MATHEMATICS AND ORIGAMI

Mathematics and Origami

8.2.8 SINGULAR POINTS IN TRIANGLES
8.2.8.1 ORTHOCENTER

It ́s the intersection point of the three altitudes of a triangle. Let ́s make these two folds:

1- A → A ; BC → BC

2- B → B ; AC → AC

This way, we get intersection point O: it will be the orthocenter if fold OC becomes perpen-
dicular to AB (starting hypothesis).
Demonstration 1
The six angles in O add up to 360º, and they are congruent (taken by pairs) as vertical angles.
Moreover, Ang.B = Ang.ROC once their sides are perpendicular (if the hypothesis is fulfilled);
Same applies to Angs. A and C.
Therefore it ́ll be: 360 = 2A + 2B + 2C ; 180 = A + B + C
which is true because it expresses the sum of the angles of the ∆ABC.
Demonstration 2
∆ACR and ∆BCS are similar (right-angled with Ang. C in common)
∆CBT and ∆ABR ,, (if the hypothesis is fulfilled)
∆BAS and ∆CAT ,, ,, ,,
From all it is born that:

BS

AR

BC

AC

= ;

AR

CT

AB

CB

= ;

CT

BS

CA

BA

=

or its equivalent:

AC×BS=BC×AR ; CB×AR=AB×CT ; BA×CT=BS×CA

There will be orthocenter if the former three equalities hold true. And they do, because each

one of them is equivalent to twice the area of ∆ABC.

8.2.8.2 CIRCUMCENTER
Is the center of the circumference passing through the three vertices of a triangle: it coincides
with the intersection of the three perpendicular bisectors of its sides.
Let these folds:
A → B (produces the perpendicular bisector c)
C → A (idem b)
B → C (idem a)
If O is the intersection of a and b, symmetry gives:
OB = OC ; OC = OA
hence:
OB = OA (fold c passes through O)

A

S

2

C

R

1

B

T

O