Heron’s Formula from a 4-Dimensional Perspective

J. Scott CarterDavid Mullens

March 10, 2011

Outline1. The Pythagorean Theorem

2. The distributive law is a scissors congruence

3. The product of two scissors congruences(general statement)

4. The product of two scissors congruences(right triangles)

5. The product of two scissors congruences(parallelograms)

6. The product of two scissors congruences (Πθγ × Πθγ )

7. Lots of pictures

8. The geometry and algebra together

Outline1. The Pythagorean Theorem

2. The distributive law is a scissors congruence

3. The product of two scissors congruences(general statement)

4. The product of two scissors congruences(right triangles)

5. The product of two scissors congruences(parallelograms)

6. The product of two scissors congruences (Πθγ × Πθγ )

7. Lots of pictures

8. The geometry and algebra together

Outline1. The Pythagorean Theorem

2. The distributive law is a scissors congruence

3. The product of two scissors congruences(general statement)

4. The product of two scissors congruences(right triangles)

5. The product of two scissors congruences(parallelograms)

6. The product of two scissors congruences (Πθγ × Πθγ )

7. Lots of pictures

8. The geometry and algebra together

Outline1. The Pythagorean Theorem

2. The distributive law is a scissors congruence

3. The product of two scissors congruences(general statement)

4. The product of two scissors congruences(right triangles)

5. The product of two scissors congruences(parallelograms)

6. The product of two scissors congruences (Πθγ × Πθγ )

7. Lots of pictures

8. The geometry and algebra together

Outline1. The Pythagorean Theorem

2. The distributive law is a scissors congruence

3. The product of two scissors congruences(general statement)

4. The product of two scissors congruences(right triangles)

5. The product of two scissors congruences(parallelograms)

6. The product of two scissors congruences (Πθγ × Πθγ )

7. Lots of pictures

8. The geometry and algebra together

Outline1. The Pythagorean Theorem

2. The distributive law is a scissors congruence

3. The product of two scissors congruences(general statement)

4. The product of two scissors congruences(right triangles)

5. The product of two scissors congruences(parallelograms)

6. The product of two scissors congruences (Πθγ × Πθγ )

7. Lots of pictures

8. The geometry and algebra together

Outline1. The Pythagorean Theorem

2. The distributive law is a scissors congruence

3. The product of two scissors congruences(general statement)

4. The product of two scissors congruences(right triangles)

5. The product of two scissors congruences(parallelograms)

6. The product of two scissors congruences (Πθγ × Πθγ )

7. Lots of pictures

8. The geometry and algebra together

Outline1. The Pythagorean Theorem

2. The distributive law is a scissors congruence

3. The product of two scissors congruences(general statement)

4. The product of two scissors congruences(right triangles)

5. The product of two scissors congruences(parallelograms)

6. The product of two scissors congruences (Πθγ × Πθγ )

7. Lots of pictures

8. The geometry and algebra together

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On the right-hand-side of Heron’s formula, theexpression

2a2b2 + 2a2c2 + 2b2c2 − a4 − b4 − c4

appears. We first consider the four terms a2b2,a2b2, (−a4), and (−b4). Recall, that we haveplaced the triangle with edge lengths a ≤ b ≤ c inthe coordinate plane with vertices (0, 0), (p, 0)and (q, r) where a2 = q2 + r2, b2 = (p− q)2 + r2),and c = p.

Since both a and b are expressed as the sum ofsquares (a2 = q2 + r2 and b2 = (p− q)2 + r2), eachof the hyper-solids (hyper-rectangles)[0, a]× [0, b]× [0, a]× [0, b],[0, b]× [0, a]× [0, b]× [0, a],[0, a]× [0, a]× [0, a]× [0, a], and[0, b]× [0, b]× [0, b]× [0, b] can be decomposed asin the preceding section.

The algebraic identities

a4 = (q2 + r2)(q2 + r2)

= q4 + 2r2q2 + r4,

b4 = ((p− q)2 + r2)((p− q)2 + r2)

= (p− q)4 + 2(r2(p− q)2) + (p− q)4,

a2b2 = (q2 + r2)((p− q)2 + r2)

= q2(p− q)2 + r2(p− q)2 + q2r2 + r4

are all realized as scissors congruences.

The expressions a2c2 and b2c2 have similar, butmuch easier decompositions as scissorscongruences. So the identities

a2c2 = (q2 + r2)p2

= q2p2 + r2p2,

b2c2 = ((p− q)2 + r2)p2

= (p− q)2p2 + r2

are also realized as scissors congruences.

2a2b2 + 2a2c2 + 2b2c2 − a4 − b4 − c4

= 2[(q2 + r2)((p− q)2 + r2)

]+2

[(q2 + r2)(p2)

]+ 2

[((p− q)2 + r2)(p2)

]−(q2 + r2)2 − ((p− q)2 + r2)2 − p4

= 2q2(p− q)2 + 2r2(p− q)2 + 2q2r2 + 2r4

+2q2p2 + 2p2r2 + 2(p− q)2p2 + 2p2r2

−[q4 + 2q2r2 + r4 + (p− q)4+

2r2(p− q)2 + r4 + p4]is a scissors congruence.

2a2b2 + 2a2c2 + 2b2c2 − a4 − b4 − c4

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

+2q2p2 + 4p2r2 − q4 − p4

+2q2r2 + 2r4 −[2q2r2 + 2r4]

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

+4p2r2 −[q4 − 2q2p2 + p4]

= (p− q)2 [2q2 + 2p2 − (p− q)2]

+4p2r2 −[p2 − q2]2

= (p− q)2 [q + p]2

−[p2 − q2]2

+ 4p2r2

= 4p2r2.