2003 Pan African MO Problems/Problem 2

Problem

The circumference of a circle is arbitrarily divided into four arcs. The midpoints of the arcs are connected by segments. Show that two of these segments are perpendicular.

Solution

Let $A, B, C, D$ in that order be the four points that divide the circle into four arcs, and let $X$ be the intersection of $AC$ and $BD$ . Let $F, G, H, J$ be the midpoints of $\overarc{AB}, \overarc{BC}, \overarc{CD}, \overarc{DA}$ respectively. Additionally, let $a = \overarc{AB}, b = \overarc{BC}, c = \overarc{CD},$ and $d = \overarc{DA}$ .

Note that $a+b+c+d = 360^\circ$ . Additionally, from the definition of midpoint, $FB = \tfrac{a}{2}$ and $BG = \tfrac{b}{2}$ . Thus, $FG = \tfrac{a+b}{2}$ . Likewise, $HD = \tfrac{c}{2}$ and $DJ = \tfrac{d}{2}$ , so $HJ = \tfrac{c+d}{2}$ . Therefore, $\angle FXG = \tfrac{a+b+c+d}{2} = \tfrac{180}{2} = 90^\circ$ , so $FH \perp GJ$ .

2003 Pan African MO (Problems)
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
All Pan African MO Problems and Solutions

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2003 Pan African MO Problems/Problem 2

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