MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

That is, O is equidistant from A, B, C; in consequence O is the center of the unique circumfer-

ence passing through the three vertices of the triangle.

If ∆ABC is a right-angled one, its circumcenter is placed at the mid-point of its hypotenuse.

Distances form the circumcenter to the mid-points of legs are equal to half of those legs

(Thales ́ theorem). Last figure shows how the triangle can be flattened.

8.2.8.3 BARICENTER
Is the intersection point of the three medians of the sides of a triangle.

That ́s the reason why it is also its center of gravity: The c.o.g. must be on each one of the me-

dians and, being unique, it has to be placed over the intersection of the three. It is easy to see

that the median divides a triangle in another two of equal area (equal bases and same height),

hence, of equal weight (in Greek, βαρος = weighty).

To get O (fig. 1), fold mid-point of each side, then mid-point / opposite vertex.

Fig. 2 demonstrates an important property of baricenter. Uniting side mid-points we

produce ∆A ́B ́C ́ which is similar to ∆ABC (Thales ́ theorem). Once demonstrated the exis-

tence of baricenter O of great ∆ABC, O is also the baricenter of small ∆A ́B ́C ́. Because of the

similarity of those triangles, it is:OC OC

2

1

́= which proves that the baricenter is distant from a

vertex twice as much as from the mid-point of the opposite side.

This remarkable property serves to divide a segment in three equal parts. Though this

matter has its natural place when dealing with division in equal parts, I prefer to develop it now

to profit of its background.

Let segment AB (fig. 3). Mark any point C and get the baricenter O of ∆ABC. Fold re-

spective parallels to BC and AC through O: O is the mid-point of DF and GE. Being similar

∆ABC and BDF, if B ́ is the mid-point of AC in the former, O will be also the mid-point of DF

in the latter. Same demonstration applies to GE. On the other hand, ∆ABC and AGE, are also

similar: D is the homologous of C ́, hence D is the mid-point of AE. A similar reasoning leads

to E as the midpoint of DB. Therefore

AD DE EB AB

3

1

= = =

C

B

a

c b

O

A

O

A

B C

A

A

B

B

C

C

A

O C

B

O

B

A

C

A ́

C ́ B ́

A

B ́

O

C ́

C

A ́

B

F

D E

(^12) G 3