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Difference between revisions of "Monge's theorem"
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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>
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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]]) ==
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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.
  
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[[Category:Theorems, Geometry]]
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[[Category:Theorems]]
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[[Category:Geometry]]

Latest revision as of 08:08, 11 October 2024

[asy] import geometry; size(6cm); point O1 = (-0.5, 0.5); point O2 = (1.0, 1.0); point O3 = (0.3, -0.5); dot(O1); dot(O2); dot(O3); real r1 = 0.2; real r2 = 0.7; real r3 = 0.3; circle omega1 = circle(O1, r1); circle omega2 = circle(O2, r2); circle omega3 = circle(O3, r3); draw(omega1); draw(omega2); draw(omega3); point X = O2 + (1 / (1 - r3 / r2)) * (O3 - O2); point Y = O3 + (1 / (1 - r1 / r3)) * (O1 - O3); point Z = O1 + (1 / (1 - r2 / r1)) * (O2 - O1); dot("$X$", X); dot("$Y$", Y, NE); dot("$Z$", Z, SE); line[] t1 = tangents(omega3, X); line[] t2 = tangents(omega1, Y); line[] t3 = tangents(omega2, Z); draw(t1, red); draw(t2, red); draw(t3, red); draw(X--Y, purple + dotted); [/asy]

Enlarge.png
A diagram of Monge's theorem; $X$, $Y$ and $Z$ are collinear.

Monge's theorem is a theorem in Euclidean geometry. It states that given three circles, $\omega_1, \omega_2$ and $\omega_3$, 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 $\omega_2$ and $\omega_3, \omega_3$ and $\omega_1, \omega_1$ and $\omega_2$ by $X, Y, Z$ 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.

Proof 1 (Menelaus' theorem)

[asy] import geometry; size(6cm); point O2 = (0, 0); point O3 = (1, 0); real r2 = 0.3; real r3 = 0.5; circle omega2 = circle(O2, r2); circle omega3 = circle(O3, r3); draw(omega2); draw(omega3); dot("$O_2$", O2, S); dot("$O_3$", O3, S); point X = O2 + (1 / (1 - r3 / r2)) * (O3 - O2); line[] t = tangents(omega2, X); draw(t, red); dot("$X$", X, S); draw(X--O3, dotted); point P = intersectionpoint(t[1], perpendicular(O2, t[1])); point Q = intersectionpoint(t[1], perpendicular(O3, t[1])); draw(O2--P, dotted); draw(O3--Q, dotted); dot("$P$", P, N); dot("$Q$", Q, N); [/asy]

Let $O_1, O_2, O_3$ be the centers and $r_1, r_2, r_3$ be the radii of $\omega_1, \omega_2, \omega_3$ respectively. First consider $\omega_2$ and $\omega_3$. $X$ lies on line $O_1O_2$. Let one of the two common tangents touch $\omega_2$ at $P$, and touch $\omega_3$ at $Q$. $\triangle XO_2P\sim \triangle XO_3Q$, hence \[\frac{O_2X}{O_3X}=\frac{O_2P}{O_3Q}=\frac{r_2}{r_3}.\] As $X$ doesn't lie on segment $O_2O_3$, we get \[\frac{\overrightarrow{O_2X}}{\overrightarrow{XO_3}}=-\frac{r_2}{r_3}.\] Similarly, \[\frac{\overrightarrow{O_3Y}}{\overrightarrow{YO_1}}=-\frac{r_3}{r_1}, \frac{\overrightarrow{O_1Z}}{\overrightarrow{ZO_2}}=-\frac{r_1}{r_2}.\] Therefore, \[\frac{\overrightarrow{O_2X}}{\overrightarrow{XO_3}}\cdot\frac{\overrightarrow{O_3Y}}{\overrightarrow{YO_1}}\cdot\frac{\overrightarrow{O_1Z}}{\overrightarrow{ZO_2}}=\left(-\frac{r_2}{r_3}\right)\left(-\frac{r_3}{r_1}\right)\left(-\frac{r_1}{r_2}\right)=-1.\] By Menelaus' theorem, $X, Y$ and $Z$ are collinear. $\square$

Proof 2 (Homothety)

Let $h_{12}$ be the positive homothety taking $\omega_1$ to $\omega_2$. Let $h_{23}$ be the positive homothety taking $\omega_2$ to $\omega_3$. $h_{12}$ and $h_{23}$ have centers $Z$ and $X$ respectively. Consider $h_{23}\circ h_{12}$. 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 $h_{23}\circ h_{12}$ is positive. As $h_{23}\circ h_{12}$ takes $\omega_1$ to $\omega_3$ we know its center is $Y$. Therefore $X, Y, Z$ are collinear. $\square$

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)

[asy] import geometry; size(10cm); point O1 = (-0.5, 0.5); point O2 = (1.0, 1.0); point O3 = (0.3, -0.5); dot(O1); dot(O2); dot(O3); real r1 = 0.2; real r2 = 0.7; real r3 = 0.3; circle omega1 = circle(O1, r1); circle omega2 = circle(O2, r2); circle omega3 = circle(O3, r3); draw(omega1); draw(omega2); draw(omega3); point X = O2 + (1 / (1 - r3 / r2)) * (O3 - O2); point Y = O3 + (1 / (1 - r1 / r3)) * (O1 - O3); point Z = O1 + (1 / (1 - r2 / r1)) * (O2 - O1); dot("$X$", X); dot("$Y$", Y, NE); dot("$Z$", Z, SE); line[] t1 = tangents(omega3, X); line[] t2 = tangents(omega1, Y); line[] t3 = tangents(omega2, Z); draw(t1[0], red); draw(t2[0], red); draw(t3[1], red); point A1 = intersectionpoint(t2[0], t3[1]); point A2 = intersectionpoint(t3[1], t1[0]); point A3 = intersectionpoint(t1[0], t2[0]); point I = intersectionpoint(line(A1, O1), line(A2, O2)); draw(X--Y, purple + dotted); draw(X--O2, dotted); draw(Y--O3, dotted); draw(Z--O2, dotted); draw(A1--I, blue + dashed); draw(A2--I, blue + dashed); draw(A3--I, blue + dashed); dot("$A_1$", A1, N); dot("$A_2$", A2); dot("$A_3$", A3); dot("$O_1$", O1, NE); dot("$O_2$", O2, NW); dot("$O_3$", O3, E); [/asy]

Let $O_1, O_2, O_3$ be the centers of $\omega_1, \omega_2, \omega_3$ respectively. Let $t_1$ be the external tangent to $\omega_2, \omega_3$ which doesn't intersect $\triangle O_1O_2O_3$. Define $t_2$ and $t_3$ similarly. Let $A_1$ be the intersection of $t_2$ and $t_3$. Define $A_2$ and $A_3$ similary. Note that $X, Y, Z$ lie on $O_2O_3, O_3O_1, O_1O_2$ respectively.

Lines $A_1O_1, A_2O_2, A_3O_3$ are the internal angle bisectors in $\triangle A_1A_2A_3,$ hence they concur in the incenter of $\triangle A_1A_2A_3.$ As $\triangle A_1A_2A_3$ and $\triangle O_1O_2O_3$ are perspective from a point, by Desargues' theorem, they are also perspective from a line, hence $A_2A_3\cap O_2O_3, A_3A_1\cap O_3O_1, A_1A_2\cap O_1O_2$ are collinear, but these are precisely the points $X, Y, Z.$ $\square$

Degenerate cases

If any two of $\omega_1, \omega_2, \omega_3$ 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 $X, Y, Z$ are all points at infinity. Hence, all of them lie on the line at infinity.

If two circles are congruent (say $\omega_2$ and $\omega_3$) and the third isn't ($\omega_1$,) then $X$ is a point at infinity of the pencil of lines parallel to $O_2O_3,$ and $Y$ and $Z$ are Euclidean (ordinary) points. By similarity (similar to proof 1,) we get \[\frac{\overrightarrow{O_1Y}}{\overrightarrow{O_1O_3}}=\frac{\overrightarrow{O_1Z}}{\overrightarrow{O_1O_2}},\] hence $YZ\parallel O_2O_3,$ so $X, Y, Z$ are collinear.

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