Given a right triangle with side lengths $3$, $4$, and $5$, a circle is drawn with each side as a diameter. Find the radius of the circle that is internally tangent to all three circles.

Would radical axes be a valid approach here? I am thinking coordinates at the moment, as well. I haven't really made any progress with this problem as it's very hard.

Any solutions not involving advanced geometry techniques?

UPDATE: Graphing isn't working either. I can't see anything that might be useful.

enter image description here

  • $\begingroup$ Please read tag descriptions before applying them: the algebraic geometry tag expressly states that "[t]his tag should not be used for elementary problems which involve both algebra and geometry." $\endgroup$ – KReiser Aug 26 at 23:41
  • $\begingroup$ I'm sorry. I will do so in the future; I'm quite new here. $\endgroup$ – Fleccerd Aug 26 at 23:49
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    $\begingroup$ This is the Problem of Apollonius, the tenth case in the table. $\endgroup$ – Ross Millikan Aug 27 at 1:46

I think the best naive approach is to place the figure on a coordinate plane. To make the numbers nice, it might be better to scale up the triangle by a factor of $2$, then when we find the circle, scale everything back down.

The most natural way to proceed, I think, is to let the midpoint of the hypotenuse be the origin, and the right angle is at $C = (4,3)$. Then the centers of the semicircles are at $C' = (0,0)$, $B' = (4,0)$, and $A' = (0,3)$. We are looking for a point $P = (x,y)$ such that $PC' + 5 = PB' + 3 = PA' + 4 = r$, where $r$ is the radius of the tangent circle. This leads to the system of equations $$\begin{align} (r-5)^2 &= x^2 + y^2 \\ (r-3)^2 &= (x-4)^2 + y^2 \\ (r-4)^2 &= x^2 + (y-3)^2. \end{align}$$ Consequently $$4(4-r) = (r-5)^2 - (r-3)^2 = x^2 - (x-4)^2 = 8(x-2),$$ or $x = \frac{8-r}{2}$, and similarly $$9-2r = (r-5)^2 - (r-4)^2 = y^2 - (y-3)^2 = 3(2y-3),$$ or $y = \frac{9-r}{3}$. Thus $$(r-5)^2 = \left(\frac{8-r}{2}\right)^2 + \left(\frac{9-r}{3}\right)^2,$$ for which the unique positive root is $r = \frac{144}{23}$, hence after undoing the scaling, the desired radius is $$\boxed{r = \frac{72}{23}}.$$ The center is located at $P = (\frac{10}{23}, \frac{21}{46})$.

Whether we can find a more elegant solution, or through purely geometric means, remains an open question, but this approach I find quite straightforward, elementary, and not computationally difficult.

Since two users have disputed my answer, I show the following Mathematica code and Geogebra figure:

First[{x, y, r}/2 /. Solve[{EuclideanDistance[{x, y}, {0, 0}] + 5 == 
 EuclideanDistance[{x, y}, {4, 0}] + 3 == 
 EuclideanDistance[{x, y}, {0, 3}] + 4 == r}, {x, y, r}]]

The output is

{10/23, 21/46, 72/23}

The figure is shown below.

enter image description here

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  • $\begingroup$ I think there's some mistake. When I tried to sketch this in Geogebra, I'm not getting the desired result. $\endgroup$ – SarGe Aug 27 at 13:52
  • $\begingroup$ With @heropup's method, I'm getting $r = 72/23$ and center $P = (36/23, 24/23)$ $\endgroup$ – cosmo5 Aug 27 at 15:30
  • $\begingroup$ @cosmo5 You have ignored the fact that I have rotated the figure. $\endgroup$ – heropup Aug 27 at 16:59

enter image description here

Let the center of the circle be $(x, y)$ and radius of the circle be $r$.

Since, the circle touches all the three circles, we get the constraints as $$\begin{align}(x-2)^2+y^2&=(2-r)^2\\ x^2+(y-1.5)^2&=(1.5-r)^2\\ (x-2)^2+(y-1.5)^2&=(2.5-r)^2 \end{align}$$

Solving gives us $\displaystyle (x,y,r)=\left(\frac{36}{23},\frac{24}{23},\frac{72}{23}\right),(0,0,0)$.

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