2003 Pan African MO Problems/Problem 2

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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.


Let $A, B, C, D$ in that order be the four points that divide the circle into four arcs. Let $F, G, H, J$ be the midpoints of $\overarc{AB}, \overarc{BC}, \overarc{CD}, \overarc{DA}$ respectively, and let $X$ be the intersection of $FH$ and $GJ$. 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 $FH \perp GJ$.

See Also

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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