G4G Celebration of Mind 2021 October

Glen Whitney

studioinfinity.org/pintegons

G4G Celebration of Mind 2021 October

Glen Whitney

studioinfinity.org/pintegons

Joint work with Alissa Crans

Mughal jali, India c. 1615 c.e.

image from the Toledo Museum of Art

instead of insisting that our pentagons are *regular*, allow them to be merely *convex*

images from ‘kjo’ on math.stackexchange.com

1918

• We’re looking for integer-sided pentagons in those families

• We’re looking for pintegons in those families

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⋮

The pintegons in Family 7 are those with sides $(a,b,b,b,b)$ for whole numbers $a$ and $b$ satisfying \[0 < a < (2\sqrt\phi)b.\]

\[\begin{split}C: & \,\left(b+(d-2a)\cos\frac{C}2, (d-2a)\sin\frac{C}2\right) = & \,\left(d\left(\sin C + \cos\frac{C}2\right), a + d\left(\cos C − \sin\frac{C}2\right)\right) \end{split}\]

\[\begin{split}b & = d \sin 2\gamma + 2a \cos \gamma 2(d − 2a) \sin \gamma & = a + d \cos 2\gamma − 2a \sin \gamma \end{split}\]

\[\sin\gamma = \frac{a-d + \sqrt{a^2+3d^2}}{2d}\]

$H = a^2+3d^2$

\[ b = \frac{(\sqrt{H} + 3a − d)\sqrt{2(d − a)(\sqrt{H} + a)}}{2d} \]

$H = a^2+3d^2$

When is \[ \sqrt{2(d − a)(\sqrt{H} + a)} \] of the form \[ r + s\sqrt{H} \] for rational numbers $r$ and $s$?

When is $\sqrt{2(d − a)(\sqrt{H} + a)}$ of the form $ r + s\sqrt{H} $ for rational numbers $r$ and $s$?

Well, when is $\sqrt{n}$ of the form $r$ for a rational number $r$?

$\sqrt1, \sqrt2, \sqrt3, \sqrt4, \sqrt5, \sqrt6, \sqrt7, \sqrt8, \sqrt9, \sqrt{10}, \ldots$

When is $\sqrt{2(d − a)(\sqrt{H} + a)}$ of the form $ r + s\sqrt{H} $ for rational numbers $r$ and $s$?

**Answer:** Only when $2(d − a)(\sqrt{H} + a) = (j + k\sqrt{H})^2$ for $j$ and $k$ each either a whole number or half of an odd number. (Dedekind, 1871)

$H = a^2+3d^2$

Turns out $H$ must be a perfect square!

So there’s some $h$ such that $a^2 + 3d^2 = h^2.$

\[ b = \frac{(\sqrt{H} + 3a − d)\sqrt{2(d − a)(\sqrt{H} + a)}}{2d} \]

$H = a^2+3d^2$

Turns out $H$ must be a perfect square!

So there’s some $h$ such that $a^2 + 3d^2 = h^2.$

\[ b = \frac{(h + 3a − d)\sqrt{2(d − a)(h + a)}}{2d} \]

Integers $a, d, h,$ and $m$ so:

$a^2 + 3d^2 = h^2$

$2(d-a)(h+a) = m^2$

$h+3a-d > 0$

\[ b = \frac{(h+3a-d)\sqrt{2(d-a)(h + a)}}{2d} \]

45

448

115

275

275

629

3750

1921

3179

3179

X

X

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✔

✔

X

X

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✔

✔

studioinfinity.org/pintegons