Golden Ratio In a 2x2 Square: Without And Within

The most straightforward construction of the Golden Ratio has been posted by Dung Thanh Nguyen at the CutTheKnotMath facebook page.

Assume $AO=OB=BC=OD\;$ and $OD\perp AB.\;$ $M\;$ is the intersection of $AB\;$ with the circle $C(D),\;$ centered at $C\;$ and passing through $D.$

Golden Ratio In a 2x2 Square: Without And Within, diagram 1

Then $M\;$ divides $AB\;$ in the Golden Ratio: $\displaystyle\frac{BM}{AM}=\varphi.$

The proof is also straightforward: assume $AO=OB=BC=OD=1.\;$ Then $CM=CD=\sqrt{5};\;$ $BM=\sqrt{5}-1;\;$ $AM=3-\sqrt{5},\;$ so that $\displaystyle\frac{BM}{AM}=\frac{\sqrt{5}-1}{3-\sqrt{5}}=\frac{\sqrt{5}+1}{2}=\varphi.$

Dung Thanh Nguyen's construction can be represented a little differently, starting with a $2\times 2\;$ square:

Golden Ratio In a 2x2 Square: Without And Within, diagram 2

This construction is responsible to the Without part in the title of the page. The Within part is illustrated below.

Golden Ratio In a 2x2 Square: Without And Within, diagram 3

In all three diagrams the ratio of the red segment to the blue one is Golden.

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