# Three Circles in a Triangle: A Solution

In my last post, I posed the problem of showing that if triangles $ABC$ and $PQR$ (as shown in the diagram below) are similar, then $\Delta ABC$ is equilateral.

We will solve this problem in two steps:

1. Show that the three circles all have the same radius.
2. Use the fact that the radiuses are equal to show that $\Delta ABC$ is equilateral.

Let’s start with the first step of showing that the radiuses are equal. It will help if we draw in some radiuses, so let’s draw the radius of each circle to each triangle side.

Since the circles are tangent to the sides of the triangle, each of the six radiuses drawn form a right angle with the side it meets. As long as we’re labeling angles, we should probably also mark the fact that $\angle BAC=\angle QPR$, $\angle ABC=\angle PQR$, and $\angle ACB=\angle PRQ$ since $\Delta ABC$ and $\Delta PQR$ are similar.

The radiuses we drew in created some quadrilaterals. Consider quadrilateral $AX_1PZ_2$, for example. We know that the angle sum in a quadrilateral is $360^{\circ}$, and this particular quadrilateral has two $90^{\circ}$-angles. Therefore, the measure of $\angle X_1PZ_2$ is $180^{\circ}$ minus the measure of $\angle X_1AZ_2$. In other words, $\angle X_1PZ_2$ and $\angle X_1AZ_2$ are supplementary angles. The same logic shows us that $\angle Y_1QX_2$ is supplementary to $\angle Y_1BX_2$ and that $\angle Y_2RZ_1$ is supplementary to $\angle Y_2CZ_1$. We mark the supplementary angles on our diagram.

The four angles around point $P$ add up to a full rotation, or $360^{\circ}$. The two marked angles are supplementary, so the remaining two angles, $\angle RPZ_2$ and $\angle QPX_1$, also add up to $180^{\circ}$; we mark them as supplementary.

Now look at quadrilateral $PQX_2x_1$. Its angle sum is $360^{\circ}$, and since it has right angles at $X_1$ and $X_2$, its angles at $P$ and $Q$ must be supplementary. But then we can repeat our last two lines of thought around the triangle: the angles $\angle PQX_2$ and $\angle RQY_1$ are supplementary due to the angle sum around point $Q$, then $\angle QRY_2$ is supplementary to $\angle RQY_1$ because of the angle sum in quadrilateral $QRY_2Y_1$, and finally $\angle Z_1RP$ is supplementary to $\angle Y_2RQ$ as a result of the angle sum around point $R$ being $360^{\circ}$.

With our diagram set up, we are finally ready to show that the radiuses of the three circles are equal. Suppose that one radius is shorter than another, say $PX_1<QX_2$. Then, because the red angles in quadrilateral $PQX_2X_1$ match those of $QRY_2Y_1$, it must be the case that $QY_1<RY_2$ (since the side touching the solid red angle must be the short one). By the same logic in quadrilateral $RPZ_2Z_1$, we would have that $RZ_1<PZ_2$. But those inequalities lead to the impossible conclusion that $$\text{(radius of circle }P) < \text{(radius of circle }Q) <\text{(radius of circle }R) <\text{(radius of circle }P) .$$ Since the initial radius cannot be less than itself, it could not have been the case that one radius was shorter than another. Therefore, all radiuses are equal.

Knowing that the radiuses are equal, we proceed to Step 2. Each side of $\Delta PQR$ is made up of two radiuses, and since the radiuses are all equal, each side of $\Delta PQR$ has the same length. Therefore, $\Delta PQR$ is equilateral, and so $\Delta ABC$, which is similar to it, is equilateral as well.