Nagel Point Meets Incircle

The Nagel point of the triangle has barycentric coordinates $s-a:s-b:s-c$ , while the incentre has barycentric coordinates $a:b:c$ . The distance between two points $P,Q$ with normalized barycentric coordinates $p_1:p_2:p_3$ and $q_1:q_2:q_3$ (so that $p_1+p_2+p_3=q_1+q_2+q_3=1$ ) is given by the formula $PQ^2 \; = \; -a^2(p_2-q_2)(p_3-q_3) - b^2(p_1-q_1)(p_3-q_3) - c^2(p_1-q_1)(p_2-q_2)$

Thus the distance between the Nagel point and the incentre is given by $NaI^2 \; = \; - \frac{a^3 + b^3 + c^3 - 2(a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2) + 9abc}{a+b+c}$ On the other hand the inradius $r$ of $ABC$ has formula $r^2 \; = \; \frac{(s-a)(s-b)(s-c)}{s} \; = \; \frac{(b+c-a)(a+c-b)(a+b-c)}{4(a+b+c)}$ and hence we deduce that $NaI^2 - r^2 \; = \; \frac{(a+b-3c)(a+c-3b)(b+c-3a)}{4(a+b+c)}$ Since $Na$ lies on the incircle we know that $NaI = r$ , and hence $(a+b-3c)(a+c-3b)(b+c-3a) \; = \; 0$ Without loss of generality we deduce that $b+c = 3a$ . Thus we can write $b = \tfrac32a - d$ , $c = \tfrac32a + d$ for some $0 < d < \tfrac32a$ . It is clear that $c$ is then the largest of the three sides of the triangle. Since $a,b,c$ form the sides of the triangle, we must have $\tfrac32a + d = c < a + b = a + \tfrac32a - d$ , so that $d < \tfrac12a$ . Thus the triangle $ABC$ has sides $a \; < \; b = \tfrac32a-d \; < \; c = \tfrac32a + d$ for any $0 < d < \tfrac12a$ . Thus the perimeter of the triangle $ABC$ is $4a = 256$ , so we deduce that the length of its shortest side is $a = \boxed{64}$ .

Let $a = BC$ , $b = AC$ , and $c = AB$ , and let the coordinates of the vertices be $B(-\frac{1}{2}a, 0)$ , $C(\frac{1}{2}a, 0)$ , and $A(p, q)$ .

The centroid $G$ has coordinates $G\bigg(\cfrac{p - \frac{1}{2}a + \frac{1}{2}a}{3}, \cfrac{q + 0 + 0}{3}\bigg) = G\bigg(\cfrac{p}{3}, \cfrac{q}{3}\bigg)$ .

The incenter $I$ has coordinates $I\bigg(\cfrac{ap + \frac{1}{2}ab + \frac{1}{2}ac}{a + b + c}, \cfrac{aq + 0 + 0}{a + b + c}\bigg) = I\bigg(\cfrac{2ap + ab + ac}{2(a + b + c)}, \cfrac{aq}{a + b + c}\bigg)$ .

Since $BC$ lies along the $x$ -axis, the radius $r$ of the incenter is the same as the $y$ -coordinate of the incenter, so $r = \cfrac{aq}{a + b + c}$ .

The centroid, incenter, and Nagel point all lie on the Nagel line such that $3 GI = r$ , so:

$3 \sqrt{\bigg(\cfrac{2ap + ab + ac}{2(a + b + c)} - \cfrac{p}{3}\bigg)^2 + \bigg(\cfrac{aq}{a + b + c} - \cfrac{q}{3}\bigg)^2} = \cfrac{aq}{a + b + c}$

By the distance formula on $AB$ , $c^2 = (p + \frac{1}{2}a)^2 + q^2$ and by the distance formula on $AC$ , $b^2 = (p - \frac{1}{2}a)^2 + q^2$ . Solving these two equations for $p$ and $q^2$ gives $p = \cfrac{c^2 - b^2}{2a}$ and $q^2 = \cfrac{2a^2b^2 + 2a^2c^2 + 2b^2c^2 - a^4 - b^4 - c^4}{4a^2}$ , and substituting these into the equation above and simplifying we obtain:

$(3a - b - c)(3b - a - c)(3c - a - b) = 0$

Without loss of generality, let $3a - b - c = 0$ , which means $b + c = 3a$ and $c = 3a - b$ .

By the triangle inequality theorem , $c < a + b$ , and substituting $c = 3a - b$ gives $3a - b < a + b$ , which solves to $a < b$ . By a similar argument, $a < c$ , so $a$ is the shortest side.

Substituting $b + c = 3a$ into the the perimeter $P = a + b + c$ and solving gives $a = \frac{1}{4}P$ .

Therefore, for a triangle with a perimeter of $256$ and a Nagel point lying on the incenter, the shortest side is $a = \frac{1}{4}P = \frac{1}{4} \cdot 256 = \boxed{64}$ .

The Nagel Point of a triangle is the point of concurrency of its Nagel cevians, which in-turn are the segments joining a triangle's vertex to the point of contact of the excircle corresponding to that vertex with the side opposite to the same vertex. Property: Let $AA_1$ be the nagel cevian corresponding to vertex $A$ and $D'$ be the point diametrically opposite to $D$ on $\odot {I}$ , then $D' \in AA_1.$ And cyclically for rest three cases. Here, ATQ, $D' \equiv N_a.$

Using the property on the nagel cevian corresponding to vertex $C$ , one gets $\angle N_aFC= 90^{\circ}$ and some angle chase would conclude that $\angle DN_aC = \angle \frac{B}{2}$ or $\angle BCN_a= 90^{\circ} - \angle \frac{B}{2}$ .

It's easily seen now that, $\Delta BI_AC \cong \Delta CN_aB$ , which implies that Exradius $I_AA_1 = N_aD = 2\times$ inradius. $\therefore AG = 2 \times AF \Rightarrow \frac{s}{2} = 2\times AF \Rightarrow AF = \frac{s}{4} = 64.$ It's apparent now that $BC= 64.$ That it indeed is the shortest side is also trivially seen.

Just For the sake of synthetic spirit:)