Mixtilinear Incircles 疑問

hbghlyj

为了找这帖用到的引理，我找到陳誼廷的A Guessing Game: Mixtilinear Incircles（PDF｜TeX）的結論5
$∠ATB = ∠CTD$ 如何證明？
import geometry;

unitsize(5cm);

pair A = dir(140);
pair B = dir(210);
pair C = dir(330);
pair M_A = dir(270);
pair M_B = dir(55);
pair M_C = dir(175);

pair I = incenter(A, B, C);
pair D = projection(line(B,C))*I;
pair E = B+C-D;

pair B_1 = extension(I, I+dir(90)*dir(A-I), A, B);
pair C_1 = extension(I, I+dir(90)*dir(A-I), A, C);

pair T = extension(M_C, B_1, M_B, C_1);
draw(C--M_C, heavygreen);
draw(B--M_B, heavygreen);

filldraw(unitcircle, opacity(0.02)+cyan, black);

draw(A--M_A);

filldraw(circumcircle(T, B_1, C_1), opacity(0.05)+lightblue, blue);
markangle(B,A,T,heavycyan);
markangle(E,A,C,heavycyan);
markangle(A,T,B,heavymagenta);
markangle(C,T,D,heavymagenta);

draw(A--B--C--cycle);

draw(T--M_C, red+dashed);
draw(T--M_B, red+dashed);

draw(A--E, blue);
pair X = dir(90);
draw(T--X, blue);

draw(B--T--C, magenta);
draw(A--T--D, magenta);

filldraw(circumcircle(T, D, M_A), opacity(0.1)+lightred, orange);

pair Z = extension(B, C, T, M_A);
draw(Z--C_1, lightgreen);
draw(Z--B, lightgreen);
draw(Z--M_A, lightgreen);

filldraw(circumcircle(B, B_1, T), opacity(0.04)+green, heavygreen);
filldraw(circumcircle(C, C_1, T), opacity(0.04)+green, heavygreen);

pair H = extension(A, D, T, M_A);
draw(A--H, grey+dashed);

dot(H);
dot(extension(A,E,X,I));
dot(extension(T,M_A,B,C));
dot(extension(A,M_A,B,C));
dot(X);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$M_A$", M_A, dir(M_A));
dot("$M_B$", M_B, dir(M_B));
dot("$M_C$", M_C, dir(M_C));
dot("$I$", I, dir(I));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$B_1$", B_1, dir(B_1));
dot("$C_1$", C_1, dir(C_1));
dot("$T$", T, dir(T));

hbghlyj

我找到陳誼廷的博文blog.evanchen.cc/2015/08/11/the-mixtilinear-incircle/

Lines $AT$ and $AE$ are isogonal. This was essentially EGMO 2012, Problem 5, and the “morally correct” solution is to do an inversion at $A$ followed by a reflection along the $\angle A$-bisector (sometimes we call this a “$\sqrt{bc}$ inversion”).

As a consequence of this, one can also show that lines $TA$ and $TD$ are isogonal (with respect to $\angle BTC$).

不懂啊……反演之后怎么得到$TA$ and $TD$ are isogonal with respect to $\angle BTC$

hbghlyj

又找到artofproblemsolving.com/community/c3016h1298152s3_mixtilinear_incircles

4. $\angle BAT = \angle CAE$, and equivalent angles.
Proof.
$\sqrt{bc}$ inversion swaps the $A$-excircle and $\omega_A$.

5. $\angle ATB = \angle CTD$, and equivalent angles.
Proof.
Equivalently, $TD$ meets $\Gamma$ at the point $A'$ so that $AA'CB$ is an isosceles trapezium. But a reflection in the perpendicular bisector of $\overline{BC}$ takes the $A$-Nagel cevian to $TD$.

$T$是 $\omega_A$ 与 $\odot ABC$ 的切点，$E$是$A$-excircle与BC的切点，
而“$\sqrt{bc}$ inversion swaps the $A$-excircle and $\omega_A$.”
并且“$\sqrt{bc}$ inversion swaps $\odot ABC$ and $BC$.”
所以$\sqrt{bc}$ inversion swaps $T$ and $E$.

4. $\angle BAT = \angle CAE$我知道这个是因为$T$的反演像仍在$AT$上，再关于$\angle BAC$角平分线对称到$E$，所以$AT,AE$是等角线。

5. $\angle ATB = \angle CTD$这个还是不懂啊……

hbghlyj

hbghlyj 发表于 2024-3-9 12:12
5. $\angle ATB = \angle CTD$这个还是不懂啊……

在§4 Sketches of Solutions中，只写了一句提示：

5. Reflect T across the perpendicular bisector of BC.

试一下：将T关于BC的中垂线反射到T’.
由4知AT、AE是等角线，所以T’在AE上。
$∠ATB=∠AT'B=∠CTD$
果然很简单就做出了于是这题就完全解决了。
import geometry;

unitsize(5cm);

pair A = dir(140);
pair B = dir(210);
pair C = dir(330);
pair M_A = dir(270);
pair M_B = dir(55);
pair M_C = dir(175);

pair I = incenter(A, B, C);
pair D = projection(line(B,C))*I;
pair E = B+C-D;

pair B_1 = extension(I, I+dir(90)*dir(A-I), A, B);
pair C_1 = extension(I, I+dir(90)*dir(A-I), A, C);

pair T = extension(M_C, B_1, M_B, C_1);
filldraw(unitcircle, opacity(0.02)+cyan, black);
pair T1 = (-T.x,T.y);
draw(A--M_A);

filldraw(circumcircle(T, B_1, C_1), opacity(0.05)+lightblue, blue);
markangle(B,A,T,red,Fill(yellow));
markangle(E,A,C,red,Fill(yellow));
markangle(radius=15,A,T,B,heavymagenta);
markangle(radius=15,C,T,D,heavymagenta);
markangle(radius=15,A,T1,B,heavymagenta);

draw(A--B--C--cycle);

draw(A--T1--B, magenta);
draw(B--T--C, magenta);
draw(A--T--D, magenta);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$M_A$", M_A, dir(M_A));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$T$", T, dir(T));
dot("$T'$", T1, dir(T1));