Dynamic Geometry: P32 Series

This note is a compilation of calculations needed in the solutions to the problems titled Dynamic Geometry: P32, 77, 89, 94, and 98 so far by @Valentin Duringer.

Centers and Radii

Let the center of the large semicircle be O(0,0)O(0,0), the origin of the xyxy-plane, its radius 11, if it is not given, and its diameter ABAB. Let the center and radius of the cyan semicircle be PP and r1r_1 and those of the green semicircle be QQ and r2r_2. Then 2r1+2r2=AB=2    r1+r2=1    r2=1r12r_1+2r_2 = AB = 2 \implies r_1 + r_2 = 1 \implies r_2 = 1 - r_1, P=(1+r1,0)=(r2,0)P=(-1+r_1, 0) = (-r_2, 0), and Q=(1r2,0)=(r1,0)Q = (1-r_2, 0) = (r_1, 0).

Let the center and radius of the left yellow circle be S(x3,y3)S(x_3, y_3) and r3r_3, those of the right yellow circle be T(x4,y4)T(x_4, y_4) and r4r_4, CDCD be the vertical dividing red segment, T2TT_2T be perpendicular to CDCD, and TNTN be perpendicular to ABAB. Then by Pythagorean theorem,

TN2=TQ2NQ2=TQ2(DQT2T)2    y42=(r2+r4)2(r2r4)2=4r2r4\begin{aligned} TN^2 & = TQ^2 - NQ^2 = TQ^2 - (DQ-T_2T)^2 \\ \implies y_4^2 & = (r_2+r_4)^2 - (r_2-r_4)^2 = 4r_2r_4 \end{aligned}

Also

TO2ON2=TN2TO2(OQNQ)2=TN2TO2(OQ(DQT2T))2=TN2(1r4)2(r1r2+r4)2=y42(r1+r2r4)2(r1r2+r4)2=4r2r42r1(2r22r4)=4r2r4r1r2r1r4=r2r4r1r2=(r1+r2)r4Note that r1+r2=1    r4=r1r2\begin{aligned} TO^2 - ON^2 & = TN^2 \\ TO^2 - (OQ-NQ)^2 & = TN^2 \\ TO^2 - (OQ-(DQ-T_2T))^2 & = TN^2 \\ (1-r_4)^2 - (r_1 - r_2 + r_4)^2 & = y_4^2 \\ (r_1 + r_2 - r_4)^2 - (r_1 - r_2 + r_4)^2 & = 4r_2r_4 \\ 2r_1(2r_2-2r_4) & = 4r_2r_4 \\ r_1r_2 - r_1r_4 & = r_2r_4 \\ r_1r_2 & = (\blue{r_1+r_2}) r_4 & \small \blue{\text{Note that }r_1+r_2 = 1} \\ \implies r_4 & = r_1r_2 \end{aligned}

Since if we flip the yellow circles and the semicircles below them horizontally about CDCD, r1r_1 becomes r2r_2 and r3r_3 becomes r4r_4. We get the equation for r3r_3 by swapping r1r_1 and r2r_2. Then r3=r2r1=r4r_3 = r_2r_1 = r_4. Similarly, y4=2r2r4=2r2r1    y3=2r1r2y_4 = 2\sqrt{r_2r_4} = 2r_2\sqrt{r_1} \implies y_3 = 2r_1\sqrt{r_2}. And x4=ON=OQ(DQT2T)=r1r2+r4=r1r2+r1r2=r1r22    x3=r2r12x_4 = ON = OQ-(DQ-T_2T) = r_1-r_2+r_4 = r_1-r_2 + r_1r_2 = r_1 - r_2^2 \implies x_3 = r_2 - r_1^2.

In summary:

{O(0,0)1P(r2,0)r1Q(r1,0)r2=1r1S(r2r12,2r1r2)r1r2T(r1r22,2r2r1)r1r2\begin{cases} O(0,0) & 1 \\ P(-r_2,0) & r_1 \\ Q(r_1, 0) & r_2 = 1 - r_1 \\ S(r_2-r_1^2, 2r_1\sqrt{r_2}) & r_1r_2 \\ T(r_1-r_2^2, 2r_2\sqrt{r_1}) & r_1r_2 \end{cases}

Vertices of Triangles

Let S1(x31.y31)S_1(x_{31}. y_{31}), S2(x32.y32)S_2(x_{32}. y_{32}), and S3(x33.y33)S_3(x_{33}. y_{33}) be the points where the left yellow circle tangent to the semicircle below it, the segment CDCD, and the unit semicircle respectively. Let the corresponding points of the right yellow circle be T1(x41.y41)T_1(x_{41}. y_{41}), T2(x42.y42)T_2(x_{42}. y_{42}), and T3(x43.y43)T_3(x_{43}. y_{43}) respectively.

Now let MT1MT_1 be perpendicular to TNTN. Then

{x41=x4+MT1=x4+r4(r1x4)r2+r4=2r1r21+r1y41=r2y4r2+r4=2r2r11+r1{x42=OD=r1r2y42=y4=2r2r1{x43=x41r4=r1r221r1r2y43=y41r4=2r2r11r1r2\begin{cases} x_{41} =x_4+MT_1 = x_4+ \dfrac {r_4(r_1-x_4)}{r_2+r_4} = \dfrac {2r_1-r_2}{1+r_1} \\ y_{41} = \dfrac {r_2y_4}{r_2+r_4} = \dfrac {2r_2\sqrt{r_1}}{1+r_1} \end{cases} \\ \begin{cases} x_{42} = OD = r_1 - r_2 \\ y_{42} = y_4 = 2r_2\sqrt{r_1} \end{cases} \\ \begin{cases} x_{43} = \dfrac {x_4}{1-r_4} = \dfrac {r_1-r_2^2}{1-r_1r_2} \\ y_{43} = \dfrac {y_4}{1-r_4} = \dfrac {2r_2\sqrt{r_1}}{1-r_1r_2} \end{cases}

Again we can find the coordinates of S1S_1, S2S_2, and S3S_3 by swapping r1r_1 and r2r_2. So in summary:

{S1(2r2r11+r2,2r1r21+r2),T1(2r1r21+r1,2r2r11+r1)S2(r2r1,2r1r2),T2(r1r2,2r2r1)S3(r2r121r1r2,2r1r21r1r2),T3(r1r221r1r2,2r2r11r1r2)\begin{cases} S_1 \left(\dfrac {2r_2-r_1}{1+r_2}, \dfrac {2r_1\sqrt{r_2}}{1+r_2}\right), & T_1 \left(\dfrac {2r_1-r_2}{1+r_1}, \dfrac {2r_2\sqrt{r_1}}{1+r_1} \right) \\ S_2 \left(r_2 - r_1, 2r_1\sqrt{r_2} \right), & T_2 \left(r_1 - r_2, 2r_2\sqrt{r_1} \right) \\ S_3 \left(\dfrac {r_2-r_1^2}{1-r_1r_2}, \dfrac {2r_1\sqrt{r_2}}{1-r_1r_2} \right), & T_3 \left(\dfrac {r_1-r_2^2}{1-r_1r_2}, \dfrac {2r_2\sqrt{r_1}}{1-r_1r_2} \right) \end{cases}

Note by Chew-Seong Cheong
3 weeks ago

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