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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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2 Answers 2

up vote 4 down vote accepted

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 $\varphi^4$.

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Thanks "André Nicolas" ! i wonder why there is such thing called "Golden Ratio" in Math . –  Frank Feb 8 '13 at 17:54
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 8 '13 at 18:20
I like this more than my answer. –  Chris Taylor Feb 8 '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 8 '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 8 '13 at 18:55

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

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

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