Geometry: prove that point M is touched by 4 circles

  • #1
Jiketz
2
0
Homework Statement
This is the problem: Given are two parallelograms ABCD and AECF with common diagonal
diagonal AC, where E and F lie inside the parallelogram ABCD.
Show:
The circumcircles of the triangles AEB, BFC, CED and DFA have a point
in common.
I've already given an answer in the pdf, but how can I prove that the point M is also on the remaining two circles? Here is another sketch I drew. Can you please finish the pdf paper ?
Relevant Equations
relevant equations can be found in the pdf solution from me
qukv4djahqia1.jpeg
 

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  • #2
Interesting exercise. I take Cartesian coordinates to look at as attached figure.
We can write the equation of circle touching three points as
[tex]x^2+y^2+lx+my+n=0[/tex]
Giving three (x,y)s of touching points, we get a set (l,m,n) . We get four (l,m,n) sets for the four circles. The system of four equations of the circles would show one degenerate solution.
That's the design of the proof, though I have not tired yet due to an expected terrible matrix calculation. The symmetry should make it easy.
1677040313458.png
 
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  • #4
Contd. from my previous post
anuttarasammyak said:
The system of four equations of the circles would show one degenerate solution.
The equation of line which connects two commom points of the cirlces i and j, is
[tex](l_i-l_j)x+(m_i-m_j)y+n_i-n_j=0[/tex]
The system of such six lines meet at a point (x,y) are recuced to three equations.
[tex]
\begin{pmatrix}
l_1-l_2 & m_1-m_2 & n_1-n_2 \\
l_2-l_3 & m_2-m_3 & n_2-n_3 \\
l_3-l_4 & m_3-m_4 & n_3-n_4 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
1 \\
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
0 \\
\end{pmatrix}
[/tex]
This 3X3 matrix in LHS shoud have determinant zero so that there is a solution.
[tex]
\begin{vmatrix}
l_1-l_2 & m_1-m_2 & n_1-n_2 \\
l_2-l_3 & m_2-m_3 & n_2-n_3 \\
l_3-l_4 & m_3-m_4 & n_3-n_4 \\
\end{vmatrix}
=\sum_{i,j,k,q=1}^4 P(ijkq)l_i m_j n_k
[/tex]
where P(ijkq)=0 when any of them equals.
P(ijkq)=1 when ijkq is even pertumation of 1234.
P(ijkq)=-1 when ijkq is odd permutation of 1234.

We expect to get coordinates of M as
[tex]
\begin{pmatrix}
l_1-l_4 & m_1-m_4 & n_1-n_4 \\
l_2-l_4 & m_2-m_4 & n_2-n_4 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
1 \\
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
\end{pmatrix}
[/tex]
or
[tex]
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
=\frac{-1}{(l_1-l_4 )(m_2-m_4 )-(l_2-l_4 )(m_1-m_4 )}
\begin{pmatrix}
m_2-m_4 & m_4-m_1 \\
l_4-l_2 & l_1-l_4 \\
\end{pmatrix}
\begin{pmatrix}
n_1-n_4 \\
n_2-n_4 \\
\end{pmatrix}
[/tex]
 
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  • #5
Supplement to my post #2

Say three points ##(a_1,a_2),(b_1,b_2),(c_1,c_2)## is on the circle which is caratterized by (l,m, n)
[tex]
\begin{pmatrix}
a_1& a_2 & 1 \\
b_1& b_2 & 1 \\
c_1& c_2 & 1 \\
\end{pmatrix}
\begin{pmatrix}
l \\
m \\
n \\
\end{pmatrix}
=-
\begin{pmatrix}
a_1^2+a_2^2 \\
b_1^2+b_2^2 \\
c_1^2+c_2^2 \\
\end{pmatrix}
[/tex]
Solving it for l,m,n
[tex]
\begin{pmatrix}
l \\
m \\
n \\
\end{pmatrix}
=\frac{-1}{det}
\begin{pmatrix}
b_2-c_2 & c_2-a_2 & a_2-b_2  \\
c_1-b_1 & a_1-c_1 & b_1-a_1 \\
b_1c_2-b_2c_1  & c_1a_2-c_2a_1 & a_1b_2-a_2b_1 \\
\end{pmatrix}
\begin{pmatrix}
a_1^2+a_2^2 \\
b_1^2+b_2^2 \\
c_1^2+c_2^2 \\
\end{pmatrix}
[/tex]
where
[tex]det=a_1b_2-a_2b_1+b_1c_2-b_2c_1+c_1a_2-c_2a_1[/tex]

Parameters in this exercise are
[tex]
\begin{matrix}
& CircleDAF & CircleDCE & CircleBCF & CircleBAE \\
a_1 & d_1 & d_1 & -d_1 & -d_1 \\
a_2 & d_2 & d_2 & -d_2 & -d_2 \\
b_1 & f_1 & -f_1 & f_1 & -f_1 \\
b_2 & f_2 & -f_2 & f_2 & -f_2 \\c_1 & -1 & 1 & 1 & -1 \\
c_2 & 0 & 0 & 0 & 0 \\
det & -d_2+f_2+d_1f_2-d_2f_1 & d_2+f_2-d_1f_2+d_2f_1 & -d_2-f_2-d_1f_2+d_2f_1 & d_2-f_2+d_1f_2-d_2f_1
\end{matrix}[/tex]
As for circle DAF
[tex]
\begin{pmatrix}
l \\
m \\
n \\
\end{pmatrix}
=\frac{-1}{det}
\begin{pmatrix}
f_2 & -d_2 & d_2-f_2 \\
-1-f_1 & d_1+1 & -d_1+f_1 \\
f_2& -d_2 & d_1f_2-d_2f_1 \\
\end{pmatrix}
\begin{pmatrix}
d_1^2+d_2^2 \\
f_1^2+f_2^2 \\
1 \\
\end{pmatrix}
[/tex]
[tex]
=\frac{-1}{det}
\begin{pmatrix}
f_2 |d|^2 -d_2 |f|^2 +d_2-f_2 \\
(-1-f_1 )|d|^2+ (d_1+1 )|f|^2 -d_1+f_1 \\
f_2|d|^2 -d_2|f|^2+d_1f_2-d_2f_1 \\
\end{pmatrix}
[/tex]
where
[tex]|d|^2=d_1^2+d_2^2[/tex]
[tex]|f|^2=f_1^2+f_2^2[/tex]
[tex]det=-d_2+f_2+d_1f_2-d_2f_1[/tex]

As for circle DCE
[tex]
\begin{pmatrix}
l \\
m \\
n \\
\end{pmatrix}
=\frac{-1}{det}
\begin{pmatrix}
-f_2 & -d_2 & d_2+f_2 \\
1+f_1 & d_1-1 & -d_1-f_1\\
f_2 & d_2 & -d_1f_2+d_2f_1 \\
\end{pmatrix}
\begin{pmatrix}
d_1^2+d_2^2 \\
f_1^2+f_2^2 \\
1 \\
\end{pmatrix}
[/tex]
[tex]=\frac{-1}{det}
\begin{pmatrix}
-f_2 |d|^2 -d_2 |f|^2+ d_2+f_2 \\
(1+f_1) |d|^2+ (d_1-1 )|f|^2 -d_1-f_1\\
f_2 |d|^2+ d_2|f|^2 -d_1f_2+d_2f_1 \\
\end{pmatrix}
[/tex]
where
[tex]det=d_2+f_2-d_1f_2+d_2f_1[/tex]

As for circle BCF
[tex]
\begin{pmatrix}
l \\
m \\
n \\
\end{pmatrix}
=\frac{-1}{det}
\begin{pmatrix}
f_2 & d_2& -d_2-f_2\\
1-f_1 & -1-d_1 & d_1+f_1\\
-f_2 & -d_2 & -d_1f_2+d_2f_1 \\
\end{pmatrix}
\begin{pmatrix}
d_1^2+d_2^2 \\
f_1^2+f_2^2 \\
1 \\
\end{pmatrix}
[/tex]
[tex]
=\frac{-1}{det}
\begin{pmatrix}
f_2 |d|^2+ d_2|f|^2 -d_2-f_2\\
(1-f_1)|d|^2 +( -1-d_1) |f|^2+ d_1+f_1\\
-f_2 |d|^2 -d_2 |f|^2 -d_1f_2+d_2f_1 \\
\end{pmatrix}
[/tex]
where
[tex]det=-d_2-f_2-d_1f_2+d_2f_1[/tex]
for circle BAE

[tex]
\begin{pmatrix}
l \\
m \\
n \\
\end{pmatrix}
=\frac{-1}{det}
\begin{pmatrix}
-f_2 & d_2 & -d_2+f_2 \\
-1+f_1 & -d_1+1 & d_1-f_1 \\
-f_2 & d_2 & d_1f_2-d_2f_1 \\
\end{pmatrix}
\begin{pmatrix}
d_1^2+d_2^2 \\
f_1^2+f_2^2 \\
1 \\
\end{pmatrix}
[/tex]
[tex]
=\frac{-1}{det}
\begin{pmatrix}
-f_2 |d|^2+ d_2 |f|^2 -d_2+f_2 \\
(-1+f_1)|d|^2+ (-d_1+1 )|f|^2+ d_1-f_1 \\
-f_2 |d|^2+ d_2 |f|^2+ d_1f_2-d_2f_1 \\
\end{pmatrix}
[/tex]
where
[tex]det=d_2-f_2+d_1f_2-d_2f_1[/tex]
I hope I have not made careless mistake.
 
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