# Geometry: prove that point M is touched by 4 circles

• Jiketz
In summary, the conversation discusses the equation for a circle touching three points and the system of equations for four circles. The design of the proof is shown, but the speaker has not tried it yet due to a potentially difficult matrix calculation. The conversation also mentions a link for additional information. The final part of the conversation discusses parameters and equations for four specific circles. The speaker provides equations for the circles and their determinants, and concludes with a note about making careless mistakes.
Jiketz
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

#### Attachments

• P1.pdf
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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.

Last edited:
Lnewqban
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.
$$\begin{pmatrix} l_1-l_2 & m_1-m_2 & n_1-n_2 \\ l_2-l_3 & m_2-m_3 & n_2-n_3 \\ l_3-l_4 & m_3-m_4 & n_3-n_4 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix}$$
This 3X3 matrix in LHS shoud have determinant zero so that there is a solution.
$$\begin{vmatrix} l_1-l_2 & m_1-m_2 & n_1-n_2 \\ l_2-l_3 & m_2-m_3 & n_2-n_3 \\ l_3-l_4 & m_3-m_4 & n_3-n_4 \\ \end{vmatrix} =\sum_{i,j,k,q=1}^4 P(ijkq)l_i m_j n_k$$
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
$$\begin{pmatrix} l_1-l_4 & m_1-m_4 & n_1-n_4 \\ l_2-l_4 & m_2-m_4 & n_2-n_4 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \end{pmatrix}$$
or
$$\begin{pmatrix} x \\ y \\ \end{pmatrix} =\frac{-1}{(l_1-l_4 )(m_2-m_4 )-(l_2-l_4 )(m_1-m_4 )} \begin{pmatrix} m_2-m_4 & m_4-m_1 \\ l_4-l_2 & l_1-l_4 \\ \end{pmatrix} \begin{pmatrix} n_1-n_4 \\ n_2-n_4 \\ \end{pmatrix}$$

Last edited:
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)
$$\begin{pmatrix} a_1& a_2 & 1 \\ b_1& b_2 & 1 \\ c_1& c_2 & 1 \\ \end{pmatrix} \begin{pmatrix} l \\ m \\ n \\ \end{pmatrix} =- \begin{pmatrix} a_1^2+a_2^2 \\ b_1^2+b_2^2 \\ c_1^2+c_2^2 \\ \end{pmatrix}$$
Solving it for l,m,n
$$\begin{pmatrix} l \\ m \\ n \\ \end{pmatrix} =\frac{-1}{det} \begin{pmatrix} b_2-c_2 & c_2-a_2 & a_2-b_2 \\ c_1-b_1 & a_1-c_1 & b_1-a_1 \\ b_1c_2-b_2c_1 & c_1a_2-c_2a_1 & a_1b_2-a_2b_1 \\ \end{pmatrix} \begin{pmatrix} a_1^2+a_2^2 \\ b_1^2+b_2^2 \\ c_1^2+c_2^2 \\ \end{pmatrix}$$
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
$$\begin{matrix} & CircleDAF & CircleDCE & CircleBCF & CircleBAE \\ a_1 & d_1 & d_1 & -d_1 & -d_1 \\ a_2 & d_2 & d_2 & -d_2 & -d_2 \\ b_1 & f_1 & -f_1 & f_1 & -f_1 \\ b_2 & f_2 & -f_2 & f_2 & -f_2 \\c_1 & -1 & 1 & 1 & -1 \\ c_2 & 0 & 0 & 0 & 0 \\ det & -d_2+f_2+d_1f_2-d_2f_1 & d_2+f_2-d_1f_2+d_2f_1 & -d_2-f_2-d_1f_2+d_2f_1 & d_2-f_2+d_1f_2-d_2f_1 \end{matrix}$$
As for circle DAF
$$\begin{pmatrix} l \\ m \\ n \\ \end{pmatrix} =\frac{-1}{det} \begin{pmatrix} f_2 & -d_2 & d_2-f_2 \\ -1-f_1 & d_1+1 & -d_1+f_1 \\ f_2& -d_2 & d_1f_2-d_2f_1 \\ \end{pmatrix} \begin{pmatrix} d_1^2+d_2^2 \\ f_1^2+f_2^2 \\ 1 \\ \end{pmatrix}$$
$$=\frac{-1}{det} \begin{pmatrix} f_2 |d|^2 -d_2 |f|^2 +d_2-f_2 \\ (-1-f_1 )|d|^2+ (d_1+1 )|f|^2 -d_1+f_1 \\ f_2|d|^2 -d_2|f|^2+d_1f_2-d_2f_1 \\ \end{pmatrix}$$
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
$$\begin{pmatrix} l \\ m \\ n \\ \end{pmatrix} =\frac{-1}{det} \begin{pmatrix} -f_2 & -d_2 & d_2+f_2 \\ 1+f_1 & d_1-1 & -d_1-f_1\\ f_2 & d_2 & -d_1f_2+d_2f_1 \\ \end{pmatrix} \begin{pmatrix} d_1^2+d_2^2 \\ f_1^2+f_2^2 \\ 1 \\ \end{pmatrix}$$
$$=\frac{-1}{det} \begin{pmatrix} -f_2 |d|^2 -d_2 |f|^2+ d_2+f_2 \\ (1+f_1) |d|^2+ (d_1-1 )|f|^2 -d_1-f_1\\ f_2 |d|^2+ d_2|f|^2 -d_1f_2+d_2f_1 \\ \end{pmatrix}$$
where
$$det=d_2+f_2-d_1f_2+d_2f_1$$

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

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

Last edited:

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