Indian National Mathematical Olympiad

Solution:

Since $E$ is the mid-point of $AC$, we have $AE = EC = b/2$. Since $BD$ bisects $\angle ABC$, we also know that $CD = ab/(a+c)$. Since $BE$ bisects $\angle ABD$, we also have
$$\frac{BD^{2}}{BA^{2}} = \frac{DE^{2}}{EA^{2}}$$
However,
$$\begin{aligned} BD^{2} & = BC^{2} + CD^{2} = a^{2} + \frac{a^{2}b^{2}}{(a+c)^{2}} \\ DE^{2} & = \left(\frac{b}{2} - \frac{ab}{a+c}\right)^{2} \end{aligned}$$

Using these in the above expression and simplifying, we get
$$a^{2}\left\{(a+c)^{2}+b^{2}\right\}=c^{2}(c-a)^{2}$$
Using $c^{2}=a^{2}+b^{2}$ and eliminating $b$, we obtain
$$c^{3}-2ac^{2}-a^{2}c-2a^{3}=0$$
Introducing $t=c/a$, this reduces to a cubic equation;
$$t^{3}-2t^{2}-t-2=0$$
Consider the function $f(t)=t^{3}-2t^{2}-t-2$ for $t>0$ (as $c/a$ is positive). For $0<t \leq 2$, we see that $f(t)=t^{2}(t-2)-t-2<0$. We also observe that $f(t)=(t-2)(t^{2}-1)-4$ is strictly increasing on $(2, \infty)$. It is easy to compute
$$f(5/2)=-\frac{11}{8}<0, \quad \text{and} \quad f(3)=4>0$$
Hence there is a unique value of $t$ in the interval $(5/2,3)$ such that $f(t)=0$. We conclude that
$$\frac{5}{2}<\frac{c}{a}<3$$

Solution:

Let us take $\angle B/4=\theta$. Then $\angle EBC=\angle DBE=\theta$ and $\angle CBD=2\theta$. Using sine rule in triangles $BEA$ and $BEC$, we get
$$\begin{aligned} \frac{BE}{\sin A} & = \frac{AE}{\sin \theta} \\ \frac{BE}{\sin 90^{\circ}} & = \frac{CE}{\sin 3\theta} \end{aligned}$$
Since $AE=CE$, we obtain $\sin 3\theta \sin A=\sin \theta$. However $A=90^{\circ}-4\theta$. Thus we get $\sin 3\theta \cos 4\theta=\sin \theta$. Note that
$$\frac{c}{a}=\frac{1}{\cos 4\theta}=\frac{\sin 3\theta}{\sin \theta}=3-4\sin^{2}\theta$$
This shows that $c/a<3$. Using $c/a=3-4\sin^{2}\theta$, it is easy to compute $\cos 2\theta=((c/a)-1)/2$. Hence
$$\frac{a}{c}=\cos 4\theta=\frac{1}{2}\left(\frac{c}{a}-1\right)^{2}-1$$
Suppose $c/a \leq 5/2$. Then $((c/a)-1)^{2} \leq 9/4$ and $a/c \geq 2/5$. Thus
$$\frac{2}{5} \leq \frac{a}{c}=\frac{1}{2}\left(\frac{c}{a}-1\right)^{2}-1 \leq \frac{9}{8}-1=\frac{1}{8}$$
which is absurd. We conclude that $c/a>5/2$.