### Golden Thoughts

Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.

### From All Corners

Straight lines are drawn from each corner of a square to the mid points of the opposite sides. Express the area of the octagon that is formed at the centre as a fraction of the area of the square.

### Star Gazing

Find the ratio of the outer shaded area to the inner area for a six pointed star and an eight pointed star.

# Weekly Problem 37 - 2007

##### Stage: 3 and 4 Challenge Level:

Each exterior angle of a regular hexagon is $60^{\circ} (360^{\circ}/6)$, so when sides HB and IC are extended to meet at A , an equilateral triangle, ABC is created.

Let the sides of this triangle be of length x. As BC, DE and FG are all parallel, triangles ABC, ADE and AFG are all equilateral. So, DE = DA = $p$ + $x$; and FG = FA = $q$ + $p$ + $x$.

The perimeter of trapezium BCED = $x + p + x + 2p = 2x + 3p$;

The perimeter of trapezium DEGF = $(p + x) + (q + p + x ) + 2q = 2x + 2p + 3q$;

The perimeter of hexagon FGIKJH = $2((q + p + x) + 2r ) = 2x + 2p + 2q + 4r$.

So, $2x$ + $3p$ = $2x$ + $2p$ + $3q$ ; hence $p$ = $3q$.
Also $2x$ + $2p$ + $3q$ = $2x$ + $2p$ + $2q$ + $4r$ ; hence $q$ = $4r$ .
So, $p$ : $q$ : $r$ = $12r$ : $4r$ : $r$ = $12$ : $4$ :$1$

This problem is taken from the UKMT Mathematical Challenges.

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