I came across this problem as I was preparing for an olympiad and it totally stumped me out. Can anyone please help me to solve it... Here it is:
In a quadrilateral $ABCD$, it is given that $AB = AD = 13$, $BC = CD = 20$, $BD = 24$. If $r$ is the radius of a circle inscribed in the quadrilateral, then what is the integer closest to $r$?
Thanks in advance.
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$\begingroup$Note that the triangles $\Delta ABD$ and $\Delta CBD$ are isosceles in $A$ and $C$. This implies that the side bisector of the given diagonal $BD$ is also the median, and angle bisector in the two triangles. It follows that the given quadrilateral has perpendicular diagonals $AC$, $BD$, they intersect in $O$, the mid point of the segment $BD$. Now we consider the triangles $\Delta AOB$ and $\Delta COB$, both with a right angle in $O$.
The first triangle has the known sides $13$, $12$, so $AO=5$.
The second triangle has the known sides $20$, $12$, so $CO=16$.
So the diagonals of the triangle are $BD=24$, and $AC=AO+OC=5+16=21$.
The area of the quadrilateral is half of the area of the rectangle with sides parallel to the diagonals $AC,BD$, where the sides are passing through $A,B,C,D$, so it is $\frac 12 24\cdot 21=12\cdot 21=252$.
Let now $K\in AC$ be the center of the circle which is tangent to all sides.
Let $r$ be the radius of this circle. We consider the four triangles built with $K$ as one vertex, and with one side of the quadrilateral, then compute their areas. We obtain:$$
\begin{aligned}
252
&=\operatorname{Area}(ABCD)
\\
&=
\operatorname{Area}(\Delta KAB) +
\operatorname{Area}(\Delta KBC) +
\operatorname{Area}(\Delta KCD) +
\operatorname{Area}(\Delta KDA)
\\
&=\frac 12 r(AB+BC+CD+DA)
\\
&=r(AB+BC)
\\
&=33r
\ .
\end{aligned}
$$
This leads to $r=252/33=7.63636363\dots$ .
The closest integer is $8$.
EDIT: Added picture:
Hint:
According to the Wikipedia page, the inradius of a tangential quadrilateral is:
$$r = \frac{K}{s}$$
where $K$ is the quadrilateral's area and $s$ is the semiperimeter, which is $\frac{13+13+20+20}{2} = 33$.
To find the area, split the quadrilateral into two smaller triangles, such as $\Delta ABD$ and $\Delta BDC$. Then you can apply Heron's formula on each of them:
$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$
where $a,b,c$ are the sides of the triangles.
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