Deciphering the Ratio Challenge: Inscribe Circle in Quadrilateral $ABCD$

Question

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.

Answer 1

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: $$ {OA\over OC}={3\sqrt{11}\over\sqrt{231}}=\sqrt{3\over7}. $$

Answer 2

I'm a week late, but here's an elegant solution using complex numbers. Taking $r$ to be the inradius of the tangential quadrilateral, consider the following representation:

Let's now imagine placing this geometry onto the $\mathbb{C}$ plane, where the incentre of the tangential quadrilateral becomes the origin and $W \to w$, $X \to x$, $Y \to y$, and $Z \to z$. It follows that $A \to 2\frac{wz}{w+z}, \ B \to 2\frac{yz}{y+z}, \ C \to 2\frac{xy}{x+y},\ D \to 2\frac{wx}{w+x}$. This is because $A,B,C,D$ are all points where two tangents of the circle intersect each other. This result can be proved by considering the condition of perpendicularity of the tangent to the line connecting it with the centre through $x, y, w, z$. A proof can be found here. With some elementary calculation it follows that $$2r^2\frac{|x-z|}{|(y+z)(y+x)|} = 7 \\ 2r^2\frac{|x-z|}{|(w+x)(w+z)|} = 9 \\ r^2\frac{|w-y|}{|(z+w)(z+y)|} = 2 \\ r^2\frac{|w-y|}{|(x+w)(x+y)|} = 6$$ by considering the given side lengths. The first two, and the last two equations on division give: $$\frac{9}{7} = \frac{|(y+z)(y+x)|}{|(w+x)(w+z)|} \\ 3 = \frac{|(z+w)(z+y)|}{|(x+w)(x+y)|}$$ Divide the first equation by the second, to immediately find that $$\sqrt{\frac{3}{7}} = \frac{|x+y|}{|w+z|} = \frac{AO}{CO}$$

We ignore the negative value since absolute value is always $\geq 0$. As a sidenote, utilizing complex numbers for geometry related to circles is helpful in reducing the difficulty of the problem.

NOTE: We are now qualified to prove a more general result, that in any tangential quadrilateral $ABCD$ with incentre $O$, then $$\frac{BC}{AD} = \frac{OB\cdot OC}{OA\cdot OD}$$ and $$\frac{CD}{AB} = \frac{OC\cdot OD}{OA \cdot OB}$$ by following the same procedure as earlier.

Answer 3

Lemma: In a tangential quadrilateral,

$$\frac{BC}{AD} = \frac{OB\cdot OC}{OA\cdot OD}, \quad \frac{CD}{AB} = \frac{OC\cdot OD}{OA \cdot OB}.$$

Corollary: $$\frac{OA}{OC} = \sqrt{ \frac{ AD \cdot AB}{BC\cdot CD} }.$$
Applying it to this problem, $ \frac{OA}{OC} = \sqrt{\frac{3}{7}}$.

Proof of lemma:
Since $\angle OBC = \angle OBA, \angle OCB = \angle OCD$, $\angle OAD = \angle OAB, \angle ODA = \angle DOC$.
So $\angle OBC + \angle OCB + \angle OAD + \angle ODA = \frac{360^\circ}{2} = 180^\circ$.
And $\angle BOC + \angle AOD = (180^\circ - \angle OBC - \angle OCB ) + (180^ \circ - \angle OAD - \angle ODA ) = 180^ \circ.$
Thus $ \sin \angle BOC = \sin \angle AOD $.

$$\frac{ BC}{AD} = \frac{ [BOC]}{[AOD]} = \frac{ OB\cdot OC \cdot \sin \angle BOC }{ OA \cdot OD \cdot \sin \angle AOD}\frac{OB\cdot OC}{OA\cdot OD} = \frac{ OB \cdot OC } { OA \cdot AD}.$$

The formula for $ \frac{ CD}{AB}$ can be shown in a similar manner.