Circling around this Circle

circlescontest-mathgeometry

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?

Best Answer

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 $$3\sqrt2+5\sqrt6-4\sqrt2-4\sqrt6=\sqrt6-\sqrt2$$

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