Spiral similarity

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 ${\displaystyle {\overline {AC}}}$ and ${\displaystyle {\overline {BD}}}$, and let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): P be the intersection of the two lines.
• Draw the circumcircles of triangles ${\displaystyle \triangle PAB}$ and ${\displaystyle \triangle PCD}$.
• The circumcircles intersect at a second point ${\displaystyle X\neq P}$. Then ${\displaystyle X}$ is the spiral center mapping ${\displaystyle {\overline {AB}}}$ to ${\displaystyle {\overline {CD}}.}$

Proof: Note that ${\displaystyle ABPX}$ and ${\displaystyle XPCD}$ are cyclic quadrilaterals. Thus, ${\displaystyle \angle XAB=180^{\circ }-\angle BPX=\angle XPD=\angle XCD}$. Similarly, ${\displaystyle \angle ABX=\angle APX=180^{\circ }-\angle XPC=\angle XDC}$. Therefore, by AA similarity, triangles ${\displaystyle XAB}$ and ${\displaystyle XCD}$ are similar. Thus, ${\displaystyle \angle AXB=\angle CXD,}$ so a rotation angle mapping ${\displaystyle A}$ to ${\displaystyle B}$ also maps ${\displaystyle C}$ to ${\displaystyle D}$. The dilation factor is then just the ratio of side lengths ${\displaystyle {\overline {CD}}}$ to ${\displaystyle {\overline {AB}}}$.[5]

If we express ${\displaystyle A,B,C,}$ and ${\displaystyle D}$ as points on the complex plane with corresponding complex numbers ${\displaystyle a,b,c,}$ and ${\displaystyle d}$, we can solve for the expression of the spiral similarity which takes ${\displaystyle A}$ to ${\displaystyle C}$ and ${\displaystyle B}$ to ${\displaystyle D}$. Note that ${\displaystyle T(a)=x_{0}+\alpha (a-x_{0})}$ and ${\displaystyle T(b)=x_{0}+\alpha (b-x_{0})}$, so ${\displaystyle {\frac {T(b)-T(a)}{b-a}}=\alpha }$. Since ${\displaystyle T(a)=c}$ and ${\displaystyle T(b)=d}$, we plug in to obtain ${\displaystyle \alpha ={\frac {d-c}{b-a}}}$, from which we obtain ${\displaystyle x_{0}={\frac {ad-bc}{a+d-b-c}}}$.[5]

For any points ${\displaystyle A,B,C,}$ and ${\displaystyle D}$, the center of the spiral similarity taking ${\displaystyle {\overline {AB}}}$ to ${\displaystyle {\overline {CD}}}$ is also the center of a spiral similarity taking ${\displaystyle {\overline {AC}}}$ to ${\displaystyle {\overline {BD}}}$.

This can be seen through the above construction. If we let ${\displaystyle X}$ be the center of spiral similarity taking ${\displaystyle {\overline {AB}}}$ to ${\displaystyle {\overline {CD}}}$, then ${\displaystyle \triangle XAB\sim \triangle XCD}$. Therefore, ${\displaystyle \angle AXC=\angle AXB+\angle BXC=\angle CXD+\angle BXC=\angle BXD}$. Also, ${\displaystyle {\frac {AX}{BX}}={\frac {CX}{DX}}}$ implies that ${\displaystyle {\frac {AX}{CX}}={\frac {BX}{DX}}}$. So, by SAS similarity, we see that ${\displaystyle \triangle AXC\sim \triangle BXD}$. Thus ${\displaystyle X}$ is also the center of the spiral similarity which takes ${\displaystyle {\overline {AC}}}$ to ${\displaystyle {\overline {BD}}}$.[5][6]

Spiral similarity can be used to prove Miquel's Quadrilateral Theorem: given four noncollinear points ${\displaystyle A,B,C,}$ and ${\displaystyle D}$, the circumcircles of the four triangles ${\displaystyle \triangle PAB,\triangle PDC,\triangle QAD,}$ and ${\displaystyle \triangle QBC}$ intersect at one point, where ${\displaystyle P}$ is the intersection of ${\displaystyle AD}$ and ${\displaystyle BC}$ and ${\displaystyle Q}$ is the intersection of ${\displaystyle AB}$ and ${\displaystyle CD}$ (see diagram).[1]

Let ${\displaystyle M}$ be the center of the spiral similarity which takes ${\displaystyle AB}$ to ${\displaystyle DC}$. By the above construction, the circumcircles of ${\displaystyle \triangle PAB}$ and ${\displaystyle \triangle PDC}$ intersect at ${\displaystyle M}$ and ${\displaystyle P}$. Since ${\displaystyle M}$ is also the center of the spiral similarity taking ${\displaystyle DA}$ to ${\displaystyle BC}$, by similar reasoning the circumcircles of ${\displaystyle \triangle QAD}$ and ${\displaystyle \triangle QBC}$ meet at ${\displaystyle Q}$ and ${\displaystyle M}$. Thus, all four circles intersect at ${\displaystyle M}$.[1]

Here is an example problem on the 2018 Japan MO Finals which can be solved using spiral similarity:

Given a scalene triangle ${\displaystyle ABC}$, let ${\displaystyle D}$ and ${\displaystyle E}$ be points on segments ${\displaystyle AB}$ and ${\displaystyle AC}$, respectively, so that ${\displaystyle CA=CD,BA=BE}$. Let ${\displaystyle \omega }$ be the circumcircle of triangle ${\displaystyle ADE}$ and ${\displaystyle P}$ the reflection of ${\displaystyle A}$ across ${\displaystyle BC}$. Lines ${\displaystyle PD}$ and ${\displaystyle PE}$ meet ${\displaystyle \omega }$ again at ${\displaystyle X}$ and ${\displaystyle Y}$, respectively. Prove that ${\displaystyle BX}$ and ${\displaystyle CY}$ intersect on ${\displaystyle \omega }$.[5]

Proof: We first prove the following claims:

Claim 1: Quadrilateral ${\displaystyle PBEC}$ is cyclic.

Proof: Since ${\displaystyle \triangle BAE}$ is isosceles, we note that ${\displaystyle \angle BPC=\angle BAC=180^{\circ }-\angle BEC,}$ thus proving that quadrilateral ${\displaystyle PBEC}$ is cyclic, as desired. By symmetry, we can prove that quadrilateral ${\displaystyle PBDC}$ is cyclic.

Claim 2: ${\displaystyle \triangle AXY\sim \triangle ABC.}$

Proof: We have that ${\displaystyle \angle AXY=180^{\circ }-\angle AEY=\angle YEC=\angle PEC=\angle PBC=\angle ABC.}$ By similar reasoning, ${\displaystyle \angle AYX=\angle ACB,}$ so by AA similarity, ${\displaystyle \triangle AXY\sim \triangle ABC,}$ as desired.

We now note that ${\displaystyle A}$ is the spiral center that maps ${\displaystyle XY}$ to ${\displaystyle BC}$. Let ${\displaystyle F}$ be the intersection of ${\displaystyle BX}$ and ${\displaystyle CY}$. By the spiral similarity construction above, the spiral center must be the intersection of the circumcircles of ${\displaystyle \triangle FXY}$ and ${\displaystyle \triangle FBC}$. However, this point is ${\displaystyle A}$, so thus points ${\displaystyle A,F,X,Y}$ must be concyclic. Hence, ${\displaystyle F}$ must lie on ${\displaystyle \omega }$, as desired.