I trust this message reaches you well. I am reaching out to request your assistance in solving an intriguing geometry problem that I came across in a recent competitive exam. Despite my best efforts, I have been unable to find a solution. I am eager to gain insights that will surely improve my understanding of this geometric challenge.
$ \textbf{Problem Description:} $
What is the ratio of the length of segment $OA$ to the length of segment $OC$ in a geometric figure where a circle is inscribed within a quadrilateral $ABCD$? The quadrilateral has vertices labeled as points $A, B, C,$ and $D$, with point $O$ as the center of the inscribed circle. Given the lengths of line segments $AB (4)$, $AD (9)$, $CD (12)$, and $BC (7)$, determine the value of $\frac{OA}{OC}$.
My Efforts:
I've tried various analytical approaches, but my results have been inconsistent. A more systematic strategy involving geometric properties, algebraic manipulation, or other mathematical tools seems necessary.
I made a note of this method, and then I tried to determine the values of $x$, $y$, $z$, and $p$, but it turned out that there are a countless number of solutions… I would be extremely grateful for any help or advice in figuring out the intricacies of this problem.
Thank you for your expertise and support.
Best Answer
There are infinitely many tangential quadrilaterals with the given sides. The question makes sense only if the ratio $OA/OC$ is always the same, irrespective of the shape of $ABCD$.
We can then choose a shape which is easy to compute, for instance when sides $AB$ and $BC$ lie on the same line, so that the quadrilateral becomes a triangle $ACD$ (figure below). In that case we can compute the area of the triangle via Heron's formula: $S=8\sqrt{35}$ and then the inradius $r=OB$: $$ r={S\over p}={\sqrt{35}\over2}, $$ where $p=16$ is the semiperimeter. We can then compute $OA$ and $OC$ by Pythagoras' theorem: $$ OA={3\sqrt{11}\over2},\quad OC={\sqrt{231}\over2} $$ so that: $$ {OA\over OC}={3\sqrt{11}\over\sqrt{231}}=\sqrt{3\over7}. $$