Triangle Perimeter

Note that $\Delta XYZ$ is the set of points in the interior of $\Delta ABC$ that are at least 2 units away from the sides of $\Delta ABC$ . In obvious sense, the incenters of the 2 triangles are the coinciding, and so the homothety centered at the common incenter W mapping $\Delta ABC$ to $\Delta XYZ$ has similitude ratio equivalent to the ratio between their inradii. Let $r$ and $s$ be the inradii of $\Delta ABC$ and $\Delta XYZ$ , respectively. Now, $r = \frac{2 \cdot [\Delta ABC]}{AB+BC+CA} = \frac{2 \cdot \frac{14 \cdot 12}{2}}{13+14+15} = 4$ So, $s = r-2 = 2$ and $Perimeter_{XYZ}=\frac{2 \cdot Perimeter_{ABC}}{4}=\boxed{21}$

Done in collaboration with Ahaan Rungta (diagram will be added in the comments soon):

Let $\alpha = \angle BAC$ and $\beta = \angle ABC$ and $\gamma = \angle ACB$ . Then, by the Law of Cosines, we get: $\begin{aligned} \cos \alpha &=& \frac {33}{65}, \\ \cos \beta &=& \frac {5}{13}, \\ \cos \gamma &=& \frac {3}{5}. \end{aligned}$ Now, drop perpendiculars from $X$ , $Y$ , and $Z$ to the corresponding sides of tangency. Let $M$ be the foot of the perpendicular from $Z$ to $BC$ . Let $N$ be the foot of the perpendicular from $Y$ to $BC$ . Let $O$ be the foot of the perpendicular from $X$ to $AC$ .

Let $MC = z$ . Let $NB = y$ . Let $OA = x$ . Then: $\begin{aligned} \tan \left( \dfrac {\alpha}{2} \right) &=& \dfrac {2}{x}, \\ \tan \left( \dfrac {\beta}{2} \right) &=& \dfrac {2}{y}, \\ \tan \left( \dfrac {\gamma}{2} \right) &=& \dfrac {2}{z}. \end{aligned}$ Then, we get $\left( x, y, z \right) = \left( \dfrac {7}{2}, 3, 4 \right).$ Since $M$ and $N$ are both foots of perpendiculars to $BC$ and the distances to the respective centers of the circles are equal to $2$ , we know that $MNZY$ is a rectangle. The same holds for the other two quadrilaterals.

Thus, $XY + YZ + ZX = 13 - \dfrac {7}{2} - 3 + 14 - 3 - 4 + 15 - \dfrac {7}{2} - 4 = \boxed {21}.$

We take $AB = 13, BC=14 , AC = 15$ . Let the circle centered at $X$ touch $BC$ at $X_1$ and AB at $X_2$ with $XX_1 = XX_2=2$ . Circle centered at $Y$ is tangent to $AC$ at $Y_2$ and $BC$ at $Y_1$ and the circle centered at $Z$ is tangential to $AB$ at $Z_2$ and $AC$ at $Z_1$ . We observe that $BX , CY , AZ$ are the internal angle bisectors of $\angle A , \angle B , \angle C$ respectively. Thus they are concurrent at $I$ , the incentre of $\triangle ABC$ .

Now, we find $r$ , the inradius of $\triangle ABC$ which is given by the formula

$r = \dfrac{\Delta}{s}$ = $\dfrac{\sqrt{s(s-a)(s-b)(s-c)}}{s}$ with symbols having their usual significance.

We get $r=4$ .

It is easy to notice that the sides $XY$ , $YZ$ , $ZX$ of $\triangle ABC$ are parallel to the sides $AB$ , $BC$ and $CA$ of $\triangle ABC$ and $ZXX_2Z_2$ , $XYY_1X_1$ and $YZZ_1Y_2$ are rectangles.

Now, let us consider $\triangle IBC$ . We have $XY \parallel BC$ . Therefore, $\triangle IXY \sim \triangle IBC$

$\implies$ $\dfrac{XY}{BC}$ = $\dfrac{4-2}{4}$ = $\dfrac{1}{2}$ . We conclude that $XY = \dfrac{1}{2} \times BC$ .

Similarly , we get $YZ = \dfrac{1}{2} \times AC$ , $XZ = \dfrac{1}{2} \times BC$

$\implies$ $\dfrac{XY + YZ + XZ}{AB + BC + AC}$ = $\dfrac{1}{2}$

$\implies$ $\dfrac{XY + YZ + ZX}{13 + 14 + 15}$ = $\dfrac{1}{2}$

$\implies$ $XY + YZ + XZ = \boxed{21}$

Done in collaboration with Sreejato Bhattacharya, who will draw a diagram for this later:

Let $\alpha = \angle BAC$ and $\beta = \angle ABC$ and $\gamma = \angle ACB$ . Then, by the Law of Cosines, we get: $\begin{aligned} \cos \alpha &=& \frac {33}{65}, \\ \cos \beta &=& \frac {5}{13}, \\ \cos \gamma &=& \frac {3}{5}. \end{aligned}$ Now, drop perpendiculars from $X$ , $Y$ , and $Z$ to the corresponding sides of tangency. Let $M$ be the foot of the perpendicular from $Z$ to $BC$ . Let $N$ be the foot of the perpendicular from $Y$ to $BC$ . Let $O$ be the foot of the perpendicular from $X$ to $AC$ .

Let $MC = z$ . Let $NB = y$ . Let $OA = x$ . Then: $\begin{aligned} \tan \left( \dfrac {\alpha}{2} \right) &=& \dfrac {2}{x}, \\ \tan \left( \dfrac {\beta}{2} \right) &=& \dfrac {2}{y}, \\ \tan \left( \dfrac {\gamma}{2} \right) &=& \dfrac {2}{z}. \end{aligned}$ Then, we get $\left( x, y, z \right) = \left( \dfrac {7}{2}, 3, 4 \right).$ Since $M$ and $N$ are both foots of perpendiculars to $BC$ and the distances to the respective centers of the circles are equal to $2$ , we know that $MNZY$ is a rectangle. The same holds for the other two quadrilaterals.

Thus, $XY + YZ + ZX = 13 - \dfrac {7}{2} - 3 + 14 - 3 - 4 + 15 - \dfrac {7}{2} - 4 = \boxed {21}.$

Instead of last but fourth line "So XY+YZ+ZX=......... Proceed as under.
Find the values of BP and QC.
YZ=BC-BP-QC=7.
Each side of ABC is external tangent to two equal circles. So they are || to line of centers. Thus ABC and XYZ are similar.
So Per XYZ:Per ABC: YZ:BC:7:14
It is long but a different approach.



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If we find the in-radius of the larger triangle, it comes out to be 4. Now, the in-center of the larger triangle and the inner triangle are same. The in-radius of the inner triangle = 4-2 =2

Also, the two triangles are similar. So, the ratio of the perimeter = ratio of the in-radius = 2:1

As the perimeter of the larger triangle = 42 perimeter of the inner triangle = 21