The geometric construction of a golden mean rectangle is shown in the figure below. It starts with a square ABCD of 1 cm side. Continuing the development of the golden mean rectangles, the derived length of IJ (in cm) is (rounded off to three decimal places).
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Each new golden rectangle forms by adding a square on the longer side of the previous one; use phi^2 = phi + 1 to track the growth, or track it stage by stage with coordinates.
Step 1: Set up coordinates for the unit square.
Place the square \(ABCD\) with \(A = (0,1)\), \(B = (1,1)\), \(C = (1,0)\), \(D = (0,0)\), so its side is \(1\) cm, matching the \(1\) cm mark shown along \(BC\). \(O\) is the midpoint of \(AB\), so \(O = (0.5, 1)\), since the figure marks \(AO = OB\).
Step 2: Find the swing radius OD.
By the distance formula,
\[ OD = \sqrt{(0.5-0)^2 + (1-0)^2} = \sqrt{0.25+1} = \sqrt{1.25} = \frac{\sqrt{5}}{2} \approx 1.11803 \]
This is the radius of the arc drawn from \(O\) through \(D\), which is swung across to locate \(F\) on the line through \(A\) and \(B\), extended to the left.
Step 3: Locate F and get the first golden rectangle FBCE.
Since \(F\) lies on the same horizontal line as \(A, O, B\) at distance \(OD\) from \(O\),
\[ OF = \frac{\sqrt5}{2}, \quad AF = OF - OA = \frac{\sqrt5}{2} - 0.5 = \frac{\sqrt5 - 1}{2} \approx 0.618 \]
So rectangle \(FBCE\) has width
\[ FB = AF + AB = 0.618 + 1 = 1.618 \]
and height \(1\). The ratio \(1.618 : 1\) is the golden ratio \(\varphi = \dfrac{1+\sqrt5}{2}\), so \(FBCE\) is itself a golden mean rectangle built from the square.
Step 4: Use the golden ratio identity to grow the rectangle again.
The defining property of \(\varphi\) is \(\varphi^2 = \varphi + 1\). Whenever a square is added on the LONGER side of an existing golden rectangle, the new combined rectangle is again a golden rectangle, with its long side equal to \(\varphi\) times the old long side.
Stage 1 (square \(ABCD\)): sides \(1\) and \(1\).
Stage 2 (rectangle \(FBCE\), a rectangle added to the left of stage 1): sides \(1\) and \(\varphi\).
Stage 3 (a square of side \(\varphi\) added on top, reaching \(G\) and \(H\)): sides \(\varphi\) and \(1+\varphi = \varphi^2\).
Stage 4 (a square of side \(\varphi^2\) added to the right, reaching \(I\) and \(J\)): sides \(\varphi^2\) and \(\varphi^2+\varphi=\varphi^3\).
Step 5: Identify IJ.
\(I\) and \(J\) are the top-right and bottom-right corners of the final rectangle, so \(IJ\) is the short (vertical) side of the final golden rectangle from Stage 4, which is \(\varphi^2\), the side length of the last square added.
\[ \varphi = \frac{1+\sqrt5}{2} = \frac{1+2.236068}{2} = 1.618034 \]
\[ IJ = \varphi^2 = \varphi + 1 = 2.618034 \]
Step 6: Final Answer.
Rounded to three decimal places,
\[ \boxed{IJ \approx 2.618 \text{ cm}} \]
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