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Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

regular pentagon
Here is some information about regular pentagons.

$ABCDE$ is a regular pentagon.

We first prove that triangles $AEX$ and $ADC$ are similar and that the ratio $AD/AE$ is equal to the golden ratio ${1+\sqrt 5\over 2}$.
As $ABCDE$ is a regular pentagon, triangles $AEX$ and $ADC$ have angles $36^o, 72^o, 72^o$ and so they are similar isosceles triangles. Label the side length of the pentagon $s$ so $AE=AX=CD=CX=s$ and the chord length $c$ so $AD=EC = c$ and $EX=c-s$. From the similar triangles $${c-s\over s}={s\over c}$$ so writing $AD/AE=c/s = x$ gives $$x - 1 = {1\over x}$$ which is the quadratic equation $x^2-x-1=0$ and hence, as it must be positive, the ratio $x=AD/AE$ is the golden ratio ${1+\sqrt 5\over 2}$.

Now we can find the exact values of $\cos 72^o$ and $\cos 36^o$.

Drawing a perpendicular from $E$ to $AD$ gives a right angled triangle from which $$\cos 36^o = {c\over 2s} ={\sqrt 5+1\over 4}.$$

Drawing a perpendicular from $A$ to $DC$ gives a right angled triangle from which $$\cos 72^o = {s\over 2c} ={1\over2\phi}={\sqrt 5-1\over 4}.$$

In explaining the construction of a regular pentagon coordinates are useful.

If the pentagon is inscribed in a unit circle with $A$ the point $(0,1)$ you can find the exact coordinates of $B$ and $C$ using the exact values of $\cos 72^o$ and $\cos 36^o$ which we have just found.

If the pentagon is inscribed in a unit circle with $A$ the point $(0,1)$ then $B=(\sin 72^o, \cos 72^o)$ and $C=(\sin 36^o, -\cos 36^o)$.

Hint: When you explain the construction focus on the $'y'$ coordinates of $B, C, D$ and $E$.