Spiral similarity is a plane transformation in mathematics composed of a rotation and a dilation.[1] It is used widely in Euclidean geometry to facilitate the proofs of many theorems and other results in geometry, especially in mathematical competitions and Olympiads. Though the origin of this idea is not known, it was documented in 1967 by Coxeter in his book Geometry Revisited.[2]
A spiral similarity taking triangle ABC to triangle A'B'C'.
A spiral similarity is composed of a rotation of the plane followed a dilation about a center with coordinates in the plane.[3] Expressing the rotation by a linear transformation and the dilation as multiplying by a scale factor , we find that a point gets mapped to
On the complex plane, any spiral similarity can be expressed in the form , where is a complex number. The magnitude is the dilation factor of the spiral similarity, and the argument is the angle of rotation. [4]
Through a dilation of a line, rotation, and translation, any line segment can be mapped into any other through the series of plane transformations. We can find the center of the spiral similarity through the following construction: [1]
Draw lines and , and let be the intersection of the two lines.
Draw the circumcircles of triangles and .
The circumcircles intersect at a second point . Then is the spiral center mapping to
Proof: Note that and are cyclic quadrilaterals. Thus, . Similarly, . Therefore, by AA similarity, triangles and are similar. Thus, so a rotation angle mapping to also maps to . The dilation factor is then just the ratio of side lengths to .[3]
If we express and as points on the complex plane with corresponding complex numbers and , we can solve for the expression of the spiral similarity which takes to and to . Note that and , so . Since and , we plug in to obtain , from which we obtain .[3]
For any points and , the center of the spiral similarity taking to is also the center of a spiral similarity taking to .
This can be seen through the above construction. If we let be the center of spiral similarity taking to , then . Therefore, . Also, implies that . So by AA similarity we see that , thus is also the center of the spiral similarity which takes to .[3][4]
Spiral similarity can be used to prove Miquel's Quadrilateral Theorem: given four noncollinear points and , the circumcircles of the four triangles and intersect at one point, where is the intersection of and and is the intersection of and (see diagram). [1]
Let be the center of the spiral similarity which takes to . By the above construction, the circumcircles of and intersect at and . Since is also the center of the spiral similarity taking to , by similar reasoning the circumcircles of and meet at and . Thus, all four circles intersect at .[1]
Here is an example problem on the 2018 Japan MO Finals which can be solved using spiral similarity:
Given a scalene triangle , let and be points on segments and , respectively, so that . Let be the circumcircle of triangle and the reflection of across . Lines and meet again at and , respectively. Prove that and intersect on .[3]
Proof: We first prove the following claims:
Claim 1: Quadrilateral is cyclic.
Proof: Since is isosceles, we note that thus proving that quadrilateral is cyclic, as desired. By symmetry, we can prove that quadrilateral is cyclic.
Claim 2:
Proof: We have that By similar reasoning, so by AA similarity, as desired.
We now note that is the spiral center that maps to . Let be the intersection of and . By the spiral similarity construction above, the spiral center must be the intersection of the circumcircles of and . However, this point is , so thus points must be concyclic. Hence, must lie on , as desired.
^ abcdChen, Evan (2016). Euclidean Geometry in Mathematical Olympiads. United States: MAA Press. pp. 196–200. ISBN978-0-88385-839-4.
^Coxeter, H.S.M. (1967). Geometry Revisited. Toronto and New York: Mathematical Association of America. pp. 95–100. ISBN978-0-88385-619-2.
^ abcdeBaca, Jafet (2019). "On a special center of spiral similarity". Mathematical Reflections. 1: 1–9.