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How to find the ratio between the area of the big regular pentagon $ABCDE$ and the small regular pentagon $PQRST$

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Frank
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2 Answers2

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By angle-chasing one can show that $EP=AP=AQ$.

Without loss of generality we can assume that $PQ=1$. Let $x=EA$. Then $EP=AP=AQ=x-1$.

Note that triangles $EQA$ and $APQ$ are similar. It follows that $\dfrac{EA}{AQ}=\dfrac{AQ}{PQ}$. Thus $$\frac{x}{x-1}=\frac{x-1}{1}.$$ This simplifies to $x^2-3x+1=0$, which has roots $\frac{3\pm\sqrt{5}}{2}$. One of these is less than $1$. So $x=\dfrac{3+\sqrt{5}}{2}$.

Now the ratio of the area of the big pentagon to the little one is $x^2$ to $1$. Compute. The number $x^2$ simplifies to $\dfrac{7+3\sqrt{5}}{2}$.

Remark: Note that $\frac{3+\sqrt{5}}{2}$ is the square of the famous "Golden Ratio" $\frac{1+\sqrt{5}}{2}$. The Golden Ratio is often denoted by $\varphi$. So the ratio of the areas is $\varphi^4$.

André Nicolas
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  • Thanks "André Nicolas" ! i wonder why there is such thing called "Golden Ratio" in Math . – Frank Feb 08 '13 at 17:54
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    The number comes up in a surprising number of places, the regular pentagon, as you saw here, but also the Fibonacci numbers, the worst case running time of some algorithms, and so on. There is a huge literature, much of it nonsensical from a scientific point of view, but there is also a fair amount of respectable stuff. – André Nicolas Feb 08 '13 at 18:20
  • I like this more than my answer. – Chris Taylor Feb 08 '13 at 18:44
  • Very interesting .I remember seeing it in the famous novel "de vinci code ".I guess what you mean by " in a surprising number of places" that it can be found in nature in unexplainable way . – Frank Feb 08 '13 at 18:46
  • There are some partial explanations. The biological examples tend to be somewhat questionable, since flowers, leaves have less regularity than Golden Ratio fans claim. – André Nicolas Feb 08 '13 at 18:55
  • That is obviously open a huge philosophical controversy about whether the numbers are existed in external world or we human have "invented" them .Seeing something like "Golden Ratio" is enough to convince me to follow the first trend ,Despite being the ratio inaccurate sometimes as you mentioned . – Frank Feb 08 '13 at 19:18
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The ratio of the areas is the same as the square of the ratios of the sides, so the ratio you want is

$$\frac{{\rm area}(ABCDE)}{{\rm area}(PQRST)} = \left( \frac{{\rm length}(AB)}{{\rm length}(PQ)}\right)^2$$

To calculate this, note that the angles $BAC$ and $CAD$ are all $\pi/5$, which you can get by noting that (a) the angle $BAE$ is $3\pi/5$ (sum of exterior angles is $2\pi$) and that $AEDB$ is a trapezium, so its interior angles sum to $2\pi$ and it two upper angles and two lower angles are equal.

Now you can use trigonometry. If we call the side length of the larger pentagon $a$, then dropping the perpendicular from point $A$ to intersect with $PQ$ at $S$ and $DC$ at $T$, leads to consideration of the triangles $ASE$ and $ATD$. The ratio of their heights is the same as the ratio of the length $AB:PQ$.

For the triangle $ASE$, you have

$$\sin(\pi/5) = {\rm length}(AS) / a$$

and for $ATD$ you have

$$\tan(2\pi/5) = 2\, {\rm length}(AT) / a$$

and combining these gives

$$\frac{{\rm length}(AT)}{{\rm length}(AS)} = \frac{\tan(2\pi/5)}{2\sin(\pi/5)}$$

and so the ratio of the areas is

$$\frac{\tan^2(2\pi/5)}{4\sin^2(\pi/5)} = \frac{1}{2}(7 + 3\sqrt{5}) \approx 6.85$$

Chris Taylor
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