An example on Given O(0, 0), A(12, 0), B( The aim of this small article i co-ordinates of the five classi Δ OAB and other related poin We choose a right-angled trian simplicity. (1) Orthocenter: The three altitudes of a tri one point called the orthoc (Altitudes are perpendicul vertices to the opposite sid triangles.) If the triangle is obtuse, th the orthocenter is the verte Conclusion : Simple, the o (2) Circum-center: The three perpendicular b a triangle meet in one poin circumcenter. It is the cent the circle circumscribed ab If the triangle is obtuse, th outside the triangle. If it is the circumcenter is the mid hypotenuse. (By the theore semi-circle as in the diagra Conclusion : the circum n five classical centres of a right angled trian (0, 5) . is to find the ical centres of the nts of interest. angle for iangle meet in center. lar lines from des of the he orthocenter is outside the triangle. If it ex which is the right angle. orthocenter = O(0, 0) . bisectors of the sides of nt called the ter of the circumcircle, bout the triangle. hen the circumcenter is s a right triangle, then dpoint of the em of angle in am.) m-centre, E = , 6, 2.51 ngle Yue Kwok Choy t is a right triangle,
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five classical centres example... · (Altitudes are perpendicular lines from vertices to the opposite sides of the triangles.) If the triangle is obtuse, the orthocenter is outside
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An example on five classical centres of a right angled triangle
Given O(0, 0), A(12, 0), B(0, 5) .
The aim of this small article is to find the
co-ordinates of the five classical centres
∆ OAB and other related points of interest.
We choose a right-angled triangle for
simplicity.
(1) Orthocenter:
The three altitudes of a triangle meet in
one point called the orthocenter.
(Altitudes are perpendicular lines from
vertices to the opposite sides of the
triangles.)
If the triangle is obtuse, the orthocenter is outside the triangle. If it is a right triangle,
the orthocenter is the vertex which is the right angle.
Conclusion : Simple, the orthocenter
(2) Circum-center:
The three perpendicular bisectors
a triangle meet in one point called the
circumcenter. It is the center of the circumcircle,
the circle circumscribed about the triangle.
If the triangle is obtuse, then the circumcenter is
outside the triangle. If it is a right triangle, then
the circumcenter is the midpoint of the
hypotenuse. (By the theorem of angle in
semi-circle as in the diagram.)
Conclusion : the circum
An example on five classical centres of a right angled triangle
O(0, 0), A(12, 0), B(0, 5) .
is to find the
classical centres of the
OAB and other related points of interest.
angled triangle for
of a triangle meet in
one point called the orthocenter.
(Altitudes are perpendicular lines from
vertices to the opposite sides of the
If the triangle is obtuse, the orthocenter is outside the triangle. If it is a right triangle,
the orthocenter is the vertex which is the right angle.
orthocenter = O(0, 0) .
perpendicular bisectors of the sides of
a triangle meet in one point called the
circumcenter. It is the center of the circumcircle,
the circle circumscribed about the triangle.
triangle is obtuse, then the circumcenter is
outside the triangle. If it is a right triangle, then
the circumcenter is the midpoint of the
(By the theorem of angle in
circle as in the diagram.)
the circum-centre, E = ������ , ���
� � 6,2.5�
1
An example on five classical centres of a right angled triangle
Yue Kwok Choy
If the triangle is obtuse, the orthocenter is outside the triangle. If it is a right triangle,
�
Exercise 1:
(a) Check that the circum
(b) Show that the area of the triangle
����� 2R� sinA sin
(3) Centroid:
The three medians (the lines drawn
from the vertices to the bisectors of
the opposite sides) meet in the
centroid or center of mass. The
centroid divides each median in a
ratio of 2 : 1.
Since OR : RA = 1 : 1, we have R =
Since BC : CR = 2 : 1, we
Conclusion : Centroid,
Exercise 2: Prove that i
coordinates of the centroid of
(4) In-center: The three angle
bisectors of a triangle meet in one
point called the in-center. It is the
center of the in-circle, the circle
inscribed in the triangle.
Let OA = a = 12,
OB = b = 5
AB = c = √12� � 5�
Semi-perimeter, s =
The radius of the incircle = r = EX = EY = EZ
Then Area of ∆OAB = Area of
��� ��
� � ��� � ��
�
∴ r ���! ��"�
�"��
that the circum-circle above is given by: x $ 6�� � he area of the triangle with sides a, b, c and angles A, B, C
sinB sin C , where R is the radius of the circum
(the lines drawn
from the vertices to the bisectors of
the opposite sides) meet in the
centroid or center of mass. The
centroid divides each median in a
Since OR : RA = 1 : 1, we have R = 6, 0�. Since BC : CR = 2 : 1, we therefore have
Centroid, C = ��"(��"���� , �"���"�.�
��� � �4, �*� Prove that if A x�, y��, B x�, y��, C x
coordinates of the centroid of ∆ABC is given by �,-�,.�,*
angle
of a triangle meet in one
center. It is the
circle, the circle
� 13
s = �����
� ������*� 15
The radius of the incircle = r = EX = EY = EZ
OAB = Area of ∆EOA + Area of ∆EOB+ Area of
��� ������
� � r sr 2
2
y $ 2.5�� 6.5�. with sides a, b, c and angles A, B, C is
where R is the radius of the circum-circle .
� . x*, y*� , then the
,/ , 0-�0.�0/* � .
EOB+ Area of ∆EAB
Conclusion : In-centre
In-circle :
Exercise 3:
Prove that the radius of a general
semi-perimeter s is given by :
(5) Ex-center: An ex-circle of the triangle is a
circle lying outside the triangle, tangent to
one of its sides and tangent to the
extensions of the other two.
Every triangle has three distinct ex
each tangent to one of the triangle's sides.
The center of an ex-circle is t
of the internal bisector of one angle and the
external bisectors of the other two.
As in the diagram,
Area of ∆OAB
= Area of ∆E1AB + ∆E1OB
ab2 cr�
2 � br�2 $
4������� �
∴ r� ���!5��= ��
���
Similarly, r� ���!5��=
Conclusion : Ex-centre
Exercise 4:
(a) Write down the equations of the ex
(b) Prove that for a general triangle (not just right) the radii of the ex
r� 6!!5��!5��!5� ,r
centre : E = (r, r) = (2, 2)
circle : x $ 2�� � y $ 2�� 2�
Prove that the radius of a general ∆ABC (not just right) with sides a, b, c and