A Pre-RMO question! -29

Let the side length of the rhombus be $AB=BC=CD=DA=a$ and the two diagonals $BD=b$ and $AC=c$ .

Circumradius of a triangle is given by $r = \dfrac {abc}{4\Delta}$ , where $a$ , $b$ , and $c$ are the side lengths of the triangle (and not the $a$ , $b$ , and $c$ of the rhombus of this problem), and $\Delta$ is the area of the triangle.

We know that the areas of $\triangle ABD$ and $\triangle ACD$ are the same and is equal to $100k$ . Then we have:

$\begin{cases} \dfrac {a^2b}{400k} = 12.5 & ...(1)\\ \dfrac {a^2c}{400k} = 25 & ...(2) \end{cases} \implies \dfrac {(2)}{(1)}: \implies \dfrac cb = 2 \implies c = 2b$ . Therefore the rhombus is as follows:

And $\left(\dfrac b2\right)^2 + b^2 = a^2 \implies b^2 = \dfrac 45 a^2$ . From Heron's formula , we have the area of $\triangle ABD$ , $\Delta = \sqrt{s(s-a)(s-a)(s-b)}$ , where $s = a + \dfrac b2$ . Then

$\begin{aligned} \frac {a^2b}{4\sqrt{\left(a+\frac b2\right)\left(\frac b2\right)^2\left(a-\frac b2\right)}} & = 12.5 \\ \frac {a^2}{\sqrt{a^2 - \blue{\frac {b^2}4}}} = \frac {a^2}{\sqrt{a^2-\blue{\frac {a^2}5}}} & = 25 & \small \blue{\text{Note that }b^2 = \frac 45 a^2} \\ \frac {\sqrt 5 a}2 & = 25 \\ \implies a & = 10 \sqrt 5 \end{aligned}$

Now $b = \dfrac 2{\sqrt 5} a = 20$ and area of the rhombus, $A = b^2 = 400 = 200k \implies k = \boxed 2$ .

Let each side of the rhombus be of length $a$ and $\angle {BAD}=α$

Then $a=50\sin (\fracα2)=25\cos (\frac α2)$

$\implies \tan (\frac α2)=\frac 12$

$\implies \sin α=\frac 45,a=10\sqrt 5$

Hence, $200k=a^2\sin α=500\times \frac 45=400\implies k=\boxed 2$ .