优雅几何问题

星奔川骛

在三角形 \( ABC \) 中，内切圆与 \( BC \) 相切于点 \( D \)。一个通过点 \( B \) 和 \( C \) 的圆与内切圆相切于点 \( P \)。构造一个平行四边形 \( BPCT \)，使得 \( T \) 和 \( P \) 位于直线 \( BC \) 的两侧。设 \( Q \) 是 \( P \) 的等角共轭点。证明：\( AD \parallel QT \)。

hbghlyj

相关：math.stackexchange.com/questions/3962904/
prase.cz/kalva/short/soln/sh02g7.html
anhngq.wordpress.com/wp-content/uploads/2010/ … tlist.pdf#page=20.17
ABC is an acute-angled triangle. The incircle touches BC at K. The altitude AD has midpoint M. The line KM meets the incircle again at N. Show that the circumcircle of BCN touches the incircle of ABC at N.

If $AB = AC$, then $D = K$ is the midpoint of $BC$ and the result is trivial (the figure is symmetrical about $AD$). So assume $AB$ and $AC$ are unequal. Let $X$ be the center of the circle $BNC$. Take $P$ to be the intersection of the line $KN$ and the perpendicular bisector of $BC$. Suppose that $P$ lies on the circle. Then
\[
\angle PNX = \angle NPX \, (XP = XN) = \angle NKI \, (IK \parallel XP) = \angle KNI \, (IN = IK) = \angle PNI \, (\text{same angle}),
\]
so $N, I, X$ are collinear. But if two circles meet at a point on the line of centers, then they must touch there. So it remains to show that $P$ lies on the circle.

Take $BC = a$, $CA = b$, $AB = c$, $s = \frac{a + b + c}{2}$ as usual. The usual argument shows that $BK = s - b$, $CK = s - c$. (If the incircle touches $AC$ at $K'$ and $AB$ at $K''$, then we have $BK = BK''$ (equal tangents) and $BK + KC = a$ etc., so we get three equations for three unknowns and can solve for $BK = s - b$, $CK = s - c$. Hence $BK \cdot KC = (s - b)(s - c)$. It is sufficient to show that $NK \cdot KP = (s - b)(s - c)$ also.

\[
BD = c \cos B = \frac{c^2 + a^2 - b^2}{2a}.
\]
Hence
\[
DK = BK - BD = (s - b) - \frac{c^2 + a^2 - b^2}{2a} = \frac{-ab + ac - c^2 + b^2}{2a} = \frac{(s - a)(b - c)}{a}.
\]
We have
\[
MD = \frac{AD}{2} = \frac{\text{area of } \triangle ABC}{a},
\]
so if $x = \angle MKD$, then
\[
\tan x = \frac{MD}{DK} = \frac{\text{area of } \triangle ABC}{(s - a)(b - c)}.
\]

But $NK = 2r \sin x$ (where $r$ is the inradius), and $KP = KY \sec x$, where $Y$ is the midpoint of $BC$. We have
\[
KY = BY - BK = \frac{a}{2} - (s - b) = \frac{b - c}{2}.
\]
Hence
\[
NK \cdot KP = r(b - c) \tan x.
\]
As usual, we have $\text{area of } \triangle ABC = rs$ (because $\text{area of } \triangle ABC = \text{area of } \triangle AIB + \text{area of } \triangle BIC + \text{area of } \triangle CIA$), so
\[
r = \frac{\text{area of } \triangle ABC}{s},
\]
and hence
\[
NK \cdot KP = \frac{(\text{area of } \triangle ABC)^2 (b - c)}{s(s - a)(b - c)} = (s - b)(s - c),
\]
since $(\text{area of } \triangle ABC)^2 = s(s - a)(s - b)(s - c)$ (Heron's formula).

hbghlyj

相关：artofproblemsolving.com/community/c2562h1674771_copy_of_2002_g7
The incircle $\Omega$ of the acute-angled triangle $ABC$ is tangent to its side $BC$ at a point $K$. Let $\overline{AD}$ be an altitude of triangle $ABC$, and let $M$ be the midpoint of the segment $AD$. If $N$ is the common point of the circle $\Omega$ and the line $KM$ (distinct from $K$), then prove that the incircle $\Omega$ and the circumcircle of triangle $BCN$ are tangent to each other at the point $N$.

Solution
Let $I_A$ be the $A$-excenter; by the “midpoint of altitudes” lemma, $\overline{NMKI_A}$ are collinear. In addition, let $L$ be the midpoint of $\overline{KI_A}$ and $P$ be the midpoint of $\overline{KN}$.

Observe that $\angle IPI_A=90^\circ$, so $P$ lies on $(BICI_A)$. Then Power of a Point at $K$ immediately implies $BLCN$ cyclic.

Since the in-touch and ex-touch points on $\overline{BC}$ are reflections over the midpoint, we know that $L$ must lie on the perpendicular bisector of $\overline{BC}$, and thus $L$ is the midpoint of arc $BC$ in $(NBC)$. Then the result follows from Archimedes' lemma.

星奔川骛

hbghlyj 发表于 2025-4-10 00:32
相关：https://artofproblemsolving.com/community/c2562h1674771_copy_of_2002_g7
The incircle $\Omega$  ...

是否有本题的解答？

hbghlyj

高联一百题解答
第二十四题 $\triangle ABC$ 内切圆 $\odot I$ 切 BC 于 $D, AE \perp BC$ 于 $E, F$ 为弧中点，DF 交 $\odot$ $I$ 于 $G$，作 $\triangle B C G$ 的外接圆 $\odot O$，求证：$\odot O, \odot I$ 相切于点 $G$.

证明方法一：如图，连接 $A D$ 交 $\odot I$ 于 $J$，延长 $D I$ 交 $\odot I$ 于 $K$，因为 $D K \px A E$，且 $D G$ 平分 AE ，所以 $D K, ~ D G, ~ D J, ~ D C$ 是一组调和线束，所以四边形 $D K J G$ 是一个调和四边形。过 $J$作 $\odot I$ 切线交 $B C$ 于 $M$，则 $M, ~ D, ~ C, ~ B$ 是一组调和点列，且 $K, ~ G, ~ M$ 三点共线，所以 $D G \perp$ $G M$，所以 $G D$ 平分 $\angle B G C$．延长 $G D$ 交 $\odot O$ 于 $N$，则 $N$ 为弧 $B C$ 中点，所以 $O N \px I D$，所以 $O, I,G$ 三点共线，所以 $\odot O, \odot I$ 相切于点 $G$.

证明方法二：如图，作 $\triangle ABC$ 的 A-旁切圆 $\odot J$ ，设 $\odot J$ 切 BC 于 T ，则 $CT=BD$。延长 AD交 $\odot J$ 于 $L$ ，根据位似关系知 $T L$ 为 $\odot J$ 直径。注意到 $J$ 为 $L T$ 中点，$F$ 为 $A E$ 中点，且 $L T\px A E$ ，所以 $J, ~ D, ~ F, ~ G$ 四点共线。

作 $I S \perp G D$ 于 $S$ ，则 $S$ 为 $G D$ 中点。因为 $\angle I S J=\angle I B J=\angle I C J$ ，所以 $I, ~ S, ~ B, ~ J, ~ C$ 五点共圆。取 $D J$ 中点为 $M$，则 $B D \cdot C D=S D \cdot J D=F D \cdot M D$，所以 $G, ~ B, ~ M, ~ C$ 四点共圆。作 $M N \perp B C$ 于 $N$，则 $N$ 为 $D T$ 中点，进而知 $N$ 为 $B C$ 中点，于是知 $M B=M C$，故 $G M$ 平分 $\angle B G C$，从而知 $\odot O,\odot I$ 相切于点 $G$。

星奔川骛

hbghlyj 发表于 2025-4-10 22:23
没有

好吧

星奔川骛

这个似乎用的上，看看能否证明？

$\triangle ABC$ 的内切圆圆 $I$ 切 $BC$ 于 $D$，过 $B$、$C$ 两点的圆与圆 $I$ 切于点 $P$，$P$、$Q$ 是 $\triangle ABC$ 的一对等角共轭点，$PF \perp BD$ 于 $F$，$M$ 为 $DF$ 中点，$PQ$ 交 $BC$ 于 $E$，求证：$PM \px AD$，$BE = CF$。

hbghlyj

星奔川骛 发表于 2025-4-10 15:29
$M$ 为 $DF$ 中点，$PQ$ 交 $BC$ 于 $E$，$PM \px AD$

2#、3#、5#是同一道题，似乎可以用来证明7#，但未想出

星奔川骛

解答:

$A$-伪外接圆切内切圆于 $P$，$D$ 为 $B C$ 上内切圆切点．作平行四边形 $P B R C$，$Q$ 为 $P$ 关于 $\triangle A B C$ 等角共轭点．求证：$R Q \px A D$．
证明：取 $P Q$ 中点 $T, B C$ 中点 $M$．则 $T M \px Q R$．只需证 $TM \px AD$．取旁心 $J$，只需证 $J, ~ M, ~ ~T$ 共线．注意 $P$ 在 $D J$ 上且 $J P$ 平分 $\angle B P C$．于是 $d(J, BQ)=d(J, BP)=d(J, CP)=d(J, CQ)$，即四边形PBQC有外切圆，圆心为J．故 J 在其牛顿线TM上．得证．