Extreme Ratios

If we reflect point $B$ in the line $\mathcal{L}$ , we obtain $B'$ where $A$ , $O$ and $B'$ are collinear, and the ratios $\frac{AX}{XB}$ and $\frac{AX}{XB'}$ are equal. The locus of the set of points $X$ for which $\frac{AX}{XB'}$ is a given constant $k$ is a circle in which $A$ and $B'$ are mutually inverse points (except when $k=1$ , in which case the locus is the perpendicular bisector of $AB'$ ). Thus, to extremize $x = \frac{AX}{XB} = \frac{AX}{XB'}$ , we need to find the two such circles which are tangent to the given line $\mathcal{L}$ . Let $\theta$ be the common angle of incidence of $AO$ and $BO$ to the line $\mathcal{L}$ .

The first of these circles has centre $C_1$ on the extended line $AB'$ , and has radius $R$ where $\begin{aligned} (R\csc\theta - a)(R\csc\theta + b) & = \; R^2 \\ R^2\cot^2\theta -(a-b)R\csc\theta - ab & = \; 0 \end{aligned}$ while the second circle has centre $C_2$ on the extended line $AB'$ and radius $S$ where $\begin{aligned} (S\csc\theta + a)(S\csc\theta - b) & = \; S^2 \\ S^2\cot^2\theta +(a-b)R\csc\theta - ab & = \; 0 \end{aligned}$ and hence $R \; = \; \sec\theta\big[X + \tfrac12(a-b)\tan\theta\big] \hspace{2cm} S \; = \; \sec\theta\big[X - \tfrac12(a-b)\tan\theta\big]$ where $X \; =\; \tfrac12\sqrt{(a-b)^2\tan^2\theta + 4ab\sin^2\theta}$ so that $RS \; = \; ab\tan^2\theta \hspace{2cm} R-S \; = \; (a-b)\tan\theta\,\sec\theta$ The extreme values of $x$ are obtained when the point $X$ is at either $X_1$ or $X_2$ , the points of tangency of these circles with the line $\mathcal{L}$ , and hence $x_1 \; = \; \frac{R\csc\theta - a}{R} \hspace{2cm} x_2 \; =\; \frac{S}{S\csc\theta-b} \; =\; \frac{S\csc\theta + a}{S}$ so that $x_1x_2 \; =\; \frac{(R\csc\theta - a)(S\csc\theta + a)}{RS} \; = \; \frac{RS\csc^2\theta + a(R-S)\csc\theta - a^2}{RS} \; = \; \boxed{\frac{a}{b}}$

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Now let's work the problem the other way. Suppose that $X_1,X_2$ are the excentres of the triangle $OAB$ opposite $B$ and $A$ respectively. Then $X_1$ , $O$ and $X_2$ are collinear, all lying on the line $\mathcal{L}$ . Let $C_1$ and $C_2$ be the intersections with the line $AOB'$ with the lines perpendicular to $\mathcal{L}$ passing through $X_1$ and $X_2$ respectively.

Now $X_2$ is on the angle bisector of $\angle OAB$ , so $\angle OAX_2 = \angle BAX_2 = \alpha$ . Moreover $X_1$ is on the angle bisector of $\angle OBA$ , so $\angle OBX_1 = \angle ABX_1 = \beta$ . Moreover $\angle OAX_1 = 90^\circ - \alpha$ and $\angle OBX_2 = 90^\circ - \beta$ , so we deduce that $\angle X_1AX_2 = \angle X_1BX_2 = 90^\circ$ . Thus we deduce that $X_1,X_2,A,B$ are concyclic, with the centre of the circumcircle lying on the line $\mathcal{L}$ , which means that $B'$ also lies on this circle. Thus we deduce that $\angle BX_1X_2 = \angle X_2X_1B' = \alpha$ and $X_1B'A = X_1X_2A = \beta$ , while $\angle B'OX_2 = \alpha +\beta$ .

Since $\angle AX_1B = 90^\circ - \alpha - \beta$ , we deduce that $\angle C_1X_1A = \beta$ . This means that triangles $C_1AX_1$ and $C_1X_1B'$ are similar, which implies that $C_1A \times C_1B' = C_1X_1^2$ , and hence that $A,B'$ are mutually inverse in the circle with centre $C_1$ that passes through $X_1$ (which is therefore tangent to $\mathcal{L}$ .

Since $\angle OB'X_2 = 90^\circ - \beta$ , we see that $\angle OX_2B' = 90^\circ - \alpha$ , so that $\angle B'X_2C_2 = \alpha$ . This means that the triangles $C_2B'X_2$ and $C_2X_2A$ are similar, which implies that $C_2A \times C_2B' = C_2X_2^2$ , and hence that $A,B'$ are mutually inverse in the circle with centre $C_2$ that passes through $X_2$ (which is therefore tangent to $\mathcal{L}$ ). Thus the points $X_1,X_2,C_1,C_2$ are the same as the ones previously defined. The Sine Rule now gives $\frac{AX_1}{\sin\beta} = \frac{BX_1}{\sin(90^\circ+\alpha)} = \frac{BX_1}{\cos\alpha} \hspace{2cm} \frac{AX_2}{\cos\beta} = \frac{AX_2}{\sin(90^\circ+\beta)} = \frac{BX_2}{\sin\alpha}$ and hence $x_1 \; = \; \frac{AX_1}{BX_1} = \frac{\sin\beta}{\cos\alpha} \hspace{2cm} x_2 \; = \; \frac{AX_2}{BX_2} = \frac{\cos\beta}{\sin\alpha}$ so that $x_1x_2 \; = \; \frac{\sin\beta\cos\beta}{\sin\alpha\cos\alpha} \; =\; \frac{\sin2\beta}{\sin2\alpha} \; = \; \frac{OA}{OB} \; =\; \frac{a}{b}$

Denote $\angle OAB$ , $\angle OBA$ , by $\angle A$ and $\angle B$ respectively.
Denote the length of $AB$ by $c$ , the directed length of $OX$ by $t$ and the measures of the noted angles by $\theta$ and $\phi$ (figure 1).

figure 1 By cosine rule on triangles $\triangle OAX$ , $\triangle OBX$ we have
$A{{X}^{2}}={{t}^{2}}-2at\cos \theta +{{a}^{2}}$ , $B{{X}^{2}}={{t}^{2}}-2bt\cos \left( \pi -\theta \right)+{{b}^{2}}\Rightarrow B{{X}^{2}}={{t}^{2}}+2bt\cos \theta +{{b}^{2}}$ .

It is sufficient to find the extrema of the function $f\left( t \right):=\dfrac{A{{X}^{2}}}{B{{X}^{2}}}=\dfrac{{{t}^{2}}-2at\cos \theta +{{a}^{2}}}{{{t}^{2}}+2bt\cos \theta +{{b}^{2}}}$ .
The derivative of $f$ is given by ${f}'\left( t \right)=\dfrac{2\left( a+b \right)\left[ \cos \theta \cdot {{t}^{2}}-\left( a-b \right)\cdot t-ab\cos \theta \right]}{{{\left( {{t}^{2}}+2bt\cos \theta +{{b}^{2}} \right)}^{2}}}$ The zeros of ${f}'$ are the roots of the quadratic $\cos \theta \cdot {{t}^{2}}-\left( a-b \right)\cdot t-ab\cos \theta$ , who’s discriminant is
$\begin{aligned} \Delta & ={{\left( a-b \right)}^{2}}+4ab{{\cos }^{2}}\theta \\ & ={{a}^{2}}+{{b}^{2}}+2ab\left( 2{{\cos }^{2}}\theta -1 \right) \\ & ={{a}^{2}}+{{b}^{2}}+2ab\cos \left( 2\theta \right) \\ & ={{a}^{2}}+{{b}^{2}}+2ab\cos \left( \pi -\varphi \right) \\ & ={{a}^{2}}+{{b}^{2}}-2ab\cos \varphi \\ & ={{c}^{2}} \\ \end{aligned}$ Hence, $f$ is extremal when $t=\dfrac{a-b+c}{2\cos \theta }=\dfrac{s-b}{\cos \theta }\text{:}={{t}_{1}} \ \ \ \ \ or \ \ \ \ \ t=\dfrac{a-b-c}{2\cos \theta }=-\dfrac{s-a}{\cos \theta }\text{:}={{t}_{2}} \ \ \ \ \ \ \ (1)$ where $s=\dfrac{a+b+c}{2}$ is the semiperimeter of $\triangle OAB$ .

figure 2

Now, consider the excircle $\left( {{I}_{B}},{{I}_{B}}{{Y}_{1}} \right)$ of $\triangle OAB$ tangent to the sides $BO$ , $BA$ at ${{Y}_{1}}$ and ${{Y}_{2}}$ respectively (figure 2).
Then, $B{{Y}_{1}}=B{{Y}_{2}}=\dfrac{1}{2}\left( a+b+c \right)=s$ Thus, $\left| \overline{O{{Y}_{1}}} \right|=s-b={{t}_{1}}\cos \theta$ Likewise, if the excircle $\left( {{I}_{A}},{{I}_{A}}{{Z}_{1}} \right)$ on $OB$ is tangent to the sides $AO$ , $AB$ at ${{Z}_{1}}$ and ${{Z}_{2}}$ respectively, then $\left| \overline{O{{Z}_{1}}} \right|=s-a=-{{t}_{2}}\cos \theta$

Therefore, according to $(1)$ , the extremal values of $f$ occur when the point $X$ coinsides with the excenters ${{I}_{B}}$ and ${{I}_{A}}$ , whose projections on the lines $BO$ and $AO$ are ${{Y}_{1}}$ and ${{Z}_{1}}$ respectively.

Now, by sine rule on triangles $\triangle {{I}_{A}}AB$ and $\triangle {{I}_{B}}AB$ , it is easy to see that $\dfrac{{{I}_{B}}A}{{{I}_{B}}B}=\dfrac{\sin \dfrac{B}{2}}{\sin \left( A+\dfrac{\pi -A}{2} \right)}=\dfrac{\sin \dfrac{B}{2}}{\cos \dfrac{A}{2}} \ \ \ \ \ and \ \ \ \ \ \dfrac{{{I}_{A}}A}{{{I}_{A}}B}=\dfrac{\sin \left( \pi -\dfrac{\pi -B}{2} \right)}{\sin \dfrac{A}{2}}=\dfrac{\cos \dfrac{B}{2}}{\sin \dfrac{A}{2}}$

Consequently, the required product of the minimum and the maximum value of the ratio $x=\dfrac{XA}{XB}$ is $\dfrac{{{I}_{B}}A}{{{I}_{B}}B}\cdot \dfrac{{{I}_{A}}A}{{{I}_{A}}B}=\dfrac{\sin \dfrac{B}{2}}{\cos \dfrac{A}{2}}\cdot \dfrac{\cos \dfrac{B}{2}}{\sin \dfrac{A}{2}}=\dfrac{\sin B}{\sin A}=\boxed{\dfrac{a}{b}}.$

Let $p = OX$ , $\theta$ be one of the blue angles, and $c = \cos \theta$ .

By the law of cosines on $\triangle OXA$ , $AX = \sqrt{p^2 + a^2 - 2ap \cos \theta} = \sqrt{p^2 + a^2 - 2apc}$ , and by the law of cosines on $\triangle OXB$ , $BX = \sqrt{p^2 + b^2 - 2bp \cos (180° - \theta)} = \sqrt{p^2 + b^2 + 2bpc}$ , so $x = \frac{AX}{BX} = \frac{\sqrt{p^2 + a^2 - 2apc}}{\sqrt{p^2 + b^2 + 2bpc}}$ .

For the maximum and minimum values of $x$ , $\frac{dx}{dp} = \frac{(a + b)(cp^2 - (a - b)p - abc)}{\sqrt{(p^2 - 2acp + a^2)(p^2 + 2bp + b^2)^3}} = 0$ , which solves to $p = \frac{a - b \pm \sqrt{(a - b)^2 + 4abc^2}}{2c}$ .

Substituting these $p$ values back into $x$ gives maximum and minimum values of $x = \sqrt{\frac{a^2 + 2abc^2 + b^2 \pm (a + b)\sqrt{(a - b)^2 + 4abc^2}}{2b^2(1 - c^2)}}$ .

The product $P$ of the maximum and minimum values of $x$ is therefore:

$P = \sqrt{\frac{a^2 + 2abc^2 + b^2 + (a + b)\sqrt{(a - b)^2 + 4abc^2}}{2b^2(1 - c^2)}} \cdot \sqrt{\frac{a^2 + 2abc^2 + b^2 - (a + b)\sqrt{(a - b)^2 + 4abc^2}}{2b^2(1 - c^2)}} = \frac{\sqrt{(a^2 + 2abc^2 + b^2)^2 - (a + b)^2((a - b)^2 + 4abc^2)}}{2b^2(1 - c^2)} = \frac{2ab(1 - c^2)}{2b^2(1 - c^2)} = \boxed{\frac{a}{b}}$ .