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
 
(
O,R
) be the circumcircle of the triangle
 ABC 
.Consider a point
 
 in the plane of the triangle. Denote by
 A
1
,B
1
,
1
 theprojections of 
 M 
 on triangle’s sides. The following relation holds
A
1
B
1
1
ABC 
=
 |
R
2
OM 
2
|
4
R
2
 .
Proof:
 First of all note that quadrilaterals
AB
1
M
1
,
 B
1
MA
1
,
 CA
1
MB
1
are cyclic. Applying the extended Law of Cosines in triangle
 AB
1
1
 we get
B
1
1
 =
 AM 
 sin
α
. Analogously
 A
1
1
 =
 B
 sin
β 
 and
 B
1
1
 =
 C
 sin
γ 
. Itfollows that
B
1
1
BC 
 =
 AM 
2
R , A
1
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
1
 =
A
1
B
1
+
MB
1
1
 =
A
1
CM 
+
MAC 
1
 =
ZYB
+
BY
 =
ZYX.
Similarly,
 
B
1
1
A
1
 =
 
YZX 
 and
 
B
1
A
1
1
 =
 
YX
. Thus triangles
A
1
B
1
1
 and
 XY
 are similar and
A
1
B
1
X
 =
 R
A
1
B
1
1
R .
Because triangles
 MAB
 and
 MY
 are also similar we have
XY AB
 =
 MMB.
Combining the above results we obtain
A
1
B
1
1
ABC 
=
 RR
A
1
B
1
1
·
A
1
B
1
·
B
1
1
·
A
1
1
AB
·
BC 
·
AC 
 =
 MMB
·
MA
2
R
 ·
MB
2
R
 =
 MA
·
M
4
R
2
 =
 |
R
2
OM 
2
|
4
R
2
 .
As we can see, the proof does not depend on the position of 
 
 (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 La-grange Theorem.
Theorem 2.
 Let
 M 
 be a point in the plane of triangle
 ABC 
 with barycen-tric coordinates (
u,v,w
). For any point
 
 in the plane of 
 ABC 
 the followingrelation holds
u
·
PA
2
+
v
·
PB
2
+
w
·
P
2
= (
u
+
v
 +
w
)
P
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
 
 coin-cides 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
 
 be a point in the plane of triangle
 ABC 
 withbarycentric coordinates (
u,v,w
). Denote by
 A
1
,B
1
,
1
 the projections of 
 
onto triangle’s sides. Then
A
1
B
1
1
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
 
 and the circle that passes through
 
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
1
 =
 O
2
,
 where
 O
 is the circumcenter of 
 AB
.
Solution:
 From the definition of the Brocard points we see that
1
AB
 =
1
BC 
 =
1
CA
 =
 w
1
,
 and similarly
 
2
BA
 =
2
AC 
 =
2
CB
 =
 w
2
. Let us prove that
 w
1
 =
 w
2
.Observe that
 
B
1
ABC 
=
 B
1
·
BC 
 sin
w
1
AB
·
BC 
 sin
β 
 = sin
2
w
1
sin
2
β 
 and, analogously,
1
A
ABC 
= sin
2
w
1
sin
2
γ  , 
A
1
B
ABC 
= sin
2
w
1
sin
2
α .
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Summing up the areas we obtain1 =
 
B
1
 +
1
A
 +
A
1
B
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
2
 lies inside the triangle. In order to provethat
 O
1
 =
 O
2
,
 we prove that their pedal areas are equal. If so, then wehave
 R
2
O
21
 =
 R
2
O
22
 and therefore
 O
1
 =
 O
2
.
Denote by
 A
1
,B
1
,
1
 the projections from Ω
1
 onto
 BC,AC,AB
, respec-tively. Then using the extended Law of Sines we get
 A
1
1
 =
 B
1
 sin
b
,because
 B
1
 is diameter in the cyclic quadrilateral
 BA
1
1
1
. Also from theLaw of Sines in triangle
 AB
1
 we have
B
1
sin
w
 =
 c
sin
b.
It follows that
A
1
1
 =
 B
1
 sin
b
 =
 c
sin
w.
Similarly we obtain
 B
1
1
 =
 b
sin
w
 and
 A
1
B
1
 =
 a
sin
w
. It is not difficultto see that triangle
 A
1
B
1
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
1
 =
 O
2
,
 and we are done.
Remark:
 The intersection of symmedians in the triangle is denoted by
 and is called Lemoine point. One can prove that
 
1
 =
 
2
.
 Moreover
O,
1
,K,
2
 lie on a circle with diameter
 O
 called Brocard circle.We continue with an interesting approach to another classical problem.
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