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    • Abstract: Forum GeometricorumVolume 2 (2002) 131–134.FORUM GEOMISSN 1534-1178The Perimeter of a Cevian TriangleNikolaos DergiadesAbstract. We show that the cevian triangles of certain triangle centers haveperimeters not exceeding the semiperimeter of the reference triangle. These in-

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Forum Geometricorum
Volume 2 (2002) 131–134.
FORUM GEOM
ISSN 1534-1178
The Perimeter of a Cevian Triangle
Nikolaos Dergiades
Abstract. We show that the cevian triangles of certain triangle centers have
perimeters not exceeding the semiperimeter of the reference triangle. These in-
clude the incenter, the centroid, the Gergonne point, and the orthocenter when
the given triangle is acute angled.
1. Perimeter of an inscribed triangle
We begin by establishing an inequality for the perimeter of a triangle inscribed
in a given triangle ABC.
Proposition 1. Consider a triangle ABC with a ≤ b ≤ c. Denote by X, Y , Z the
midpoints of the sides BC, CA, and AB respectively. Let D, E, F be points on
the sides BC, CA, AB satisfying the following two conditions:
(1.1) D is between X and C, E is between Y and C, and F is between Z and
B.
(1.2) ∠CDE ≤ ∠BDF , ∠CED ≤ ∠AEF , and ∠BF D ≤ ∠AF E.
Then the perimeter of triangle DEF does not exceed the semiperimeter of triangle
ABC.
B
F
Z
X β
D
γ
C E Y A
Figure 1
Proof. Denote by i, j, k the unit vectors along EF, FD, DE. See Figure 1. Since
∠BF D ≤ ∠AF E, we have i · ZF ≤ j · ZF. Similarly, since ∠CDE ≤ ∠BDF
and ∠CED ≤ ∠AEF , we have j · XD ≤ k · XD and i · EY ≤ k · EY. Now,
we have
EF + F D + DE = i · EF + j · FD + k · DE
= i · (EY + YZ + ZF) + j · (FZ + ZX + XD) + k · DE
≤ (k · EY + i · YZ + j · ZF)
+(j · FZ + j · ZX + k · XD) + k · DE
Publication Date: November 1, 2002. Communicating Editor: Paul Yiu.
132 N. Dergiades
= i · YZ + j · ZX + k · XY
≤ |i||YZ| + |j||ZX| + |k||XY| (1)
= Y Z + ZX + XY
1
= (AB + BC + CA).
2
Equality holds in (1) only when the triangles DEF and XY Z have parallel sides,
i.e., when the points D, E, F coincide with the midpoints X, Y , Z respectively, as
is easily seen.
2. Cevian triangles
Proposition 2. Suppose the side lengths of triangle ABC satisfy a ≤ b ≤ c.
Let P be an interior point with (positive) homogeneous barycentric coordinates
(x : y : z) satisfying
(2.1) x ≤ y ≤ z,
(2.2) x cot A ≥ y cot B ≥ z cot C.
Then the perimeter of the cevian triangle of P does not exceed the perimeter of the
medial triangle of ABC, i.e., the cevian triangle of the centroid.
az ay cx
Proof. In Figure 1, BD = y+z , DC = y+z , and BF = x+y . Since y ≤ z, it is
clear that BD ≥ DC. Similarly, AE ≥ EC, and AF ≥ F B. Condition (1.1) is
satisfied. Applying the law of sines to triangle BDF , we have sin(B+β) = BD . It
sin β BF
follows that
sin(B + β) sin(B + C) z(x + y)
= · .
sin B sin β sin B sin C x(y + z)
z(x+y)
From this, cot β + cot B = (cot B + cot C) · x(y+z) . Similarly, cot γ + cot C =
y(z+x)
(cot B + cot C) · x(y+z) . Consequently,
2(y cot B − z cot C)
cot γ − cot β = ,
y+z
so that β ≥ γ provided y cot B ≥ z cot C. The other two inequalities in (1.2) can
be similarly established. The result now follows from Proposition 1.
This applies, for example, to the following triangle centers. For the case of the
orthocenter, we require the triangle to be acute-angled. 1 It is easy to see that the
barycentrics of each of these points satisfy condition (2.1).
P (x : y : z) x cot A ≥ y cot B ≥ z cot C
Incenter (a : b : c) cos A ≥ cos B ≥ cos C
Centroid (1 : 1 : 1) cot A ≥ cot B ≥ cot C
Orthocenter (tan A : tan B : tan C) 1≥1≥1
Gergonne point tan A : tan B : tan C
2 2 2
1
2
(1 − tan2 A ) ≥ 1 (1 − tan2
2 2
B
2
)
≥ 1 (1 − tan2 C )
2 2
1For the homogeneous barycentric coordinates of triangle centers, see [1].
Perimeter of a cevian triangle 133
The perimeter of the cevian triangle of each of these points does not exceed the
semiperimeter of ABC.2 The case of the incenter can be found in [2].
3. Another example
The triangle center with homogeneous barycentric coordinates (sin A : sin B :
2 2
sin C ) provides another example of a point P the perimeter of whose cevian tri-
2
angle not exceeding the semiperimeter of ABC. It clearly satisfies (2.1). Since
sin A cot A = cos A − 2 cos A , it also satisfies condition (2.2). In [1], this point
2 2
1
2
appears as X174 and is called the Yff center of congruence. Here is another de-
scription of this triangle center [3]:
The tangents to the incircle at the intersections with the angle bisectors farther
from the vertices intersect the corresponding sides at the traces of the point with
homogeneous barycentric coordinates (sin A : sin B : sin C ).
2 2 2
B
F
X Z
D
I
P
C E Y A
Figure 2
2The Nagel point, with homogeneous barycentric coordinates (cot A : cot B
: cot C ), also
2 2 2
satisfies (2.2). However, it does not satisfy (2.1) so that the conclusion of Proposition 2 does not
apply. The same is true for the circumcenter.
134 N. Dergiades
References
[1] C. Kimberling, Encyclopedia of Triangle Centers, 2000
http://www2.evansville.edu/ck6/encyclopedia/.
[2] T. Seimiya and M. Bataille, Problem 2502 and solution, Crux Math., 26 (2000) 45; 27 (2001)
53–54.
[3] P. Yiu, Hyacinthos message 2114, http://groups.yahoo.com/group/Hyacinthos/message/2114,
December 18, 2000.
Nikolaos Dergiades: I. Zanna 27, Thessaloniki 54643, Greece
E-mail address: [email protected]


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