Difference between revisions of "2003 Pan African MO Problems/Problem 2"
Rockmanex3 (talk | contribs) (Solution to Problem 2 -- VERY EASY circle angle problem) |
Rockmanex3 (talk | contribs) (Fix) |
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==Solution== | ==Solution== | ||
− | Let <math>A, B, C, D</math> in that order be the four points that divide the circle into four arcs | + | Let <math>A, B, C, D</math> in that order be the four points that divide the circle into four arcs. Let <math>F, G, H, J</math> be the midpoints of <math>\overarc{AB}, \overarc{BC}, \overarc{CD}, \overarc{DA}</math> respectively, and let <math>X</math> be the intersection of <math>FH</math> and <math>GJ</math>. Additionally, let <math>a = \overarc{AB}, b = \overarc{BC}, c = \overarc{CD},</math> and <math>d = \overarc{DA}</math>. |
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Latest revision as of 16:25, 28 January 2020
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 in that order be the four points that divide the circle into four arcs. Let be the midpoints of respectively, and let be the intersection of and . Additionally, let and .
Note that . Additionally, from the definition of midpoint, and . Thus, . Likewise, and , so . Therefore, , so .
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 |