Geometry: prove that point M is touched by 4 circles

AI Thread Summary
The discussion revolves around proving that point M is touched by four circles using Cartesian coordinates. The equations of the circles are derived from the coordinates of three touching points, leading to a system of equations that reveals a degenerate solution. The participants emphasize the importance of matrix calculations to find the coordinates of point M, with a focus on the determinant of a specific 3x3 matrix being zero for a solution to exist. Additional equations and parameters related to the circles are provided to support the proof. The complexity of the calculations is acknowledged, but the symmetry in the problem is expected to simplify the process.
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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
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Interesting exercise. I take Cartesian coordinates to look at as attached figure.
We can write the equation of circle touching three points as
x^2+y^2+lx+my+n=0
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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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
(l_i-l_j)x+(m_i-m_j)y+n_i-n_j=0
The system of such six lines meet at a point (x,y) are recuced to three equations.
<br /> \begin{pmatrix}<br /> l_1-l_2 &amp; m_1-m_2 &amp; n_1-n_2 \\<br /> l_2-l_3 &amp; m_2-m_3 &amp; n_2-n_3 \\<br /> l_3-l_4 &amp; m_3-m_4 &amp; n_3-n_4 \\<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> x \\<br /> y \\<br /> 1 \\<br /> \end{pmatrix}<br /> =<br /> \begin{pmatrix}<br /> 0 \\<br /> 0 \\<br /> 0 \\<br /> \end{pmatrix}<br />
This 3X3 matrix in LHS shoud have determinant zero so that there is a solution.
<br /> \begin{vmatrix}<br /> l_1-l_2 &amp; m_1-m_2 &amp; n_1-n_2 \\<br /> l_2-l_3 &amp; m_2-m_3 &amp; n_2-n_3 \\<br /> l_3-l_4 &amp; m_3-m_4 &amp; n_3-n_4 \\<br /> \end{vmatrix}<br /> =\sum_{i,j,k,q=1}^4 P(ijkq)l_i m_j n_k<br />
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
<br /> \begin{pmatrix}<br /> l_1-l_4 &amp; m_1-m_4 &amp; n_1-n_4 \\<br /> l_2-l_4 &amp; m_2-m_4 &amp; n_2-n_4 \\<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> x \\<br /> y \\<br /> 1 \\<br /> \end{pmatrix}<br /> =<br /> \begin{pmatrix}<br /> 0 \\<br /> 0 \\<br /> \end{pmatrix}<br />
or
<br /> \begin{pmatrix}<br /> x \\<br /> y \\<br /> \end{pmatrix}<br /> =\frac{-1}{(l_1-l_4 )(m_2-m_4 )-(l_2-l_4 )(m_1-m_4 )}<br /> \begin{pmatrix}<br /> m_2-m_4 &amp; m_4-m_1 \\<br /> l_4-l_2 &amp; l_1-l_4 \\<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> n_1-n_4 \\<br /> n_2-n_4 \\<br /> \end{pmatrix}<br />
 
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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)
<br /> \begin{pmatrix}<br /> a_1&amp; a_2 &amp; 1 \\<br /> b_1&amp; b_2 &amp; 1 \\<br /> c_1&amp; c_2 &amp; 1 \\<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> l \\<br /> m \\<br /> n \\<br /> \end{pmatrix}<br /> =-<br /> \begin{pmatrix}<br /> a_1^2+a_2^2 \\<br /> b_1^2+b_2^2 \\<br /> c_1^2+c_2^2 \\<br /> \end{pmatrix}<br />
Solving it for l,m,n
<br /> \begin{pmatrix}<br /> l \\<br /> m \\<br /> n \\<br /> \end{pmatrix}<br /> =\frac{-1}{det}<br /> \begin{pmatrix}<br /> b_2-c_2 &amp; c_2-a_2 &amp; a_2-b_2  \\<br /> c_1-b_1 &amp; a_1-c_1 &amp; b_1-a_1 \\<br /> b_1c_2-b_2c_1  &amp; c_1a_2-c_2a_1 &amp; a_1b_2-a_2b_1 \\<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> a_1^2+a_2^2 \\<br /> b_1^2+b_2^2 \\<br /> c_1^2+c_2^2 \\<br /> \end{pmatrix}<br />
where
det=a_1b_2-a_2b_1+b_1c_2-b_2c_1+c_1a_2-c_2a_1

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

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

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

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