Pentabuild
Copy this straight edge and compass construction. Can you explain why it produces a regular pentagon?
The description of the construction below, and the information in the notes, should help you to explain the construction.
1. Draw a circle $C_1$ centre $O$ diameter $PQ$.
The circle $C_1$ has radius 1 unit; what is its equation?
2. Draw the perpendicular bisector of $PQ$ cutting $PQ$ at $O$ and $C_1$ at $A$ and $Y$.
3. Draw perpendicular bisectors of $PO$ and $OQ$ cutting $PQ$ at $R$ and $S$.
Find the length $YS$
4. Draw circles $C_2$ and $C_3$ centres $R$ and $S$ and radii $RO$ and $SO$.
5. Join $R$ and $S$ to the point $Y$ cutting $C_2$ at $T$ and $U$ and $C_3$ at $V$ and $W$.
6. Draw circle $C_4$ centre $Y$ radius $YW=YU$ cutting $C_1$ at $D$ and $C$.
What is the equation of $C_4$? Find the value of $y$ at the intersection of $C_1$ and $C_4$ .
7. Draw circle $C_5$ centre $Y$ radius $YT=YV$ cutting $C_1$ at $E$ and $B$.
What is the equation of $C_5$ ?
Find the value of $y$ at the intersection of $C_1$ and $C_5$.
At $B$ and $E$ $x^2 + y^2 +2y +1 = 2y + 2 = (3 + \sqrt 5)/2$ so
8. Join $AB$, $BC$, $CD$, $DE$, $EA$.
How could you adapt this construction to produce a regular decagon?
In explaining the construction look for $\sqrt 5$ appearing from triangle $YOS$ and the golden ratio appearing as the radius of circle $C_5$ (see Note). Concentrate on cosines which, in this case, are easier than the sines to find exactly.
Harry explained why we get a Pentagon:
I firstly calculated the length of $YS$ and the equations of the circle using Pythagoras' theorem on the triangle $YOS$ (and by symmetry $YOR$) to get:
$$YS=\frac{\sqrt{5}}{2}$$ Equation of $C_1$: $$x^2+y^2=1$$ Equation of $C_4$: $$x^2+(y+1)^2=\frac{(\sqrt{5}-1)^2}{4}$$ Equation of $C_5$: $$x^2+(y+1)^2=\frac{\sqrt{5}+1)^2}{4}$$
At the points of intersection of two circles, the points satisfy the equations both both circles and so we have a set of two simultaneous equations to solve. Solving these gives at the intersection of $C_1$ and $C_4$ (so for the points $C$ and $D$), $y=\frac{-(\sqrt{5}+1)}{4}$ and at the intersection of $C_1$and $C_5$, $y=\frac{\sqrt{5}-1}{4}$.
But we can now look at the hint, and find we have the same y coordinates for all our points as those of a regular hexagon. But since we are on the circle, we can work out the $x$ coordinates from the $y$ coordinates. So we have the same points as the regular pentagon in the notes section, and so this is a regular pentagon.
Tom from Bristol Grammar School then suggested a method to construct a regular decagon using the pentagon that we've already constructed.
- Construct the regular pentagon using the prescribed technique.
- Bisect the angle $\angle A C E$ by drawing a circle centre $A$ and a circle of the same radius (perhaps $E C$) centre $E$ and drawing a straight line between one of the points at which the circles intersect and point $C$. (This works because $A C=E C$, as the pentagon is regular - it is a fact that is obvious and easily proven using SAS congruence, and therefore it is equivalent to the classroom-taught angle bisection technique.)
- Let us call the point (other than $C$) at which this line crosses the pentagon's circumcircle $P$. Join $A$ to $P$, and join $E$ to $P$. Essential to our method is that now $A P=P E$, which is clearly true by SAS congruence of the triangles $A C P$ and $E C P$ ($A C=E C$ was used above, $\angle A C P= \angle P C E$ holds because $P C$ is an angle bisector, $C P$ is common).
- Construct the circle centre $A$ through $P$ and label the point (other than $P$) at which it crosses the circumcircle of the pentagon $Q$. Draw in the line segments $A Q$ and $Q B$. Similarly construct the circle centre $B$ through $Q$, join up the line segments, and repeat this process for $C$ and $D$.
- The 10-sided shape we now have inscribed in the circle is a regular decagon.
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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 ${1+\sqrt
5\over 2}$.
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