Wanted: A Pair of Friends

Geometry Level 3

The angle between the lines is $60^\circ$ . The lonely unit circle should be joined by a pair of circles.

These should be equal to each other in size, tangent to each other, and tangent to the unit circle. Each of them should also be tangent to one of the lines.

Two sets of such circles exist. Find the difference in radius between the pairs.

The answer is 0.78461.

1 solution

Jeremy Galvagni
Jul 13, 2018

$BE=1, \angle BOE=30, OE=\sqrt{3}$

Let the radius is the small circle be $AD=r$ so $BA=r+1$ and construct rectangle $ACED$ so $BC=1-r$

$\angle AOD=15$ so $OD=\frac{r}{\tan{15}}=\frac{r}{2-\sqrt{3}}=r(2+\sqrt{3})$

$DE=\sqrt{3}-r(2+\sqrt{3})=AC$

Now we can use the Pythagorean theorem on $\triangle BAC$ and solve for r

$(1-r)^{2}+(r(2+\sqrt{3})-\sqrt{3})^{2}=(r+1)^{2}$

$(7+4\sqrt{3})r^{2}+(-10-4\sqrt{3})r+3=0$

The quadratic formula gives two solutions. The second is not extraneous because it's the radius of the larger circles!

$r=21-12\sqrt{3}$ or $r=1$

The difference is $1-(21-12\sqrt{3})=12\sqrt{3}-20\approx \boxed{.78460969}$

Wanted: A Pair of Friends

The answer is 0.78461.

1 solution

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