Let $ABC$ be a triangle with incenter $I$ and $DEF$ is the contact triangle of $ABC$ (Edited, the previous definition of $D,E,F$ was wrong). Consider an inversion with respect to the incircle of $\triangle ABC$. Why does the circumcircle of $ABC$ maps to the nine point circle of $DEF$ by this inversion? I need to show that if $A^{*}$ ,$B^{*}$, $C^{*}$ are the images of $A,B,C$ then $A^{*}$ ,$B^{*}$, $C^{*}$ lie on the nine point circle of $DEF$, but why is that true?
Why does inverting with respect to the incircle sends the circumcircle to the nine-point circle of the contact triangle
contest-mathgeometrytriangles
Related Solutions
The statement is exactly the one in https://mathworld.wolfram.com/PrasolovPoint.html - in OP notations, the ortic triangle is $\Delta H_aH_bH_c$, and its reflection in the nine-point circle is $\Delta N_aN_bN_c$.
For a structural proof, the simplest way to go is to compute points using barycentric coordinates. This is not only showing the required property, but also computes the barycentric coordinates of the point of concurrence in question, from the formula we can see its complexity, recover the result claimed in ETC, and further exhibit interesting other properties - like seeing it is on the line $X(5)X(6)$. I am giving the human solution, complemented for the calculus with computer algebra assistance, easy digestible code, which may be useful for some readers (in this and similar situations).
Let $a,b,c$ be the lengths of the sides in $\Delta ABC$. A point $P$ has normed barycentric coordinates $(x,y,z)$ if $P=xA+yB+zC$, with the norming condition $1=x+y+z$. (The relation can be considered using affixes, or vectorially for some / any chosen origin point $\Omega$ in the affine plane so that the vectors relation holds $\Omega P=x\;\Omega A +y\; \Omega B + z\;\Omega C$.) Sometimes it is better to use "projective" coordinates $[x:y:z]$ instead, with $x+y+z\ne 0$, and we pass to the corresponding normed coordinates by dividing by the sum $s=x+y+z$. We often write $\frac 1s(x,y,z)$ instead of $(x/s,\ y/s,\ z/s)$ for this normed version.
Let $S$ being the area of $\Delta ABC$, so $S^2 =s(s-a)(s-b)(s-c)$. The barycentric coordinates of the nine-point center $X(5)$ are $[f(a,b,c)\ :\ f(b,c,a)\ :\ f(c,a,b)]$, see ETC, where $$ f(a,b,c)=a^2(b^2 + c^2) - (b^2 -c^2)^2\ . $$ The norming factor for this representation of $X(5)$ is $$ \begin{aligned} &\sum f(a,b,c):=f(a,b,c) + f(b,c,a)+f(c,a,b) \\ &\qquad =\sum a^2(b^2 + c^2) - (b^2 -c^2)^2 =\sum a^2b^2 + a^2c^2 + 2b^2 c^2 - b^4 - c^4 \\ &\qquad=4a^2b^2 + 4b^2c^2 + 4 c^2 a^2 - 2a^4 - 2b^4 - 2c^4 \\ &\qquad=2(a+b+c)(-a+b+c)(a-b+c)(a+b-c)=32s(s-a)(s-b)(s-c) \\ &\qquad=32S^2\ . \end{aligned} $$
Our aim is to show explicitly:
Proposition: There exists an intersection point $X(68)=AN_a\cap BN_b\cap CN_c$, and its barycentric coordinates are $$ \bbox[yellow]{\qquad X(68)= [ \ g(a,b,c)\ :\ g(b,c,a)\ :\ g(c,a,b)\ ] \qquad} $$ with $$ \bbox[yellow]{\qquad g(a,b,c) = \frac{-a^2+b^2+c^2}{a^4 + b^4 + c^4 - 2a^2(b^2+c^2))}\ . \qquad} $$
Bonus: The point $X(68)$ also lies on $X(5)X(6)$, the line through the nine-point center $X(5)$, and the symmedian point $X(6)=[a^2:b^2:c^2]$.
Proof: We compute: $$ \begin{aligned} A &= (1,0,0)\ ,\\[2mm] \frac{H_aC}{BC} &=\frac 1a\cdot b\cos C=\frac 1a\cdot b\cdot \frac{a^2 + b^2 -c^2}{2ab} =\frac 1{2a^2}(a^2+b^2-c^2)\ ,\\ H_a&=\left(0,\frac{H_aC}{BC},\frac{BH_a}{BC}\right)= \frac 1{2a^2}(0,\ a^2 + b^2 -c^2,\ a^2 - b^2 + c^2)\ , \\[2mm] X(5) &= (\ x_{X(5)},\ y_{X(5)},\ z_{X(5)}\ )\\ &= \phantom{\frac 1{8S^2}} [\ a^2(b^2+c^2) - (b^2-c^2)^2\ :\ b^2(a^2+c^2) - (a^2-c^2)^2\ :\ c^2(a^2+b^2) - (a^2-b^2)^2\ ]\ ,\\ &= \frac 1{32S^2}(\ a^2(b^2+c^2) - (b^2-c^2)^2\ ,\ b^2(a^2+c^2) - (a^2-c^2)^2\ ,\ c^2(a^2+b^2) - (a^2-b^2)^2\ )\ , \\[2mm] N_a &= 2X(5)-H_a=(\ x_{Na},\ y_{Na},\ z_{Na}\ )\ , \\[2mm] y_{Na} &=2\cdot \frac 1{32S^2}\cdot(b^2(a^2+c^2) - (a^2-c^2)^2) -\frac 1{2a^2}\cdot (a^2 + b^2 -c^2) \\ &=\frac 1{32a^2S^2}\Big(\ 2a^2b^2(a^2+c^2)-2a^2(a^2-c^2)^2 - 16S^2(a^2 + b^2 -c^2)\ \Big)\ . \\[2mm] 32a^2S^2\; y_{Na} &=2a^2b^2(a^2+c^2)-2a^2(a^2-c^2)^2 - 16S^2(a^2 + b^2 -c^2) \\ &=2a^2b^2(a^2-b^2+c^2)+2a^2 b^4-2a^2(a^2-c^2)^2 - 16S^2(a^2 + b^2 -c^2) \\ &=2a^2b^2(a^2-b^2+c^2) + 2a^2 (a^2 + b^2 -c^2)(-a^2+b^2+c^2) - 16S^2(-a^2 + b^2 -c^2) - 32a^2S^2 \\ &=(a^2-b^2+c^2)(2a^2b^2 +16S^2) + 2a^2 (a^2 + b^2 -c^2)(-a^2+b^2+c^2) - 32a^2S^2 \\[2mm] &\qquad (a^2 + b^2 -c^2)(-a^2+b^2+c^2) - 16S^2 \\ &\qquad\qquad=(b^4 - a^4 - c^4 + 2a^2c^2) - (-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2c^2a^2) \\ &\qquad\qquad=-2b^2(a^2-b^2+c^2)\ , \\[2mm] 32a^2S^2\; y_{Na} &=(a^2-b^2+c^2)(2a^2b^2 +16S^2)-4a^2b^2(a^2-b^2+c^2) \\ &=(a^2-b^2+c^2)(-2a^2b^2+16S^2) \\ &=(a^2-b^2+c^2)(-a^4-b^4-c^4+2a^2c^2+2b^2c^2)\ . \\ 32a^2S^2\; z_{Na} &=(a^2+b^2-c^2)(-a^4-b^4-c^4+2a^2b^2+2b^2c^2)\text{ similarly .} \end{aligned} $$ The line $AN_a$ has the equation in barycentric coordinates for a point $P=(x,y,z)$ given by assembly of the coordinates for $A,N_a$, and $P$ in matrix format, than building the determinant: $$ 0= \begin{vmatrix} 1 & 0 & 0\\ x_{N_a} & y_{Na} & z_{N_a}\\ x & y & z \end{vmatrix}\ , $$ so the equation involves only $y,z$ with constant proportionality: $$ \begin{aligned} \frac yz &= \frac{y_{Na}}{z_{Na}} = \frac {(a^2-b^2+c^2)(-a^4-b^4-c^4+2a^2c^2+2b^2c^2)} {(a^2+b^2-c^2)(-a^4-b^4-c^4+2a^2b^2+2b^2c^2)} \\ &= \frac {(a^2-b^2+c^2):(-a^4-b^4-c^4+2a^2b^2+2b^2c^2)} {(a^2+b^2-c^2):(-a^4-b^4-c^4+2a^2c^2+2b^2c^2)} \\ &=\frac{g(b,c,a)}{g(c,a,b)} \ . \end{aligned} $$ Similar equations are written for $BN_b$, and $CN_c$, and the proof is finished by observing that the point $X(68)$ from the proposition satisfies the above equation of $AN_a$, $y:z=g(b,c,a):g(c,a,b)$, as by symmetriy it also the case for $BN_b$, and $CN_c$.
$\square$
Sage code:
var('a,b,c');
Ha = 1/2/a^2 * vector([0, a^2 + b^2 - c^2, a^2 - b^2 + c^2])
f(a, b, c) = a^2*(b^2 + c^2) - (b^2 - c^2)^2
X5 = vector([f(a, b, c), f(b, c, a), f(c, a, b)])
X5 = 1/sum(X5) * X5 # thus norming it
Na = 2*X5 - Ha
xNa, yNa, zNa = Na
Then:
print(factor(yNa / zNa))
delivers immediately the result of our longer computation:
(a^4 + b^4 - 2*a^2*c^2 - 2*b^2*c^2 + c^4)*(a^2 - b^2 + c^2)
/
((a^4 - 2*a^2*b^2 + b^4 - 2*b^2*c^2 + c^4)*(a^2 + b^2 - c^2))
(Result was manually rearranged to fit in the MSE line.)
We can now also show the bonus point, by computing the determinant: with lines proportional to the barycentric coordinates of respectively $X(6)$, $X(5)$, and $X(68)$, and showing it vanishes: $$ \begin{vmatrix} a^2 & b^2 & c^2\\ f(a,b,c) & f(b,c,a) & f(c,a,b)\\ g(a,b,c) & g(b,c,a) & g(c,a,b) \end{vmatrix} $$
g(a,b,c) = (-a^2 + b^2 + c^2) / (a^4 + b^4 + c^4 - 2*a^2*(b^2 + c^2))
matrix([ [a^2, b^2, c^2],
[f(a, b, c), f(b, c, a), f(c, a, b)],
[g(a, b, c), g(b, c, a), g(c, a, b)], ]).det().simplify_full()
And the code delivers rapidly the zero.
This is actually known and follows from the statements:
- The radius of a triangle's circumcircle is twice the radius of that triangle's nine-point circle (see).
- If a circle passes through the center of a circumcircle and tangents to it, then its radius is twice the radius of the circumcircle.
Best Answer
We will use the known fact that given an inversion $*$ with respect to a circle $\omega$, then for any point $A$ outside $\omega$, $A^{*}$ is the midpoint of the tangency points from $A$. Using this fact, we get that in our case $A^{*}$ is the midpoint of $EF$ (as $E,F$ are the tangency points from $A$ to the incircle). Similarly, $B^{*}$,$C^{*}$ are the midpoints of $DE,DF$ respectively. Therefore, by inverting with respect to the incircle, the circumcircle of $ABC$ maps to the circumcircle of $A^{*}B^{*}C^{*}$, which is the circle passing through the midpoints of $EF,DF,DE$ or in other words the nine-point circle of the triangle $DEF$.