A geometry problem regarding inscribed circle.

contest-mathgeometry

I don't know the origin of the problem. I got it from my brother who got it from his Call Of Duty WhatsApp group. None of them could solve it nor can I. I think it would not take very high level mathematics to solve it, atmost it could take some trignometry but not higher.

The problem:-enter image description here

For those people who cannot see the picture, it asks the following:

Let there be a circle centred at $O$ whose diameter is $AB$, let there be a point $C$ on the circumference such that $AC=12$ and $BC=16$. Now draw a line from $C$ to $AB$ such it is perpendicular to $AB$. The perpendicular divides an half of the circle into $2$ pieces. Inscribe a circle in the smaller piece and name the point $P$ where it touches the diameter and inscribe a circle in the larger piece and name the point $Q$ where it touches the diameter. What is the value of $PQ$?

My trial:-

First I wanted to find the diameter, which is very easily doable by using pythagorean theorem.

So,
$12^2+16^2=AB^2$

$\boxed{AB=20}$

Now I wanted to find $AE$ and $EB$. To do that I needed to know $CE$. ($E$ is the intersection of the perpendicular with the diameter)

So the area of the of the triangle will be $\frac{1}{2}AB\cdot CE$

Making $AC$ the base of the triangle we get the area to be $\frac{1}{2}AC\cdot BC$

So,
$\frac{1}{2}AB\cdot CE=\frac{1}{2}AC\cdot BC$

$20\cdot CE=12×16$

$\boxed{CE=9.8}$

Now, $AE^2+CE^2=12^2$

$\begin{align} AE & =\sqrt{12^2-9.8^2}\\
& =7.2\end{align}$

$\boxed{AE=7.2}$

And, $AE+EB=AB$

$7.2+EB=20$

$\boxed{EB=12.8}$

My question:-

Now I am stuck, I think I have gained all the information needed about the triangles. But I can't figure out about the circles. I have no idea how to proceed.

So I am asking for some help. You can give me some hint or you can just give me the entire answer. (I have no problem with getting the answer)

Simple problems can be solved using very advanced level math. I would request the person who is answering to not use very advance math.

Best Answer

Note that $|OE| = 10 - 7.2 = 2.8$. Let $X$ be the incenter of the left circle. Then triangle $XPO$ is right-angled and if we let $r$ be the radius of the left circle, by Pythagoras we find $$ (10-r)^2 = r^2 + (r+2.8)^2. $$ Solving this equation gives $r = 3.2$. We can do the exact same thing for the other circle to find $$ (10-R)^2 = (R-2.8)^2 + R^2. $$ This gives $R = 4.8$. Hence $|PQ| = r + R = 8$.

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