Circles with diameters $3$, $4$, $6$ are packed in a rectangle of width $6$. Find a particular length.

Question

I'm trying to solve a simple problem that someone sent me.

From $A$ to $A^\prime$, I calculated the length to be $\sqrt{6}\text{ cm}$; and from $A^\prime$ to $B$, I got $\sqrt{24}\text{ cm}$. I'm not entirely sure these are correct.

Answer 1

This can be done by the application of two Pythagorean triangles.

Let $M_A$ be the center of the circle at level $A$, and others accordingly.
Let furthermore $F_A$ be the foot of $M_A$ onto level $A'$, and $F_B$ accordingly.

Thus we have to consider the right triangles $M_AF_AM_{A'}$ and $M_BF_BM_{A'}$.

The given distances are
$M_AM_{A'}=2+1.5=3.5$
$M_{A'}F_A=6-1.5-2=2.5$
$M_BM_{A'}=2+3=5$
$M_{A'}F_B=6-3-2=1$

Thus
${M_AF_A}^2={M_AM_{A'}}^2-{M_{A'}F_A}^2=12.25-6.25=6$
${M_BF_B}^2={M_BM_{A'}}^2-{M_{A'}F_B}^2=25-1=24$

Therefore
$AB=AA'+A'B=M_AF_A+M_BF_B=\sqrt6+\sqrt{24}=\sqrt6+2\sqrt6=3\sqrt6$,
i.e. your values were completely correct.

--- rk