Difference between revisions of "Concurrence"

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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.
  
Several [[line]]s (or [[curve]]s) are said to '''concur''' at a [[point]] if they all contain that point.
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== Proving concurrence ==
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In analytical geometry, one can find the point of concurrency of any two lines by solving the system of equations of the lines. To see if it shares the point of concurrency with other lines/curves requires only to test that point.
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[[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.
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Concurrence of lines can occasionally be proved by showing that a certain point is a center of a particular [[homothecy]].
  
== Proving concurrence ==
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==Problems==
In analytical geometry, one can find the points of concurrency of any two lines by solving the system of equations of the lines.
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===Introductory===
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*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]])
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===Intermediate===
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*In triangle <math>ABC^{}_{}</math>, <math>A'</math>, <math>B'</math>, and <math>C'</math> are on the sides <math>BC</math>, <math>AC^{}_{}</math>, and <math>AB^{}_{}</math>, respectively. Given that <math>AA'</math>, <math>BB'</math>, and <math>CC'</math> are concurrent at the point <math>O^{}_{}</math>, and that <math>\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92</math>, find <math>\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}</math>.
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([[1992 AIME Problems/Problem 14|Source]])
  
[[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 bisectors (which is not a cevian) of any triangle are concurrent, at the [[orthocenter]], [[incenter]], [[centroid]], [[circumcenter]], and [[Lemoine point]] respectively.
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===Olympiad===
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*Hallie is teaching geometry to Warren. She tells him that the three medians, the three angle bisectors, and the three altitudes of a triangle each meet at a point (the centroid, incenter, and orthocenter respectively). Warren gets a little confused and draws a certain triangle ABC along with the median from vertex A, the altitude from vertex B, and the angle bisector from vertex C. Hallie is surprised to see that the three segments meet at a point anyway! She notices that all three sides measure an integer number of inches, that the side lengths are all distinct, and that the side across from vertex C is 13 inches in length. How long are the other two sides? ([[Mathcamp]] Qualifying Quiz, #10, also see the [[Concurrence/Problems#Olympiad|Solution]])
  
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[[Category:Definition]]
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[[Category:Geometry]]
 
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Latest revision as of 22:57, 15 February 2008

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.

Proving concurrence

In analytical geometry, one can find the point of concurrency of any two lines by solving the system of equations of the lines. To see if it shares the point of concurrency with other lines/curves requires only to test that point.

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.

Concurrence of lines can occasionally be proved by showing that a certain point is a center of a particular homothecy.

Problems

Introductory

  • Are the lines $y=2x+2$, $y=3x+1$, and $y=5x-1$ concurrent? If so, find the the point of concurrency. (Source)

Intermediate

  • In triangle $ABC^{}_{}$, $A'$, $B'$, and $C'$ are on the sides $BC$, $AC^{}_{}$, and $AB^{}_{}$, respectively. Given that $AA'$, $BB'$, and $CC'$ are concurrent at the point $O^{}_{}$, and that $\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92$, find $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}$.

(Source)

Olympiad

  • Hallie is teaching geometry to Warren. She tells him that the three medians, the three angle bisectors, and the three altitudes of a triangle each meet at a point (the centroid, incenter, and orthocenter respectively). Warren gets a little confused and draws a certain triangle ABC along with the median from vertex A, the altitude from vertex B, and the angle bisector from vertex C. Hallie is surprised to see that the three segments meet at a point anyway! She notices that all three sides measure an integer number of inches, that the side lengths are all distinct, and that the side across from vertex C is 13 inches in length. How long are the other two sides? (Mathcamp Qualifying Quiz, #10, also see the Solution)

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