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I think this is solved using the Pythagorean theorem but cant figure out how to get the length of PQ. Any help?

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$PQ$ is the base of an isosceles triangle with height $\frac{1}{2}(XY) = 4$. The sides of the isosceles triangle are radii of the circles. The radius of each circle is $QR = PS = PQ + 1$, since we know $RP = QS = 1$.

Cutting the isosceles triangle along its height $XZ$, we have the right triangle $PXZ$. Its base is half of $PQ$, its height is $XZ = 4$, and its hypotenuse has length $PQ + 1$, so the Pythagorean theorem gives $$ \bigg(\frac{1}{2}PQ\bigg)^2 + 4^2 = (PQ + 1)^2.$$

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    $\begingroup$ Thanks for this explanation!! $\endgroup$
    – keji
    Feb 27, 2013 at 3:27
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Let $PQ=2x$. Note that $PS=2x+1$ is a radius. Consider right $\triangle PXZ$. It has hypotenuse $2x+1$ and legs $x$ and $\frac{8}{2}=4$. Thus by the Pythagorean Theorem we have $x^2+4^2=(2x+1)^2$. That simplifies to $3x^2+4x-15=0$. Conveniently, the quadratic factors as $(3x-5)(x+3)$. That gives $x=\frac{5}{3}$, and therefore $PQ=\frac{10}{3}$.

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