analytic geometry
I got this question on a grade 11 multiple choice test and had no idea how to solve it.
Q: In the diagram, the circle with centre X is tangent to the largest circle and passes through the centre of the largest circle. The circles with centres Y and Z are each tangent to the other three circles, as shown. The circle with centre X has radius 1. The circles with centres Y and Z each have radius r. The value of r is? Note: the diagram isn't to scale.
I thought I'd just update everyone by saying that the comments precisely sorted out my problem: I had been projecting the vertices correctly and then for some reason assuming I could just draw straight lines between them. Instead, of course, the straight lines are in general projected to circular arcs and having sorted this out I got the interactive simulation to work. Here is an image:
and here is a link to the website where this projection is shown live where the controls are:
On the blue plane are projected the edges and vertices (in orange) of the dodecahedron but also the projections of the incircles of each pentagonal face (in red) giving a live solution to the problem of finding a finite collection of circles in the plane such that each touch 5 others (discussed, for example, here). The red dots are the projections of the centres of the incircles on the sphere, showing how the centres of the circles are not preserved by the projection.
Best Answer
First, as shown in the diagram above, draw a horizontal line through $X$. By symmetry, it will go from the tangent point on the left and also through the tangent point of the circles $Y$ and $Z$. Since this horizontal line is perpendicular to the circle's tangent line on the left side, and the line from the larger circle's center to that point is also perpendicular to that same line, this means these $2$ lines overlap. As the largest circle's center is on circle $X$, it must be on its right side. Call this point $O$. Since the radius of the circle $X$ is $1$, and the radius of the outside circle is the diameter of $X$, this means it's $2$.
Draw a line from $O$ through $Y$ to the edge of the outside circle and call the point where it first crosses the $Y$ circle $A$ and where it touches the outside circle $B$. The length of $OB$ is $2$ and the length of $YB$ is $r$ so
$$|OY| = 2 - r \tag{1}\label{eq1A}$$
Draw a line from $Y$ to where it meets the drawn horizontal line, and call the point where it meets $C$. Note it's at a right angle to the horizontal line. Also, since it's on the circumference of circle $Y$, you have
$$|CY| = r \tag{2}\label{eq2A}$$
Next, draw a line joining $X$ and $Y$. As it goes through the circles' $X$ and $Y$ tangent point, its length is $1 + r$. Now, $\triangle XYC$ is a right-angled triangle, so by the Pythagorean theorem, you get
$$\begin{equation}\begin{aligned} |XC|^2 + r^2 & = (r+1)^2 \\ |XC|^2 & = (r+1)^2 - r^2 \\ |XC| & = \sqrt{(r+1)^2 - r^2} \\ & = \sqrt{2r + 1} \end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Since $|XO| = 1$, you have
$$|OC| = |XC| - |XO| = \sqrt{2r + 1} - 1 \tag{4}\label{eq4A}$$
Since $\triangle OYC$ is a right-angled triangle, you can use the Pythagorean theorem again to relate the lengths of the $3$ sides using the values in \eqref{eq1A}, \eqref{eq2A} and \eqref{eq4A} as
$$\begin{equation}\begin{aligned} r^2 + (\sqrt{2r + 1} - 1)^2 & = (2 - r)^2 \\ r^2 + (2r + 1 - 2\sqrt{2r + 1} + 1) & = r^2 - 4r + 4 \\ 2r + 2 - 2\sqrt{2r + 1} & = - 4r + 4 \\ - 2\sqrt{2r + 1} & = -6r + 2 \\ \sqrt{2r + 1} & = 3r - 1 \\ 2r + 1 & = 9r^2 - 6r + 1 \\ 0 & = 9r^2 - 8r \\ 0 & = r(9r - 8) \end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Since $r \gt 0$, this means that
$$9r - 8 = 0 \implies r = \frac{8}{9} \tag{6}\label{eq6A}$$