A 4-dimensional proof of Heron's Formula - University of South

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.



Purpose




Purpose




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



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



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



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





Scissors Congruence





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



A = P1 ∪ · · · ∪ Pn and A′ =⋃

Pi

B = Q1 ∪ · · · ∪Qm and B′ =⋃

Qj

Scissors Congruence



A = P1 ∪ · · · ∪ Pn and A′ =⋃

Pi

B = Q1 ∪ · · · ∪Qm and B′ =⋃

Qj

Scissors Congruence



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



x

y

z

zx

zy



x

y

z

zx

zy



.

.

.

.

.

.

.

.

.

.x

y

z

zx

zy y

z

zy

zy

zy

y

y

n

2

1

n

2

1



.

.

.

.

.

.

.

.

.

.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



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


⋃Qj.

Hence,⋃m


⋃ni=1 Pi ×Qj = A′ ×B′


Scissors Congruence



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


⋃Qj.

Hence,⋃m


⋃ni=1 Pi ×Qj = A′ ×B′


Scissors Congruence



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


⋃Qj.

Hence,⋃m


⋃ni=1 Pi ×Qj = A′ ×B′


Scissors Congruence



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


⋃Qj.

Hence,⋃m

j=1 Pi ×Qj = Pi ×B′ and finally

⋃mj=1

⋃ni=1 Pi ×Qj = A′ ×B′


Scissors Congruence



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


⋃Qj.

Hence,⋃m


⋃ni=1 Pi ×Qj = A′ ×B′


Scissors Congruence



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


⋃Qj.

Hence,⋃m


⋃ni=1 Pi ×Qj = 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


Rectangle x Rectangle


Triangle x Triangle


Trapezoid x Triangle


Triangle x Trapezoid


Trapezoid x Trapezoid


Parallelogram x Parallelogram


move tri x tri to ...


... here.


Move trap x tri...


...here.


trap x trap stays in place.


Move tri x trap ...


...here.


Scissors Congruence

A Scissors Congruence proof of the Pythagorean Theoremlooks as follows.

a2 + b2 = c2

a

a

b

b

Scissors Congruence


a2 + b2 = c2

~sca

a

b

ba

a

a

b-a

b

bc

c

b-a

Scissors Congruence


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


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

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

= x2t2 + x2u2 + y2t2 + y2u2


Scissors Congruence


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

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

= x2t2 + x2u2 + y2t2 + y2u2


Scissors Congruence


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

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

= x2t2 + x2u2 + y2t2 + y2u2


Scissors Congruence


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

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

= x2t2 + x2u2 + y2t2 + y2u2


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


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

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


z

z

v

vx

y

t

u


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


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

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


z

z

v

vx

y

t

u


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

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


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


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


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


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


t = 14r2p2

A2t =

14r

2p2

In fact Heron’s Formula tells us....

A2t =

14r

2p2


x ≤ y w ≤ z

A2t =

14r

2p2


( 0 , 0 ) ( 0 , 0 )

x ≤ y z ≤ w

x ≤ y w ≤ z

A2t =

14r

2p2


( 0 , 0 ) ( 0 , 0 )

x ≤ y z ≤ w

y ≤ x w ≤ z

x ≤ y w ≤ z

A2t =

14r

2p2


( 0 , 0 ) ( 0 , 0 )

x ≤ y z ≤ w

y ≤ x w ≤ z

x ≤ y w ≤ z

y ≤ x z ≤ w

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A 4-dimensional proof of Heron's Formula - University of South