A Singular Triangle

Add $CD$ perpendicular to $AF$ and label the points of tangency $P$ , $Q$ , and $R$ , as shown below:

Let $x = AG = GH = HF$ , $y = FQ$ , and $z = CF = FB$ .

By the tangent-secant theorem , $FQ^2 = HF \cdot GF = AG \cdot GF = AP^2$ , so $AP = FQ = y$ and $y^2 = x \cdot 2x = 2x^2$ .

Also, since the segments from a point to two tangent points on the same circle are equal, $CP = CQ$ , which means $CA = CP + AP = CQ + FQ = CF = z$ , and $AR = AP = y$ and $RB = QB = FQ + FB = y + z$ , which makes $AB = AR + RB = 2y + z$ .

$\triangle CDF \cong \triangle CDA$ by HL congruence, so $DF = \frac{1}{2}AF = \frac{1}{2}(AG + GH + HF) = \frac{3}{2}x$ .

From $\triangle CFD$ , $\cos \angle CFD = \cfrac{DF}{CF} = \cfrac{3x}{2z}$ , which means $\cos \angle AFB = \cos (180° - \angle CFD) = -\cos \angle CFD = -\cfrac{3x}{2z}$ .

By the law of cosines on $\triangle AFB$ , $\cos \angle AFB = \cfrac{FB^2 + AF^2 - AB^2}{2 \cdot FB \cdot AF}$ , or $-\cfrac{3x}{2z} = \cfrac{z^2 + (3x)^2 - (2y + z)^2}{2 \cdot z \cdot 3x}$ , which after substituting $y^2 = 2x^2$ from above solves to $y = \frac{4}{5}z$ .

That means the sides of the triangle have a ratio of $AC : BC : AB = z : 2z : (2y + z) = z : 2z : \frac{13}{5}z$ , and the smallest $z$ for all integer sides is $z = 5$ , making the perimeter $z + 2z + \frac{13}{5}z = 5 + 10 + 13 = \boxed{28}$ .

Because of symmetry, $\triangle CAF$ is isosceles with $CA=CF$ . Let $CA=CF=1$ , this means that $CB=2$ . Add in the inradii of length $r$ $OD$ , $OE$ , and $OI$ . Then we have:

$\begin{cases} CA: & r\cot \dfrac A2 + r \cot \dfrac C2 = 1 \\ CB: & r\cot \dfrac B2 + r \cot \dfrac C2 = 2 \\ AB: & r\cot \dfrac A2 + r \cot \dfrac B2 = c \end{cases}$

$CA+CB - AB: \quad 2 r \cot \dfrac C2 = 3-c \implies c = 3- 2r \cot \dfrac C2$ .

We note that $AG=GH=HF = \dfrac 23 \sin \dfrac C2$ . By tangent-secant theorem of circles we have $AD^2 = AG \cdot AH = \dfrac 83 \sin^2 \dfrac C2$ , $\implies AD = \dfrac {2\sqrt 2}3 \sin \dfrac C2$ . We also note that $AD+DC = \dfrac {2\sqrt 2}3 \sin \dfrac C2 + r\cot \dfrac C2 = 1$ , $\implies r\cot \dfrac C2 = 1 - \dfrac {2\sqrt 2}3 \sin \dfrac C2$ and $c = 1 + \dfrac {4\sqrt 2}3 \sin \dfrac C2$ .

By cosine rule ,

$\begin{aligned} c^2 & = 1^2 + 2^2 - 2(1)(2) \cos C \\ \left(1 + \dfrac {4\sqrt 2}3 \sin \dfrac C2\right)^2 & = 5 - 4\left(1-2 \sin^2 \dfrac C2 \right) \\ 1 + \dfrac {8\sqrt 2}3 \sin \dfrac C2 + \dfrac {32}9 \sin^2 \dfrac C2 & = 1 + 8\sin^2 \dfrac C2 \\ \dfrac {8\sqrt 2}3 \sin \dfrac C2 & = \frac {40}9 \sin^2 \dfrac C2 & \small \blue{\text{For }\sin \dfrac C2 \ne 0} \\ \implies \frac {4\sqrt 2}3 \sin \dfrac C2 & = \dfrac {8\sqrt 2}3 \times \frac 9{40} \times \frac {4\sqrt 2}3 = \frac 85 \\ \implies c & = 1 + \frac 85 = \frac {13}5 \end{aligned}$

Therefore $a:b:c = 2:1:\frac {13}5$ and the smallest integer side lengths triangle is $10:5:13$ with a perimeter of $\boxed{28}$ .

Hints: Consider the power of A and F to the incircle in different ways in terms of AB,BC,CA, then the you'll find the ratio.

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