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Two circles, green and red, are placed in a regular pentagon as shown below. The circles are vertically aligned and touch each other. The base of the pentagon is the diameter of the red circle and the apex of the pentagon is on the boundary of the green circle. Which circle is bigger?

two circles in pentagon

I would prefer a solution that doesn’t involve any calculation.

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8 Answers 8

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Here's a solution using minor calculation.

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

pentagon

Let the radius of the red circles be 1.

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

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 circle.

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    $\begingroup$ +1. Cool. This is the same as the answer to the other question except for the figure. $\endgroup$ Commented Apr 6 at 21:38
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    $\begingroup$ The connection with proofs using radical expressions idls thst once the golden rstio between diagonal and sudebis observed, the radicals for the relevant trig functions follow. This method uses the radicals in a particularly simple way. $\endgroup$ Commented Apr 7 at 1:24
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Not a rigorous proof, but "it's obvious"

The radius of the arc is equal to the diameter of the red circle, and this is clearly less than the diameter of the green circle. Therefore we can conclude that the green circle is bigger.
Circular arc added to original diagram

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  • $\begingroup$ I was going to answer in the same way as you, as this is the only possible way to prove that green circle is greater without any calculation. Euclid performed several demonstrations just using compass without using tan() or sin() $\endgroup$ Commented Apr 6 at 16:25
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    $\begingroup$ While I see it, as you said, this isn’t a rigorous proof. But +1 anyway. $\endgroup$ Commented Apr 6 at 17:32
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    $\begingroup$ One limitation is that we cannot expect hand-drawn compass/straightedge constructions to be implemented without error, and that could be important when the difference in radii is less than 4%. Using a computerized version eliminates this problem but is a borderline "calculator" solution. $\endgroup$ Commented Apr 7 at 1:16
  • $\begingroup$ It can be made rigorous. If $r_g = r_r = r \implies \tan{72^\circ} = \dfrac{2r + r}{r} = 3$. Knowing that $\sec{72^\circ} = 2\phi = 1 + \sqrt{5}$ we can see that actually $\tan{72^\circ} = \sqrt{5 + 2\sqrt{5}} \; > \; \sqrt{5 + 2\sqrt{4}} \; (i.e. 3)$. So $\; r_g \;>\; r_r $. $\endgroup$ Commented Jun 23 at 18:04
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The calculations are straightforward and easy.

Let $r$ and $g$ be the radius of the red and the green circle, respectively. The triangle whose vertices are the vertices of the pentagon lying on the circles is isosceles, with the angles $\frac {\pi}{5}$, $\frac {2\pi}{5}$, and $\frac {2\pi}{5}$. The side opposite to the first angle has length $2r$ and the altitude to it has length $r+2g$. Thus $r<g$, because $\frac r{r+2g}=\tan \frac \pi{10}<\frac 13$. We can justify the last inequality without a calculator, observing that $\tan \frac \pi{10}<\frac 13$ iff $\frac \pi{10}<\operatorname{arctan}\frac 13$. The Taylor series for the function $\operatorname{arctan} x$ shows that when $0<x<1$ then $\operatorname{arctan} x>x-\frac {x^3}3$. So $$\frac\pi{10}<0.315<0.333-0.015<\frac 1{3}-\frac 1{81}<\operatorname{arctan}\frac 13.$$

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    $\begingroup$ Is there a way to see the trigonometric value is less than 1/3 without plugging it into an calculator? $\endgroup$ Commented Apr 5 at 18:08
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    $\begingroup$ Alternatively: the tangent of 18° has the exact radical form $\sqrt{\ frac15(5-2\sqrt5)}$. Through algebra this is less than $1/3$ iff $\sqrt5>20/9$. Then $5=400/80$ and $(20/9)^2=400/81$. $\endgroup$ Commented Apr 6 at 4:09
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    $\begingroup$ +1. This is essentially the same as the answer I had. I want to wait a bit more to see if there’s an answer with no calculations, but if none appears, I’ll accept this one. $\endgroup$ Commented Apr 6 at 17:35
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    $\begingroup$ @Oscar Lanzi Tidy your MathJax a bit. $\tan{18} = 1 / \sqrt{5 + 2\sqrt{5}}$ $\endgroup$ Commented Apr 11 at 14:41
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    $\begingroup$ @OscarLanzi A simple solution of this problem which usually fits is to delete an old comment and post a fixed or updated comment instead. $\endgroup$ Commented Apr 11 at 15:30
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Construct a pentagram centered at one end of the base of the pentagon with a vertex at the apex of the pentagon. Inscribe an orange circle inside the pentagram and construct three congruent, tangent, blue circles centered along an edge of the pentagram.

pentagram

The orange circle is congruent to the red circle. Based on Four circles and a star, the blue circles are bigger than the orange circle. Therefore, the middle blue circle is bigger than the red circle. Therefore, the green circle is bigger than the top blue circle. Therefore, the green circle is bigger than the red circle.

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    $\begingroup$ +1 I like what you did with connecting to the other problem. This was indeed how I came up with the current problem. $\endgroup$ Commented Apr 6 at 17:34
  • $\begingroup$ @Pranay I think the two problems are essentially equivalent because you can reduce from either one to the other in this manner. Also, the proof without using trig from the other problem can basically be copy-pasted to answer this one. $\endgroup$ Commented Apr 6 at 17:42
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    $\begingroup$ I completely agree with you. I was trying to prove that problem without any calculations and reduced it to this problem. But still no luck on a proof without calculations. $\endgroup$ Commented Apr 6 at 18:11
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Playing around with numbers:

Let the side of the pentagon be one unit, which then is also the red circle diameter. Then the height of the pentagon is determined by standard trigonometric methods as

$h=\cos(18°)+\cos(54°)=\sqrt{\frac{5+\sqrt5}{8}}+\sqrt{\frac{5-\sqrt5}{8}}.$

That is just math, so I did not put a spoiler, but now comes the creative part. We

square the height to obtain

$h^2=(\sqrt{\frac{5+\sqrt5}{8}}+\sqrt{\frac{5-\sqrt5}{8}})^2=\frac{5+\sqrt5}{8}+2\sqrt{(\frac{5+\sqrt5}{8})(\frac{5-\sqrt5}{8})}+\frac{5-\sqrt5}{8}$

The first and third terms in the binomial power expansion are the sum of two conjugate quadratic residues, while the square root radicand in the middle contains the product of those same residues. Both the sum and the product are rational numbers and the squared height is thus greatly simplified:

$h^2=\frac54+2\sqrt{\frac5{16}}=\frac54+\sqrt{\frac54}.$

Well, then ...

$\sqrt{\frac54}>1\implies h^2>\frac94\implies h>\frac32,$

and

since the diameter of the green circle is clearly $h-\frac12>1$ the green circle wins.

The calculated ratio of the green circle diameter to the red circle diameter is about

1.0388.

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  • $\begingroup$ Some continued fraction convergents: sagecell.sagemath.org/… $\endgroup$ Commented Apr 8 at 3:14
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    $\begingroup$ @PM2Ring See this question for an approximation of the ratio of the height to the red-circle radius using periodic continued fractions with digits in $\mathbb{Z}[(1+\sqrt5)/2]$. $\endgroup$ Commented Apr 8 at 7:14
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enter image description here

Radii of enclosing circle for pentagon, red and green circles being $R $, $ R_1 $ and $R_2$, we have:

$$ R_1 = R \sin{A} $$

$$ 2 R_2 = R + R \cos{A} - R_1 $$

$$ \Rightarrow R_2 = \dfrac {R(1 + \cos{A} - \sin{A})}{2}$$

Looking at the limiting case for $R_2 > R_1 $:

$$ \Rightarrow \dfrac {R(1 + \cos{A} - \sin{A})}{2} > R \sin{A}$$

$$ 1 + \cos{A} - \sin{A} > 2 \sin{A} $$

$$ 1 + \cos{A} > 3 \sin{A} $$

$$ \sec{A} + 1 > 3 \tan{A} $$

$$ \text{Squaring:} \;\;\;\;\; \sec^2{A} + 2 \sec{A} + 1 > 9 \tan^2{A} $$

$$ \text{Hence: } \;\;\;\;\; 8 \sec^2{A} - 2 \sec{A} - 10 < 0 $$

$$ \text{Or: } \;\;\;\;\; 4 \sec^2{A} - \sec{A} - 5 < 0 $$

$$ \text{Factorizing:} \;\;\;\;\; (4 \sec{A} - 5)\;(\sec{A} + 1)\; < \;0 $$

For a physical pentagon only a first quadrant angle, i.e. $\sec{A} < \dfrac{5}{4} $, is possible.

Thus $\cos{A} > \dfrac{4}{5} $ and $\sin{A} < \dfrac{3}{5} $. (Very interesting how the 3-4-5 triangle comes into this situation.)

Now the question remaining is whether $ \cos{A} > \dfrac{4}{5} $.

From the regular pentagon's geometry, we have $ A = \dfrac{\pi}{5} = 36^\circ $

Oscar Lanzi says that $ \cos{A} = \dfrac{(1 + \sqrt{5})}{4} $.

Those unfamiliar with the latter expression may explore its derivation in the following spoiler.

[ DIGRESSION . . .

Pentagon & Phi

Let the pentagon edge length be $1$.

From the regular pentagon's symmetry: $$|BEG| \parallel |FA| \;\;\;\; |AEC| \parallel |FB| $$
Hence $AFBE$ is a rhombus with
$$ |BE| = |AF| = |AE| = |FB| = 1 $$
Let pentagon diagonal length = $D$.

Similar triangles, $\triangle ABC$ and $\triangle BCE$:
$$ \implies \dfrac{|AB|}{|BC|} = \dfrac{|BC|}{|CE|} $$
$$ \implies \dfrac{D}{1} = \dfrac{1}{D-1} $$ $$ \implies D^2 - D - 1 = 0 $$
Leaving $ D = \dfrac{(1 + \sqrt{5})}{2} $ as the only positive solution to the ratio of pentagon diagonal to edge length.

$ \dfrac{(1 + \sqrt{5})}{2} $ is referred to as the golden mean, $\phi$.

We also see that $|FI|$ bisects $|AB|$, $|CG|$ and $\angle AEB $.

Thus in $\triangle AHC $:
$$\cos{36^\circ} = \dfrac{|AH|}{|AE|} = \dfrac{\dfrac{D}{2}}{1} $$ $$ \boldsymbol{ \therefore \;\; \cos{36^\circ} \; = \; \frac{1 + \sqrt{5}}{4} } $$

. . . END OF DIGRESSION ]

$$ \text{If} \;\;\; \cos{A} > \dfrac{4}{5} $$

$$ \Rightarrow \dfrac{1 + \sqrt{5}}{4} > \dfrac{4}{5} $$

$$ \Rightarrow \sqrt{5} > \dfrac{11}{5} $$

$$ \text{Squaring, this means that: } \;\;\;\;\; 5 = \dfrac{125}{25} > \dfrac{121}{25} $$

And it is.

So the size of the regular pentagon semi-angle ($\dfrac{\pi}{5}$) means that the green circle will be marginally bigger than the red one.

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    $\begingroup$ There are other answers with fewer calculations that don’t involve calculators. $\endgroup$ Commented Apr 6 at 17:33
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    $\begingroup$ Reposted due to typos. I would try to incorporate exact trigonometric values derivable from the geometric properties of the regular pentagon. You get $\sin A=\sqrt{(5−\sqrt5)/8}$ and $\cos A=(\sqrt5+1)/4$. You then must prove with these radicals that $R_2>R_1$. $\endgroup$ Commented Apr 6 at 19:37
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    $\begingroup$ Much better. There are many derivations of the trigonometric functiobs of multiple if $\pi/10$ (or multiples of 18°, if you will) on Math SE and they are also given in Wikipedia. You should familiarize yourself with them. $\endgroup$ Commented Apr 6 at 22:42
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    $\begingroup$ So many things a man should do . . . ! $\endgroup$ Commented Apr 7 at 1:18
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    $\begingroup$ as in my answer to PSE#131103 : funny to spot 8/5 here (there 21/13) as rational approximation for golden ratio Φ where numerator and denominator are two consecutive fibonacci numbers $\endgroup$ Commented Apr 10 at 23:13
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enter image description here

The green circle is bigger.

1. Find the center of the pentagon.

2. Draw the radii from the center to the vertices.

3. Notice that the distance from the green circle vertex to the midpoint of the red diameter $V = R + b$ where b < R.

4. The length of a side of a pentagon a < 2R.

For the two circles to be equal, they would have to intersect at a point above the midpoint. Your diagram clearly shows the intersection point below so the green circle is bigger.

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  • $\begingroup$ You depend on the precision of the diagram to prove $r_g \; > \; r_r $ . . . Drawings can be massaged. $\endgroup$ Commented Jun 23 at 18:20
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Here's one without doing much math, although it hinges on one bit of knowledge (sorta), so already having it might be considered calculating anything or not...

Here's is the logic:

The half circle and the circle are stacked on top of each other. The bottom of the stack is the bottom of the pentagon and the top of the stack is the apex.

If the circles are exactly equal, that means the distance from the top to the bottom is 1.5, given that it is made out of half a circle and a whole circle, and the diameter of the red circle is 1. That distance would be larger if the unknown green circle would be larger than the 1.0 in diameter, and smaller if the green circle would be smaller.

The same distance is also the height of the pentagon, which is known to be 1.53... something, does not matter exactly. It is bigger than 1.5. That makes the diameter of the green circle somewhat more than 1.0.

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  • $\begingroup$ All true but this forum needs to know where you got the 1.53 . . . figure from. Refer to the similar answer in geometric form above and the proof I offered for it. $\endgroup$ Commented Jun 23 at 18:14

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