Difference between revisions of "Excircle"

Revision as of 09:13, 5 November 2007

An excircle is a circle tangent to the extensions of two sides of a triangle and the third side.

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Properties

For any triangle, there are three unique excircles. This follows from the fact that there is one, if any, circle such that three given distinct lines are tangent to it.

Related Formulas

If the circle is tangent to side $a$ of the circle, the radius is $\frac{2K}{-a+b+c}$ , where $K$ is the triangle's area, and $a,b,c$ are side lengths.
If the circle is tangent to side $b$ of the circle, the radius is $\frac{2K}{a-b+c}$ , where $K$ is the triangle's area, and $a,b,c$ are side lengths.
If the circle is tangent to side $c$ of the circle, the radius is $\frac{2K}{a+b-c}$ , where $K$ is the triangle's area, and $a,b,c$ are side lengths.

Problems

Introductory

Intermediate

Olympiad

$\triangle ABC$ is a triangle. Take points $D, E, F$ on the perpendicular bisectors of $BC, CA, AB$ respectively. Show that the lines through $A, B, C$ perpendicular to $EF, FD, DE$ respectively are concurrent. (Source)
Let $ABC$ be a triangle with circumcircle $\omega.$ Point $D$ lies on side $BC$ such that $\angle BAD = \angle CAD.$ Let $I_{A}$ denote the excenter of triangle $ABC$ opposite $A,$ and let $\omega_{A}$ denote the circle with $AI_{A}$ as its diameter. Circles $\omega$ and $\omega_{A}$ meet at $P$ other than $A.$ The circumcle of triangle $APD$ meet line $BC$ again at $Q\, ($ other than $D).$ Prove that $Q$ lies on the excircle of triangle $ABC$ opposite $A$ . (Source: Problem 13.2 - MOSP 2007)

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 ===Olympiad===
 *<math>\triangle ABC</math> is a triangle. Take points <math>D, E, F</math> on the perpendicular bisectors of <math>BC, CA, AB</math> respectively. Show that the lines through <math>A, B, C</math> perpendicular to <math>EF, FD, DE</math> respectively are concurrent. ([[1997 USAMO Problems/Problem 2|Source]])
+*Let <math>ABC</math> be a triangle with circumcircle <math>\omega.</math> Point <math>D</math> lies on side <math>BC</math> such that <math>\angle BAD = \angle CAD.</math> Let <math>I_{A}</math> denote the excenter of triangle <math>ABC</math> opposite <math>A,</math> and let <math>\omega_{A}</math> denote the circle with <math>AI_{A}</math> as its diameter. Circles <math>\omega</math> and <math>\omega_{A}</math> meet at <math>P</math> other than <math>A.</math> The circumcle of triangle <math>APD</math> meet line <math>BC</math> again at <math>Q\, (</math>other than <math>D).</math> Prove that <math>Q</math> lies on the excircle of triangle <math>ABC</math> opposite <math>A</math>. (Source: Problem 13.2 - MOSP 2007)
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