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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.

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Building Tetrahedra

Can you make a tetrahedron whose faces all have the same perimeter?

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Ladder and Cube

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?


Stage: 4 Short Challenge Level: Challenge Level:2 Challenge Level:2
The diagram shows the top-right-hand portion of the square.
The shaded trapezium is labelled QXYZ and W is the point at which ZY produced meets PQ.
As QXYZ is an isosceles trapezium, ∠QZY = ∠ZQX = 45°.
Also, as YX is parallel to ZQ, ∠XYW = ∠WXY = 45°. So WYX and WZQ are both isosceles right-angled triangles. As ∠ZWQ = 90° and Z is at centre of square PQRS, we deduce that W is the midpoint of PQ. Hence WX = XQ = $\frac{1}{4}$PQ. So the ratio of the side-lengths of similar triangles WYX and WZQ is 1:2 and hence the ratio of their areas 1:4.
Therefore the area of trapezium QXYZ = $\frac{3}{4}$ x area of triangle ZWQ = $\frac{3}{32}$ x area PQRS since triangle ZWQ is one-eighth of PQRS. So the fraction of the square which is shaded is 4 x $\frac{3}{32}$ = $\frac{3}{8}$.

This problem is taken from the UKMT Mathematical Challenges.
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