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？

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/c3016h12981 … ixtilinear_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。