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 OO, its radius 11 and diameter ABAB, the center and radius of the purple circle be PP and rr, the segment OPOP meets the unit semicircle at CC, PNPN be perpendicular to ABAB, COB=θ\angle COB = \theta and t=tanθ2t = \tan \frac \theta 2 (refer to half-angle tangent substitution). Then

OP+PC=OCPNcscθ+PC=OCrcscθ+r=1    r=11+cscθ=2t(1+t)2\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 RR and r2r_2. By Pythagorean theorem,

PN2+NR2=PR2PN2+(OBONRB)2=PR2r2+(1rcotθr2)2=(r+r2)2r2+(12t(1+t)21t22tr2)2=r2+2rr2+r22(11t1+tr2)2=2rr2+r22(2t1+tr2)2=2rr2+r224t2(1+t)24t1+tr2+r22=4t(1+t)2r2+r224t24t(1+t)r2=4tr2    r2=t2+t\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 QQ and r1r_1. We note that the green and cyan semicircles are interchangeable with θ\theta replaced with πθ\pi - \theta, then tt replaced with tan(πθ2)=cotθ2=1t\tan \left(\frac {\pi - \theta}2 \right) = \cot \frac \theta 2 = \frac 1t. Therefore,

r1=1t2+1t=11+2tr_1 = \frac {\frac 1t}{2+\frac 1t} = \frac 1{1+2t}

Let the center and radius of the orange circle be TT and r4r_4. By Decartes' theorem,

1r4=1r+1r211+21rr21r1r2=(1+t)22t+2+tt1+2(1+t)2(2+t)2t2(1+t)22t2+tt=t2+2t+52t+2t=t2+2t+92t    r4=2tt2+2t+9\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 SS and r3r_3. Again r3r_3 and r4r_4 are interchangeable, we have:

r3=2t1t2+2t+9=2t9t2+2t+1r_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 OO of the unit semicircle to be (0,0)(0,0) of the xyxy-plane, the center P(x0,y0)P(x_0,y_0) of the purple circle is given by:

{x0=PNcscθ=r1t22t=2t(1+t)21t22t=1t1+ty0=r=2t(1+t)2\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(x1,y1)=(r11,0)=(11+2t1,0)=(2t1+2t,0)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(x2,y2)=(1r2,0)=(1t2+t,0)=(22+t,0)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(x4,y4)T(x_4, y_4). By Pythagorean theorem, OT2OM2=TM2OT^2-OM^2 = TM^2 and TR2MR2=TM2TR^2 - MR^2 = TM^2. Therefore,

OT2OM2=TR2MR2(1r4)2x42=(r2+r4)2(1r2x)212r4=2r2r41+2r2+2(1r2)x4x4=1r2r4r2r41r2=11+r21r2r4=12t(1+t)t2+2t+9=9t2t2+2t+9\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 TM2=OT2OM2TM^2 = OT^2 - OM^2:

y42=(1r4)2x42=(9+t2t2+2t+9)2(9t2t2+2t+9)2=36t2(t2+2t+9)2    y4=6tt2+2t+9\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 tt with 1t\dfrac 1t, we have the coordinates of the center of the red circle S(x3,y3)S(x_3,y_3):

x3=9t219t2+2t+1,y3=6t9t2+2t+1x_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 T1(x41,y41)T_1 (x_{41},y_{41}) and T1UT_1U be perpendicular to TMTM. Note that TT1U\triangle TT_1U and TRM\triangle TRM are similar. Then

UT1MR=TT1TR=r4r2+r4    r4r2+r4MRUMTM=T1RTR=r2r2+r4    UM=r2r2+r4TM\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 T1T_1 are

{x41=OM+UT1=x4+r4(1r2x4)r2+r4=13t2t2+4t+13y41=UM=r2r2+r4y4=6tt2+4t+13\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 tt with 1t\dfrac 1t and negative direction for xx-coordinate, the corresponding point S1(x31,y31)S_1(x_{31}, y_{31}) of the red circle is:

x31=113t213t2+4t+1,y31=6t13t2+4t+1x_{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 T2(x42,y42)T_2 (x_{42},y_{42}). By interpolation,

{x42=rx4+r4x0r+r4=5t2t2+2t+5y42=ry4+r4y0r+r4=4tt2+2t+5\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 tt with 1t\dfrac 1t and negative direction for xx-coordinate, the corresponding point S1(x32,y32)S_1(x_{32}, y_{32}) of the red circle is:

x32=15t25t2+2t+1,y32=4t5t2+2t+1x_{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 T3(x43,y43)T_3 (x_{43},y_{43}). By extrapolation,

{x43=x41r4=9t2t2+9y43=y41r4=6tt2+9\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 tt with 1t\dfrac 1t and negative direction for xx-coordinate, the corresponding point S1(x33,y33)S_1(x_{33}, y_{33}) of the red circle is:

x33=19t29t2+1y33=6t9t2+1x_{33} = \frac {1-9t^2}{9t^2+1} \quad \quad y_{33} = \frac {6t}{9t^2+1}

Note by Chew-Seong Cheong
3 months, 3 weeks ago

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Great work !

Valentin Duringer - 3 months, 2 weeks ago

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That was why it took me so long.

Chew-Seong Cheong - 3 months, 2 weeks ago

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

Valentin Duringer - 3 months, 2 weeks ago

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

Chew-Seong Cheong - 3 months, 2 weeks ago

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