We have drawn the inscribed and excircles of a triangle, as well as the six other lines, each of which touches two of these circles, but not the sides of the triangle. In how many ways can we select from these six lines

a) the side lines of a trapezoid,

b) the side lines of a parallelogram, if no two sides of the original triangle are equal in length?
1. Let the center of the incircle $k$ of triangle $ABC$ be $O$, the center of the excircle $k_a$ (which is tangent to the extension of sides $AB$ and $AC$ and the side $BC$ externally) be $O_a$, and similarly, the centers of the excircles $k_b$ and $k_c$ (tangent to sides $CA$ and $AB$ respectively) be $O_b$ and $O_c$. Any two non-coincident circles, such as any pair from $k, k_a, k_b, k_c$, have four common tangents: two external and two internal, and the pairs are reflections of each other across the single common axis of symmetry of the two circles, which is the line connecting their centers, known as the central line. In other words: the central line bisects both the angle between the common external tangents and the angle between the common internal tangents.

For any pair of our circles, three common tangents are the three sides of the triangle, so we need to draw one more common tangent for each pair. Consider, for example, the pair of circles $k$ and $k_c$. For these, the sides $CA$ and $CB$ are the common external tangents, the central line $OO_c$ is the angle bisector $f_c$ of angle $BCA$, so the fourth common tangent is the reflection of the line $AB$ across $f_c$, the line $A_cB_c = e_c$, where $A_c$ is on the line $CB$ and $B_c$ is on the line $CA$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_c94505aa295d06c02352g-1.jpg?height=949&width=696&top_left_y=596&top_left_x=693)

Figure 1

Similarly, for the pair of circles $k_a$ and $k_b$, $CA$ and $CB$ are the common internal tangents, the central line $O_aO_b$ is the bisector $f_c'$ of the external angles at vertex $C$, so the fourth common tangent is again the reflection of the line $AB$, this time across $f_c'$, the line $A_c'B_c' = g_c$, where $A_c'$ is on the line $CB$ and $B_c'$ is on the line $CA$.

We claim that the $180^\circ$ rotation around point $C$ (or reflection across point $C$) maps $e_c$ to $g_c$, so they are parallel to each other. Indeed, $A_c$ and $A_c'$ are on the line $CB$ and due to the two reflections, $A_c'C = AC = A_cC$, and similarly, $B_c'C = B_cC$, which proves our claim.

Similarly, the fourth tangents $e_a$ and $g_a$ of the pairs $k, k_a$ and $k_b, k_c$, and the fourth tangents $e_b$ and $g_b$ of the pairs $k, k_b$ and $k_a, k_c$ are also parallel. (We cannot get more fourth tangents, as we have already paired $k$ with each of the other three circles and always included the two remaining circles.)

2. Since the problem is looking for the sides of trapezoids and parallelograms among the 6 drawn lines, we need to check if any additional parallelism can occur between one line from each of the 3 pairs, such as between $e_a, e_b, e_c$.

Choose the labeling of the triangle such that the angles, using the usual notation, satisfy $\alpha < \beta < \gamma$, or $\beta < 90^\circ$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_c94505aa295d06c02352g-2.jpg?height=849&width=729&top_left_y=123&top_left_x=687)

Then, by reflection and the theorem of external angles, the lines $e_a, e_b, e_c$ from the direction of $BA$ are respectively

$$
\gamma, \quad 180^\circ - \gamma, \quad \beta - \alpha
$$

oriented. Thus, $e_a \| e_b$ if $\gamma = 90^\circ$; however, $e_a \| e_c$ is impossible because $\beta - \alpha = \gamma = 180^\circ - \beta - \alpha$ implies $\beta = 90^\circ$; finally, $e_b \| e_c$ is also impossible because it would imply $\beta - \alpha = 180^\circ - \gamma = \beta + \alpha$, or $\alpha = 0^\circ$. Therefore, in the case of a right triangle, 4 of the 6 lines are parallel, and these are distinct (since, for example, $C_a$ and $C_b$ coinciding would mean $CA + CB = BC$, which is impossible, Figure 3).

![](https://cdn.mathpix.com/cropped/2024_05_02_c94505aa295d06c02352g-2.jpg?height=749&width=661&top_left_y=1390&top_left_x=713)

Figure 3

If the triangle has no right angle, the directions of the parallel tangents are different.

3. Returning to the question, if triangle $ABC$ is not a right triangle, among the six tangent lines, any four chosen will always have two parallel lines, but not more than two. Thus, a quadrilateral is formed, and it is always a trapezoid. How many ways can we choose the four lines? As many ways as we can "omit" the remaining two lines: $\frac{6 \cdot 5}{2} = 15$ (we divide by 2 due to symmetry).

Therefore, 15 trapezoids are formed. These can be convex, crossed (where the non-parallel sides intersect between the two parallel lines), or degenerate, and also parallelograms if both pairs of their sides are parallel. We get a parallelogram when the four chosen lines are pairwise parallel, i.e., we leave out a complete pair from the three pairs. This is 3 out of the 15 possibilities.

If the triangle is a right triangle, we can only choose 2 out of the 4 tangents perpendicular to the hypotenuse, so the two tangents in the other direction must always be among the sides of the quadrilateral. We can choose 2 out of 4 in 6 ways (as we saw above for the circles), so here 6 selections are possible, and they always form a parallelogram.

Note. We did not explicitly mention in the solution the constraint of the problem that no two sides of the triangle are equal in length. Nevertheless, we used this fact when we stated that $e_c$ is different from the side $AB = c$. Indeed, $e_c$ is only identical to $c$ if the angle bisector $f_c$ is perpendicular to $c$, i.e., if $CA = CB$. By excluding such cases, we preempted the examination of situations where the number of drawn tangent lines is less than 6.