圆内接四边形有Brocard点⇔调和四边形

hbghlyj

圆内接四边形ABCD和点P使得
∠PAB=∠PBC=∠PCD=∠PDA
求证：AB·CD=BC·AD

hbghlyj

Given a cyclic quadrilateral $ABCD$ and a point $P$ such that $\angle PAB = \angle PBC = \angle PCD = \angle PDA = \omega$. Prove that $AB \cdot CD = BC \cdot AD$.
jstor.org/stable/3603502
Let the internal angles of the quadrilateral be $A = \angle DAB$, $B = \angle ABC$, $C = \angle BCD$, $D = \angle CDA$.
The problem states $\angle PAB = \angle PBC = \angle PCD = \angle PDA = \omega$.
Consider the angles:
$\angle PBA = \angle ABC - \angle PBC = B - \omega$ (assuming P is positioned such that $\omega < B$, etc.)
$\angle PCB = C - \omega$
$\angle PDC = D - \omega$
$\angle PAD = A - \omega$

From the sine rule applied to triangles formed by P and vertices (e.g., $\triangle APB, \triangle BPC$, etc.):
\begin{align*}
    \frac{BP}{\sin(\angle PAB)} &= \frac{AP}{\sin(\angle PBA)} \implies \frac{BP}{\sin \omega} = \frac{AP}{\sin(B-\omega)} \\
    \frac{CP}{\sin(\angle PBC)} &= \frac{BP}{\sin(\angle PCB)} \implies \frac{CP}{\sin \omega} = \frac{BP}{\sin(C-\omega)} \\
    \frac{DP}{\sin(\angle PCD)} &= \frac{CP}{\sin(\angle PDC)} \implies \frac{DP}{\sin \omega} = \frac{CP}{\sin(D-\omega)} \\
    \frac{AP}{\sin(\angle PDA)} &= \frac{DP}{\sin(\angle PAD)} \implies \frac{AP}{\sin \omega} = \frac{DP}{\sin(A-\omega)}
\end{align*}
Rearranging these gives:
\begin{align*}
    AP \sin \omega &= BP \sin(B-\omega) \\
    BP \sin \omega &= CP \sin(C-\omega) \\
    CP \sin \omega &= DP \sin(D-\omega) \\
    DP \sin \omega &= AP \sin(A-\omega)
\end{align*}
Multiplying these four equations together:
\[ (AP \cdot BP \cdot CP \cdot DP) (\sin \omega)^4 = (AP \cdot BP \cdot CP \cdot DP) \sin(A-\omega)\sin(B-\omega)\sin(C-\omega)\sin(D-\omega) \]
Dividing by $AP\cdot BP\cdot CP\cdot DP$
\[ (\sin \omega)^4 = \sin(A-\omega)\sin(B-\omega)\sin(C-\omega)\sin(D-\omega) \]
Since $ABCD$ is cyclic, $C = 180^\circ - A$ and $D = 180^\circ - B$.
So, $\sin(C-\omega) = \sin(180^\circ - A - \omega) = \sin(A+\omega)$.
And $\sin(D-\omega) = \sin(180^\circ - B - \omega) = \sin(B+\omega)$.
Substituting these into the equation:
\[ (\sin \omega)^4 = \sin(A-\omega)\sin(B-\omega)\sin(A+\omega)\sin(B+\omega) \]
Using the identity $\sin(x-y)\sin(x+y) = \sin^2 x - \sin^2 y$:
\[ (\sin \omega)^4 = (\sin^2 A - \sin^2 \omega)(\sin^2 B - \sin^2 \omega) \]
Divide by $(\sin\omega)^4 \sin^2 A \sin^2 B$
\[ \frac{1}{\sin^2 A \sin^2 B} = \frac{\sin^2 A - \sin^2 \omega}{\sin^2\omega \sin^2 A} \cdot \frac{\sin^2 B - \sin^2 \omega}{\sin^2\omega \sin^2 B} \]
\[ \csc^2 A \csc^2 B = \left(\frac{1}{\sin^2\omega} - \frac{1}{\sin^2 A}\right) \left(\frac{1}{\sin^2\omega} - \frac{1}{\sin^2 B}\right) \]
\[ \csc^2 A \csc^2 B = (\csc^2\omega - \csc^2 A)(\csc^2\omega - \csc^2 B) \]
which implies
\begin{equation} \label{eq:1}
\csc^2\omega = \csc^2 A + \csc^2 B \quad
\end{equation}
Let $AB=a, BC=b, CD=c, DA=d$.
The angles subtended by the sides at point $P$ are given as:
$\angle APB = 180^\circ - B = D$ (where $B$ is $\angle ABC$, etc.)
$\angle BPC = 180^\circ - C = A$
$\angle CPD = 180^\circ - D = B$
$\angle DPA = 180^\circ - A = C$
Using the sine rule:
In $\triangle PAB$: $\frac{BP}{\sin\omega} = \frac{a}{\sin(\angle APB)} = \frac{a}{\sin D}$. So $BP = \frac{a \sin\omega}{\sin D}$.
In $\triangle PBC$: $\frac{BP}{\sin(C-\omega)} = \frac{b}{\sin(\angle BPC)} = \frac{b}{\sin A}$. So $BP = \frac{b \sin(C-\omega)}{\sin A}$.
Equating the expressions for $BP$:
\[ \frac{a \sin\omega}{\sin D} = \frac{b \sin(C-\omega)}{\sin A} \]
\[ \frac{a}{b} = \frac{\sin(C-\omega)\sin D}{\sin A \sin\omega} \]
Since $ABCD$ is cyclic, $C = 180^\circ - A \implies \sin(C-\omega) = \sin(180^\circ - A - \omega) = \sin(A+\omega)$.
And $D = 180^\circ - B \implies \sin D = \sin B$.
So,
\[ \frac{a}{b} = \frac{\sin(A+\omega)\sin B}{\sin A \sin\omega} = \frac{(\sin A \cos\omega + \cos A \sin\omega)\sin B}{\sin A \sin\omega} = (\cot\omega + \cot A)\sin B \]
Similarly, for $DP$:
In $\triangle PCD$: $\frac{DP}{\sin\omega} = \frac{c}{\sin(\angle CPD)} = \frac{c}{\sin B}$. So $DP = \frac{c \sin\omega}{\sin B}$.
In $\triangle PDA$: $\frac{DP}{\sin(A-\omega)} = \frac{d}{\sin(\angle DPA)} = \frac{d}{\sin C}$. So $DP = \frac{d \sin(A-\omega)}{\sin C}$.
Equating the expressions for $DP$:
\[ \frac{c \sin\omega}{\sin B} = \frac{d \sin(A-\omega)}{\sin C} \]
\[ \frac{c}{d} = \frac{\sin(A-\omega)\sin B}{\sin C \sin\omega} \]
Since $ABCD$ is cyclic, $\sin C = \sin A$.
So,
\[ \frac{c}{d} = \frac{\sin(A-\omega)\sin B}{\sin A \sin\omega} = \frac{(\sin A \cos\omega - \cos A \sin\omega)\sin B}{\sin A \sin\omega} = (\cot\omega - \cot A)\sin B \]
Now, multiply $\frac{a}{b}$ and $\frac{c}{d}$:
\begin{align*} \frac{ac}{bd} &= [(\cot\omega + \cot A)\sin B] \cdot [(\cot\omega - \cot A)\sin B] \\ &= (\cot^2\omega - \cot^2 A)\sin^2 B \end{align*}
Using the identity $\cot^2 x - \cot^2 y = (\csc^2 x - 1) - (\csc^2 y - 1) = \csc^2 x - \csc^2 y$:
\[ \frac{ac}{bd} = (\csc^2\omega - \csc^2 A)\sin^2 B \]
From relation \eqref{eq:1}, $\csc^2\omega = \csc^2 A + \csc^2 B$.
So, $\csc^2\omega - \csc^2 A = \csc^2 B$.
Substituting this into the expression for $\frac{ac}{bd}$:
\[ \frac{ac}{bd} = (\csc^2 B) \sin^2 B = \frac{1}{\sin^2 B} \cdot \sin^2 B = 1 \]
Therefore, $ac = bd$. This means $AB \cdot CD = BC \cdot AD$.

lxz2336831534

托勒密是否有用

hejoseph

一些量：设 $AB=a$，$BC=b$，$CD=c$，$DA=d$，$ac=bd=s$，四边形 $ABCD$ 的外接圆圆心是 $O$，点 $P$ 关于四边形 $ABCD$ 的等角共轭点是 $Q$，$AC$ 中点 $M$、$BD$ 中点 $N$、点 $O$、点 $P$、点 $Q$ 共圆，这个圆的半径是 $r$，则
\begin{align*}
\angle POQ&=2\omega\\
\cos\omega&=\frac{(a^2+b^2)(a^2b^2+s^2)}{2ab\sqrt{a^4b^4+(a^4+4a^2b^2+b^4)s^2+s^4}}\\
OP=OQ&=\frac{s\sqrt{(a^2+b^2)(a^2b^2+s^2)((a^4+b^4)(a^4b^4+s^4)-4a^4b^4s^2)}}{2ab(a^4b^4+(a^4+4a^2b^2+b^4)s^2+s^4)\sin\omega}\\
PQ&=\frac{s\sqrt{(a^2+b^2)(a^2b^2+s^2)((a^4+b^4)(a^4b^4+s^4)-4a^4b^4s^2)}}{ab(a^4b^4+(a^4+4a^2b^2+b^4)s^2+s^4)}\\
r&=\frac{s}{2\sin\omega}\sqrt{\frac{(a^4+b^4)(a^4b^4+s^4)-4a^4b^4s^2}{(a^2+b^2)(a^2b^2+s^2)(a^4b^4+(a^4+4a^2b^2+b^4)s^2+s^4)}}
\end{align*}

hbghlyj

guanmo1 wrote at 2018-3-27 08:52
现在关于勃罗卡点的研究的一个最新方向是一般多边形存在勃罗卡点的充要条件。

一般多边形存在勃罗卡点的充要条件是