CTK Insights

23 May

A square in parallel lines

An a\times a square intersects two parallel lines located at distance a from each other. The points of intersection are cross joined. Prove that the angle so obtained is 45^{\circ}.

The problem is credited to V. V. Proizvolov (Kvant 1996 (2)) and may serve an example where using dynamical software to sketch a diagram proves to be a distraction. I put together a GeoGebra applet and looked at the possible properties of the configuration with several elements added. Meanwhile, Hubert Shutrick pointed out its obvious, salient feature. The solution is a one liner.

Solution

The angle formed by a transversal with two parallel lines is uniquely determined by its length between the lines - and vice versa.

The rest is easy:

\gamma+\delta=90^{\circ}

2\alpha + 2\beta +\gamma+\delta=360^{\circ}

\alpha + \beta =135^{\circ}

\epsilon =45^{\circ}

4 Responses to “A square in parallel lines”

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