From 3e838d56acbcbf68f1859e57f707f5cce4f324a0 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Thu, 23 Apr 2026 17:46:50 +0800
Subject: [PATCH 01/34] color 4 fdh change.md
---
content/categories/FDH/_index.md | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/content/categories/FDH/_index.md b/content/categories/FDH/_index.md
index a005294..1210cfe 100644
--- a/content/categories/FDH/_index.md
+++ b/content/categories/FDH/_index.md
@@ -5,6 +5,6 @@ description:
style:
background: "#ff348f8f"
- color: "#fff"
+ color: "#ff66ccff"
---
From eb8184a0a824242b3fb8bd999bece0232a234ab0 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Sun, 3 May 2026 12:58:08 +0800
Subject: [PATCH 02/34] color
---
content/categories/FDH/_index.md | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/content/categories/FDH/_index.md b/content/categories/FDH/_index.md
index f58c3f3..aa0d969 100644
--- a/content/categories/FDH/_index.md
+++ b/content/categories/FDH/_index.md
@@ -5,6 +5,6 @@ description:
style:
background: "#ff348f8f"
- color: "#ff66ccff"
+ color: "#8ef"
---
From 14e7437cdf1aa37d299dcb081cb2f5572d472222 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Thu, 7 May 2026 18:12:59 +0800
Subject: [PATCH 03/34] Add files via upload
---
.../post/articles/cute/cutsandwichinhalf.md | 80 +++++++++++++++++++
1 file changed, 80 insertions(+)
create mode 100644 content/post/articles/cute/cutsandwichinhalf.md
diff --git a/content/post/articles/cute/cutsandwichinhalf.md b/content/post/articles/cute/cutsandwichinhalf.md
new file mode 100644
index 0000000..7a8b1a9
--- /dev/null
+++ b/content/post/articles/cute/cutsandwichinhalf.md
@@ -0,0 +1,80 @@
+---
+title: "呜呜呜,小南梁吃不起饭了😢"
+description: 三角形面积的平分线
+date: 2026-03-28
+image:
+math: true
+license: All Right Reserved, 河源中学数学研究协会
+hidden: false
+comments: true
+darft: false
+categories:
+ - Femboy's Dessert House
+ - Articles
+tags:
+ - Femboy's Dessert House
+ - 圆锥曲线
+---
+
+## 导入
+
+小南梁卡洛特是古早动漫《幸运星》的大粉丝,她花钱买了里面陵樱学校的夏季款水手服,接着就发现钱包已然空空,连饭钱都要没了😥😥😥.于是,她只好用剩下的钱买了一块三明治,想着把它切成面积相等的两块,一块中午吃,一块晚上吃(同学们要节制消费,不要模仿哦).
+
+问题:请问卡洛特应该怎么切?能找到一个直线系[^1]囊括所有的切法吗?
+
+[^1]: 一堆直线构成的集合.
+
+## 先上解答
+
+容易想到,卡洛特只要沿着一条中线切,或者根据三角形相似找出一条平行于一边的直线(条件如下图所示)去切,就可以完成任务.
+
+要找满足题意的所有直线,需要如下步骤:
+1.在$\bigtriangleup ABC$中,作出三条中线,设为$AD, BE, CF$.
+2.选取其中一角,这里选$\angle A$,取**这个角的两边上的中线**$BE, CF$的中点分别为$P, Q$.
+3.以$AC, AB$所在直线为渐近线,作与$CF, BE$切于$P,Q$两点的**双曲线段**[^2].
+4.对$\angle B, \angle C$重复操作2,3.
+5.能够平分$\bigtriangleup ABC$面积的直线构成的直线系,就是上面作出来的三条双曲线段的所有切线的集合.
+
+[^2]: 这里仿照直线和线段,用“双曲线段”表示双曲线上的一部分.
+
+完成一条双曲线段的图大致如下.
+
+## 关于双曲线是怎么来的
+
+在初中的学习中,我们知道经反比例函数图象上的一点,作坐标轴的垂线,所围成的矩形面积一定.在高中,我们知道反比例函数的图象是双曲线,坐标轴就是它的两条渐近线,实际上,类似的与双曲线的渐近线有关的面积结论还有很多.在推导这个问题时,我们用到如下结论:
+
+>双曲线上任意一点的切线与它的渐近线围成的三角形面积为一个定值.
+
+该定理的证明涉及复杂的字母运算,我们放到文章末尾.
+
+三角形的中线平分三角形的面积,这是显而易见的.以上图为例,我们希望这条直线由$BE$向$CF$“过渡”的过程中,不要改变$\angle A$那部分三角形的面积.而根据上述定理,我们构造如上双曲线段,让这条直线作为该双曲线段的切线,就能保证这部分三角形是“双曲线上一点的切线与它的渐近线围成的三角形”,从而保证面积为定值,换个角也是同理的.
+
+## 为何一定切于中线的中点
+
+如下图,我们用一种微元的思想来解释.
+
+设上面第3步作出的双曲线(假装我们还不知道过哪个点)与$CF$切于点$X$.
+
+接着,作出该双曲线的另一条切线$l$,这条切线与$CF$的夹角很小,为$\theta$.$l$分别交$AC, AB$于$G,H$.
+
+由于偏转角度相当小,可以认为$l$仍过点$X$.[^3]同时也可以认为$XG=XC, XH=XF$.
+
+[^3]: 这步有点奇妙,希望有大佬解释一下.
+
+由三角形面积不变,有
+$$
+S_{\bigtriangleup XCG}=\frac{1}{2} XC\cdot XG=\frac{1}{2} XC^2=S_{\bigtriangleup XFH}=\frac{1}{2} XF\cdot XH=\frac{1}{2} XF^2
+$$
+
+因此
+$$
+XC=XF
+$$
+
+$X$为$CF$的中点,证毕!!!
+
+## 上面那个结论的证明
+
+待补充...
+
+更多疑问可致信河中数协官方邮箱或者在下方评论区留言,小南梁会耐心为您解答♡
\ No newline at end of file
From 234fc2e227447f4f672c933880bb70ab30d1c683 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Thu, 7 May 2026 18:18:45 +0800
Subject: [PATCH 04/34] =?UTF-8?q?=E8=BE=BE=E7=89=B9=E5=AB=A6=E5=A8=A5.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
content/post/articles/cute/cutsandwichinhalf.md | 4 ++--
1 file changed, 2 insertions(+), 2 deletions(-)
diff --git a/content/post/articles/cute/cutsandwichinhalf.md b/content/post/articles/cute/cutsandwichinhalf.md
index 7a8b1a9..c3eeef0 100644
--- a/content/post/articles/cute/cutsandwichinhalf.md
+++ b/content/post/articles/cute/cutsandwichinhalf.md
@@ -1,7 +1,7 @@
---
title: "呜呜呜,小南梁吃不起饭了😢"
description: 三角形面积的平分线
-date: 2026-03-28
+date: 2026-05-07
image:
math: true
license: All Right Reserved, 河源中学数学研究协会
@@ -77,4 +77,4 @@ $X$为$CF$的中点,证毕!!!
待补充...
-更多疑问可致信河中数协官方邮箱或者在下方评论区留言,小南梁会耐心为您解答♡
\ No newline at end of file
+更多疑问可致信河中数协官方邮箱或者在下方评论区留言,小南梁会耐心为您解答♡
From f49f3df09bb0792439007a9a76f89eabee2a38d2 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Fri, 8 May 2026 09:25:32 +0800
Subject: [PATCH 05/34] filenamechangetocheckifcategoriescanbeapplycorrectly
---
content/categories/{FDH => femboy's-dessert-house}/_index.md | 0
1 file changed, 0 insertions(+), 0 deletions(-)
rename content/categories/{FDH => femboy's-dessert-house}/_index.md (100%)
diff --git a/content/categories/FDH/_index.md b/content/categories/femboy's-dessert-house/_index.md
similarity index 100%
rename from content/categories/FDH/_index.md
rename to content/categories/femboy's-dessert-house/_index.md
From 18b4ae53c0b99f20970f71f6d10f2bf41ef09a12 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Fri, 8 May 2026 09:31:26 +0800
Subject: [PATCH 06/34] =?UTF-8?q?tag=E5=BE=AE=E8=B0=83?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
content/post/articles/cute/derange.md | 1 +
1 file changed, 1 insertion(+)
diff --git a/content/post/articles/cute/derange.md b/content/post/articles/cute/derange.md
index f5bab64..928fa17 100644
--- a/content/post/articles/cute/derange.md
+++ b/content/post/articles/cute/derange.md
@@ -14,6 +14,7 @@ categories:
tags:
- Femboy's Dessert House
- 全错排
+ - 排列组合
---
## 导入
From 10cf56489faa86d8f310f5933136c2d42ac1af19 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Fri, 8 May 2026 09:35:49 +0800
Subject: [PATCH 07/34] =?UTF-8?q?=E6=9C=89=E4=B8=AA=E5=9C=B0=E6=96=B9?=
=?UTF-8?q?=E4=B8=8D=E5=AF=B9.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
content/post/articles/cute/cutsandwichinhalf.md | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/content/post/articles/cute/cutsandwichinhalf.md b/content/post/articles/cute/cutsandwichinhalf.md
index c3eeef0..1f73ca5 100644
--- a/content/post/articles/cute/cutsandwichinhalf.md
+++ b/content/post/articles/cute/cutsandwichinhalf.md
@@ -63,7 +63,7 @@ tags:
由三角形面积不变,有
$$
-S_{\bigtriangleup XCG}=\frac{1}{2} XC\cdot XG=\frac{1}{2} XC^2=S_{\bigtriangleup XFH}=\frac{1}{2} XF\cdot XH=\frac{1}{2} XF^2
+S_{\bigtriangleup XCG}=\frac{1}{2} XC\cdot XGsin{\theta}=\frac{1}{2} XC^{2}sin{\theta}=S_{\bigtriangleup XFH}=\frac{1}{2} XF\cdot XHsin{\theta}=\frac{1}{2} XF^{2}sin{\theta}
$$
因此
From 523df950b0ec9832822bfc5d15b745571bb4af85 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Fri, 8 May 2026 17:27:55 +0800
Subject: [PATCH 08/34] Update content/post/articles/cute/cutsandwichinhalf.md
Co-authored-by: Website-xieyuen <131447547+Website-xieyuen@users.noreply.github.com>
---
content/post/articles/cute/cutsandwichinhalf.md | 10 +++++-----
1 file changed, 5 insertions(+), 5 deletions(-)
diff --git a/content/post/articles/cute/cutsandwichinhalf.md b/content/post/articles/cute/cutsandwichinhalf.md
index 1f73ca5..8f93ec6 100644
--- a/content/post/articles/cute/cutsandwichinhalf.md
+++ b/content/post/articles/cute/cutsandwichinhalf.md
@@ -29,11 +29,11 @@ tags:
容易想到,卡洛特只要沿着一条中线切,或者根据三角形相似找出一条平行于一边的直线(条件如下图所示)去切,就可以完成任务.
要找满足题意的所有直线,需要如下步骤:
-1.在$\bigtriangleup ABC$中,作出三条中线,设为$AD, BE, CF$.
-2.选取其中一角,这里选$\angle A$,取**这个角的两边上的中线**$BE, CF$的中点分别为$P, Q$.
-3.以$AC, AB$所在直线为渐近线,作与$CF, BE$切于$P,Q$两点的**双曲线段**[^2].
-4.对$\angle B, \angle C$重复操作2,3.
-5.能够平分$\bigtriangleup ABC$面积的直线构成的直线系,就是上面作出来的三条双曲线段的所有切线的集合.
+1. 在 $\bigtriangleup ABC$ 中,作出三条中线,设为 $AD, BE, CF$.
+2. 选取其中一角,这里选 $\angle A$,取**这个角的两边上的中线** $BE, CF$ 的中点分别为 $P, Q$.
+3. 以 $AC, AB$ 所在直线为渐近线,作与 $CF, BE$ 切于 $P,Q$ 两点的**双曲线段**[^2].
+4. 对 $\angle B, \angle C$ 重复操作2,3.
+5. 能够平分 $\bigtriangleup ABC$ 面积的直线构成的直线系,就是上面作出来的三条双曲线段的所有切线的集合.
[^2]: 这里仿照直线和线段,用“双曲线段”表示双曲线上的一部分.
From a7e41d6c927914daa3eff671d4e9bcad31f7842f Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Fri, 8 May 2026 17:33:20 +0800
Subject: [PATCH 09/34] kongge
---
content/post/articles/cute/cutsandwichinhalf.md | 17 ++++++++++-------
1 file changed, 10 insertions(+), 7 deletions(-)
diff --git a/content/post/articles/cute/cutsandwichinhalf.md b/content/post/articles/cute/cutsandwichinhalf.md
index 8f93ec6..0971c77 100644
--- a/content/post/articles/cute/cutsandwichinhalf.md
+++ b/content/post/articles/cute/cutsandwichinhalf.md
@@ -28,7 +28,8 @@ tags:
容易想到,卡洛特只要沿着一条中线切,或者根据三角形相似找出一条平行于一边的直线(条件如下图所示)去切,就可以完成任务.
-要找满足题意的所有直线,需要如下步骤:
+要找满足题意的所有直线,需要如下步骤:
+
1. 在 $\bigtriangleup ABC$ 中,作出三条中线,设为 $AD, BE, CF$.
2. 选取其中一角,这里选 $\angle A$,取**这个角的两边上的中线** $BE, CF$ 的中点分别为 $P, Q$.
3. 以 $AC, AB$ 所在直线为渐近线,作与 $CF, BE$ 切于 $P,Q$ 两点的**双曲线段**[^2].
@@ -43,35 +44,37 @@ tags:
在初中的学习中,我们知道经反比例函数图象上的一点,作坐标轴的垂线,所围成的矩形面积一定.在高中,我们知道反比例函数的图象是双曲线,坐标轴就是它的两条渐近线,实际上,类似的与双曲线的渐近线有关的面积结论还有很多.在推导这个问题时,我们用到如下结论:
->双曲线上任意一点的切线与它的渐近线围成的三角形面积为一个定值.
+> 双曲线上任意一点的切线与它的渐近线围成的三角形面积为一个定值.
该定理的证明涉及复杂的字母运算,我们放到文章末尾.
-三角形的中线平分三角形的面积,这是显而易见的.以上图为例,我们希望这条直线由$BE$向$CF$“过渡”的过程中,不要改变$\angle A$那部分三角形的面积.而根据上述定理,我们构造如上双曲线段,让这条直线作为该双曲线段的切线,就能保证这部分三角形是“双曲线上一点的切线与它的渐近线围成的三角形”,从而保证面积为定值,换个角也是同理的.
+三角形的中线平分三角形的面积,这是显而易见的.以上图为例,我们希望这条直线由 $BE$ 向 $CF$ “过渡”的过程中,不要改变 $\angle A$ 那部分三角形的面积.而根据上述定理,我们构造如上双曲线段,让这条直线作为该双曲线段的切线,就能保证这部分三角形是“双曲线上一点的切线与它的渐近线围成的三角形”,从而保证面积为定值,换个角也是同理的.
## 为何一定切于中线的中点
如下图,我们用一种微元的思想来解释.
-设上面第3步作出的双曲线(假装我们还不知道过哪个点)与$CF$切于点$X$.
+设上面第3步作出的双曲线(假装我们还不知道过哪个点)与 $CF$ 切于点 $X$ .
-接着,作出该双曲线的另一条切线$l$,这条切线与$CF$的夹角很小,为$\theta$.$l$分别交$AC, AB$于$G,H$.
+接着,作出该双曲线的另一条切线 $l$ ,这条切线与 $CF$ 的夹角很小,为 $\theta$ . $l$ 分别交 $AC, AB$ 于 $G,H$ .
-由于偏转角度相当小,可以认为$l$仍过点$X$.[^3]同时也可以认为$XG=XC, XH=XF$.
+由于偏转角度相当小,可以认为 $l$ 仍过点 $X$ .[^3]同时也可以认为 $XG=XC, XH=XF$ .
[^3]: 这步有点奇妙,希望有大佬解释一下.
由三角形面积不变,有
+
$$
S_{\bigtriangleup XCG}=\frac{1}{2} XC\cdot XGsin{\theta}=\frac{1}{2} XC^{2}sin{\theta}=S_{\bigtriangleup XFH}=\frac{1}{2} XF\cdot XHsin{\theta}=\frac{1}{2} XF^{2}sin{\theta}
$$
因此
+
$$
XC=XF
$$
-$X$为$CF$的中点,证毕!!!
+$X$ 为 $CF$ 的中点,证毕!!!
## 上面那个结论的证明
From 3c3137f4cee40d715b6b46fbb14fd9ffdadda45c Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Fri, 8 May 2026 18:08:10 +0800
Subject: [PATCH 10/34] pf
---
.../post/articles/cute/cutsandwichinhalf.md | 63 ++++++++++++++++++-
1 file changed, 61 insertions(+), 2 deletions(-)
diff --git a/content/post/articles/cute/cutsandwichinhalf.md b/content/post/articles/cute/cutsandwichinhalf.md
index 0971c77..48d426b 100644
--- a/content/post/articles/cute/cutsandwichinhalf.md
+++ b/content/post/articles/cute/cutsandwichinhalf.md
@@ -65,7 +65,7 @@ tags:
由三角形面积不变,有
$$
-S_{\bigtriangleup XCG}=\frac{1}{2} XC\cdot XGsin{\theta}=\frac{1}{2} XC^{2}sin{\theta}=S_{\bigtriangleup XFH}=\frac{1}{2} XF\cdot XHsin{\theta}=\frac{1}{2} XF^{2}sin{\theta}
+S_{\bigtriangleup XCG}=\frac{1}{2} XC\cdot XG\sin{\theta}=\frac{1}{2} XC^{2}\sin{\theta}=S_{\bigtriangleup XFH}=\frac{1}{2} XF\cdot XH\sin{\theta}=\frac{1}{2} XF^{2}\sin{\theta}
$$
因此
@@ -78,6 +78,65 @@ $X$ 为 $CF$ 的中点,证毕!!!
## 上面那个结论的证明
-待补充...
+考虑平面直角坐标系 $xOy$ 中的双曲线 $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ ,作它在点 $P(x_0,y_0)$ 处的切线 $l$ ,分别交双曲线的渐近线 $m: y=\frac{b}{a} x$ 与 $n: y=-\frac{b}{a} x$ 于点 $A(x_1,y_1), B(x_2, y_2)$ .
+
+由“代一半”法(证明略),得切线方程为:
+
+$$
+\frac{x_{0}x}{a^{2}}-\frac{y_{0}y}{b^{2}}=1
+$$
+
+接着,直接联立 $l$ 与 $m$, $l$ 与 $n$ 的方程,解得
+
+$$
+\left\{\begin{matrix}
+ x_1=\frac{a^2 b}{bx_0-y_0} \\
+ y_1=\frac{ab^2}{bx_0-y_0}
+\end{matrix}\right.
+
+\left\{\begin{matrix}
+ x_2=\frac{a^2 b}{bx_0+y_0} \\
+ y_2=-\frac{ab^2}{bx_0+y_0}
+\end{matrix}\right.
+$$
+
+因此
+
+$$
+\left | OA \right | \left | OB \right | =\sqrt{x_1^2+y_1^2} \sqrt{x_2^2+y_2^2}=\frac{a^4b^2+a^2b^4}{b^2x_0^2+a^2y_0^2}=\frac{a^2+b^2}{\frac{x_0^{2}}{a^{2}}-\frac{y_0^{2}}{b^{2}}}
+$$
+
+根据 $P(x_0,y_0)$ 在双曲线上,有
+
+$$
+\frac{x_0^{2}}{a^{2}}-\frac{y_0^{2}}{b^{2}}=1
+$$
+
+因此
+
+$$
+\left | OA \right | \left | OB \right | = \frac{a^2+b^2}
+$$
+
+
+渐近线与 $x$ 轴所成角满足
+
+$$
+\sin{\angle AOx}=\frac{a}{\sqrt{a^2+b^2} }, \cos{\angle AOx}=\frac{b}{\sqrt{a^2+b^2} }
+$$
+
+所以由对称性
+
+$$
+\sin{\angle AOB}=\sin{2\angle AOx}=2\sin{\angle AOx}\cos{\angle AOx}=\frac {2ab}{a^2+b^2}
+$$
+
+因此
+
+$$
+S_{\triangle AOB}=\frac{1}{2} \left | OA \right | \left | OB \right | \sin{\angle AOB}=ab
+$$
+
+所以 $\triangle AOB$ 的面积为定值 $ab$ .证毕!!!!!!!!!!
更多疑问可致信河中数协官方邮箱或者在下方评论区留言,小南梁会耐心为您解答♡
From 56d9f13099d4a55ad48fee63a0989fad1a57f050 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Fri, 8 May 2026 18:13:27 +0800
Subject: [PATCH 11/34] debugcutsandwichinhalf.md
---
content/post/articles/cute/cutsandwichinhalf.md | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/content/post/articles/cute/cutsandwichinhalf.md b/content/post/articles/cute/cutsandwichinhalf.md
index 48d426b..5e27498 100644
--- a/content/post/articles/cute/cutsandwichinhalf.md
+++ b/content/post/articles/cute/cutsandwichinhalf.md
@@ -115,7 +115,7 @@ $$
因此
$$
-\left | OA \right | \left | OB \right | = \frac{a^2+b^2}
+\left | OA \right | \left | OB \right | = {a^2+b^2}
$$
From ff7d1873cbb64625802ab0884f43d2e828c45c03 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Fri, 8 May 2026 18:16:29 +0800
Subject: [PATCH 12/34] gubed.md
---
content/post/articles/cute/cutsandwichinhalf.md | 2 ++
1 file changed, 2 insertions(+)
diff --git a/content/post/articles/cute/cutsandwichinhalf.md b/content/post/articles/cute/cutsandwichinhalf.md
index 5e27498..52720a3 100644
--- a/content/post/articles/cute/cutsandwichinhalf.md
+++ b/content/post/articles/cute/cutsandwichinhalf.md
@@ -93,7 +93,9 @@ $$
x_1=\frac{a^2 b}{bx_0-y_0} \\
y_1=\frac{ab^2}{bx_0-y_0}
\end{matrix}\right.
+$$
+$$
\left\{\begin{matrix}
x_2=\frac{a^2 b}{bx_0+y_0} \\
y_2=-\frac{ab^2}{bx_0+y_0}
From 2b33f188e67692309b4bd165e6fb3d9690c5fb05 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Fri, 8 May 2026 18:17:12 +0800
Subject: [PATCH 13/34] Update content/post/articles/cute/derange.md
Co-authored-by: Website-xieyuen <131447547+Website-xieyuen@users.noreply.github.com>
---
content/post/articles/cute/derange.md | 1 -
1 file changed, 1 deletion(-)
diff --git a/content/post/articles/cute/derange.md b/content/post/articles/cute/derange.md
index 928fa17..f5bab64 100644
--- a/content/post/articles/cute/derange.md
+++ b/content/post/articles/cute/derange.md
@@ -14,7 +14,6 @@ categories:
tags:
- Femboy's Dessert House
- 全错排
- - 排列组合
---
## 导入
From 475c74441d3f52a199b505f15a097b940d415a4b Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Fri, 8 May 2026 18:22:35 +0800
Subject: [PATCH 14/34] bugde.md
---
content/post/articles/cute/cutsandwichinhalf.md | 12 +-----------
1 file changed, 1 insertion(+), 11 deletions(-)
diff --git a/content/post/articles/cute/cutsandwichinhalf.md b/content/post/articles/cute/cutsandwichinhalf.md
index 52720a3..f14ff25 100644
--- a/content/post/articles/cute/cutsandwichinhalf.md
+++ b/content/post/articles/cute/cutsandwichinhalf.md
@@ -89,17 +89,7 @@ $$
接着,直接联立 $l$ 与 $m$, $l$ 与 $n$ 的方程,解得
$$
-\left\{\begin{matrix}
- x_1=\frac{a^2 b}{bx_0-y_0} \\
- y_1=\frac{ab^2}{bx_0-y_0}
-\end{matrix}\right.
-$$
-
-$$
-\left\{\begin{matrix}
- x_2=\frac{a^2 b}{bx_0+y_0} \\
- y_2=-\frac{ab^2}{bx_0+y_0}
-\end{matrix}\right.
+x_1=\frac{a^2 b}{bx_0-y_0}, y_1=\frac{ab^2}{bx_0-y_0}, x_2=\frac{a^2 b}{bx_0+y_0}, y_2=-\frac{ab^2}{bx_0+y_0}
$$
因此
From 0a970a680166c7cbf32b6465fc3e1f504c2bd009 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Sat, 9 May 2026 09:22:42 +0800
Subject: [PATCH 15/34] casesenvironment.md
---
content/post/articles/cute/cutsandwichinhalf.md | 12 +++++++++++-
1 file changed, 11 insertions(+), 1 deletion(-)
diff --git a/content/post/articles/cute/cutsandwichinhalf.md b/content/post/articles/cute/cutsandwichinhalf.md
index f14ff25..b99b08b 100644
--- a/content/post/articles/cute/cutsandwichinhalf.md
+++ b/content/post/articles/cute/cutsandwichinhalf.md
@@ -89,7 +89,17 @@ $$
接着,直接联立 $l$ 与 $m$, $l$ 与 $n$ 的方程,解得
$$
-x_1=\frac{a^2 b}{bx_0-y_0}, y_1=\frac{ab^2}{bx_0-y_0}, x_2=\frac{a^2 b}{bx_0+y_0}, y_2=-\frac{ab^2}{bx_0+y_0}
+\begin{cases}
+x_1=\frac{a^2 b}{bx_0-y_0}, \\
+y_1=\frac{ab^2}{bx_0-y_0},
+\end{cases}
+$$
+
+$$
+\begin{cases}
+x_2=\frac{a^2 b}{bx_0+y_0}, \\
+y_2=-\frac{ab^2}{bx_0+y_0}.
+\end{cases}
$$
因此
From e2066ea06dc05a156f64350aff38032e77164c37 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Sat, 9 May 2026 09:26:04 +0800
Subject: [PATCH 16/34] Update content/post/articles/cute/cutsandwichinhalf.md
Co-authored-by: Website-xieyuen <131447547+Website-xieyuen@users.noreply.github.com>
---
content/post/articles/cute/cutsandwichinhalf.md | 2 ++
1 file changed, 2 insertions(+)
diff --git a/content/post/articles/cute/cutsandwichinhalf.md b/content/post/articles/cute/cutsandwichinhalf.md
index b99b08b..7382b93 100644
--- a/content/post/articles/cute/cutsandwichinhalf.md
+++ b/content/post/articles/cute/cutsandwichinhalf.md
@@ -44,6 +44,8 @@ tags:
在初中的学习中,我们知道经反比例函数图象上的一点,作坐标轴的垂线,所围成的矩形面积一定.在高中,我们知道反比例函数的图象是双曲线,坐标轴就是它的两条渐近线,实际上,类似的与双曲线的渐近线有关的面积结论还有很多.在推导这个问题时,我们用到如下结论:
+> 引理:
+>
> 双曲线上任意一点的切线与它的渐近线围成的三角形面积为一个定值.
该定理的证明涉及复杂的字母运算,我们放到文章末尾.
From 6cb9d24f6d7d9f879731bb52fddd43f4857b7ab9 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Sat, 9 May 2026 09:26:16 +0800
Subject: [PATCH 17/34] Update content/post/articles/cute/cutsandwichinhalf.md
Co-authored-by: Website-xieyuen <131447547+Website-xieyuen@users.noreply.github.com>
---
content/post/articles/cute/cutsandwichinhalf.md | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/content/post/articles/cute/cutsandwichinhalf.md b/content/post/articles/cute/cutsandwichinhalf.md
index 7382b93..720bac5 100644
--- a/content/post/articles/cute/cutsandwichinhalf.md
+++ b/content/post/articles/cute/cutsandwichinhalf.md
@@ -78,7 +78,7 @@ $$
$X$ 为 $CF$ 的中点,证毕!!!
-## 上面那个结论的证明
+## 引理的证明
考虑平面直角坐标系 $xOy$ 中的双曲线 $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ ,作它在点 $P(x_0,y_0)$ 处的切线 $l$ ,分别交双曲线的渐近线 $m: y=\frac{b}{a} x$ 与 $n: y=-\frac{b}{a} x$ 于点 $A(x_1,y_1), B(x_2, y_2)$ .
From 369eaa40b42f03dc6d18379172d483aec44775d6 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Sat, 9 May 2026 09:28:05 +0800
Subject: [PATCH 18/34] pf
---
.../cute/{cutsandwichinhalf.md => cut-sandwich-in-half.md} | 0
1 file changed, 0 insertions(+), 0 deletions(-)
rename content/post/articles/cute/{cutsandwichinhalf.md => cut-sandwich-in-half.md} (100%)
diff --git a/content/post/articles/cute/cutsandwichinhalf.md b/content/post/articles/cute/cut-sandwich-in-half.md
similarity index 100%
rename from content/post/articles/cute/cutsandwichinhalf.md
rename to content/post/articles/cute/cut-sandwich-in-half.md
From a85f945f21d35b9ec8469a0a8f679913e9453d98 Mon Sep 17 00:00:00 2001
From: ZFCarrotFDH <3504574822@qq.com>
Date: Sat, 9 May 2026 20:16:52 +0800
Subject: [PATCH 19/34] Add files via upload
---
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.../cute/81da59a72952073f4241cbf14aa57ef6.jpg | Bin 0 -> 288982 bytes
2 files changed, 0 insertions(+), 0 deletions(-)
create mode 100644 content/post/articles/cute/12c65f7fba740405b2ca1e596afeda62.jpg
create mode 100644 content/post/articles/cute/81da59a72952073f4241cbf14aa57ef6.jpg
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