A 4-dimensional proof of Heron’s Formula

J. Scott CarterDavid Mullens

University of South Alabama

November 2010MAA Regional meeting, Pensacola, FL

Purpose

Give 4-d proof of Heron’s formula

using ScissorsCongruences.

Thanks to organizers

Thanks to my undergraduate advisor J. Scott Carter

Purpose

Give 4-d proof of Heron’s formula using ScissorsCongruences.

Thanks to organizers

Thanks to my undergraduate advisor J. Scott Carter

Purpose

Give 4-d proof of Heron’s formula using ScissorsCongruences.

Thanks to organizers

Thanks to my undergraduate advisor J. Scott Carter

Purpose

Give 4-d proof of Heron’s formula using ScissorsCongruences.

Thanks to organizers

Thanks to my undergraduate advisor J. Scott Carter

Area of Polygons

Ayt = 12pr

( q , r )

( 0 , 0 ) ( p , 0 )

r

( p - q )

a b

c

Area of Polygons

Ayt = 12pr, Ap = pr

( q , r )

( 0 , 0 ) ( p , 0 )

( p+q , r )

r

( p - q )

a b

c

Area of Polygons

Ayt = 12pr, Ap = pr

( q , r )

( 0 , 0 ) ( p , 0 )

( p+q , r )

r

( p - q )

a b

c

Area of Polygons

Ayt = 12pr, Ap = pr, Ar = pr

( q , r )

( 0 , 0 ) ( p , 0 )

( p , r )

r

( p - q )

a b

c

( 0 , r )

Area of Polygons

Ayt = 12pr, Ap = pr, Ar = pr, Abt = 1

2pr

( q , r )

( 0 , 0 ) ( p , 0 )

( p , r )

r

( p - q )

a b

c

( 0 , r )

Heron’s Formula

At =√s(s− a)(s− b)(s− c) where the semiperimeter,

s = (a + b + c)/2. We will keep a < b < c.

a b

c

Heron’s Formula can be written strictly in terms of its edgesBy squaring both sides and clearing the denominator, i.e.,f(a, b, c) = 16A2

t = (a+b+c)(−a+b+c)(a−b+c)(a+b−c)

Heron’s Formula

At =√s(s− a)(s− b)(s− c) where the semiperimeter,

s = (a + b + c)/2. We will keep a < b < c.

a b

c

Heron’s Formula can be written strictly in terms of its edges

By squaring both sides and clearing the denominator, i.e.,f(a, b, c) = 16A2

t = (a+b+c)(−a+b+c)(a−b+c)(a+b−c)

Heron’s Formula

At =√s(s− a)(s− b)(s− c) where the semiperimeter,

s = (a + b + c)/2. We will keep a < b < c.

a b

c

Heron’s Formula can be written strictly in terms of its edgesBy squaring both sides and clearing the denominator, i.e.,

f(a, b, c) = 16A2t = (a+b+c)(−a+b+c)(a−b+c)(a+b−c)

Heron’s Formula

At =√s(s− a)(s− b)(s− c) where the semiperimeter,

s = (a + b + c)/2. We will keep a < b < c.

a b

c

Heron’s Formula can be written strictly in terms of its edgesBy squaring both sides and clearing the denominator, i.e.,f(a, b, c) = 16A2

t = (a+b+c)(−a+b+c)(a−b+c)(a+b−c)

Scissors Congruence

Polygons A and A′ are scissors congruent if there existspolygon decomposition P1, P2, ..., Pn and Q1, Q2, ..., Qn of Aand A′ respectively such that Pi is congruent to Qj.

In short two polygons are scissors congruent if one can becut up and reassembled into the other. Let us denotescissors congruence by A ∼ sc A′

If we take the cross product of two polygons A and B in R2

then the resulting figure will exist in R4, i.e., A×B ∈ R4 isa 4-d hypersolid.

Scissors Congruence

Polygons A and A′ are scissors congruent if there existspolygon decomposition P1, P2, ..., Pn and Q1, Q2, ..., Qn of Aand A′ respectively such that Pi is congruent to Qj.

In short two polygons are scissors congruent if one can becut up and reassembled into the other. Let us denotescissors congruence by A ∼ sc A′

If we take the cross product of two polygons A and B in R2

then the resulting figure will exist in R4, i.e., A×B ∈ R4 isa 4-d hypersolid.

Scissors Congruence

Polygons A and A′ are scissors congruent if there existspolygon decomposition P1, P2, ..., Pn and Q1, Q2, ..., Qn of Aand A′ respectively such that Pi is congruent to Qj.

In short two polygons are scissors congruent if one can becut up and reassembled into the other. Let us denotescissors congruence by A ∼ sc A′

If we take the cross product of two polygons A and B in R2

then the resulting figure will exist in R4, i.e., A×B ∈ R4 isa 4-d hypersolid.

Higher Dimensional Polytopes

Higher Dimensional Polytopes

Higher Dimensional Polytopes

Scissors Congruence

Let A,A′, B, and B′ be polygons in R2, let A ∼ sc A′, andlet B ∼ sc B′.

Then ∃P1, ..., Pn, Q1, ..., Qm such that,

A = P1 ∪ · · · ∪ Pn and A′ =⋃

Pi

B = Q1 ∪ · · · ∪Qm and B′ =⋃

Qj

Scissors Congruence

Let A,A′, B, and B′ be polygons in R2, let A ∼ sc A′, andlet B ∼ sc B′.

Then ∃P1, ..., Pn, Q1, ..., Qm such that,

A = P1 ∪ · · · ∪ Pn and A′ =⋃

Pi

B = Q1 ∪ · · · ∪Qm and B′ =⋃

Qj

Scissors Congruence

Let A,A′, B, and B′ be polygons in R2, let A ∼ sc A′, andlet B ∼ sc B′.

Then ∃P1, ..., Pn, Q1, ..., Qm such that,

A = P1 ∪ · · · ∪ Pn and A′ =⋃

Pi

B = Q1 ∪ · · · ∪Qm and B′ =⋃

Qj

Scissors Congruence

Let A,A′, B, and B′ be polygons in R2, let A ∼ sc A′, andlet B ∼ sc B′.

Then ∃P1, ..., Pn, Q1, ..., Qm such that,

A = P1 ∪ · · · ∪ Pn and A′ =⋃

Pi

B = Q1 ∪ · · · ∪Qm and B′ =⋃

Qj

Scissors CongruenceLemma 1: A×B ∼ sc A′ ×B′

Lemma 2: The distributive law is a scissors congruence.

x

y

z

zx

zy

Scissors CongruenceLemma 1: A×B ∼ sc A′ ×B′

Lemma 2: The distributive law is a scissors congruence.

x

y

z

zx

zy

Scissors CongruenceLemma 1: A×B ∼ sc A′ ×B′

Lemma 2: The distributive law is a scissors congruence.

x

y

z

zx

zy

Scissors CongruenceLemma 1: A×B ∼ sc A′ ×B′

Lemma 2: The distributive law is a scissors congruence.

.

.

.

.

.

.

.

.

.

.x

y

z

zx

zy y

z

zy

zy

zy

y

y

n

2

1

n

2

1

Scissors CongruenceLemma 1: A×B ∼ sc A′ ×B′

Lemma 2: The distributive law is a scissors congruence.

.

.

.

.

.

.

.

.

.

.x

y

z

zx

zy

v

w

u

uv

uw

tv

tw

( t + u )( v + w )

y

z

zy

zy

zy

y

y

n

2

1

n

2

1

t

Scissors CongruenceDistributive Law in R2

c

(a+b)c = ac + bc

a b

Scissors CongruenceDistributive Law in R2, R3

c

(a+b)c = ac + bc

b

(a+b)c = ac + bc 22 2

a

a

b

c

Scissors CongruenceDistributive Law in R2, R3, R4

c

(a+b)c = ac + bc

b

(a+b)c = ac + bc (a+b)c = ac + bc33 3

22 2

a

a

a

b b

c

c

Scissors Congruence

Lemma 1: A×B ∼ sc A′ ×B′

Since A = P1 ∪ · · · ∪ Pn and B = Q1 ∪ · · · ∪Qm

then A×B = (P1 ∪ · · · ∪ Pn)× (Q1 ∪ · · · ∪Qm).

We also have that A′ =⋃Pi and B′ =

⋃Qj.

Hence,⋃m

j=1 Pi ×Qj = Pi ×B′ and finally⋃mj=1

⋃ni=1 Pi ×Qj = A′ ×B′

Ergo, A×B ∼ sc A′ ×B′.

Scissors Congruence

Lemma 1: A×B ∼ sc A′ ×B′

Since A = P1 ∪ · · · ∪ Pn and B = Q1 ∪ · · · ∪Qm

then A×B = (P1 ∪ · · · ∪ Pn)× (Q1 ∪ · · · ∪Qm).

We also have that A′ =⋃Pi and B′ =

⋃Qj.

Hence,⋃m

j=1 Pi ×Qj = Pi ×B′ and finally⋃mj=1

⋃ni=1 Pi ×Qj = A′ ×B′

Ergo, A×B ∼ sc A′ ×B′.

Scissors Congruence

Lemma 1: A×B ∼ sc A′ ×B′

Since A = P1 ∪ · · · ∪ Pn and B = Q1 ∪ · · · ∪Qm

then A×B = (P1 ∪ · · · ∪ Pn)× (Q1 ∪ · · · ∪Qm).

We also have that A′ =⋃Pi and B′ =

⋃Qj.

Hence,⋃m

j=1 Pi ×Qj = Pi ×B′ and finally⋃mj=1

⋃ni=1 Pi ×Qj = A′ ×B′

Ergo, A×B ∼ sc A′ ×B′.

Scissors Congruence

Lemma 1: A×B ∼ sc A′ ×B′

Since A = P1 ∪ · · · ∪ Pn and B = Q1 ∪ · · · ∪Qm

then A×B = (P1 ∪ · · · ∪ Pn)× (Q1 ∪ · · · ∪Qm).

We also have that A′ =⋃Pi and B′ =

⋃Qj.

Hence,⋃m

j=1 Pi ×Qj = Pi ×B′ and finally⋃mj=1

⋃ni=1 Pi ×Qj = A′ ×B′

Ergo, A×B ∼ sc A′ ×B′.

Scissors Congruence

Lemma 1: A×B ∼ sc A′ ×B′

Since A = P1 ∪ · · · ∪ Pn and B = Q1 ∪ · · · ∪Qm

then A×B = (P1 ∪ · · · ∪ Pn)× (Q1 ∪ · · · ∪Qm).

We also have that A′ =⋃Pi and B′ =

⋃Qj.

Hence,⋃m

j=1 Pi ×Qj = Pi ×B′ and finally

⋃mj=1

⋃ni=1 Pi ×Qj = A′ ×B′

Ergo, A×B ∼ sc A′ ×B′.

Scissors Congruence

Lemma 1: A×B ∼ sc A′ ×B′

Since A = P1 ∪ · · · ∪ Pn and B = Q1 ∪ · · · ∪Qm

then A×B = (P1 ∪ · · · ∪ Pn)× (Q1 ∪ · · · ∪Qm).

We also have that A′ =⋃Pi and B′ =

⋃Qj.

Hence,⋃m

j=1 Pi ×Qj = Pi ×B′ and finally⋃mj=1

⋃ni=1 Pi ×Qj = A′ ×B′

Ergo, A×B ∼ sc A′ ×B′.

Scissors Congruence

Lemma 1: A×B ∼ sc A′ ×B′

Since A = P1 ∪ · · · ∪ Pn and B = Q1 ∪ · · · ∪Qm

then A×B = (P1 ∪ · · · ∪ Pn)× (Q1 ∪ · · · ∪Qm).

We also have that A′ =⋃Pi and B′ =

⋃Qj.

Hence,⋃m

j=1 Pi ×Qj = Pi ×B′ and finally⋃mj=1

⋃ni=1 Pi ×Qj = A′ ×B′

Ergo, A×B ∼ sc A′ ×B′.

Scissors Congruence

Remember...

( q , r )

( 0 , 0 ) ( p , 0 )

( p , r )

r

( p - q )

a b

c

( 0 , r )( q , r )

( 0 , 0 ) ( p , 0 )

( p , r )

r

( p - q )

a b

c

( 0 , r )

Scissors Congruence

Remember...

( q , r )

( 0 , 0 ) ( p , 0 )

( p+q , r )

r

( p - q )

a b

c

( q , r )

( 0 , 0 ) ( p , 0 )

( p+q , r )

r

( p - q )

a b

c

Scissors CongruenceExample in R4 looks as follows.

Rectangle x Rectangle

Scissors CongruenceExample in R4 looks as follows.

Rectangle x Rectangle

Scissors CongruenceExample in R4 looks as follows.

Triangle x Triangle

Scissors CongruenceExample in R4 looks as follows.

Trapezoid x Triangle

Scissors CongruenceExample in R4 looks as follows.

Triangle x Trapezoid

Scissors CongruenceExample in R4 looks as follows.

Trapezoid x Trapezoid

Scissors CongruenceExample in R4 looks as follows.

Parallelogram x Parallelogram

Scissors CongruenceExample in R4 looks as follows.

move tri x tri to ...

Scissors CongruenceExample in R4 looks as follows.

... here.

Scissors CongruenceExample in R4 looks as follows.

Move trap x tri...

Scissors CongruenceExample in R4 looks as follows.

...here.

Scissors CongruenceExample in R4 looks as follows.

trap x trap stays in place.

Scissors CongruenceExample in R4 looks as follows.

Move tri x trap ...

Scissors CongruenceExample in R4 looks as follows.

...here.

Scissors CongruenceExample in R4 looks as follows.

Scissors Congruence

A Scissors Congruence proof of the Pythagorean Theoremlooks as follows.

a2 + b2 = c2

a

a

b

b

Scissors Congruence

A Scissors Congruence proof of the Pythagorean Theoremlooks as follows.

a2 + b2 = c2

~sca

a

b

ba

a

a

b-a

b

bc

c

b-a

Scissors Congruence

A Scissors Congruence proof of the Pythagorean Theoremlooks as follows.

a2 + b2 = c2

~sc ~sca

a

b

ba

a

a

b-a

b

bc

c

b-a

c

c

c

c

ba

b-a

Scissors Congruence

Why is this important?

Because each time we see the sum of squares,

z

z

x

y

z2 = x2 + y2

Scissors Congruence

Why is this important?Because each time we see the sum of squares,

z

z

x

y

z2 = x2 + y2

Scissors Congruence

Why is this important?Because each time we see the sum of squares,

z

z

v

vx

y

t

u

z2 = x2 + y2 and v2 = t2 + u2.

Scissors Congruence

Now it is important to note,

z2v2 = (x2 + y2)(t2 + u2)

= x2(t2 + u2) + y2(t2 + u2)

= x2t2 + x2u2 + y2t2 + y2u2

is a scissor congruence by Lemma 2.

Scissors Congruence

Now it is important to note,

z2v2 = (x2 + y2)(t2 + u2)

= x2(t2 + u2) + y2(t2 + u2)

= x2t2 + x2u2 + y2t2 + y2u2

is a scissor congruence by Lemma 2.

Scissors Congruence

Now it is important to note,

z2v2 = (x2 + y2)(t2 + u2)

= x2(t2 + u2) + y2(t2 + u2)

= x2t2 + x2u2 + y2t2 + y2u2

is a scissor congruence by Lemma 2.

Scissors Congruence

Now it is important to note,

z2v2 = (x2 + y2)(t2 + u2)

= x2(t2 + u2) + y2(t2 + u2)

= x2t2 + x2u2 + y2t2 + y2u2

is a scissor congruence by Lemma 2.

Scissors Congruence

Now it is important to note,

z2v2 = (x2 + y2)(t2 + u2)

= x2(t2 + u2) + y2(t2 + u2)

= x2t2 + x2u2 + y2t2 + y2u2

is a scissor congruence by Lemma 2.

Scissors Congruence

Once again recall...

( q , r )

( 0 , 0 ) ( p , 0 )

( p+q , r )

r

( p - q )

a b

c

which tells us that we can writea2 = (q2 + r2), b2 = ((p− q)2 + r2), and c2 = p2.

Back to Heron’s Formula

16A2t = (a + b + c)(−a + b + c)(a− b + c)(a + b− c)

= 2a2b2 + 2a2c2 + 2b2c2 − a4 − b4 − c4

Since a2 = (q2 + r2), b2 = ((p− q)2 + r2), and c2 = p2, everypiece of Heron’s Formula left over that contains an a and bcan be thought of as

z

z

v

vx

y

t

u

Back to Heron’s Formula

16A2t = (a + b + c)(−a + b + c)(a− b + c)(a + b− c)

= 2a2b2 + 2a2c2 + 2b2c2 − a4 − b4 − c4

Since a2 = (q2 + r2), b2 = ((p− q)2 + r2), and c2 = p2, everypiece of Heron’s Formula left over that contains an a and bcan be thought of as

z

z

v

vx

y

t

u

Back to Heron’s Formula

16A2t = (a + b + c)(−a + b + c)(a− b + c)(a + b− c)

= 2a2b2 + 2a2c2 + 2b2c2 − a4 − b4 − c4

Since a2 = (q2 + r2), b2 = ((p− q)2 + r2), and c2 = p2,

everypiece of Heron’s Formula left over that contains an a and bcan be thought of as

z

z

v

vx

y

t

u

Back to Heron’s Formula

16A2t = (a + b + c)(−a + b + c)(a− b + c)(a + b− c)

= 2a2b2 + 2a2c2 + 2b2c2 − a4 − b4 − c4

Since a2 = (q2 + r2), b2 = ((p− q)2 + r2), and c2 = p2, everypiece of Heron’s Formula left over that contains an a and bcan be thought of as

z

z

v

vx

y

t

u

Back to Heron’s Formula

16A2t = (a + b + c)(−a + b + c)(a− b + c)(a + b− c)

= 2a2b2 + 2a2c2 + 2b2c2 − a4 − b4 − c4

Since a2 = (q2 + r2), b2 = ((p− q)2 + r2), and c2 = p2, everypiece of Heron’s Formula left over that contains an a and bcan be thought of as

z

z

v

vx

y

t

u

Algebraic Proof

2a2b2 + 2a2c2 + 2b2c2 − a4 − b4 − c4

= 2[(q2 + r2)((p− q)2 + r2)

]+ 2[((p− q)2 + r2)(p2)

]+2[(q2 + r2)(p2)

]−(q2 + r2)2 − ((p− q)2 + r2)2 − p4

=[2[(q2 + r2)((p− q)2 + r2)

]+ 2[((p− q)2 + r2)(p2)

]−((p− q)2 + r2)2

]+2[(q2 + r2)(p2)

]− ((q2 + r2)2 − p4

= 4r2p2 + (p− q)2[2(q2 + r2) + 2p2 − (p− q)2 − 2r2

]+2(q2 + r2)r2 − r4 + 2q2p2 − (q2 + r2)− p4

Algebraic Proof

2a2b2 + 2a2c2 + 2b2c2 − a4 − b4 − c4

= 2[(q2 + r2)((p− q)2 + r2)

]+ 2[((p− q)2 + r2)(p2)

]+2[(q2 + r2)(p2)

]−(q2 + r2)2 − ((p− q)2 + r2)2 − p4

=[2[(q2 + r2)((p− q)2 + r2)

]+ 2[((p− q)2 + r2)(p2)

]−((p− q)2 + r2)2

]+2[(q2 + r2)(p2)

]− ((q2 + r2)2 − p4

= 4r2p2 + (p− q)2[2(q2 + r2) + 2p2 − (p− q)2 − 2r2

]+2(q2 + r2)r2 − r4 + 2q2p2 − (q2 + r2)− p4

Algebraic Proof

2a2b2 + 2a2c2 + 2b2c2 − a4 − b4 − c4

= 2[(q2 + r2)((p− q)2 + r2)

]+ 2[((p− q)2 + r2)(p2)

]+2[(q2 + r2)(p2)

]−(q2 + r2)2 − ((p− q)2 + r2)2 − p4

=[2[(q2 + r2)((p− q)2 + r2)

]+ 2[((p− q)2 + r2)(p2)

]−((p− q)2 + r2)2

]+2[(q2 + r2)(p2)

]− ((q2 + r2)2 − p4

= 4r2p2 + (p− q)2[2(q2 + r2) + 2p2 − (p− q)2 − 2r2

]+2(q2 + r2)r2 − r4 + 2q2p2 − (q2 + r2)− p4

Algebraic Proof

2a2b2 + 2a2c2 + 2b2c2 − a4 − b4 − c4

= 2[(q2 + r2)((p− q)2 + r2)

]+ 2[((p− q)2 + r2)(p2)

]+2[(q2 + r2)(p2)

]−(q2 + r2)2 − ((p− q)2 + r2)2 − p4

=[2[(q2 + r2)((p− q)2 + r2)

]+ 2[((p− q)2 + r2)(p2)

]−((p− q)2 + r2)2

]+2[(q2 + r2)(p2)

]− ((q2 + r2)2 − p4

= 4r2p2 + (p− q)2[2(q2 + r2) + 2p2 − (p− q)2 − 2r2

]+2(q2 + r2)r2 − r4 + 2q2p2 − (q2 + r2)− p4

Algebraic Proof

= 4r2p2 + (p− q)2[q2 + 2pq + p2

]+ 2q2p2 − q4 − p4

= 4r2p2 + (p− q)2[(q + p)2

]−[q4 − 2q2p2 + p4

]= 4r2p2 +

[(p− q)2(q + p)2

]−[(p− q)2(q + p)2

]= 4r2p2

So now we can see that 16A2t = 4r2p2 ⇒ A2

t = 14r2p2

Algebraic Proof

= 4r2p2 + (p− q)2[q2 + 2pq + p2

]+ 2q2p2 − q4 − p4

= 4r2p2 + (p− q)2[(q + p)2

]−[q4 − 2q2p2 + p4

]

= 4r2p2 +[(p− q)2(q + p)2

]−[(p− q)2(q + p)2

]= 4r2p2

So now we can see that 16A2t = 4r2p2 ⇒ A2

t = 14r2p2

Algebraic Proof

= 4r2p2 + (p− q)2[q2 + 2pq + p2

]+ 2q2p2 − q4 − p4

= 4r2p2 + (p− q)2[(q + p)2

]−[q4 − 2q2p2 + p4

]= 4r2p2 +

[(p− q)2(q + p)2

]−[(p− q)2(q + p)2

]

= 4r2p2

So now we can see that 16A2t = 4r2p2 ⇒ A2

t = 14r2p2

Algebraic Proof

= 4r2p2 + (p− q)2[q2 + 2pq + p2

]+ 2q2p2 − q4 − p4

= 4r2p2 + (p− q)2[(q + p)2

]−[q4 − 2q2p2 + p4

]= 4r2p2 +

[(p− q)2(q + p)2

]−[(p− q)2(q + p)2

]= 4r2p2

So now we can see that 16A2t = 4r2p2 ⇒ A2

t = 14r2p2

Algebraic Proof

= 4r2p2 + (p− q)2[q2 + 2pq + p2

]+ 2q2p2 − q4 − p4

= 4r2p2 + (p− q)2[(q + p)2

]−[q4 − 2q2p2 + p4

]= 4r2p2 +

[(p− q)2(q + p)2

]−[(p− q)2(q + p)2

]= 4r2p2

So now we can see that 16A2t = 4r2p2 ⇒ A2

t = 14r2p2

A2t =

14r

2p2

In fact Heron’s Formula tells us....

A2t =

14r

2p2

In fact Heron’s Formula tells us....

x ≤ y w ≤ z

A2t =

14r

2p2

In fact Heron’s Formula tells us....

( 0 , 0 ) ( 0 , 0 )

x ≤ y z ≤ w

x ≤ y w ≤ z

A2t =

14r

2p2

In fact Heron’s Formula tells us....

( 0 , 0 ) ( 0 , 0 )

x ≤ y z ≤ w

y ≤ x w ≤ z

x ≤ y w ≤ z

A2t =

14r

2p2

In fact Heron’s Formula tells us....

( 0 , 0 ) ( 0 , 0 )

x ≤ y z ≤ w

y ≤ x w ≤ z

x ≤ y w ≤ z

y ≤ x z ≤ w