Vinent Pantaloni (Orléans, Frane)
Problem :Two straight non-parallel lines are drawn on a gymnasium floor as shown, for example, in the figure below. You are to use a compass and straightedge to construct the portion of the bisector of the angle that would be formed by these two lines if they extended outside the gymnasium. Your construction must take place entirely within the gymnasium.
Solution :
My solution uses the following property : The center of a parallelogram which is the midpoint of the diagonals is a center of symmetry of the parallelogram. So bisectors of opposite angles are parallel and symmetrical with respect to that center.
There are two cases : either one of the sides of the gymnasium is cut by the two lines, or the two lines cut all four sides once. The latter case can be solved in quite the same way as the first one which I will start with.
We know that it is possible with straight edge and compass to draw : 1. the line parallel to a given line passing through a given point.
2. the midpoint of a segment line by drawing its perpendicular bisector.
3. the bisectors of two non parallel lines if we have the point where they intersect.
Here’s a figure showing the gymnasiumABC D and in green and blue the two straight non-parallel lines.
The goal is to construct the red bisector.
① Draw the two dashed blue and green paral- lels. Let’s suppose that they intersect inJ inside the gymnasium.
EH J I is a parallelogram.
② Draw the red dashed bisector of∠H J Ithat intersects[AD]in L.
③ Find the midpointKof[I H], center of sym- metry ofEH J I.
④ The circle whose center isKand radiusKL cuts[I H]in a second pointM.
⑤ Draw the parallel to (J L) passing through M. That is the bisector we were looking for.
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Now ifJ is not inside the gymnasium we have a problem, but here is a workaround shown on the figure below. The idea is to construct a parallelogram that is twice as small as EH J I by drawing the green and blue dotted parallels passing through the midpointK of[I H]. Suppose they meet inO inside the gymnasium. Repeat the previous construction with angle∠KOI giving the black bisector (OP)and small black circle. Since the parallelogram is twice as small, the small black circle (cyan diameter) whose center is N has a radius one half of the one we need to draw with centerK. We can then haveM and draw the parallel to(OP)passing throughM to get the desired bisector.
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If O is still out of the gymnasium we can repeat this process, building each time a parallelogram that is half the size of the previous one, so even- tually we will have a third vertex of the parallelo- gram inside the gymnasium as seen on this figure : after four series of parallels we get the point V.
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Finally, here is a last figure showing that the other case where the non parallel lines do not intersect a common side of the gymnasium is actually trea- ted in the same manner. Again you might need to draw some more parallels going through succes- sive midpoints of the pink diagonal in order to have your third vertex of a parallelogram inside the gymanasium.
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