I am trying to solve this geometry problem from an exam. The exam is supposed to be 3 hours long, and this is supposed to be 1 out of 10 problems. So given that, the solution should be something quick. However, the only solution I've been able to find is extremely convoluted and realistically I'd never be able to give an answer in less than 3 hours. So, I'm looking for a simpler solution ( the intended solution ) which I should have been able to do in the exam. Here's the problem:
Consider this illustration:
In the image, the outer shape is a square with side length $8$, a circle is drawn so that is is tangent to two sides of the square at points $G$ and $F$, then the line $AH$ is drawn so that it is tangent to the circle at point $I$ and also such that the length of the segment $AI$ is also equal to $8$. Lastly the line $BJ$ is drawn such that it is tangent to the circle at point $K$. Find the area of the blue triangle, that is the area of $\triangle{AJB}$.
My approach so far is to divide the square into 9 different areas :
I also added the segment $EL$ such that it is the continuation of the line $GE$. From here on out when I say $a_{i}$ , I am referring to the area of region $a_{i}$ , $r$ will be the radius of the circle, $x$ will be the length of segment $JI$ and $y$ will be the length of segment $IH$.
So we want to find $a_{9}$
Notice then that:
$$
a_{1} = a_{2} = \frac{r(8-r)}{2} \\
a_{3} = r^2 \\
a_{4} = a_{5} = \frac{ry}{2} \\
a_{6} = a_{7} = \frac{rx}{2} \\
a_{8} = \frac{8(8-r-y)}{2} = 4(8-r-y) \\
a_{1} + \cdots + a_{9} = 8*8 = 64 \\
a_{1} + \cdots + a_{8} = 4r + ry + rx + 32 – 4y = 32 + r(4 + x) + y(r – 4) \\
a_{9} = 64 – (32 + r(4 + x) + y(r – 4)) = 32 – r(4 + x) + y(4 – r)
$$
So from here really all we need to do is find $x, y, r$ and problem solved, right? Finding $r$ is pretty easy, notice that line segment $AE$ can be thought of as the hypotenuse of $\triangle{AIE}$ and the hypotenuse of $\triangle{AEL}$. Using Pythagoras, we have $8^2 + r^2 = (8-r)^2 + (8-r)^2$. Solving for $r$ we get two solutions: $16+8\sqrt{3}, 16-8\sqrt{3}$ and since $r$ is lesser than $8$, then the second one is our value of $r$.
To find the value of $y$, notice that by Pythagoras on $\triangle{ACH}$, we have: $(8 + y)^2 = 8^2 + (8-r-y)^2$, and since we know the value of $r$ we can solve for $y$ and we get $y = \frac{16\sqrt{3}-24}{3}$.
From here, I don't actually know how to find the value of $x$, so I went back to our expression for the desired area $a_{9} = 32 – r(4 + x) + y(4 – r)$, plugged in the values of $r$ and $y$ to make the area a function of $x$, then I found another function of $x$ for the area, using Heron's formula. Notice the perimeter of the desired triangle is equal to $8 + (8 – x) + (8 – r + x)$, and when we plugged for $r$ we get that the semiperimeter $S$ should be equal to $S = 4 \sqrt{3} + 4$. Since the sides of the triangle have length $8, 8-x, 8-r+x$ and since we have precise values for the semiperimeter and $r$, we can sort of make a function for the area based on values of $x$. If we make these two functions equal to each other and solve the equation, we get two values which are possible candidates for $x$.
Problem is, these values are super convoluted, involve really large numbers and a bunch of radical expressions, which is why I gave up on this solution, realistically if it wasn't for Symbolab solving the equation for me, I probably wouldn't have time to do so in the exam.
So that's why I am writing this post : I would like for you guys to help me come up with a concrete value for $x$ in an easier way, or maybe a completely different way of computing the area that doesn't involve $x$, anything that is simpler than what I've been doing.
Thank you so much for reading all of this, I know that it's kind of a long post.
Best Answer
In such issues, you can "deconvolute" the difficulties, first of all by using the power of point $A$ with respect to circle (C) in 2 different ways :
$$P(A,(C))=AE^2-r^2=PI^2 \tag{1}$$
Knowing that $AE=AD-ED=\sqrt{2}(8-r).$
(1) gives :
$$2(8-r)^2-r^2=64$$
This quadratic equation has the unique compatible solution
$$r=8(2-\sqrt{3})\tag{2}$$ (as you have obtained).
Edit : Here is the end of the computation.
Let us compute the tangent of the two angles at the base of triangle $ABJ$ :
$$\tan(\angle BAJ)=\tan(\alpha)=\tan(\frac{\pi}{4}+\arctan \frac{r}{8})=\frac{1+\frac{r}{8}}{1-\frac{r}{8}}=\frac{8+r}{8-r}\tag{1}$$
$$\tan(\angle JBA)=\tan(\frac{\pi}{2}-2\beta)=\cot(2 \beta)=\frac{1}{\tan(2 \beta)}=\frac{1- (\tan \beta)^2}{2 \tan \beta}=\frac{8(4-r)}{r(8-r)},\tag{3}$$
the last equality being a consequence of relationship :
$$\tan(\beta)=\frac{r}{8-r}\tag{3'}$$
If we denote by $P$ the foot of the altitude issued from $J$, and $h=JP$, we can write :
$$\begin{cases}AP&=&\frac{h}{\tan(\angle BJA)}\\ PB&=&\frac{h}{\tan(\angle JBA)}\end{cases}$$
adding them, we get, by dividing by $h$ :
$$\frac{8}{h}=\frac{r(8-r)}{8(4-r)}+\frac{8-r}{8+r}$$
from which (using (2))
$$h=\frac{8}{11}(6+\sqrt{3})\approx 5.623309678$$
is obtained and from there the area :
$$\frac{32}{11}(6+\sqrt{3})$$
of triangle $JAB$.