synthetic proof of X(12)

https://artofproblemsolving.com/community/c374081h2712819

Author: auntyellow

diagrams/monge.ggb

@@ -51,14 +51,16 @@ Other proofs can be found [here](https://imomath.com/index.php?options=323) (Pro
 
 **Theorem 1** The circle through the [feet of the internal bisectors](https://mathworld.wolfram.com/IncentralTriangle.html) *I*<sub>A</sub>*I*<sub>B</sub>*I*<sub>C</sub> of a triangle *ABC* passes through the Feuerbach point.
 
-[Here](pythagoras/feuerbach-1.py) is a computational proof *starting from incircle*.
+[Here](pythagoras/feuerbach-1.py) is the computational proof *starting from incircle*.
 
 **Theorem 2** The Feuerbach point of a triangle *ABC* is the [anti-Steiner point](https://artofproblemsolving.com/community/c1646h1025320s3_antisteiner_point) of the Euler line of the [intouch triangle](https://mathworld.wolfram.com/ContactTriangle.html) *C*<sub>A</sub>*C*<sub>B</sub>*C*<sub>C</sub> with respect to the same intouch triangle *C*<sub>A</sub>*C*<sub>B</sub>*C*<sub>C</sub>.
 
-[Here](pythagoras/feuerbach-2.py) is a computational proof *starting from incircle*.
+[Here](pythagoras/feuerbach-2.py) is the computational proof *starting from incircle*.
 
 Other proofs of the two theorems can be found [here](http://blancosilva.github.io/post/2013/07/15/some-results-related-to-the-feuerbach-point.html) (computational) and [here](https://forumgeom.fau.edu/FG2012volume12/FG201205.pdf) (synthetic).
 
 **Theorem 3** Let *F*<sub>A</sub>, *F*<sub>B</sub>, *F*<sub>C</sub> be the touch points of the nine-point circle with the *A*-, *B*-, *C*- excircles, respectively. The lines *AF*<sub>A</sub>, *BF*<sub>B</sub>, *CF*<sub>C</sub> meet at *X*(12), the harmonic conjugate of the Feuerbach point *F*<sub>I</sub> with respect to the incenter *I* and the nine-point center *N*. <sup>[1](https://www.cut-the-knot.org/Curriculum/Geometry/FeuerbachIncidence.shtml)</sup>
 
-[Here](pythagoras/feuerbach-3.py) is a computational proof *starting from incircle*.
+[Here](pythagoras/feuerbach-3.py) is the computational proof *starting from incircle*.
+
+This theorem can also be proved by [Monge's theorem](monge.md#monges-theorem) (Theorem 2).

diagrams/monge.png

@@ -93,31 +93,41 @@ which is between *a* and *b*. So it should be the internal intersection. Denote
 
 <img src="https://latex.codecogs.com/gif.latex?\frac{\overrightarrow{H'A}}{\overrightarrow{BH'}}=\frac{a-h'}{h'-b}=\frac{R}r\quad\text{(Eq.\,2)}">
 
+The external and internal intersections, are the external and internal **[homothetic centers](https://en.wikipedia.org/wiki/Homothetic_center#Circles)** of two circles, and they are harmonic conjugate with respect to the two circle centers:
+
+<img src="https://latex.codecogs.com/gif.latex?(A,B;H,H')=-1">
+
 For example, one circle is inside the other:
 
 <img src="https://latex.codecogs.com/gif.latex?\begin{cases}x^2+y^2=9\\x^2+(y+1)^2=1\end{cases}">
 
-The two external tangent lines are <img src="https://latex.codecogs.com/gif.latex?y=\pm\sqrt{3}ix/2-3/2"> (tangent points are <img src="https://latex.codecogs.com/gif.latex?(\pm{3}\sqrt{3}i,-6)"> and <img src="https://latex.codecogs.com/gif.latex?(\pm\sqrt{3}i,-3)">), and their intersection is <img src="https://latex.codecogs.com/gif.latex?(0,-3/2)">.
+The two external tangent lines are <img src="https://latex.codecogs.com/gif.latex?y=\pm\sqrt{3}ix/2-3/2"> (tangent points are <img src="https://latex.codecogs.com/gif.latex?(\pm{3}\sqrt{3}i,-6)"> and <img src="https://latex.codecogs.com/gif.latex?(\pm\sqrt{3}i,-3)">), and their intersection (internal homothetic center) is <img src="https://latex.codecogs.com/gif.latex?(0,-3/2)">.
 
-The two internal tangent lines are <img src="https://latex.codecogs.com/gif.latex?y=\pm\sqrt{15}ix/4-3/4"> (tangent points are <img src="https://latex.codecogs.com/gif.latex?(\pm{3}\sqrt{15}i,-12)"> and <img src="https://latex.codecogs.com/gif.latex?(\pm\sqrt{15}i,3)">), and their intersection is <img src="https://latex.codecogs.com/gif.latex?(0,-3/4)">.
+The two internal tangent lines are <img src="https://latex.codecogs.com/gif.latex?y=\pm\sqrt{15}ix/4-3/4"> (tangent points are <img src="https://latex.codecogs.com/gif.latex?(\pm{3}\sqrt{15}i,-12)"> and <img src="https://latex.codecogs.com/gif.latex?(\pm\sqrt{15}i,3)">), and their intersection (external homothetic center) is <img src="https://latex.codecogs.com/gif.latex?(0,-3/4)">.
 
 ### Monge's theorem
 
 <img src="diagrams/monge.png">
 
-**Monge's theorem** states that for any three circles in a plane, the intersection points of each of the three pairs of external tangent lines are collinear. This still holds even if one circle is completely inside the other.
+**Theorem 1** For any three circles in a plane, the intersection points of each of the three pairs of external tangent lines (i.e. three external homothetic centers) are collinear. This still holds even if one circle is completely inside the other.
+
+This can be proved by applying Eq. 1 onto *D*, *E* and *F* and Menelaus's theorem:
+
+<img src="https://latex.codecogs.com/gif.latex?\frac{\overrightarrow{DB}}{\overrightarrow{DC}}\cdot\frac{\overrightarrow{EC}}{\overrightarrow{EA}}\cdot\frac{\overrightarrow{FA}}{\overrightarrow{FB}}=\frac{r_B}{r_C}\cdot\frac{r_C}{r_A}\cdot\frac{r_A}{r_B}=1">
+
+**Theorem 2** For three circles (*A*), (*B*) and (*C*) in a plane, the external homothetic center *D* of (*B*) and (*C*), the internal homothetic center *E'* of (*C*) and (*A*), and the internal homothetic center *F'* of (*A*) and (*B*), are collinear.
 
-Let's apply Eq. 1 onto three external intersections *D*, *E* and *F*:
+This can be proved by applying Eq. 1 onto *D*, Eq. 2 onto *E'* and *F'* and Menelaus's theorem:
 
-<img src="https://latex.codecogs.com/gif.latex?\frac{\overrightarrow{DA}}{\overrightarrow{DB}}\cdot\frac{\overrightarrow{EC}}{\overrightarrow{EA}}\cdot\frac{\overrightarrow{FB}}{\overrightarrow{FC}}=\frac{r_A}{r_B}\cdot\frac{r_C}{r_A}\cdot\frac{r_B}{r_C}=1">
+<img src="https://latex.codecogs.com/gif.latex?\frac{\overrightarrow{DB}}{\overrightarrow{DC}}\cdot\frac{\overrightarrow{CE'}}{\overrightarrow{E'A}}\cdot\frac{\overrightarrow{AF'}}{\overrightarrow{F'B}}=\frac{r_B}{r_C}\cdot\frac{r_C}{r_A}\cdot\frac{r_A}{r_B}=1">
 
-According to Menelaus's theorem, *D*, *E* and *F* are collinear.
+A famous triangle center *X*(12) mentioned [here](feuerbach.md#theorems-related-to-the-feuerbach-point) (Theorem 3) can be proved by this theorem.
 
-The three internal intersections *D'*, *E'* and *F'* have a similar property. Let's apply Eq. 2 onto them:
+**Theorem 3** For three circles (*A*), (*B*) and (*C*) in a plane, let *D'*, *E'* and *F'* be the internal homothetic centers of (*B*)(*C*), (*C*)(*A*) and (*A*)(*B*), respectively, then *AD'*, *BE'* and *CF'* are concurrent.
 
-<img src="https://latex.codecogs.com/gif.latex?\frac{\overrightarrow{D'A}}{\overrightarrow{BD'}}\cdot\frac{\overrightarrow{E'C}}{\overrightarrow{AE'}}\cdot\frac{\overrightarrow{F'B}}{\overrightarrow{CF'}}=\frac{r_A}{r_B}\cdot\frac{r_C}{r_A}\cdot\frac{r_B}{r_C}=1">
+This can be proved by applying Eq. 2 onto *D'*, *E'* and *F'* and Ceva's theorem:
 
-According to Ceva's theorem, *CD'*, *BE'* and *AF'* are concurrent.
+<img src="https://latex.codecogs.com/gif.latex?\frac{\overrightarrow{BD'}}{\overrightarrow{D'C}}\cdot\frac{\overrightarrow{CE'}}{\overrightarrow{E'A}}\cdot\frac{\overrightarrow{AF'}}{\overrightarrow{F'B}}=\frac{r_B}{r_C}\cdot\frac{r_C}{r_A}\cdot\frac{r_A}{r_B}=1">
 
 ### Note
 

feuerbach.md

@@ -30,7 +30,7 @@ Then it's easy to calculate the intersection *O* of the three tangent planes, an
 
 ### Eight vertices of a hexahedron lie on a sphere
 
-> If seven vertices of a (quadrilaterally-faced) hexahedron lie on a sphere, then so does the eighth vertex. <sup>[2](https://imomath.com/index.php?options=323) (Problem 11)</sup>
+> If seven vertices of a (quadrilaterally-faced) hexahedron lie on a sphere, then so does the eighth vertex. <sup>[1](https://imomath.com/index.php?options=323) (Problem 11)</sup>
 
 Just like the previous problem, we define arbitrary points *O*(0,0,0), *A*(*a*,0,0), *B*(*b*,*c*,0) and *C*(*d*,*e*,*f*), then calculate the sphere passing through these four points: