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Origami Geometry Projects for Math Fairs
Robert Geretschläger
Graz, Austria
6 Problems from 1 Fold
G
D’
E
F
C’ BA
CD
1. Prove that C‘D‘ is a tangent of the circle with center C. passing through B and D.
2. Prove that the perimeter of triangle GAC‘ is equal to half the perimeter of ABCD.
3. Prove the identity AG = C‘B + GD‘
4. Prove that the sum of the perimeters of triangles C‘BE and GD‘F is equal to the perimeter of triangle GAC‘.
5. Prove that the perimeter of triangle GD‘F is equal to the length of line segment AC‘.
6. Prove that the inradius of GAC‘ is equal to the length of line segment GD‘.
1. More Mathematical Morsels; Ross Honsberger
2. VIII Nordic Mathematical Contest 1994
4. 37th Slovenian Mathematical Olympiad 1993
6. classic Sangaku problem
6 Problems from 1 Fold
Problem 1
Prove that C‘D‘ is a tangent
of the circle with center C.
passing through
B and D.
P
G
D’
E
F
C’ BA
CD
6 Problems from 1 Fold
Problem 2
Prove that the perimeter of triangle GAC‘ is equal to half
the perimeter of ABCD.
P
G
D’
E
F
C’ BA
CD
AC‘ + C‘G + GA
= AC‘ + C‘P + GP + GA
= AC‘ + C‘B + GD + GA
= AB + DA
6 Problems from 1 Fold
Problem 3
Prove the identity AG =
C‘B + GD‘
P
G
D’
E
F
C’ BA
CD
AC‘ + C‘G + GA
= AB + C‘D‘
= AC‘ + C‘B + C‘G + GD‘
AG = C‘B + GD‘
6 Problems from 1 Fold
Problem 4
Prove that the sum of the perimeters of triangles C‘BE
and GD‘F is equal to the perimeter of triangle GAC‘.
G
D’
E
F
C’ BA
CD
GAC‘ ~ C’BE ~ GD’F
AG = C’B + GD’ AC’ = BE + D’F C’G = EC’ + FG
AG + AC’ + C’G = (C’B + BE + EC’) + (GD’ + D’F + FG)
6 Problems from 1 Fold
Problem 5
Prove that the perimeter of triangle GD‘F is equal to the length of line segment AC‘.
P
G
D’
E
F
C’ BA
CD
AC‘
= D‘P
= D‘G + GP
= D‘G + GD
= D‘G + GF + FD
= D‘G + GD + FD‘
6 Problems from 1 Fold
Problem 6
Prove that the inradius of GAC‘ is equal to the length of line
segment GD‘.
II I
II
I
M
G
D’
E
F
C’ BA
CDC‘I = C‘III = x, GII = GIII = y, AI = AII = r
2C‘D‘ = AC‘ + AG + GC‘
= (r + x) + (r + y) + (x + y)
= 2(x + y + r)
2(x + y + GD‘) = 2(x + y + r)
GD‘ = r
The Pentagon Project
The Golden Ratio
a
1
1
1
a-1
1
a : 1 = 1 : (a-1)
a² - a = 1
a² - a – 1 = 0
a =
a : 1
The Pentagon Project
Angles in a regular pentagon
d d
1
1
108°
36°
36°
72° 72°
36°
The Pentagon Project
The Golden Triangle
d : 1 = 1 : (d-1)
d = =
1
d
36°
36°72°
36°
72°1
1
d-1
The Pentagon Project
Placing the
Pentagon on the
Paper
d=1
a
1
1
The Pentagon Project
1D C
BA
Step 1
The Pentagon ProjectStep 2
1
The Pentagon ProjectStep 3
1
The Pentagon ProjectStep 4
I
2
1
The Pentagon ProjectStep 5
5 2
1
The Pentagon ProjectStep 6
2
1
5
The Pentagon ProjectStep 7
The Pentagon ProjectStep 8
The Pentagon Project
Additional challenges for advanced pentagonists:
+++ Can a regular pentagon with sides a longer than 1/ be placed in the interior of a unit square?
+++ Determine a folding sequence for a larger regular pentagon.
+++ Determine the largest possible value of a. Prove that your value is the largest possible.
The Pentagon Project
Folding a
pentagram
Part 1Folding method by Shuzo Fujimoto; from „The New Origami“ by Steve and Migumi Biddle
The Pentagon Project
Folding a
pentagram
Part 2Folding method by Shuzo Fujimoto; from „The New Origami“ by Steve and Migumi Biddle
The Pentagon Project
Challenge Question:
Decide whether the pentagram folded by this method is regular or not and prove your assertion.
Answer:
The pentagram is not regular!
Why?
Have a closer look at step 6!
The Pentagon Project
2
3
123d
33
33a
62,123
33
23
Axioms of Construction
Straight-edge and Compass: P1
P2
rP1
P2
P3point on object
point at intersection
P point at random
P
P
Axioms of Construction
Paper folding:
P
(O1)
Axioms of Construction
a
l 2
1l
a'
l1
l2
m
(O2) (O3)
b
Q
P
l Q
P
(O4) (O5)
Axioms of Constructionl P
l
Q
P
l
Y
XP
P'
c
P
l
P
(O6)
(O7)
Axioms of Construction
2l
1P
P2
l1(O7*)
Folding Roots
Linear equation ax = b
Solution: x = ab
a
b
slope of the crease is ab
y
x
5
5
-5
-5
Folding Roots
op :x²=2uy
O
y
x
oP (v,w)
Quadratic Equation x²+px+q = 0
x² - 2usx + 2uvs – 2uw = 0
u²s² - 2uvs + 2uw = 0
Parabola:x² = 2uy
Tangent:y = s(x - v) + w
0222 uw
uv ss
u = 2, v = -p, w = q
Parabola: x² = 4y
(Focus F(0,1), directrix y = -1)
P0(-p,q)
Folding Roots
1p :y²=2ax
2p :x²=2by
111P (x ,y )
t:y=cx+d
222P (x ,y )
x
y t: y = cx + d
p1: yy1 = ax + ax1
p2: xx2 = by + by2
3
b
ac
Folding Roots
1
1
2
F
F
2l
t
1l
x
y
ab
3
Folding Roots
_
_
H
C
Ea
b
c PB
A
E
G
B C
H
F
DA
CB
HG
FE
DA
AC : CB =
3 2 _ _
Folding Roots
111
2 2 2
P (x ,y )
P (x ,y )
1p :(y-n)²=2a(x-m)2
p :x²= 2by
1A (m ,n)
t: y=cx+d
y
x
t: y = cx + d
p1: (y-n)(y1–n) = a(x-m) + a(x1–m)
p2: xx2 = by + by2 02223 ba
bn
bm ccc
x³ + px² + qx + r = 0
p = -2m, q = 2n,
r = a, b = 1
222221 :;, rpqrp xlF
Folding Roots
D E C
F_
_G
BY
A'
F'
A
G
F
(A)
(D) D' E C
_F
_G
BY
A'G'
F'
X
D
F
G
A Y_A B
_G
_F
CE
F'
G'
ZA’
angle trisection:
cos 34cos³ - 3cos
or:
Thanks for listening!
robert.geretschlaeger@brgkepler.at
http://geretschlaeger.brgkepler.at
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