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'Pentabuild' printed from https://nrich.maths.org/
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.