*Solution by Anthony Roitman, 9*^{th} grade.

**Problem:**

*Two spheres, which are tangent to each other, are also tangent to a plane and to a line perpendicular to that plane. What is the ratio of the spheres' radii if the points where the spheres touch the plane form: *

*(i) a right triangle with the point at which the line intersects the plane?*

*(ii) an isosceles triangle with the point at which the line intersects the plane?*

* *

**Solution:**

In order to approach this problem, let us look at this problem in two ways. One will be the projection of the line and two spheres onto a plane parallel to the given plane, and the other way will be the projection of the spheres and the line onto a plane perpendicular to the given plane. In other words, let us look at the top view and the side view of our problem, in Figures 1 and 2. In Fig. 2, the intersection of the segments $a$ and $b$ is the projection of the tangent line $T$ onto this parallel plane, since $T$ is perpendicular to the given plane, and therefore this parallel plane. Also, the segment $c$ in Figure 2 is the same length as the segment $c$ in Figure 1.

Now, let us consider the first case, where the triangle formed by $a$, $b$ and $c$ is a right triangle. But then, there are two further cases. Namely, these are when

1. The right angle is contained between $a$ and $b$, or

2. when it is contained between $b$ and $c$.

There is of course a third case, where the right angle is contained between $c$ and $a$, but in this case, the ratio of the radii will simply be the reciprocal of what it would be in the previous case. Therefore, this case can be disregarded.

Now, let us try to create a system of equations to solve this problem. For both subcases, the equation $a^2+ (a-b)^2=(a+b)^2$ holds true (we can derive this from Fig. 1). For subcase 1, the other equation is $a^2+ b^2=c^2$. For subcase 2, the other equation is $c^2+b^2=a^2$. Upon solving these equations for $\frac{a}{b}$ (which is the ratio of the radii), we get that for subcase 1, the ratio is $\sqrt{3} + 2$. For subcase 2, we get $\sqrt{5} + 2$.

The other case, where $a$, $b$ and $c$ form an isosceles triangle, can be treated similarly. We again have two subcases:

1. The two equal sides of the triangle are $a$ and $b$

2. The two equal sides of the triangle are $b$ and $c$

The first case is obvious. Since $a$ and $b$ are the radii of the spheres, their ratio is 1, since they are equal. The second case is more complex. Again we have the equation, $a^2+ (a-b)^2=(a+b)^2$. We also have the equation $b=c$. Solving for $\frac{a}{b}$, we get that $\frac{a}{b} = 4$.

Overall, we have two possible ratios for two cases:

1. $\sqrt{3} + 2$ or $\sqrt{5}+ 2$

2. $1$ or $4$