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Age 16 to 18 Challenge Level:

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.

  1. Construct the regular pentagon using the prescribed technique.
  2. 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.)
  3. 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).
  4. 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 $D$.
  5. The 10-sided shape we now have inscribed in the circle is a regular decagon.