Tam giác Pedal

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On the area of a pedal triangle
Ivan Borsenco

Geometry has been always the area of mathematics that attracted problem
solvers with its exactness and intriguing results. The article presents one of
such beautiful results - the Euler’s Theorem for the pedal triangle and its
applications. We start with the proof of this theorem and then we discuss
Olympiad 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 A1, B1, C1 the

projections of M on triangle’s sides. The following relation holds
SA1B1C1

SABC

= |R

2_{− OM}2_|

4R2 .

Proof: First of all note that quadrilaterals AB1M C1, BC1M A1, CA1M B1

are cyclic. Applying the extended Law of Cosines in triangle AB1C1 we get

B1C1 = AM sin α. Analogously A1C1 = BM sin β and B1C1 = CM sin γ. It

follows that

B1C1

BC =
AM

2R ,
A1C1

AC =
BM

2R ,
A1B1

BC =
CM

2R .

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Angle chasing yields

∠A1B1C1 = ∠A1B1M +∠M B1C1 = ∠A1CM +∠M AC1 = ∠ZY B+∠BY X = ∠ZY X.

Similarly, ∠B1C1A1 = ∠Y ZX and ∠B1A1C1 = ∠Y XZ. Thus triangles

A1B1C1 and XY Z are similar and

A1B1

XY =

RA1B1C1

R .

Because triangles M AB and M Y X are also similar we have
XY

AB =
M X
M B.
Combining the above results we obtain

SA1B1C1

SABC

= R

RA1B1C1

·A1B1· B1C1· A1C1
AB · BC · AC =

M X
M B·

M A
2R ·

M B
2R =

M A · M X
4R2 =

|R2_{− OM}2_|

4R2 .

As we can see, the proof does not depend on the position of M (inside or
outside the circle).

Corollary 1. If M lies on the circle, the projections of M onto triangle’s
sides 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 P in the plane of ABC the following
relation holds

u · P A2+ v · P B2+ w · P C2 = (u + v + w)P M2+vwa

2_{+ uwb}2_{+ uvc}2

u + v + w .
The proof can be found after applying Stewart’s Theorem a few times. The
corollary of this theorem in which we are interested is the case when P

coin-cides with the circumcenter O. We get

R2− OM2 = vwa

2_{+ uwb}2_{+ uvc}2

(u + v + w)2 .

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Theorem 3. Let M be a point in the plane of triangle ABC with
barycentric coordinates (u, v, w). Denote by A1, B1, C1 the projections of M

onto triangle’s sides. Then
SA1B1C1

SABC

= vwa

2_{+ uwb}2_{+ uvc}2

4R2_{(u + v + w)}2 .

From the above results we can see that Theorem 1 and Theorem 3 give us
insight 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 one
application we present is about Brocard points.

Let us introduce the definition of a Brocard point. In triangle ABC the
circle that passes through A and is tangent to BC at B, circle that passes
through 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 concurrency
is called Brocard point. In general, we have two Brocard points, the second
one being obtained by reversing clockwise or counterclockwise the tangency of
circles.

Problem 1. Let Ω1 and Ω2 be the two Brocard points of triangle ABC.

Prove that OΩ1 = OΩ2, where O is the circumcenter of ABC.

Solution: From the definition of the Brocard points we see that

∠Ω1AB = Ω1BC = Ω1CA = w1, and similarly ∠Ω2BA = Ω2AC =

Ω2CB = w2. Let us prove that w1 = w2.

Observe that SBΩ1C

SABC

= BΩ1· BC sin w1
AB · BC sin β =

sin2w1

sin2β and, analogously,
SCΩ1A

SABC

= sin

2_w
1

sin2γ ,

SAΩ1B

SABC

= sin

2_w
1

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Summing up the areas we obtain

1 = SBΩ1C + SCΩ1A+ SAΩ1B

SABC

= sin w

2
1

sin2α +
sin w2₁
sin2β +

sin w2₁
sin2γ or
1

sin w2
1

= 1

sin2α +
1
sin2β +

1
sin2γ.
The same expression we get summing the areas for Ω2:

1
sin w2₂ =

1
sin2α +

1
sin2β +

1
sin2γ
From two equalities we conclude that w1= w2 = w.

The idea for the solution comes from Euler’ Theorem for pedal triangles.
Observe that Ω1 and Ω2 always lie inside triangle ABC, because every circle

lies in that half of the plane that contains triangle ABC. It follows that the
point of their intersection Ω1 or Ω2 lies inside the triangle. In order to prove

that OΩ1 = OΩ2, we prove that their pedal areas are equal. If so, then we

have R2− OΩ2

1 = R2− OΩ22 and therefore OΩ1 = OΩ2.

Denote by A1, B1, C1 the projections from Ω1 onto BC, AC, AB,

respec-tively. Then using the extended Law of Sines we get A1C1 = BΩ1sin b,

because BΩ1 is diameter in the cyclic quadrilateral BA1Ω1C1. Also from the

Law of Sines in triangle ABΩ1 we have

BΩ1

sin w =
c
sin b.
It follows that

A1C1= BΩ1sin b = c sin w.

Similarly we obtain B1C1 = b sin w and A1B1 = a sin w. It is not difficult

to see that triangle A1B1C1 is similar to ABC with ratio of similarity sin w.

From the fact w1 = w2 = w we conclude that pedal triangles of Ω1 and Ω2

have the same area. Thus OΩ1 = OΩ2, 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 KΩ1 = KΩ2. Moreover

O, Ω1, K, Ω2 lie on a circle with diameter OK called Brocard circle.

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Problem 2. Consider a triangle ABC and denote by O, I, H its
circum-center, incircum-center, and orthocircum-center, respectively. Prove that

OI ≤ OH.

Solution: Many solvers will start to play with formulas for OI and OH.
We choose another way, we consider the pedal areas for I and H. Clearly,
the incenter is always inside the triangle ABC and if OH ≥ R, then we are
done. Therefore assume that OH < R. It follows that in this case we have an
acute-angled triangle. In order to prove that OI ≤ OH, we can use Euler’s
Theorem for pedal triangles and prove that the area of pedal triangle I, SI

is greater than the area of the pedal triangle of H, SH. This is because if

SI ≥ SH then R2− OI2 ≥ R2− OH2 and therefore OH ≥ OI.

Let us find both areas. Denote by A1, B1, C1 the projections of H and

A2, B2, C2 the projections of I onto BC, CA, AB, respectively. Recalling the

fact that A1, B1, C1 lie on the Euler circle of radius R/2 we get

SA1B1C1 =

A1B1· B1C1· A1C1

4 · R/2 .

Using the Law of Sines it is not difficult to see that B1C1 = a cos α, A1C1 =

b cos β, A1B1= c cos γ and therefore

SA1B1C1 =

abc · cos α cos β cos γ

2R = 2S cos α cos β cos γ.
To calculate the area of triangle A2B2C2, observe that

SA2B2C2 = SA2IB2+SA2IC2+SB2IC2 =

1
2r

2_{(sin α+sin β+sin γ) =} r2(a + b + c)

4R .

Recalling the formula r
4R = sin

α
2 sin

β
2sin

γ

2 we obtain
SA2B2C2 = 2S ·

r

4R = 2S sin
α
2 sin

β
2 sin

γ
2.
It follows that it is enough to prove that

sinα
2 sin

β
2 sin

γ

2 ≥ cos α cos β cos γ.

Now we use the classical Jensen Inequality. Consider the concave function
f : (0,π₂) → R, f (x) = ln cos x. Then

f (a + b
2 ) + f (

b + c
2 ) + f (

a + c

2 ) ≥ f (a) + f (b) + f (c).
Thus ln


sinα

2 sin
β
2sin

γ
2



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Problem 3. (Balkan Mathematical Olympiad )

Let ABC be an acute triangle. Denote by A1, B1, C1 the projections of

centroid G onto triangle’s sides. Prove that
2

9 ≤

SA1B1C1

SABC

≤ 1
4.

Solution: As we will see this problem requires a direct application of
Euler’s Theorem for pedal triangles. Applying it we have to prove

2
9 ≤

R2− OG2

4R2 ≤

1

4.

The right-hand side immediately follows. To prove the left-hand side recall
the well known formula

9OG2 = OH2= R2(1 − 8 cos α cos β cos γ).

Because triangle ABC is acute-angled we have cos α, cos β, cos γ ≥ 0, hence
8 cos α cos β cos γ ≥ 0, and thus 9OG2 ≤ R2 _{or OG}2 _{≤ R}2_/9.

The conclusion now follows.

Problem 4. (Mathematical Reflections, proposed by Ivan Borsenco)
For every point M inside triangle ABC we define a triplet (d1, d2, d3),

where d1, d2, d3 are distances from M to the sides BC, AC, AB. Prove that

the set of points M , satisfing condition d1· d2· d3≥ r3, where r is the incircle’s

radius, lies inside the circle with center O and radius OI.

Solution: Let A1, B1, C1 be projections from M onto sides BC, AC, AB.

Consider pedal triangle A1B1C1 for point M , we have ∠B1M C1 = 180 − α,

∠A1M C1= 180 − β, ∠A1M B1 = 180 − γ. It follows that

2 · AreaA1B1C1 = 2S1 = d2· d3· sinα + d1· d3· sinβ + d1· d2· sinγ

and rewriting it we get 2S1 =

d1· d2· d3

2R ·

 a
d1

+ b
d2

+ c
d3


.
Also we know that 2 · AreaABC = 2S = a · d1+ b · d2+ c · d3.

Applying the Cauchy-Schwarz inequality we obtain
4S·S1 =

d1· d2· d3

2R (ad1+ bd2+ cd3)
 a

d1

+ b
d2

+ c
d3



≥ d1· d2· d3(a + b + c)

2

2R .

Using Euler’s Theorem for pedal triangles we have
4S2(R2− OM2₎

4R2 ≥

d1· d2· d3(a + b + c)2

2R ≥

r3(a + b + c)2

2R .

From here it follows that R2− OM2_{≥ 2Rr or OI}2 _{= R}2_{− 2Rr ≥ OM}2_.

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Finally, we want to direct the reader’s attention to the result that finds
unexpectedly simple approximation for the circumradius of the pedal triangle.

Problem 5. (proposed by Ivan Borsenco)

Let M be a point inside triangle ABC, which has barycentric coordinates
(u, v, w). Denote by RM the circumradius of the pedal triangle of M . Prove

that

s

(u + v + w) a

2

u +
b2

v +
c2

w


≥ 6√3 · RM.

Solution: Denote by N the isogonal point of M . We need the following
two lemmas

Lemma 1. If A1B1C1 and A2B2C2 are the pedal triangles of two isogonal

points M and N then these six points lie on a circle.

Proof: Let us prove that B1B2C1C2is a cyclic quadrilateral. Let ∠BAM =

∠CAN = φ. Then because AB1M C1 is cyclic, we have

∠AB1C1= 90◦− ∠C1B1C2 = 90◦− ∠C1AM = 90◦− φ.

Similarly, because AB2N C2 is cyclic, we have

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Hence ∠AB1C1 = ∠AC2B2, and so B1B2C2C1 is a cyclic quadrilateral.

Analogously, we obtain that A1A2B2B1 and A1A2C2C1 are also cyclic.

Con-sider the three circles that circumscribe our quadrilaterals. If they do not
coincide on a common circle, they should have a radical center, which is the
intersection of their radical axes. However, we can see that radical axes, that
are A1A2, B1B2, C1C2, form a triangle, namely ABC, a contradiction. It

follows that A1, A2, B1, B2, C1, C2 lie on the same circle.

Lemma 2. If M and N are isogonal points, then the following equality
holds

AM · AN

bc +

BM · BN

ac +

CM · CN
ab = 1.

Proof: Let A1, B1, C1 be the pedal triangle of M . It is not difficult to

prove that B1C1 ⊥ AN . Thus the area of quadrilateral AB1N C1 is equal to

1

2B1C1· AN . Because B1C1 = AM · sin α, we get SAB1N C1 =

1

2AM · AN sin α.
Analogously, we find that SBC1N A1 =

1

2BM · BN sin β and SCA1N B1 =

1
2CM ·
CN sin γ.

Summing them up we obtain

SABC= SAB1N C1 + SBC1N A1+ SCA1N B1

or

SABC =

1

2(AM · AN sin α + BM · BN sin β + CM · CN sin γ) .
Using the Extended Law of Sines we obtain the desired result

AM · AN

bc +

BM · BN

ac +

CM · CN
ab = 1.

Now let us return to the problem. Applying the AM-GM inequality in
Lemma 2 we get

1
27 =

1
27

 AM · AN

bc +

BM · BN

ac +

CM · CN
ab

3

≥ AM · AN · BM · BN · CM · CN
a2_b2_c2 .

Using the fact that AM1 =

B1C1

sinα, BM1=
A1C1

sinβ, CM1 =
A1B1

sinγ and
AN = B2C2

sinα, BN =
A2C2

sinβ , CN =
A2B2

sinγ , we obtain
1

27 ≥

A1B1· B1C1· A1C1· A2B2· B2C2· A2C2

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As a2b2c2· sin2α · sin2β · sin2γ = 4S

4

R2 it follows that

1
27 ≥

R2

4S4 · A1B1· B1C1· A1C1· A2B2· B2C2· A2C2.

Recalling Lemma 1 and Euler’s Theorem for areas of pedal triangles, we have

A1B1·B1C1·A1C1 = 4RM·

R2− OM2

4R2 ·S, A2B2·B2C2·A2C2 = 4RM·

R2− ON2

4R2 ·S.

Thus

1
27 ≥

R_M2 (R2− OM2_)(R2_{− O}2_{N )}

4R2_{· S}2 .

The next step is to use Theorem 3 for point M and its isogonal conjugate

R2− OM2₌ vwa2+ uwb2+ uvc2

(u + v + w)2 =

uvw(a2/u + b2/v + c2/w)
(u + v + w)2

The isogonal point of M , N has a

2

u,
b2

v,
c2
w



as barycentric coordinates and
thus

R2− ON2 ₌ a2b2c2· uvw(u + v + w)

(vwa2_{+ uwb}2_{+ uvc}2₎2 =

a2b2c2(u + v + w)
uvw(a2_{/u + b}2_{/v + c}2_/w)2.

Combining these two results with the inequality we have
1

27 ≥

R2_M · a2_b2_c2

4R2_{· S}2_{· (u + v + w)(a}2_{/u + b}2_{/v + c}2_/w).

Finally, we obtain
s

(u + v + w) a

2

u +
b2

v +
c2

w


≥ 6√3 · RM.

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