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Four circles are placed in a regular pentagram as in the figure below. The three green circles are congruent, and their centres are on the horizontal segment. Wherever things look tangent, they are tangent.

Is the red circle bigger than, smaller than, or congruent to the green circles?

four circles star

Disclaimer: I don't yet have a proof that doesn't involve calculations, but I would prefer if someone can come up with such a proof.

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Here's a solution using minor calculation.

Consider the following diagram with red circles in place of green circles.

pentagram with circles

Let the radius of the red circles be 1.

By the Pythagorean Theorem, the length of $OA'$ is $\sqrt{1^2+3^2}=\sqrt{10}$.

On the other hand, $OA$ is a diagonal of the regular pentagon with side $OB$ opposite vertex $A$. Using similar triangles, the ratio of the lengths of a diagonal and a side of a regular pentagon is the golden ratio, so the length of $OA$ is $2\phi=1+\sqrt5$.

Because $\sqrt{10}=\sqrt{6+2\sqrt4}<\sqrt{6+2\sqrt5}=1+\sqrt5$, the red circle is smaller than the green circles.

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    $\begingroup$ +1 this is a very minimal geometric creative approach $\endgroup$ Commented Mar 28 at 0:22
  • $\begingroup$ The diagram as you've drawn it doesn't make clear how $ |OA| = 2\; \phi $. It's clearer to see this when you construct an outer circle of radius $R = |OA|$ touching the star points and also radial lines showing that $R \sin{18}^\text{o} = \;r =\;1$. $\endgroup$ Commented Apr 14 at 20:10
  • $\begingroup$ $|OA|$ isn't a diagonal to either the inner or outer pentagon in this situation. It looks to me that you simply used the fact that $\Delta OCA$ (where C is the mid-point of $|OB|$) is an 18-72-90 right angle triangle or one half of the 36-72-72 triangle involved in the pentagon $\phi$ relation. Of course you get the correct answer (and very efficiently) but saying OA is a diagonal of the regular pentagon with side OB opposite vertex A demands the reader to imagine a new pentagon that isn't drawn. $\endgroup$ Commented Apr 14 at 20:35
  • $\begingroup$ I would say it as: OA is a diagonal of a regular pentagon with side OB opposite vertex A. But again, all you are doing is using the proportions of the 36-72-72 triangle that are found in regular pentagons. Generally, unless someone has pentagons on their mind they are unlikely to envision them emerge ahead of seeing the triangles characterizing them, I think. But each to their own, of course. $\endgroup$ Commented Apr 14 at 20:48
  • $\begingroup$ I would say a pentagon rather than the pentagon as this one is putative and different to either the one in the red circle or the outer one formed by joining the star points. This I would do because when you have people looking at a puzzle involving explicit pentagons, a solution referring to "the pentagon" would likely be read as referring to one of the two visible pentagons. $\endgroup$ Commented Apr 14 at 20:53
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Let's have a labelled picture (source):

pentagram with vertices labelled and angles shown

I'm going to fix $\overline{CD}=1$, i.e. we have a regular pentagon of side-length $1$ in the middle. Let $\omega=\overline{CH}$, so that we have an isosceles triangle with side-lengths $1,\omega,\omega$ and angles $36^{\circ},72^{\circ},72^{\circ}$. That means

by the cosine rule, $\omega=\dfrac{1}{\sqrt{2(1-\cos36^{\circ})}}$.

The green radius is $\dfrac{1+2\omega}{6}$.

Meanwhile, by constructing a right-angled triangle inside the regular pentagon, we get immediately that

the red radius is $\dfrac{1}{2\tan36^{\circ}}$.

By calculation, these numbers are

(red) $0.68819...$ and (green) $0.70611...$, so the answer is that green is larger.

I'm also trying to find a neat trick to show this without needing a calculator. Here's what I have so far, which looks pretty cool and might help someone to get further:

$$\text{ green }<\text{ red }\Leftrightarrow\frac{1+2\omega}{6}<\frac{1}{2\tan36^{\circ}}\Leftrightarrow(1+2\omega)\tan36^{\circ}<3$$ $$\Leftrightarrow\left(1+\frac{2}{\sqrt{2(1-\cos36^{\circ})}}\right)\frac{\sqrt{1-\cos^236^{\circ}}}{\cos36^{\circ}}<3$$ $$\Leftrightarrow\frac{\sin36^{\circ}+\sqrt{2(1+\cos36^{\circ})}}{\cos36^{\circ}}<3$$

The exact values of $\cos36^{\circ}$ and $\sin36^{\circ}$ are known:

$$\cos36^{\circ}=\frac{1+\sqrt{5}}{4}\qquad\text{ and }\qquad\sin36^{\circ}=\frac{\sqrt{10-2\sqrt{5}}}{4}$$

so we have

$$\text{ green }<\text{ red }\Leftrightarrow\frac{\sqrt{10-2\sqrt{5}}+2\sqrt{10+2\sqrt{5}}}{1+\sqrt{5}}<3\Leftrightarrow4\sqrt{5}<4$$

which resolves the question without need of a calculator (although it's still quite computational in a different way).

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  • $\begingroup$ Re: no-calculator solution: Multiplying out the denominator, squaring, and canceling like terms yields $\text{green}<\text{red}\leftrightarrow 4\sqrt 5<4$. $\endgroup$ Commented Mar 25 at 18:37
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    $\begingroup$ This is correct, but if we allow calculations, there is a much simpler proof. $\endgroup$ Commented Mar 25 at 19:30
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pentagon & star

This solution evaluates the exact distance of the diagonal $FG$ in the hexagon so as to see if it is equal to, greater or less than three times the diameter of the inner circle.
Taking the inner pentagon's enclosed circle to be of unity radius, we have: $$|FG| \; = \; 2 \, |FJ| \; = \; 2 \left(\dfrac{1}{\tan{18^\circ}} \right)$$ Referring to the proportions of the regular pentagon and the $\phi$ relation between a diagonal and an edge of unit length, we can deduce: $$ \begin{align}\tan{18^\circ} \; &= \; \dfrac{1}{2}{\sqrt{\phi^2 - \left(\dfrac{1}{2}\right)^2}} \\ &= \dfrac{1}{\sqrt{4\phi^2 - 1}} \\ &= \dfrac{1}{\sqrt{4\; \cdot \left[\left(\dfrac{1 + \sqrt{5}}{2}\right)^2 \right] - 1}} \\ &= \dfrac{1}{\sqrt{4 \; \cdot \left(\dfrac{1 + 2\sqrt{5} + 5}{4}\right) - 1 }} \\ &= \dfrac{1}{\sqrt{5 + 2\sqrt{5}}} \\ \\ \text{Hence} \;\;\;\;\;\;\;\; |FG| \; &= \; 2 \sqrt{5 + 2\sqrt{5}} \\ \\ \text{Since} \;\;\;\;\;\;\;\; 2 \sqrt{5 + 2\sqrt{5}} \;\; &> \;\;2 \sqrt{5 + 2\sqrt{4}} \;\;\;\;\;(\text{i.e.} \,\, 6) \end{align}$$
we can thus say that $|FG|$ is greater than the width of three unit radius circles laid along that line.
This means that a circle enclosed by the inner pentagon (e.g. the red circle) will not be as large as either of three similar circles which if laid end to end along $|FG|$ would exactly cover it, e.g. the green circles.

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Here's a geometric approach with no calculation needed, using only a compass and straightedge - you'll need to know how to find the center of a circle, and how to make a perpendicular bisector of a line segment.

Draw lines tangent to the green circles which pass through the points where they touch, which can be accomplished by constructing perpendicular bisectors of the segments connecting adjacent green circles' centers. By construction, these lines (shown here in yellow) are vertical, parallel, and one green-circle diameter apart, yet the red circle fits visibly in between them. The red circle is smaller.
enter image description here

Alternatively, note that if the red and green circles were indeed the same size

the fact that the middle green circle's center lies on the red circle's perimeter would mean that the red circle's center was also on the green circle's perimeter. If instead the red circle is smaller, its center lies inside the green circle, and if instead the red circle is larger, its center lies outside the green circle. Use the compass and straightedge to find the red circle's center - we find that it lies inside the middle green circle, allowing us to conclude that the red circle is smaller.

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    $\begingroup$ It's not clear to me that the yellow lines are vertical (in other words, this is not a proof). $\endgroup$ Commented Mar 27 at 18:40
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    $\begingroup$ @Pranay The yellow lines are perpendicular bisectors of the segments connecting the centers of the green circles. By construction, they are perpendicular to the line upon which the green circles' centers lie. The tangent line must be perpendicular to the radius. $\endgroup$ Commented Mar 27 at 19:04
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Following answer does not use clever geometrical constructs or explicit trigonometric formulas like in the excellent nice answers already given by others, it rather simply calculates symbolically and uses some known properties of golden ratio and regular pentagon.

Straightforward, but certainly rather long, calculations and simplifications.

let green regular pentagon have side $ s=1 $
write $\phi$ for golden ratio $\dfrac{1+\sqrt{5}}{2}$
then blue diagonal has well known length $d = \phi$

ratio between red and green regular pentagons is $2 -\phi$
(this is easy to prove)

green inscribed circle has well known radius $\dfrac{1}{2\sqrt{5-2\sqrt{5}}}$
(see e.g. Wikipedia regular pentagon page)

so red inscribed circle has radius $r = \dfrac{2-\phi}{2\sqrt{5-2\sqrt{5}}} = \dfrac{2-\phi}{2\sqrt{7-4\phi}}$

and the three green circles have given radius $g = \dfrac{\phi}{6}$

we can show g > r

i.e. $$\dfrac{\phi}{6} > \dfrac{2-\phi}{2\sqrt{7-4\phi}}$$ <=> (simplify and square) $$\dfrac{\phi^2}{9} > \dfrac{(2-\phi)^2}{7-4\phi}$$ <=> ($\phi^2 = 1+\phi$) $$\dfrac{1+\phi}{9} > \dfrac{\phi^2-4\phi+4}{7-4\phi}$$ <=> ($\phi^2=1+\phi$) $$\dfrac{1+\phi}{9} > \dfrac{5-3\phi}{7-4\phi}$$ <=> (multiply) $$7+3\phi - 4\phi^2 > 45-27\phi$$ <=> ($\phi^2=1+\phi$) $$3-\phi > 45-27\phi$$ <=> $$26\phi > 42$$ <=> $$\phi > \dfrac{21}{13}$$ <=> $$1 + \sqrt{5} > \dfrac{42}{13}$$ <=> $$\sqrt{5} > \dfrac{29}{13}$$ <=> (square) $$5 > \dfrac{841}{169}$$ <=> $$845 > 841$$

rather unimportant notes:

As visual confirmation, white center effectively just crosses green circle border.

By luck 21/13 appears during deduction as reasonably good rational approximation for golden ratio $\phi$ where, relatively small, numerator and denominator are two consecutive Fibonacci numbers.

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    $\begingroup$ Aren't you using trigonometry when you say "green inscribed circle has well known radius"? $\endgroup$ Commented Mar 27 at 18:24
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    $\begingroup$ @Pranay : thanks. implicitly most probably yes, I took result from Wikipedia ... $\endgroup$ Commented Mar 27 at 19:24

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