α
+ 1sin
2
β
+ 1sin
2
γ
From two equalities we conclude that
w
1
=
w
2
=
w
.The idea for the solution comes from Euler’ Theorem for pedal triangles.Observe that Ω
1
and Ω
2
always lie inside triangle
ABC
, because every circlelies in that half of the plane that contains triangle
ABC
. It follows that thepoint of their intersection Ω
1
or Ω
2
lies inside the triangle. In order to provethat
O
Ω
1
=
O
Ω
2
,
we prove that their pedal areas are equal. If so, then wehave
R
2
−
O
Ω
21
=
R
2
−
O
Ω
22
and therefore
O
Ω
1
=
O
Ω
2
.
Denote by
A
1
,B
1
,C
1
the projections from Ω
1
onto
BC,AC,AB
, respectively. Then using the extended Law of Sines we get
A
1
C
1
=
B
Ω
1
sin
b
,because
B
Ω
1
is diameter in the cyclic quadrilateral
BA
1
Ω
1
C
1
. Also from theLaw of Sines in triangle
AB
Ω
1
we have
B
Ω
1
sin
w
=
c
sin
b.
It follows that
A
1
C
1
=
B
Ω
1
sin
b
=
c
sin
w.
Similarly we obtain
B
1
C
1
=
b
sin
w
and
A
1
B
1
=
a
sin
w
. It is not diﬃcultto see that triangle
A
1
B
1
C
1
is similar to
ABC
with ratio of similarity sin
w
.From the fact
w
1
=
w
2
=
w
we conclude that pedal triangles of Ω
1
and Ω
2
have the same area. Thus
O
Ω
1
=
O
Ω
2
,
and we are done.
Remark:
The intersection of symmedians in the triangle is denoted by
K
and is called Lemoine point. One can prove that
K
Ω
1
=
K
Ω
2
.
Moreover
O,
Ω
1
,K,
Ω
2
lie on a circle with diameter
OK
called Brocard circle.We continue with an interesting approach to another classical problem.
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(2007)
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