Geometry by multiplication
Tonight's idea
Treat each point of the plane as a complex number. In Multiplying is rotating you saw that multiplying by a number \(w\) of length 1 turns every point about 0. To turn about some other point \(c\): slide \(c\) to 0, turn, slide back.
Rotation about \(c\). \(z \mapsto c + w(z - c)\). The arrow \(\mapsto\) means "is sent to".
Use \(w = i\) for a quarter turn anticlockwise. For a third of a turn (120°) use \(\omega = -\tfrac12 + \tfrac{\sqrt3}{2}\,i\). Three thirds make a full turn, so \(\omega^3 = 1\). And \(1, \omega, \omega^2\) are three arrows of length 1 set 120° apart, which cancel: \(1 + \omega + \omega^2 = 0\).
Equilateral test. A triangle with corners \(p, q, r\), named anticlockwise, is equilateral exactly when \(p + \omega q + \omega^2 r = 0\).
Why: write the \(p\) as \((-\omega - \omega^2)p\) and divide by \(\omega\). The equation becomes \(q - p = -\omega\,(r - p)\). Multiplying by \(-\omega\) is a turn of 60° clockwise. So side \(pq\) is side \(pr\) turned through 60°: two equal sides with 60° between them.
Watch one: Napoleon's theorem
Take any triangle \(a, b, c\), named anticlockwise. On each side build an equilateral triangle pointing outwards. Call their centres \(p, q, r\). Claim: \(pqr\) is equilateral.
A third of a turn about \(p\) carries \(b\) to \(a\). So \(a = p + \omega(b - p)\), which rearranges to \((1 - \omega)p = a - \omega b\). The other sides give \((1 - \omega)q = b - \omega c\) and \((1 - \omega)r = c - \omega a\). Feed the test:
\[(1 - \omega)(p + \omega q + \omega^2 r) = a - \omega b + \omega b - \omega^2 c + \omega^2 c - \omega^3 a = a - \omega^3 a = 0.\]
Everything cancels because \(\omega^3 = 1\). Since \(1 - \omega \neq 0\), the test passes. \(\square\)
Drag \(z\) and watch \(iz\), its quarter turn about 0: mathdetours.com/demos/complex.html.
Your turn
- Rotate \(z = 5 + 2i\) a quarter turn anticlockwise about \(c = 1 + i\). (Use \(w = i\).)
- A square has corners \(a, b, c, d\) in anticlockwise order, with \(a = 1 + i\) and \(b = 4 + 2i\). Find \(c\) and \(d\). Hint: side \(bc\) is side \(ab\) turned a quarter turn.
- Multiply out \(\omega^2\) and check that \(1 + \omega + \omega^2 = 0\). Then use the equilateral test on \(p = 0\), \(q = 2\), \(r = 1 + \sqrt3\,i\).
- A treasure map says: walk from the gallows to the oak, turn right 90°, walk the same distance again and drive in a stake. Go back to the gallows, walk to the pine, turn left 90°, walk the same distance again and drive in a second stake. Dig halfway between the stakes. The gallows has rotted away. Put the oak at 0, the pine at 10 and the gallows at an unknown \(g\). Where is the treasure?
Stretch: a shared centre
The centroid of a triangle is the average of its three corners. Add the three equations for \(p\), \(q\) and \(r\) in the proof of Napoleon's theorem. What does the sum tell you about the centroids of \(abc\) and \(pqr\)?
Answers
- \(c + i(z - c) = 1 + i + i(4 + i) = 1 + i + 4i - 1 = 5i\).
- \(b - a = 3 + i\), and a quarter turn of that is \(i(3 + i) = -1 + 3i\). So \(c = b + (-1 + 3i) = 3 + 5i\) and \(d = a + (-1 + 3i) = 4i\).
- \(\omega^2 = \tfrac14 - \tfrac34 - 2 \cdot \tfrac12 \cdot \tfrac{\sqrt3}{2}\,i = -\tfrac12 - \tfrac{\sqrt3}{2}\,i\). Adding, \(1 + \omega + \omega^2 = 1 - \tfrac12 - \tfrac12 + 0i = 0\). Test: \(\omega q = -1 + \sqrt3\,i\) and \(\omega^2 r = \left(-\tfrac12 - \tfrac{\sqrt3}{2}i\right)(1 + \sqrt3\,i) = 1 - \sqrt3\,i\). So \(p + \omega q + \omega^2 r = 0 - 1 + \sqrt3\,i + 1 - \sqrt3\,i = 0\). Equilateral, with side 2.
- Turning right multiplies your direction by \(-i\), turning left by \(i\). First stake: \(0 + (-i)(0 - g) = ig\). Second stake: \(10 + i(10 - g) = 10 + 10i - ig\). Halfway: \(\tfrac12(ig + 10 + 10i - ig) = 5 + 5i\). The \(g\) cancels, so the gallows is not needed. Stand at the middle of the line from oak to pine, face the pine, and the treasure is 5 units to your left.
Stretch. Adding gives \((1 - \omega)(p + q + r) = (a + b + c) - \omega(a + b + c) = (1 - \omega)(a + b + c)\). Divide by \(1 - \omega\): \(p + q + r = a + b + c\). Divide by 3 and the two centroids are the same point. In the figure \(p = 3 - \sqrt3\,i\), \(q = (3 + \sqrt3)(1 + i)\) and \(r = -\sqrt3 + 3i\), each side of \(pqr\) is \(\sqrt{24 + 12\sqrt3} \approx 6.69\), and both centroids are \(2 + 2i\).