You may also like

Darts and Kites

Explore the geometry of these dart and kite shapes!

Pent

The diagram shows a regular pentagon with sides of unit length. Find all the angles in the diagram. Prove that the quadrilateral shown in red is a rhombus.

Pentakite

ABCDE is a regular pentagon of side length one unit. BC produced meets ED produced at F. Show that triangle CDF is congruent to triangle EDB. Find the length of BE.

Golden Thoughts

Age 14 to 16
Challenge Level

Sue from Madras College, St Andrew's, Scotland sent in a good solution to this problem.



Let $SX = a$ and $XR= b$ then $PQ=a+b$ and, since the areas of the triangles $SPX, XYR$ and $PQY$ are equal, we know that $YR= ax/b$ and $QY = ax/(a+b)$. Because $PS = QR$ we now have $$ \frac{ax}{a+b} + \frac{ax}{b} = x$$ and simplifying this we get $b^2 -ab - a^2 = 0$. So, writing $t$ for $\frac{b}{a}$, we get the quadratic equation $$ t^2 - t - 1 = 0$$ which has roots $$ t = \frac{1 \pm \sqrt 5}{2}$$ but we know that $t$ is positive so $$ t =\frac {1 + \sqrt 5}{2}.$$ Also, writing the ratio $RY/YQ$ in terms of $t$ we get $$ \frac{RY}{YQ} = \frac{a+b}{b} = \frac {1}{t} + 1.$$ We already know that $t=\frac{1}{t} + 1$ because $t^2 = 1 + t$. So $$ \frac{RX}{XS} = \frac {RY}{YQ} = \frac {1 + \sqrt 5}{2}.$$