# Dynamic Geometry: P80 Series

This note provides the general calculations in the solutions to problems titled Dynamic Geometry P80, 85, 90, 95, 99, and 105 so far by @Valentin Duringer.

## Radii of Circles and Semicircles

Let the center of the big semicircle be $O$, its radius $1$ and diameter $AB$, the center and radius of the purple circle be $P$ and $r$, the segment $OP$ meets the unit semicircle at $C$, $PN$ be perpendicular to $AB$, $\angle COB = \theta$ and $t = \tan \frac \theta 2$ (refer to half-angle tangent substitution). Then

\begin{aligned} OP + PC & = OC \\ PN \cdot \csc \theta + PC & = OC \\ r \csc \theta + r & = 1 \\ \implies r & = \frac 1{1+\csc \theta} = \frac {2t}{(1+t)^2} \end{aligned}

Let the center and radius of the cyan semicircle be $R$ and $r_2$. By Pythagorean theorem,

\begin{aligned} PN^2 + NR^2 & = PR^2 \\ PN^2 + (OB-ON-RB)^2 & = PR^2 \\ r^2 + (1-r \cot \theta - r_2)^2 & = (r+r_2)^2 \\ r^2 + \left(1 - \frac {2t}{(1+t)^2} \cdot \frac {1-t^2}{2t} - r_2 \right)^2 & = r^2 + 2rr_2 + r_2^2 \\ \left(1 - \frac {1-t}{1+t} - r_2 \right)^2 & = 2rr_2 + r_2^2 \\ \left(\frac {2t}{1+t} - r_2 \right)^2 & = 2rr_2 + r_2^2 \\ \frac {4t^2}{(1+t)^2} - \frac {4t}{1+t} r_2 + r_2^2 & = \frac {4t}{(1+t)^2} r_2 + r_2^2 \\ 4t^2 - 4t(1+t) r_2 & = 4t r_2 \\ \implies r_2 & = \frac t{2+t} \end{aligned}

Let the center and radius of the green semicircle be $Q$ and $r_1$. We note that the green and cyan semicircles are interchangeable with $\theta$ replaced with $\pi - \theta$, then $t$ replaced with $\tan \left(\frac {\pi - \theta}2 \right) = \cot \frac \theta 2 = \frac 1t$. Therefore,

$r_1 = \frac {\frac 1t}{2+\frac 1t} = \frac 1{1+2t}$

Let the center and radius of the orange circle be $T$ and $r_4$. By Decartes' theorem,

\begin{aligned} \frac 1{r_4} & = \frac 1r + \frac 1{r_2} - \frac 11 + 2 \sqrt{\frac 1{rr_2}-\frac 1r - \frac 1{r_2}} \\ & = \frac {(1+t)^2}{2t} + \frac {2+t}t - 1 + 2\sqrt{\frac {(1+t)^2(2+t)}{2t^2}-\frac {(1+t)^2}{2t} - \frac {2+t}t } \\ & = \frac {t^2+2t+5}{2t} + \frac 2t = \frac {t^2 + 2t + 9}{2t} \\ \implies r_4 & = \frac {2t}{t^2+2t+9} \end{aligned}

Let the center and radius of the red circle be $S$ and $r_3$. Again $r_3$ and $r_4$ are interchangeable, we have:

$r_3 = \frac {\frac 2t}{\frac 1{t^2} + \frac 2t + 9} = \frac {2t}{9t^2 + 2t + 1}$

## Centers of Circles and Semicircles

Setting the center $O$ of the unit semicircle to be $(0,0)$ of the $xy$-plane, the center $P(x_0,y_0)$ of the purple circle is given by:

$\begin{cases} x_0 = PN \cdot \csc \theta = r \cdot \dfrac {1-t^2}{2t} = \dfrac {2t}{(1+t)^2} \cdot \dfrac {1-t^2}{2t} = \dfrac {1-t}{1+t} \\ y_0 = r = \dfrac {2t}{(1+t)^2} \end{cases}$

The center of the green semicircle $Q(x_1,y_1) = (r_1-1, 0) = \left(\dfrac 1{1+2t}-1, 0 \right) = \left(- \dfrac {2t}{1+2t}, 0 \right)$.
The center of the cyan semicircle $R(x_2, y_2) = (1-r_2, 0) = \left(1 - \dfrac t{2+t},0 \right) = \left(\dfrac 2{2+t}, 0 \right)$.

Let the center of the orange circle be $T(x_4, y_4)$. By Pythagorean theorem, $OT^2-OM^2 = TM^2$ and $TR^2 - MR^2 = TM^2$. Therefore,

\begin{aligned} OT^2 - OM^2 & = TR^2 - MR^2 \\ (1-r_4)^2 - x_4^2 & = (r_2+r_4)^2 - (1-r_2 - x)^2 \\ 1 - 2r_4 & = 2r_2r_4 - 1 + 2r_2 + 2(1-r_2)x_4 \\ x_4 & = \frac {1-r_2 - r_4 - r_2r_4}{1-r_2} = 1 - \frac {1+r_2}{1-r_2}r_4 \\ & = 1 - \frac {2t(1+t)}{t^2+2t+9} = \frac {9-t^2}{t^2+2t+9} \end{aligned}

From $TM^2 = OT^2 - OM^2$:

\begin{aligned} y_4^2 & = (1-r_4)^2 - x_4^2 = \left(\frac {9+t^2}{t^2+2t+9} \right)^2 - \left(\frac {9-t^2}{t^2+2t+9} \right)^2 = \frac {36t^2}{(t^2+2t+9)^2} \\ \implies y_4 & = \frac {6t}{t^2+2t+9} \end{aligned}

Replacing $t$ with $\dfrac 1t$, we have the coordinates of the center of the red circle $S(x_3,y_3)$:

$x_3 = \dfrac {9t^2-1}{9t^2 + 2t + 1}, \quad \quad y_3 = \dfrac {6t}{9t^2+2t+1}$

## Vertices of Triangles

Let the point where the orange circle is tangent to the cyan semicircle be $T_1 (x_{41},y_{41})$ and $T_1U$ be perpendicular to $TM$. Note that $\triangle TT_1U$ and $\triangle TRM$ are similar. Then

$\frac {UT_1}{MR} = \frac {TT_1}{TR} = \frac {r_4}{r_2+r_4} \implies \frac {r_4}{r_2+r_4} \cdot MR \quad \frac {UM}{TM} = \frac {T_1R}{TR} = \frac {r_2}{r_2+r_4} \implies UM = \frac {r_2}{r_2+r_4} \cdot TM$

Then the coordinates of $T_1$ are

$\begin{cases} x_{41} = OM + UT_1 = x_4 + \dfrac {r_4(1-r_2-x_4)}{r_2+r_4} = \dfrac {13-t^2}{t^2+4t+13} \\ y_{41} = UM = \dfrac {r_2}{r_2+r_4} y_4 = \dfrac {6t}{t^2+4t+13} \end{cases}$

Replacing $t$ with $\dfrac 1t$ and negative direction for $x$-coordinate, the corresponding point $S_1(x_{31}, y_{31})$ of the red circle is:

$x_{31} = \frac {1-13t^2}{13t^2+4t+1}, \quad \quad y_{31} = \frac {6t}{13t^2+4t+1}$

Let the point where the orange circle is tangent to the purple circle be $T_2 (x_{42},y_{42})$. By interpolation,

$\begin{cases} x_{42} = \dfrac {rx_4+r_4x_0}{r+r_4} = \dfrac {5-t^2}{t^2+2t+5} \\ y_{42} = \dfrac {ry_4+r_4y_0}{r+r_4} = \dfrac {4t}{t^2+2t+5} \end{cases}$

Replacing $t$ with $\dfrac 1t$ and negative direction for $x$-coordinate, the corresponding point $S_1(x_{32}, y_{32})$ of the red circle is:

$x_{32} = \frac {1-5t^2}{5t^2+2t+1}, \quad \quad y_{32} = \frac {4t}{5t^2+2t+1}$

Let the point where the orange circle is tangent to the unit semicircle be $T_3 (x_{43},y_{43})$. By extrapolation,

$\begin{cases} x_{43} = \dfrac {x_4}{1-r_4} = \dfrac {9-t^2}{t^2+9} \\ y_{43} = \dfrac {y_4}{1-r_4} = \dfrac {6t}{t^2+9} \end{cases}$

Replacing $t$ with $\dfrac 1t$ and negative direction for $x$-coordinate, the corresponding point $S_1(x_{33}, y_{33})$ of the red circle is:

$x_{33} = \frac {1-9t^2}{9t^2+1} \quad \quad y_{33} = \frac {6t}{9t^2+1}$

Note by Chew-Seong Cheong
5 months, 2 weeks ago

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Great work !

- 5 months, 1 week ago

That was why it took me so long.

- 5 months, 1 week ago

I get it now ! Are going to post a solution to P107, 101 and 98?

- 5 months, 1 week ago

I will write solution of 98 first. Need same treatment as P80 series

- 5 months, 1 week ago