Difference between revisions of "Monge's theorem"
m |
m |
||
(2 intermediate revisions by the same user not shown) | |||
Line 72: | Line 72: | ||
Let <math>h_{12}</math> be the positive [[homothety]] taking <math>\omega_1</math> to <math>\omega_2</math>. Let <math>h_{23}</math> be the positive homothety taking <math>\omega_2</math> to <math>\omega_3</math>. <math>h_{12}</math> and <math>h_{23}</math> have centers <math>Z</math> and <math>X</math> respectively. Consider <math>h_{23}\circ h_{12}</math>. It is well known that the composition of two homotheties is also a homothety, with center lying on the line joining the centers of the two original homotheties. Also, its coefficient of homothety equals the product of the coefficients of the original two homotheties, so we know that the coefficient of <math>h_{23}\circ h_{12}</math> is positive. As <math>h_{23}\circ h_{12}</math> takes <math>\omega_1</math> to <math>\omega_3</math> we know its center is <math>Y</math>. Therefore <math>X, Y, Z</math> are collinear. <math>\square</math> | Let <math>h_{12}</math> be the positive [[homothety]] taking <math>\omega_1</math> to <math>\omega_2</math>. Let <math>h_{23}</math> be the positive homothety taking <math>\omega_2</math> to <math>\omega_3</math>. <math>h_{12}</math> and <math>h_{23}</math> have centers <math>Z</math> and <math>X</math> respectively. Consider <math>h_{23}\circ h_{12}</math>. It is well known that the composition of two homotheties is also a homothety, with center lying on the line joining the centers of the two original homotheties. Also, its coefficient of homothety equals the product of the coefficients of the original two homotheties, so we know that the coefficient of <math>h_{23}\circ h_{12}</math> is positive. As <math>h_{23}\circ h_{12}</math> takes <math>\omega_1</math> to <math>\omega_3</math> we know its center is <math>Y</math>. Therefore <math>X, Y, Z</math> are collinear. <math>\square</math> | ||
+ | |||
+ | This proof is particularly useful because it shows a more general claim: For any three circles, their pairwise exsimilicenters are collinear. | ||
== Proof 3 ([[Desargues' theorem]]) == | == Proof 3 ([[Desargues' theorem]]) == | ||
Line 142: | Line 144: | ||
If two circles are congruent (say <math>\omega_2</math> and <math>\omega_3</math>) and the third isn't (<math>\omega_1</math>,) then <math>X</math> is a point at infinity of the pencil of lines parallel to <math>O_2O_3,</math> and <math>Y</math> and <math>Z</math> are Euclidean (ordinary) points. By similarity (similar to proof 1,) we get <cmath>\frac{\overrightarrow{O_1Y}}{\overrightarrow{O_1O_3}}=\frac{\overrightarrow{O_1Z}}{\overrightarrow{O_1O_2}},</cmath> hence <math>YZ\parallel O_2O_3,</math> so <math>X, Y, Z</math> are collinear. | If two circles are congruent (say <math>\omega_2</math> and <math>\omega_3</math>) and the third isn't (<math>\omega_1</math>,) then <math>X</math> is a point at infinity of the pencil of lines parallel to <math>O_2O_3,</math> and <math>Y</math> and <math>Z</math> are Euclidean (ordinary) points. By similarity (similar to proof 1,) we get <cmath>\frac{\overrightarrow{O_1Y}}{\overrightarrow{O_1O_3}}=\frac{\overrightarrow{O_1Z}}{\overrightarrow{O_1O_2}},</cmath> hence <math>YZ\parallel O_2O_3,</math> so <math>X, Y, Z</math> are collinear. | ||
+ | |||
+ | [[Category:Theorems]] | ||
+ | [[Category:Geometry]] |
Latest revision as of 08:08, 11 October 2024
|
A diagram of Monge's theorem; , and are collinear. |
Monge's theorem is a theorem in Euclidean geometry. It states that given three circles, and , none of which lies completely inside one of the others, if we construct the intersections of their common external tangents, then these intersections are collinear.
For the rest of this article, we will denote the intersections of the external tangents to and and and by respectively. Note that the external tangent to two circles is a line tangent to both of them which doesn't pass through the segment connecting the circles' centers. Exactly two of these tangents exist for any two circles not contained in each other. Internal tangents are defined similarly, but they pass through the segment joining the centers of the circles. If we replace two of the external tangent intersections with internal tangent intersections, the statement still holds. Each of the proofs below may be modified to show this. Also note that the degenerate cases with parallel tangents may need to be handled seperately.
Contents
Proof 1 (Menelaus' theorem)
Let be the centers and be the radii of respectively. First consider and . lies on line . Let one of the two common tangents touch at , and touch at . , hence As doesn't lie on segment , we get Similarly, Therefore, By Menelaus' theorem, and are collinear.
Proof 2 (Homothety)
Let be the positive homothety taking to . Let be the positive homothety taking to . and have centers and respectively. Consider . It is well known that the composition of two homotheties is also a homothety, with center lying on the line joining the centers of the two original homotheties. Also, its coefficient of homothety equals the product of the coefficients of the original two homotheties, so we know that the coefficient of is positive. As takes to we know its center is . Therefore are collinear.
This proof is particularly useful because it shows a more general claim: For any three circles, their pairwise exsimilicenters are collinear.
Proof 3 (Desargues' theorem)
Let be the centers of respectively. Let be the external tangent to which doesn't intersect . Define and similarly. Let be the intersection of and . Define and similary. Note that lie on respectively.
Lines are the internal angle bisectors in hence they concur in the incenter of As and are perspective from a point, by Desargues' theorem, they are also perspective from a line, hence are collinear, but these are precisely the points
Degenerate cases
If any two of are congruent, their external tangents are parallel. The non-projective proofs fail in these cases, so they need to be handled seperately.
If all three circles are congruent, all three tangent pairs are parallel, so are all points at infinity. Hence, all of them lie on the line at infinity.
If two circles are congruent (say and ) and the third isn't (,) then is a point at infinity of the pencil of lines parallel to and and are Euclidean (ordinary) points. By similarity (similar to proof 1,) we get hence so are collinear.