Circling around this Circle Circles with centers $P, Q$ and $R$, having radii $1, 2$ and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQR$?
On this problem, I was able to find the side lengths of PQ and QR (pretty obvious). I am not able to figure out how to find PR though. Can someone please help?
 A: Assuming without loss of generality that $l$ is horizontal:


*

*the line $PQ$ is 3 units long and rises up 1 unit, so its projection onto $l$ or $P'Q'$ is $\sqrt{3^2-1^2}=2\sqrt2$ units

*$QR$ is 5 units long and rises up 1 unit, so $Q'R'$ is $\sqrt{5^2-1^2}=2\sqrt6$ units


This allows calculating the areas of three trapeziums:


*

*$A(PP'Q'Q)=\frac12\cdot2\sqrt2(1+2)=3\sqrt2$

*$A(QQ'R'R)=\frac12\cdot2\sqrt6(2+3)=5\sqrt6$

*$A(PP'R'R)=\frac12\cdot(2\sqrt2+2\sqrt6)(1+3)=4(\sqrt2+\sqrt6)$


$A(\triangle PQR)$ is then $|A(PP'Q'Q)+A(QQ'R'R)-A(PP'R'R)|$, the absolute difference between the areas of the two small trapeziums and the large one – in other words
$$3\sqrt2+5\sqrt6-4\sqrt2-4\sqrt6=\sqrt6-\sqrt2$$
A: Another way of solving this is to find $\angle PQR$. Your area will then be:
$$A_{\triangle PQR}=\frac12\cdot PQ\cdot PR\cdot\arcsin PQR$$

You can think of $\angle PQR$ as the sum of the following angles:
$$\angle PQR = \angle PQG +\angle RQK+\frac\pi2$$

It's very easy to see that $PQ=3$ and $QG=1$, therefore $PG=2\sqrt2$, thus:
$$\angle PQG=\arccos \left(\frac{1}{3}\right)$$
Since, $QR=5$ and $RK=1$, then $QK=2\sqrt6$, then:
$$\angle RQK=\arcsin\left(\frac1{5}\right)$$
From here, you get that the area:
$$\begin{align}
A_{\triangle PQR}&=\frac12\cdot3\cdot5\sin\left(\arccos\left(\frac1{3}\right)+\arcsin\left(\frac1{5}\right)+\frac\pi2\right)\\
&=\frac12\cdot15\cdot\frac{2}{15} \sqrt{2} \left(\sqrt{3}-1\right)\\
&=\sqrt{2} \left(\sqrt{3}-1\right)
\end{align}$$
