No Measures Given

Let $BC = 1$ and $\frac{AB}{BC} = \alpha \Rightarrow AB = \alpha$ . We see that $\triangle IAC \sim \triangle ABC \sim \triangle IBA \sim \triangle CBH.$

So $\large \frac{IB}{BA} = \frac{AB}{BC}$ $\Rightarrow IB = \alpha^2$ . Also, $\large \frac{BH}{BC} = \frac{BC}{BA}$ $\Rightarrow BH = \frac{1}{\alpha}$

From $IB$ and $BH$ we get $IH^2 = IB^2 + BH^2 = \large\alpha^4 + \frac{1}{\alpha^2}$

$AH = AB + BH = \alpha + \large\frac{1}{\alpha}$ $\Rightarrow AH^2 = \alpha^2 + \large\frac{1}{\alpha^2}$ $+ 2$

Since $\triangle IAK$ is right angled at $A$ and $H$ is the mid point of hypotenuse we get that $H$ is the circumcentre of the $\triangle IAK \Rightarrow IH = AH \Rightarrow IH^2 = AH^2$ Now putting the $IH$ and $AH$ from above we get

$\alpha^4 + \large\frac{1}{\alpha^2}$ $= \alpha^2 + \large\frac{1}{\alpha^2}$ $+2 \newline \alpha^4 - \alpha^2 - 2 = 0 \newline \Rightarrow (\alpha^2 - 2)(\alpha^2 + 1) = 0 \newline \Rightarrow \alpha^2 = 2 \newline \Rightarrow \alpha = \sqrt{2}$ .

Hence $\large \frac{AB}{BC}$ $= \sqrt{2} = 1.414$

Let $BC=1$ and $AB=x$ . Then $\dfrac {AB}{BC} = x$ . Since $\triangle ABC$ and $\triangle ABI$ are similar, $BI = x^2$ . Let $JK$ be perpendicular to line $t$ . Then $\triangle IJK$ and $\triangle IBH$ are similar. If $IH=HK$ , then $IB=BJ=x^2$ and $CJ = x^2 - 1$ . Since $\triangle ABC$ and $\triangle CJK$ are similar, $JK = x(x^2-1)$ .

By tangent-secant theorem , we have:

$\begin{aligned} BK^2 & = CK \cdot AK \\ BJ^2 + JK^2 & = CK(AC+CK) \\ x^4 + x^2(x^2-1)^2 & = \sqrt{x^2+1}(x^2-1)\left(\sqrt{x^2+1}+\sqrt{x^2+1}(x^2-1)\right) \\ x^4 + x^6-2x^4 + x^2 & = (x^2+1)(x^2-1)x^2 \\ x^6 - x^4 + x^2 & = x^6-x^2 \\ x^4 = 2x^2 \\ x & = \sqrt 2 \approx \boxed{1.414} \end{aligned}$

Draw $JK$ perpendicular to line $r$ , and let $AB = b$ and $BC = 1$ .

$\triangle IBA \sim \triangle ABC$ by AA similarity, so $\frac{IB}{AB} = \frac{AB}{BC}$ , or $\frac{IB}{b} = \frac{b}{1}$ , which solves to $IB = b^2$ .

$\triangle CBH \sim \triangle ABC$ by AA similarity, so $\frac{BH}{BC} = \frac{BC}{AB}$ , or $\frac{BH}{1} = \frac{1}{b}$ , which solves to $BH = \frac{1}{b}$ .

Since $\angle ABC$ is a right angle, $AC$ is the diameter of circle $\gamma$ , and its diameter is at the midpoint of $AC$ which is $(\frac{1}{2}, \frac{1}{2}b)$ . The slope of the radius through $B$ is therefore $b$ , which means $BK$ has an equation of $y = -\frac{1}{b}x$ . $AK$ has an equation of $y = -bx + b$ , so $K$ has coordinates $(\frac{b^2}{b^2 - 1}, -\frac{b}{b^2 - 1})$ . Since $IB = b^2$ and $BH = \frac{1}{b}$ , $I$ has coordinates $(-b^2, 0)$ and $H$ has coordinates $(0, -\frac{1}{b})$ , and both the slopes of $IH$ and $HK$ calculate to $-\frac{1}{b^3}$ . Therefore, $I$ , $H$ , and $K$ are collinear.

$\triangle HJK \cong \triangle HBI$ by AAS congruence, so $HJ = BH = \frac{1}{b}$ and $JK = IB = b^2$ .

$\triangle AJK \sim \triangle ABC$ by AA similarity, so $\frac{AJ}{JK} = \frac{AB}{BC}$ , or $\frac{b + \frac{1}{b} + \frac{1}{b}}{b^2} = \frac{b}{1}$ , which rearranges to $(b^2 + 1)(b^2 - 2) = 0$ , and solves to $b = \sqrt{2}$ for positive real $b$ .

Therefore, the ratio $\frac{AB}{BC} = \frac{b}{1} = \sqrt{2} \approx \boxed{1.414}$ .

I solved this using similar triangle proportionality. Since we know that IH = HK then we know that triangle AIK has double the side lengths of triangle CHK. That makes AI = 2(CH). The tangent lines passing through line S are parallel making triangle AIB similar to triangle HCB with a scale factor of 2. Both of these triangles are similar to triangle ABC by angle-angle similarity. We can establish the proportional relationship as $\frac{IB}{AB}$ = $\frac{AB}{BC}$ . After cross multiplying we get that AB^2 = IB(BC). Because triangle AIB is twice the size of HCB, IB = 2(BC). By substitution AB^2 = 2BC^2. That makes AB/BC = root 2 or 1.414.

First thing first, $I, H, K$ are collinear for any $\triangle ABC$ . It is a direct consequence of the Pascal's theorem applied to the degenerate hexagon $(AABBCC).$ Maybe inversion provides better proof anyway.

To the prob. Since, $AI, CH$ both are tangents to the circle and $AC$ is a diameter so, we have $AI || CH \Rightarrow AC= CK$ Let $N$ denote the midpoint of $AB$ then $CN || KH \Rightarrow \angle BCN = \angle KBC = \angle A \Rightarrow BC^{2} = BN \times BA = \dfrac{1}{2} AB^2$ $\Rightarrow \dfrac{AB}{BC}= \sqrt{2} = 1.4144 \square$

Let KB and AI intersect at D.Then using Newton-gauss lemma on the complete quadrilateral ABCD we find that,the midpoint of CD lie on AB since H is the midpoint of IK.Let that midpoint be X.Now,using some trig we get BD= $\frac{ACsin(C)}{2cos(C)}$ . But again from trinagle BCD we get , $\frac{CX}{DX}=\frac{BDsinC}{ACcosC}=1$ . Then, using these two eqaution we get,cosC= $\frac{1}{\sqrt{3}}$ .The rest is simple.

Let the position coordinates of $A, B$ and $C$ be $(0,c), (0,0)$ and $(a, 0)$ respectively. Then the equation of the circle is $x^2+y^2-ax-cy=0$ . Slope of the tangent to this circle at any point $(h, k)$ is $\dfrac{2h-a}{c-2k}$ . Using this, the equation of $\overline {AI}$ is $y=\dfrac{ax+c^2}{c}$ , and the position coordinates of $I$ are $(-\dfrac{c^2}{a},0)$ . Similarly, the position coordinates of $H$ and $K$ are $(0,-\dfrac{a^2}{c})$ and $(\dfrac{ac^2}{c^2-a^2}, \dfrac{-a^2c}{c^2-a^2})$ respectively. Hence the condition $|\overline {IH}|=|\overline {HK}|$ yields $c^2=2a^2\implies \dfrac{c}{a}=\sqrt 2\approx \boxed {1.414}$

Let O be the centre of circle that will be on side AC. Let OA = OB = OC = R Side CH and AI both will be parallel. So, (KH/KI = KC/AK = 1/2) So, (AK = 2KC) So, (AK = 2(AK-AC) So, (2AC = AK) As, (AC = 2R) So, (AK= 4R) Hence, KC = 2R. Let angle ACB = X So, Angle OBC = OCB = X Hence Angle BOC = 180-2X In Triangle OBK Cos(180-2X) = R/3R = 1/3 -Cos(2X) = 1/3 Cos(X) = 1/1.732 Tan(X) = 1.414 Also, Tan(X) = AB/BC = 1.414