Download Proofs and Problems without Words - CMC3

CMC3–S Spring 2016 Conference
Kellogg West Conference Center, Pomona, CA
March 4th – 5th, 2016
Proofs and Problems
without Words
But, pictures are not proofs in themselves,
but may offer inspiration and direction.
Mathematical proofs require rigor, but
mathematical ideas benefit from insight.
Speaker: Karl Ting,
Mission College, Santa Clara
Going Beyond Computations,
Looking at Rectangles and Triangles.
1 + 3 + 5 + . . . + (2n – 1) = n2
1
4
+
1 2
1 3
+
+
4
4
1
...=
3
CMC3 – South Conference
Proofs and Problems without Words
-adapted from the Chou Pei Suan Ching
(author unknown, circa B.C. 200?)
1
CMC3 – South Conference
Proofs and Problems without Words
1
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem I
a2 + b 2 = c 2
-adapted from the Chou Pei Suan Ching
(author unknown, circa B.C. 200?)
1
CMC3 – South Conference
Proofs and Problems without Words
The Challenge
The Picture
Prove the Pythagorean Theorem I
a2 + b 2 = c 2
-adapted from the Chou Pei Suan Ching
(author unknown, circa B.C. 200?)
Reassembling dissection . . .
1
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem I
a2 + b 2 = c 2
-adapted from the Chou Pei Suan Ching
(author unknown, circa B.C. 200?)
1
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem I
a2 + b 2 = c 2
-adapted from the Chou Pei Suan Ching
(author unknown, circa B.C. 200?)
END slide
2
CMC3 – South Conference
Proofs and Problems without Words
Another Proof Within
2
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem II
a 2 + b 2 = c2
-Bhāskara (12th Century)
2
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem II
a 2 + b 2 = c2
-Bhāskara (12th Century)
2
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem II
a 2 + b 2 = c2
-Bhāskara (12th Century)
Reassembling dissection . . .
2
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem II
a 2 + b 2 = c2
-Bhāskara (12th Century)
2
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem II
a 2 + b 2 = c2
-Bhāskara (12th Century)
2
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem II
a 2 + b 2 = c2
-Bhāskara (12th Century)
END slide
3
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem III
a2 + b2 = c2
-Euclid
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
3
CMC3 – South Conference
Proofs and Problems without Words
Euclidean
Transitions
END slide
4
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem IV
a 2 + b 2 = c2
-Euclid
Adapted from the transitional
approach attributed to Euclid and
appears in Proofs Without Words.
4
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem IV
a 2 + b 2 = c2
-Euclid
4
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem IV
a 2 + b 2 = c2
-Euclid
4
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem IV
a 2 + b 2 = c2
-Euclid
4
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem IV
a 2 + b 2 = c2
-Euclid
4
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem IV
a 2 + b 2 = c2
-Euclid
END slide
5
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem V
2
a
+
2
b
=
2
c
-Leonardo da Vinci – (1452-1519)
5
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem V
a 2 + b 2 = c2
-Leonardo da Vinci – (1452-1519)
5
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem V
a 2 + b 2 = c2
-Leonardo da Vinci – (1452-1519)
5
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem V
a 2 + b 2 = c2
-Leonardo da Vinci – (1452-1519)
5
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem V
a 2 + b 2 = c2
-Leonardo da Vinci – (1452-1519)
5
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem V
a 2 + b 2 = c2
-Leonardo da Vinci – (1452-1519)
5
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem V
a 2 + b 2 = c2
-Leonardo da Vinci – (1452-1519)
5
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem V
a 2 + b 2 = c2
-Leonardo da Vinci – (1452-1519)
5
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem V
a 2 + b 2 = c2
-Leonardo da Vinci – (1452-1519)
5
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Prove the Pythagorean Theorem V
a 2 + b 2 = c2
-Leonardo da Vinci – (1452-1519)
END slide
3
CMC
– South
Conference
Proofs and Problems
without Words
6
CMC3 – South Conference
Proofs and Problems without Words
The Picture
Vertical Tangent Line on the x-axis.
The Challenge
Verify the Pythagorean Trig Identities
sin2Ɵ + cos2Ɵ = 1
1 + tan2Ɵ = sec2Ɵ
cot2Ɵ + 1 = csc2Ɵ
6
CMC3 – South Conference
Proofs and Problems without Words
The Picture
Vertical Tangent Line on the x-axis.
The Challenge
Verify the Pythagorean Trig Identities
sin2Ɵ + cos2Ɵ = 1
1 + tan2Ɵ = sec2Ɵ
cot2Ɵ + 1 = csc2Ɵ
6
CMC3 – South Conference
Proofs and Problems without Words
The Picture
Vertical Tangent Line on the x-axis.
The Challenge
Verify the Pythagorean Trig Identities
sin2Ɵ + cos2Ɵ = 1
1 + tan2Ɵ = sec2Ɵ
cot2Ɵ + 1 = csc2Ɵ
6
CMC3 – South Conference
Proofs and Problems without Words
The Picture
Vertical Tangent Line on the x-axis.
The Challenge
Verify the Pythagorean Trig Identities
sin2Ɵ + cos2Ɵ = 1
1 + tan2Ɵ = sec2Ɵ
cot2Ɵ + 1 = csc2Ɵ
END slide
7
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Verify the Pythagorean Trig Identities
sin2Ɵ + cos2Ɵ = 1
1 + tan2Ɵ = sec2Ɵ
cot2Ɵ + 1 = csc2Ɵ
Tangent Line at the point
(cos Ɵ, sin Ɵ).
7
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Verify the Pythagorean Trig Identities
sin2Ɵ + cos2Ɵ = 1
1 + tan2Ɵ = sec2Ɵ
cot2Ɵ + 1 = csc2Ɵ
Tangent Line at the point
(cos Ɵ, sin Ɵ).
7
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Verify the Pythagorean Trig Identities
sin2Ɵ + cos2Ɵ = 1
1 + tan2Ɵ = sec2Ɵ
cot2Ɵ + 1 = csc2Ɵ
Tangent Line at the point
(cos Ɵ, sin Ɵ).
7
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Verify the Pythagorean Trig Identities
sin2Ɵ + cos2Ɵ = 1
1 + tan2Ɵ = sec2Ɵ
cot2Ɵ + 1 = csc2Ɵ
Tangent Line at the point
(cos Ɵ, sin Ɵ).
END slide
8
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Verify the following trig
identities:
sin2Ɵ + cos2Ɵ = 1
tan2Ɵ + 1 = sec2Ɵ
cot2Ɵ + 1 = csc2Ɵ
(tan Ɵ + 1)2 + (cot Ɵ + 1)2
= (sec Ɵ + csc Ɵ)2
tan 𝜃+1
Collinear sec Ɵ and csc Ɵ.
tan Ɵ = cot 𝜃+1
8
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Verify the following trig
identities:
sin2Ɵ + cos2Ɵ = 1
tan2Ɵ + 1 = sec2Ɵ
cot2Ɵ + 1 = csc2Ɵ
(tan Ɵ + 1)2 + (cot Ɵ + 1)2
= (sec Ɵ + csc Ɵ)2
tan 𝜃+1
Collinear sec Ɵ and csc Ɵ.
tan Ɵ = cot 𝜃+1
8
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Verify the following trig
identities:
sin2Ɵ + cos2Ɵ = 1
tan2Ɵ + 1 = sec2Ɵ
cot2Ɵ + 1 = csc2Ɵ
(tan Ɵ + 1)2 + (cot Ɵ + 1)2
= (sec Ɵ + csc Ɵ)2
tan 𝜃+1
Collinear sec Ɵ and csc Ɵ.
tan Ɵ = cot 𝜃+1
8
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Verify the following trig
identities:
sin2Ɵ + cos2Ɵ = 1
tan2Ɵ + 1 = sec2Ɵ
cot2Ɵ + 1 = csc2Ɵ
(tan Ɵ + 1)2 + (cot Ɵ + 1)2
= (sec Ɵ + csc Ɵ)2
tan 𝜃+1
Collinear sec Ɵ and csc Ɵ.
tan Ɵ = cot 𝜃+1
8
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Verify the following trig
identities:
sin2Ɵ + cos2Ɵ = 1
tan2Ɵ + 1 = sec2Ɵ
cot2Ɵ + 1 = csc2Ɵ
(tan Ɵ + 1)2 + (cot Ɵ + 1)2
= (sec Ɵ + csc Ɵ)2
tan 𝜃+1
Collinear sec Ɵ and csc Ɵ.
tan Ɵ = cot 𝜃+1
8
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Verify the following trig
identities:
sin2Ɵ + cos2Ɵ = 1
tan2Ɵ + 1 = sec2Ɵ
cot2Ɵ + 1 = csc2Ɵ
(tan Ɵ + 1)2 + (cot Ɵ + 1)2
= (sec Ɵ + csc Ɵ)2
tan 𝜃+1
Collinear sec Ɵ and csc Ɵ.
END slide
tan Ɵ = cot 𝜃+1
9
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Geometric
Mean and
Similar Triangles
The Challenge
sin2Ɵ + cos2Ɵ = 1
sin2Ɵ + cos2Ɵ = 1
Determine:
The lengths of the sides
of DOAC.
The lengths of the sides
of DABC.
Verify the trig identity:
sin2Ɵ + cos2Ɵ = 1
9
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Geometric
Mean and
Similar Triangles
The Challenge
sin2Ɵ + cos2Ɵ = 1
sin2Ɵ + cos2Ɵ = 1
Determine:
The lengths of the sides
of DOAC.
The lengths of the sides
of DABC.
Verify the trig identity:
sin2Ɵ + cos2Ɵ = 1
9
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Geometric
Mean and
Similar Triangles
The Challenge
sin2Ɵ + cos2Ɵ = 1
sin2Ɵ + cos2Ɵ = 1
Determine:
The lengths of the sides
of DOAC.
The lengths of the sides
of DABC.
Verify the trig identity:
sin2Ɵ + cos2Ɵ = 1
9
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Geometric
Mean and
Similar Triangles
The Challenge
sin2Ɵ + cos2Ɵ = 1
sin2Ɵ + cos2Ɵ = 1
Determine:
The lengths of the sides
of DOAC.
The lengths of the sides
of DABC.
Verify the trig identity:
sin2Ɵ + cos2Ɵ = 1
9
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Geometric
Mean and
Similar Triangles
The Challenge
sin2Ɵ + cos2Ɵ = 1
sin2Ɵ + cos2Ɵ = 1
Determine:
The lengths of the sides
of DOAC.
The lengths of the sides
of DABC.
Verify the trig identity:
sin2Ɵ + cos2Ɵ = 1
9
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Geometric
Mean and
Similar Triangles
The Challenge
sin2Ɵ + cos2Ɵ = 1
sin2Ɵ + cos2Ɵ = 1
Determine:
The lengths of the sides
of DOAC.
The lengths of the sides
of DABC.
Verify the trig identity:
sin2Ɵ + cos2Ɵ = 1
END slide
3
CMC
– South
Conference
END slide
Proofs and Problems
without Words
10
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Verify the Law of Cosines
c2 = a2 + b2 – 2ab cos ϴ
-Timothy A. Sipka
10
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Verify the Law of Cosines
c2 = a2 + b2 – 2ab cos ϴ
-Timothy A. Sipka
10
The Picture
CMC3 – South
Conference
Proofs and Problems
without Words
The Challenge
Verify the Law of Cosines
c2 = a2 + b2 – 2ab cos ϴ
-Timothy A. Sipka
10
The Picture
CMC3 – South
Conference
Proofs and Problems
without Words
The Challenge
Verify the Law of Cosines
c2 = a2 + b2 – 2ab cos ϴ
-Timothy A. Sipka
10
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Verify the Law of Cosines
-Timothy A. Sipka
The Pythagorean Theorem states that
(b sin ϴ)2 + (a – b cos ϴ) 2 = c2
b2 sin2ϴ + a2 – 2ab cos ϴ + b2 cos2ϴ = c2
a2 + b2 (sin2ϴ + cos2ϴ ) – 2ab cos ϴ = c2
a2 + b2 – 2ab cos ϴ = c2
c2 = a2 + b2 – 2ab cos ϴ
END slide
Triangular Numbers
and Other Figurate Numbers
CMC3 – South Conference
Proofs and Problems without Words
We will now proceed onto Triangular Numbers
The Kid
Genius.
CMC3 – South Conference
Proofs and Problems without Words
Triangular Numbers
𝑘 (𝑘 + 1)
𝑇𝑘 =
2
CMC3 – South Conference
Proofs and Problems without Words
Triangular Numbers
Tk = 1 + 2 + 3 + … + k =
𝑘 (𝑘 + 1)
2
Square Numbers (sum of consecutive odd natural #s)
Sk = 1 + 3 + 5 + … + (2k – 1) = 𝑘
2
11
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Determine the formula for
the kth pentagonal number.
Pentagonal Numbers
11
CMC3 – South Conference
Proofs and Problems without Words
The Challenge
The Picture
Tk =
𝑘 (𝑘+1)
2
Pk = 2Tk – 1 + Tk-2
=
=
=
END slide
2𝑘 𝑘+1 −2 +(𝑘−2)(𝑘−1)
2
2
2𝑘 +2𝑘−2+𝑘 2 −3𝑘+2
2
2
3𝑘 − 𝑘 𝑘(3𝑘−1)
=
2
2
12
CMC3 – South Conference
Proofs and Problems without Words
Here is the formula for the kth
hexagonal number.
Hk = 2Tk – 1 + Sk-1
=
2𝑘 𝑘+1
2
- 1 + (k – 1)2
= k2 + k - 1 + k2 – 2k + 1
= 2k2 – k = k (2k – 1)
END slide
13
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Determine the kth n-gonal number, Nk.
where n is the number of sides of the
n-gon and k represent the length
number of elements along one of the
sides.
13
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
The kth n-gonal number is
Nk = 1 + (n – 1)(k – 1) + (𝑛 – 2)
(𝑘 –2)(𝑘 –1)
.
2
where n is the number of sides of the
n-gon and k represent the length
number of elements along one of the
sides.
-David Logothetti
END slide
1 + (n – 1) splines of length (k – 1) + (n -2) Tk-2.
14
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Determining a formula for the kth
HEX number?
14
CMC3 – South Conference
Proofs and Problems without Words
“FLATTENING A CUBE”
h1 = 1
h2 = 1 + (6 x 2) – 6
= 1+6=7
h3 = 7 + (6 x 3) – 6
= 7 + 12 = 19
14
CMC3 – South Conference
Proofs and Problems without Words
h4 = 19 + (6 x 4) – 6
= 37
h5 = 61
Justify the claim:
END slide
hn = n3 – (n – 1)3
or
hn = 3n2 – 3n + 1
Sums of Squares and Cubes
15
CMC3 – South Conference
Proofs and Problems without Words
The Challenge
Verify the sum of the squares of the natural numbers formula.
𝑛 2
1𝑖
=
=
2
(1
+
2
2
+
2
3
+...+
𝑛(𝑛+1)(2𝑛+1)
6
2
n)
15
CMC3 – South Conference
Proofs and Problems without Words
The Picture
𝑛 2
1𝑖
= (12 + 22 + 32 + . . . + n2)
=
=
𝑛(𝑛+1)(2𝑛+1)
6
𝑛 𝑛+1 (2𝑛+1)
2
×
3
15
CMC3 – South Conference
Proofs and Problems without Words
The Picture
Three towers forms a
rectangle with height, Tn,
and base, (2n+1); so
𝑛 2
1𝑖
=
In general, (2n – 1) + 2 = 2n + 1 (base)
END slide
16
CMC3 – South Conference
Proofs and Problems without Words
The Challenge
Verify the sum of the cubes of the natural numbers formula.
𝑛 3
1𝑖
=
=
=
3
(1
+
3
2
+
3
3
+...+
(𝑛2 +𝑛)(𝑛2 +𝑛)
4
𝑛 (𝑛+1) 2
2
3
n)
16
3
CMC
– South
Conference
The Picture
Proofs and Problems
without Words
The Challenge
Sum of Cubes I
-J. Barry Love
13 + 23 + 33 + 43 + 53 + … + n3
=
𝑛 3
1𝑖 =
(𝑛2 +𝑛)(𝑛2 +𝑛)
4
16
CMC3 –
South
Conference
Proofs and Problems
without Words
𝑛 3
1𝑖 =
(𝑛2 +𝑛)(𝑛2 +𝑛)
4
END slide
n2
n
17
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Sum of Cubs II
-Alan L. Fry
𝑛 3
1𝑖 =
𝑛(𝑛+1) 2
2
𝑇𝑛 =
2
17
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Sum of Cubs II
-Alan L. Fry
𝑛 3
1𝑖 =
𝑛(𝑛+1) 2
2
𝑇𝑛 =
2
17
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Sum of Cubs II
-Alan L. Fry
𝑛 3
1𝑖 =
𝑛(𝑛+1) 2
2
𝑇𝑛 =
2
17
CMC3 –
South
Conference
Proofs and
Problems
without Words
Sum of Cubs II
𝑛 3
1𝑖 =
-Alan L. Fry
𝑛(𝑛+1) 2
2
𝑇𝑛 =
2
17
CMC3 –
South
Conference
Proofs and
Problems
without Words
Sum of Cubs II
𝑛 3
1𝑖 =
END slide
-Alan L. Fry
𝑛(𝑛+1) 2
2
𝑇𝑛 =
2
Fibonacci, da Vinci,
and the Golden Rectangle
Really? Of course, I
have to check this out!
18
CMC3 – South Conference
Proofs and Problems without Words
The Challenge
Prove that the sum of the squares of the first n terms of
the Fibonacci sequence is the product of the last term and
its successor.
Fn = Fn-1 + Fn-2
where F1 = 1, F2 = 1
Fibonacci Sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
𝑛 2
1 𝐹𝑖 =
(12 + 12 + 22 + 32 + 52 + . . . + Fn2)
= Fn∙ Fn+1.
18
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
The Fibonacci Sequence is a sequence that
starts with the terms, F1 = 1 and F2 = 1, and
each terms that follows are constructed by
adding together the two previous terms;
that is, Fn = Fn-1 + Fn-2.
The first 6 terms are:
1, 1, 2, 3, 5, 8, . . .
The first 5 squared terms are:
1, 1, 4, 9, 25, . . .
The first 5 partial sums are:
1, 2, 6, 15, 40, . . .
Prove:
F12 + F22 + F32 + . . . + Fn2 = Fn ∙ Fn+1.
18
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
The Fibonacci Sequence is a sequence that
starts with the terms, F1 = 1 and F2 = 1, and
each terms that follows are constructed by
adding together the two previous terms;
that is, Fn = Fn-1 + Fn-2.
The first 6 terms are:
1, 1, 2, 3, 5, 8, . . .
The first 5 squared terms are:
1, 1, 4, 9, 25, . . .
The first 5 partial sums are:
1, 2, 6, 15, 40, . . .
Prove:
F12 + F22 + F32 + . . . + Fn2 = Fn ∙ Fn+1.
18
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
The Fibonacci Sequence is a sequence that
starts with the terms, F1 = 1 and F2 = 1, and
each terms that follows are constructed by
adding together the two previous terms;
that is, Fn = Fn-1 + Fn-2.
The first 6 terms are:
1, 1, 2, 3, 5, 8, . . .
The first 5 squared terms are:
1, 1, 4, 9, 25, . . .
The first 5 partial sums are:
1, 2, 6, 15, 40, . . .
Prove:
F12 + F22 + F32 + . . . + Fn2 = Fn ∙ Fn+1.
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CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
The Fibonacci Sequence is a sequence that
starts with the terms, F1 = 1 and F2 = 1, and
each terms that follows are constructed by
adding together the two previous terms;
that is, Fn = Fn-1 + Fn-2.
The first 6 terms are:
1, 1, 2, 3, 5, 8, . . .
The first 5 squared terms are:
1, 1, 4, 9, 25, . . .
The first 5 partial sums are:
1, 2, 6, 15, 40, . . .
Prove:
F12 + F22 + F32 + . . . + Fn2 = Fn ∙ Fn+1.
18
CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
The Fibonacci Sequence is a sequence that
starts with the terms, F1 = 1 and F2 = 1, and
each terms that follows are constructed by
adding together the two previous terms;
that is, Fn = Fn-1 + Fn-2.
The first 6 terms are:
1, 1, 2, 3, 5, 8, . . .
The first 5 squared terms are:
1, 1, 4, 9, 25, . . .
The first 5 partial sums are:
1, 2, 6, 15, 40, . . .
Prove:
F12 + F22 + F32 + . . . + Fn2 = Fn ∙ Fn+1.
18
CMC3 – South Conference
Proofs and Problems without Words
So what is: 12 + 12 + 22 + 52 + 82 + 132 + 212 + 342?
18
CMC3 – South Conference
Proofs and Problems without Words
So what is: 12 + 12 + 22 + 52 + 82 + 132 + 212 + 342?
1:1.618 ≈ 1:ф
Mona, what kind of answer is that?
34 (55) = 1870
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CMC3 – South Conference
Proofs and Problems without Words
So what is: 12 + 12 + 22 + 52 + 82 + 132 + 212 + 342?
Mona responded: The rectangles
are “almost”
GOLDEN rectangles
with a golden spiral
so the dimension is approximately
34 by (1.618 x 34) and having an
area of 1870.408 sq. units or ≈ 1870.
Leo was a good teacher!
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CMC3 – South
Conference
The Golden Rectangle is “approximated”
by rectangles whose dimensions are
consecutive Fibonacci numbers. In the
triangle below: 55/34 = 1.617647059…
END slide
Apple used my sequence in their logo,
and all I got was this chameleon shirt.
I would rather have had Apple stock.
Well, ”All that glitters is not gold.”
The Fibonacci Challenge
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CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
Why is there a difference in the areas?
Note:
In light of the two
examples, is dissecting a square, a valid
indicator of the
truthfulness of the
Pythagorean
Theorem?
(In the first case, the rectangle is larger, but in the
second case, the rectangle is smaller. It’s a puzzlement!)
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CMC3 – South Conference
Proofs and Problems without Words
The Picture
The Challenge
It is called the
Principle of Concealed Distribution.
The larger the rectangle, the greater
the concealment of the unit area.
𝐹
(ratio 𝑛 → ф + 1)
𝐹𝑛−2
Problems to Elicit Thinking.
I have left two problems for you. Problem #1: Take a square
and locate the center. Draw two lines through the center that intersect
at right angles. The four puzzle pieces are congruent. Reposition the
four pieces as pictured in the second square. A fifth square would
appear. Where did it come from?
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CMC3 – South Conference
Proofs and Problems
without Words
Challenge
Problem #2: The flapper is
made from one single piece of
paper with only the cuts that
appear in the illustration.
END slide
CMC3 – South Conference
When there may be NO words,
ENCOURAGE
PERSISTENCE.