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Brazilian Mathematical Olympiad

Geometry Difficulty 5.4 AIME, harder Prove it Brazil

Line r passes through the corner $A$ of a sheet of paper and makes an angle $\alpha$ with the horizontal border, as shown in figure 1. In order to divide $\alpha$ into three equal parts we proceed as follows:

a) initially we mark two points $B$ and $C$ on the vertical border such that $AB = BC$ ; through $B$ we draw a line $s$ parallel to the border (figure 2);

b) after that, we fold the sheet so as to make $C$ coincide with a point $C'$ on the line $r$ and $A$ with a point $A'$ on line $s$ (figure 3); we call $B'$ the point which coincides with $B$ .

Figure 1

Figure 2

Figure 3

Show that lines $AA'$ and $AB'$ divides angle $\alpha$ into three equal parts.

Solution

Let $P$ and $X$ be the points determined by the crease in the lower horizontal border of the sheet and on the line $s$ , respectively. Let $\beta = \angle PAA'$ . Since $AP = AP'$ , $\angle PA'A = \angle AA'B = \beta$ , thus $\angle PA'X = 2\beta$ . Moreover $\triangle APX \cong \triangle A'PX$ , so that $\angle XAA' = \angle XAP - \beta = \beta$ .

Now observe that $\angle BXA = 2\beta$ and, since $A'P \parallel B'P$ , $\angle B'XA' = 2\beta$ also. Thus $A, X$ and $B'$ are collinear. Therefore $AB'$ is a median and height of $\triangle C'AA'$ , so $AB'$ is also an angle bisector of $\angle C'AA'$ . Thus $\angle C'AB' = \angle B'AA' = \angle A'AP = \beta$ .

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