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The red, blue, and green circles have diameters 3, 4, and 5, respectively.

What is the radius of the black circle tangent to all three of these circles?

I just figured out the radius is exactly $\dfrac{72}{23}$ but I don't know how to do the solution.

enter image description here

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    $\begingroup$ If your radius is correct, then whatever you did to determine that radius is the solution. $\endgroup$
    – John Douma
    Aug 26, 2018 at 21:05
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    $\begingroup$ Did you just measure it or how did you find out? $\endgroup$
    – Creator
    Aug 26, 2018 at 21:15
  • $\begingroup$ Does this generalize to an arbitrary Pythagorean triangle? $\endgroup$ Aug 28, 2018 at 18:14

4 Answers 4

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Radius is exactly $\dfrac{72}{23}$, pretty neat. See the figure below, where dotted lines from the center of the circumcribing circle passes through the midpoints of the sides of the triangle.

enter image description here

Let : $B$ be the origin $(0,0)$, and the center of circle of unknown radius $r$ be $(x,y)$. Then we solve the following three equations to find $r$

$\dfrac{3}{2}+\sqrt{x^2+(\dfrac{3}{2}-y)^2}=r$

$2+\sqrt{(2-x)^2+y^2}=r$

$\dfrac{5}{2}+\sqrt{(2-x)^2+(\dfrac{3}{2}-y)^2}=r$

so that $(r, x, y)= \left(\dfrac{72}{23}, \dfrac{36}{23}, \dfrac{24}{23}\right)$

Note: The midpoints of the sides $(0,\dfrac{3}{2})$, $(2,0)$, and $(2,\dfrac{3}{2})$ are centers of circles of radii $\dfrac{3}{2}, 2, \dfrac{5}{2}$. Given any two tangent circles, their centres and point of tangency are colinear.

Bonus: If right triangle : $\triangle ABC$ has rational sides, then $(r,x,y)$ are also all rationals.

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    $\begingroup$ Excuse me, i´m confused about the first equotation. So how do you get to this equation? $\endgroup$
    – X X
    Aug 26, 2018 at 21:41
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    $\begingroup$ Look at the dotted line going from center of large circle though the midpoint of side that has length $3$. The dotted line has length $r$, while the radius of the circle centered at $(0,\dfrac{3}{2})$ has radius $\dfrac{3}{2}$. $\endgroup$
    – user585825
    Aug 26, 2018 at 21:43
  • $\begingroup$ I cannot understand whence you got your system of equations. I understand that the constants correspond to the radii of the coloured circles, but I don’t understand the meaning of the equations. $\endgroup$ Aug 27, 2018 at 6:02
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    $\begingroup$ @ChaseRyanTaylor each equation corresponds to a dotted line. The constant corresponds to the segment from the midpoint of the triangles edge to the edge of the large circle, and the rooted part is just the distance formula from the midpoint of an edge to the center of the large circle. $\endgroup$
    – Vaelus
    Aug 27, 2018 at 6:06
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    $\begingroup$ @Vaelus Ah; the radius of the black circle is the radius of the coloured circle plus the distance from the coloured circle’s centre to the centre of the black circle. $\endgroup$ Aug 27, 2018 at 6:33
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Let's find tangent circle by applying inversion transformation, or equivalently working in complex domain and applying reciprocal transformation $w=1/\overline{z}$.

Let (0,0), or the inversion centre, be the intersection of 3 circles. Other circle intersections are $(0,3)$, $(4,0)$ and $\frac{12}{25}\left(3,4\right)$.

Inversion transformation transform a circle passing through centre into a line, and other circles into circles. Thus, given three circles are transformed into lines, intersecting at 3 points; while big (still unknown) tangent circle transforms into circle tangent on those three lines. Transformed intersection coordinates are:

$ \begin{array}{lcr} (0,3) && \longrightarrow&& \left(0,\frac{1}{3}\right)\\ (4,0) && \longrightarrow && \left(\frac{1}{4},0\right)\\ \frac{12}{25}\left(3,4\right) && \longrightarrow && \left(\frac{1}{4},\frac{1}{3}\right)\end{array} $ enter image description here There are 4 circles that are tangent on all 3 lines: 1 inscribed and 3 escribed circle. The correct one has (0,0) in its interior. Finding coordinates of the centre and radius (solving quadratic equation) gives following circle:

$c=\left(-\frac{1}{6},-\frac{1}{4}\right)\quad r=\frac{1}{2}$

Transforming back escribed circle gives the required tangent circle. Take notice that, while points on a circle transform into a circle, a circle centre does not transform to a corresponding circle centre. Radius of transformed circle can be deduce by working on the line that connects origin and centre of the circle. Transforming 2 points of the circle that are collinear with circle centre and the origin gives.

$2R= \left|\frac{1}{|c|-r}-\frac{1}{|c|+r}\right|\\ R= \frac{r}{\left||c|^2-r^2\right|}\\ R=\frac{72}{23}$

Here is the picture with requested tangent circle, as well as one other tangent circle corresponding to the inscribed circle of the reciprocal space.

enter image description here

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    $\begingroup$ What a beautiful application of inversion...+1 $\endgroup$
    – user279540
    Aug 27, 2018 at 2:10
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Thank you for your answers that helped me make a solution

$M_{red}=\dbinom{0}{3/2}$ , $r_{red}=\dfrac{3}{2}$ , $M_{blue}=\dbinom{0}{2}$ , $r_{blue}=2$, $M_{green}=\dbinom{2}{3/2}$ , $r_{green}=\dfrac{5}{2}$

$M_{black}=M=\dbinom{x}{y}$ , $r_{black}=r$

$length(v)=\dfrac{v}{\sqrt{v_{x}^{2}+v_{y}^{2}}}$ , $unitvec(v)=\dfrac{v}{length(v)}$

$d_{red}=length[M_{red}+unitvec(M_{red}-M)\times r_{red}-M]$

$d_{blue}=length[M_{blue}+unitvec(M_{blue}-M)\times r_{blue}-M]$

$d_{green}=length[M_{green}+unitvec(M_{green}-M)\times r_{green}-M]$

$\dbinom{0}{3/2}+\dfrac{\dbinom{0}{3/2}-\dbinom{x}{y}}{\sqrt{(0-x)^{2}+(3/2-y)^{2}}}\times \dfrac{3}{2}-M$ $\Rightarrow $ $d_{red}=\sqrt{\left(-x-\dfrac{3x}{\sqrt{4x^{2}+(3-2y)^{2}}}\right)^{2}+\left(3/2-y+\dfrac{9-6y}{2\sqrt{4x^{2}+(3-2y)^{2}}}\right)^{2}}$

$\dbinom{2}{0}+\dfrac{\dbinom{2}{0}-\dbinom{x}{y}}{\sqrt{(2-x)^{2}+(0-y)^{2}}}\times 2-M$ $\Rightarrow $ $d_{blue}=\sqrt{\left(2-x+\dfrac{4-2x}{\sqrt{(-2+x)^{2}+y^{2}}}\right)^{2}+\left(-y-\dfrac{2y}{\sqrt{(-2+x)^{2}+y^{2}}}\right)^{2}}$

$\dbinom{2}{3/2}+\dfrac{\dbinom{2}{3/2}-\dbinom{x}{y}}{\sqrt{(2-x)^{2}+(3/2-y)^{2}}}\times \dfrac{5}{2}-M$ $\Rightarrow $ $d_{red}=\sqrt{\left(2-x-\dfrac{5(-2+x)}{\sqrt{25+4x(-4+x)+4y(-3+y)}}\right)^{2}+\left(3/2-y+\dfrac{5(3/2-y)}{2\sqrt{(-2+x)^{2}+(-3/2+y)^{2}}}\right)^{2}}$

Setting these 3 equations to r and solving for x,y,r of the black circle results in:

$x=\dfrac{36}{23} , y=\dfrac{24}{23} , r=\dfrac{72}{23}$

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It is simply the Apollonius Problem and solved by the following three quadratic equations using WolframAlpha for center $ (x,y) $ and radius $r$ of the biggest circle, which is tangent to other three circles;

(1) $ x^2 + (y - \frac{3}{2})^2 = (r - \frac{3}{2})^2 $, circle on side of length $ = 3 $,

(2) $ (x - 2)^2 + y ^ 2 = (r - 2)^2 $, circle on side of length $ = 4 $,

(3) $ (x - 2)^2 + (y - \frac{3}{2})^2 = (r - \frac{5}{2})^2 $, circle on side of length $ = 5 $.

The solution is $ x = \frac{36}{23}, y = \frac{24}{23} $ and radius $ r $ is the Answer $ = \frac{72}{23} $.

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    $\begingroup$ i don´t get why (r-3/2)^2 ; (r-2)^2 and (r-5/2)^2. can you explain it to me? thank you. $\endgroup$
    – X X
    Aug 26, 2018 at 21:50
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    $\begingroup$ If you study the Apollonius problem of circles tangent to three circles as given in en.wikipedia.org/wiki/Problem%20of%20Apollonius, you will find that to get the largest circle, which is encircling all the three circles, we need to choose negative sign in RHS of all the three quadratic equations i.e. (r-ri)^2, r1=3/2, r2=4/2 and r3=5/2. Here, 3, 4 and 5 are the diameters of the three semi-circles on the three sides of the triangle. I hope it clears the doubt. $\endgroup$
    – user585308
    Aug 26, 2018 at 21:51
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    $\begingroup$ @MarkHennings Here's how to use MathJax. You can type fractions like this: \$ \frac{m}{n} \$. $\endgroup$
    – Toby Mak
    Aug 26, 2018 at 22:16
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    $\begingroup$ Beside that i solved the equations myself....yeah i did the same $\endgroup$
    – user585547
    Aug 26, 2018 at 22:21
  • $\begingroup$ @XX: Here we have the same equations as in the top voted answer. You might find the explanation there helpful. The radius $r$ of the wanted circle is the length of the radius of one of the small circles plus the distance from the center of the wanted circle to the center of the corresponding small circle. $\endgroup$
    – epi163sqrt
    Aug 28, 2018 at 9:13