### 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.

A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?

### Days and Dates

Investigate how you can work out what day of the week your birthday will be on next year, and the year after...

# Weekly Problem 3 - 2011

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

Let $r$ be the radius of each of the larger circles.
The sides of the square are equal to $r+1$, the sum of the two radii.
The diagonal of the square is $2r$.

By Pythagoras,         $$(r+1)^2+(r+1)^2 = (2r)^2$$Simplifying gives:      $$2(r+1)^2 = 4 r^2$$ i.e.                          $$(r+1)^2 = 2r^2$$
so$$r+1 = \sqrt{2}r$$

[$-\sqrt{2}r$ is not possible since $r+1> 0$].

Therefore $(\sqrt{2} - 1)r = 1$.
Hence $r= \frac {1}{\sqrt{2}-1} = \sqrt{2} + 1$.

This problem is taken from the UKMT Mathematical Challenges.

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