Difference between revisions of "Concurrence"
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− | Several [[line]]s (or [[curve]]s) are said to ''' | + | Several (that is, three or more) [[line]]s (or [[curve]]s) are said to be '''concurrent''' at a [[point]] if they all contain that point. The point is said to be the point of concurrence. |
== Proving concurrence == | == Proving concurrence == | ||
In analytical geometry, one can find the points of concurrency of any two lines by solving the system of equations of the lines. | In analytical geometry, one can find the points of concurrency of any two lines by solving the system of equations of the lines. | ||
− | [[Ceva's Theorem]] gives a criteria for three [[cevian]]s of a triangle to be concurrent. In particular, the three [[altitude]]s, [[angle bisector]]s, [[median]]s, [[symmedian]]s, and [[perpendicular bisector]]s (which are not cevians) of any triangle are concurrent, at the [[orthocenter]], [[incenter]], [[centroid]], [[Lemoine point]], and [[circumcenter]], respectively. | + | ==Related Theorems== |
+ | *[[Ceva's Theorem]] gives a criteria for three [[cevian]]s of a triangle to be concurrent. In particular, the three [[altitude]]s, [[angle bisector]]s, [[median]]s, [[symmedian]]s, and [[perpendicular bisector]]s (which are not cevians) of any triangle are concurrent, at the [[orthocenter]], [[incenter]], [[centroid]], [[Lemoine point]], and [[circumcenter]], respectively. | ||
+ | ==Problems== | ||
+ | ===Introductory=== | ||
+ | *Are the lines <math>y=2x+2</math>, <math>y=3x+1</math>, and <math>y=5x-1</math> concurrent? If so, find the the point of concurrency. ([[Concurrence/Problems#Introductory|Source]]) | ||
+ | ===Intermediate=== | ||
+ | ===Olympiad=== | ||
+ | |||
+ | [[Category:Definition]] | ||
+ | [[Category:Geometry]] | ||
{{stub}} | {{stub}} |
Revision as of 20:29, 23 November 2007
This is an AoPSWiki Word of the Week for Nov 22-28 |
Several (that is, three or more) lines (or curves) are said to be concurrent at a point if they all contain that point. The point is said to be the point of concurrence.
Contents
Proving concurrence
In analytical geometry, one can find the points of concurrency of any two lines by solving the system of equations of the lines.
Related Theorems
- Ceva's Theorem gives a criteria for three cevians of a triangle to be concurrent. In particular, the three altitudes, angle bisectors, medians, symmedians, and perpendicular bisectors (which are not cevians) of any triangle are concurrent, at the orthocenter, incenter, centroid, Lemoine point, and circumcenter, respectively.
Problems
Introductory
- Are the lines , , and concurrent? If so, find the the point of concurrency. (Source)
Intermediate
Olympiad
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