# Extended General Malfatti's Problem
## 3 Definition and Theorem for the Tri(T\({}_{n}\)) Problem
### The Tri(T\({}_{2}\)) Problem  
We aim to find three circles inside a triangle that are tangent to each other pairwise, and every circle is tangent to two sides of the triangle. There are formulae to solve this problem. These formulae were discovered by [PERSON] and published in 1811 [8]. **Theorem 2** ([3,8]). : _The radii of the Malfatti circles are_  
\[\mathrm{r}_{1} = \frac{r}{2(s-a)}(s-r+d-e-f) \tag{1}\] \[\mathrm{r}_{2} = \frac{r}{2(s-b)}(s-r-d+e-f)\] (2) \[\mathrm{r}_{3} = \frac{r}{2(s-c)}(s-r-d-e+f) \tag{3}\]  
_where \(s=(a+b+c)/2\), \(r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}\) is the radius of the inscribed circle, and \(d\), \(e\), and \(f\) is the distance from the center of the inscribed circle to \(A\), \(B\), and \(C\), respectively._  
The values \(\mathrm{r}_{1}\), \(\mathrm{r}_{2}\), and \(\mathrm{r}_{3}\) are the radii of the circles near vertices A, B, and C, respectively. Notice that when we find the radii of these three circles (for example, \(\mathrm{r}_{1}\)), the center of these circles is easy to find from the location of the vertex (A), angle bisectors (of \(\angle\)A), and radius (\(\mathrm{r}_{1}\)). We provide a simple example for finding the inscribed circle and [PERSON]'s circle. **Example 1**. : _Consider a triangle whose vertices are \(A=(-8,-7)\), \(B=(6,-2)\), and \(C=(9,8)\). From theorem 1, we find the inscribed center (x,y) = (3.88, 0.01), with a radius of 2.61, as shown in Figure 3(a). The centers of three Malfatti circles located at (1.99, \(-1.01\)), (5.18, \(-1.22\)), and (5.03, 1.80) with radii 2.29, 1.01, and 2.02, respectively, are shown in Figure 3(b)._  
Using a = b = c or a = b, we establish the following corollaries (proof by Theorem 2):  
**Corollary 3**. : _The radii of the three Malfatti circles for the equilateral triangle with side length a have the same value \(r_{1}=r_{2}=r_{3}=\frac{3-\sqrt{3}}{2}\ r=\frac{\sqrt{3}-1}{4}a\)._  
**Corollary 4**. : _Consider an isosceles triangle whose leg and base lengths are \(a\) and \(c\), respectively. Its inscribed circle is centered at (x,y) = \([a(A+B)+cC]/(2a+c)\) with radius \(r=\frac{c}{2}w\), where \(w=\sqrt{\frac{2a-c}{2a+c}}\). The three Malfatti circles for this triangle have radii \(r_{1}=r_{2}=(2a+c)w(1-w)/4\ r_{3}=c\{\frac{1}{4w}+\frac{w^{2}}{4}-\frac{c}{2a +c}\sqrt{\frac{a}{2a-c}}\}\)._  
Note that the variables r, d, e, f can be represented by the constant w where r = \(\frac{c}{2}\)w, d = e = \(\sqrt{\frac{a}{2a-c}}\)cw, f = aw, and when a = c, the result for Corollary 4 is the same as that for Corollary 3. When the angle opposite to the base is \(2\theta\), we have \(\sin(\theta)=c/2\)a. Assume that T = \(\tan(\frac{\theta}{2})\), Corollary 4 can be rewritten as  
**Corollary 5**. : _Consider an isosceles triangle whose leg and base lengths are \(a\) and \(c\), respectively. The angle opposite to the side with length c is \(2\theta\). The three Malfatti circles for this triangle have radii \(r_{1}=r_{2}=s\frac{2T(1-T)}{1+T^{2}}\), \(r_{3}=\frac{c}{2(1+T)^{2}(1-T)}\Big{(}3T^{2}-2\sqrt{2}T\sqrt{1+T^{2}}+1\Big{)}\) where \(s=(2a+c)/2\), \(T=tan(\frac{\theta}{2})\)._  
Let us give an example of finding the Malfatti circles of an isosceles triangle and observe the result. **Example 2**. : _Consider an isosceles triangle whose leg and base lengths are \(l=10\) and \(b=12\), respectively. The radii for the Malfatti circles are \(r_{1}=r_{2}=2.0\), \(r_{3}=1.718\), as shown in Figure 4(a). Let (l,b) = (10, 8), the radii for the Malfatti circles are \(r_{1}=r_{2}=1.583\), \(r_{3}=1.827\), as shown in Figure 4(b). We observe that when the length of leg \(l=\sqrt{\left(\frac{b}{2}\right)^{2}+h^{2}}<b\) (as in Figure 4(a)), \(r_{1}=r_{2}<r_{3}\). Additionally, when \(l>b\), \(r_{1}=r_{2}>r_{3}\). It is easy to guess that when \(l=b\), the isosceles triangle becomes an equilateral triangle, so that \(r_{1}=r_{2}=r_{3}\). Let \(l=b=10\), \(r_{1}=r_{2}=r_{3}=1.830\)._  
Figure 4: Solution for the Tri(T\({}_{1}\)) and Tri(T\({}_{2}\)) problems. (a) Inscribed circle. (b) Malfatti circle.