Here's a problem I just came up with :
A semicircle (O) is inscribed in a quadrilateral ABCD , as shown in the figure.
If sides AD , DC , CB measure 17 ; 16 and 14 respectively, what is the length of side AB ?
Using GeoGebra, I found an answer to be very close to 31 , but I wonder if it's possible to calculate the exact solution(s) ?
Your ideas are welcome.
Problem:
To find $AB$ in the specific scenario.
Findings:
($OZ=OY=OX=r$)
Theorem: Tangents drawn from a single external point to a circle are always equal in length.
By using the above theorem:
$$DY+CY=16; DZ+AZ=17; CX+XB=14$$
which transforms to:
$$DY+CY=16; DY+\sqrt{AO^2-r^2}=17; CY+\sqrt{BO^2-r^2}=14$$
Now,
$$DY+\sqrt{AO^2-r^2}+CY+\sqrt{BO^2-r^2}=14+17 \implies \sqrt{AO^2-r^2}+\sqrt{BO^2-r^2}=15$$
Now, we can further solve and find the values of $AO+BO$ according to the radius given to us.
(The value of $AB$ comes out in a range of $[15,16]$ if the radius was assumed to be $1$, but I doubt that finding $AB$ without the radius given would be possible.)
Case-wise discussion and analysis:
(Consider $AO=x, BO=y$)
$r=8$ case:
$$\text{Starting equation: } \alpha+\beta=15$$
$$\text{If we assume }\alpha=\sqrt{x^2-64},\beta=\sqrt{y^2-64} \implies x=\sqrt{64+\alpha^2}, y=\sqrt{(15-\alpha)^2+64}$$
$\text{ for } \alpha \in [0,15]$
Now, if $$\alpha=0 \implies x=8,\beta=8; y=\sqrt{(15-0)^2+8^2}=17 \implies x+y=25 $$
Now, if $$\alpha=7.5 \implies x=y=\sqrt{7.5^2+64}=\sqrt{120.25} \approx 10.966 \implies x+y \approx 21.93$$
Again, if we look at the boundary case, when $\alpha=15$, we obtain:
$$\alpha=15 \implies \beta=0, x=\sqrt{15^2+8^2}=17,y=\sqrt{(15-15)^2+8^2}=8 \implies x+y=25$$
Hence,
$$\boxed{\text{The range of $x+y$ comes out as } [\sqrt{481},25]}$$
Look further into this for seeing why the range of $x+y$ comes out as $[\sqrt{481},25]$.
Thank You.
Let $E$, $F$, $G$ be the tangency points, as in figure below. If we set $x=AG$, then: $$ GD=DF=17-x,\quad FC=CE=x-1,\quad EB=15-x. $$ Let $r$ be the radius of the circle. As the angles at $O$ sum up to $180°$ we also have: $$ \arctan{x\over r}+2\arctan{17-x\over r}+2\arctan{x-1\over r} +\arctan{15-x\over r}=\pi. $$
EDIT.
In a first attempt, I solved this equation for $x$ with Mathematica, obtaining four explicit but very long algebraic expressions. I then realized that solving instead for $r$ is much better: the above equation, in fact, can be rewritten as: $$ 47 r^4+\left(94 x^2-1596 x-2786\right)r^2 +(47 x^4-1596 x^3+14554 x^2-17340 x+4335)=0 $$ and solving for $r^2$ is then straightforward: $$ r^2=\frac{1}{47} \left(-47 x^2+798 x+1393 \pm 32 \sqrt{-174 x^2+2967 x+1696}\right) $$ Only the larger solution gives results consistent with the limiting cases $x=1$ and $x=15$, which can be computed by hand. Hence the smaller solution is spurious and I'll describe here only the larger one. Here's a plot of $r$ versus $x$: I chose $x\in[1,15]$ because outside that range one or more sides of the quadrangle become negative.
And here's a plot of $AB=\sqrt{x^2+r^2}+\sqrt{(15-x)^2+r^2}$.
The maximum value of $AB$ is attained at $x=227/29$ and is $$ AB_\max =\sqrt{\frac{27965}{29}}\approx 31.0533 $$
