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PrintXXII OBM
Brazil geometry
Problem
Line r passes through the corner of a sheet of paper and makes an angle with the horizontal border, as shown in figure 1. In order to divide into three equal parts we proceed as follows:
a) initially we mark two points and on the vertical border such that ; through we draw a line parallel to the border (figure 2);
b) after that, we fold the sheet so as to make coincide with a point on the line and with a point on line (figure 3); we call the point which coincides with .
Figure 1
Figure 2
Figure 3
Show that lines and divides angle into three equal parts.
a) initially we mark two points and on the vertical border such that ; through we draw a line parallel to the border (figure 2);
b) after that, we fold the sheet so as to make coincide with a point on the line and with a point on line (figure 3); we call the point which coincides with .
Show that lines and divides angle into three equal parts.
Solution
Let and be the points determined by the crease in the lower horizontal border of the sheet and on the line , respectively. Let . Since , , thus . Moreover , so that .
Now observe that and, since , also. Thus and are collinear. Therefore is a median and height of , so is also an angle bisector of . Thus .
Now observe that and, since , also. Thus and are collinear. Therefore is a median and height of , so is also an angle bisector of . Thus .
Techniques
Angle chasingConstructions and loci