This is a follow up to my solution to the question of peta arantes. In that solution, i am showing a proposition, which solves the cited question in a more general situation, starting with points $A,B,D$ in general position. However, during the process of searching for a solution in given special case of a pentagon $ABCDE$, a coincidence was catching my eye. I have a proof for this "coincidence", but not one inside synthetic geometry. Below i am stating a proposition - which involves two pentagons, three squares, and some few related points - using notations close to those in loc. cit.. The question is how to show this proposition using (only) geometric constructions, and it will be stated after the proposition. Finally i have some notes on how my present proof works.
Result: Let $ABCDE$ be a regular pentagon. The mid point of $AD$ is denoted by $H$. The reflection in the line $BC$ is denoted by a prime, i.e. $P\to P'$. Reflecting the regular pentagon and the point $H$, we thus obtain the regular pentagon $A'BCD'E'$ and the mid point $H'$ of $A'D'$. Let us construct the squares $ABTU$ and $DBKJ$ in the exterior of $\Delta ABD$. Let $ALDY$ be the square with diagonal $AD$. Let $X$ be the intersection of the diagonals in the orthogonal quadrilateral $ADKT$. The mid points of its sides $AD$, $DK$, $KT$, $TA$ are denoted by $H$, $\Xi$, $S$, $\Sigma$. The mid point of $BC$ is $N$. Then we have the following properties:
- $\bf(1)$ The five points $U,X,L,C,J$ are colinear.
- $\bf(2)$ The five lines are related by parallelism: $$ UXLCJ\ \| \ \Sigma N\Xi\ \| \ TD'\ \| \ SH'\ \| \ A'K\ . $$
Question: Is there a (more or less) purely geometric argument for $C$ being on the line $UXLJ$, for $N$ being on the line $\Sigma\Xi$, and for the above parallelism?
Here are some notes in the context of the above Result. The colinearity of $U,X,L,J$ was shown in the general case of an arbitrary triangle $\Delta ABD$ by using (rather) geometric arguments in my solution of the already mentioned question of peta arantes. (That question went in an other direction, however, and it was solved by the relation $CH=BH=SL=TK/2$, since $T,L,K$ are three vertices of a square - just in case somebody needs one more square in the picture.)
As by-product of that proposition, the quadrilateral $ADKT$ is orthodiagonal, so its Varignon parallelogram $H\Xi S\Sigma$ is a rectangle. The line $\Xi\Sigma$ is also in general parallel to the line $UXLJ$. Now the special case of a regular pentagon $ABCDE$ enters the scene. (Although we will never need the point $E$...) It turns out that $C$ is also a point on the line $UXLJ$. My proof so far "destroys the geometry", and uses complex number towards an almost algorithmic proof. Here is this argument explicitly. Affixes of points denoted by upper case letters, are denoted by the corresponding lower case letters. We can arrange (by rescaling, translation, and rotation) that the points of the regular pentagon have the affixes $c=1$, $b=\zeta=\zeta_5$, $a=\zeta^2$, $e=\zeta^3$, $d=\zeta^4=1/\zeta$, where $\zeta=\cos\frac{2\pi}5+i\sin\frac{2\pi}5$ is this choice of a primitive root of order five of the unity. Then $$ \begin{aligned} u &= a+i(b-a)=\zeta^2 +i(\zeta-\zeta^2)\ ,\\ \bar u &= \bar a-i(\bar b-\bar a)=\zeta^3 -i(\zeta^4-\zeta^3)\ ,\\[2mm] j &= d-i(b-d)=\zeta^4 -i(\zeta-\zeta^4)\ ,\\ \bar j &= \bar d + i(\bar b-\bar d)=\zeta +i(\zeta^4-\zeta)\ ,\\ \end{aligned} $$ and we have to compute the following determinant, and show it vanishes: $$ \begin{aligned} \begin{vmatrix} 1 & c & \bar c\\ 1 & u & \bar u\\ 1 & j & \bar j \end{vmatrix} &= \begin{vmatrix} 1 & 1 & 1\\ 1 & u & \bar u\\ 1 & j & \bar j \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1\\ 0 & u-1 & \bar u-1\\ 0 & j-1 & \bar j-1 \end{vmatrix} \\ &\qquad=(u-1)(\bar j-1) -\overline{ (u-1)(\bar j-1) }\ . \\ (u-1)(\bar j-1) &=\Big(\ (\zeta^2-1) +i(\zeta-\zeta^2) \ \Big)\Big(\ (\zeta-1) + i (\zeta^4-\zeta)\ \Big) \\ &=-i\zeta\Big(\ i(\zeta-\zeta^4) + (\zeta-1) \ \Big)\Big(\ (\zeta-1) + i (\zeta^4-\zeta)\ \Big) \\ &=-i\zeta^6\Big(\ (\zeta-\zeta^{-1})^2 + (\zeta-1)^2 \ \Big) \\ &=-i\Big(\ (\zeta^4-\zeta^2)^2 + (\zeta^4-\zeta^3)^2 \ \Big) \\ &=-i\Big(\ \zeta^8-2\zeta^6+\zeta^4 + \zeta^8-2\zeta^7+\zeta^6 \ \Big) \\ &=-i\Big(\ \zeta^4 + 2\zeta^3 - 2\zeta^2 -\zeta \ \Big) \ ,\\[2mm] \overline{(u-1)(\bar j-1)} &=\overline{-i\Big(\ \zeta^4 + 2\zeta^3 - 2\zeta^2 -\zeta \ \Big)} \\ &=i\Big(\ \zeta + 2\zeta^2 - 2\zeta^3 -\zeta^4 \ \Big)\ , \end{aligned} $$ and the difference $(u-1)(\bar j-1) - (\bar u-1)(j-1)$ vanishes. The calculus may seem messy, but is straightforward. To destroy even more, note that the computer does the job in a few lines...
F.<ZETA20> = CyclotomicField(20)
z, I = ZETA20^4, ZETA20^5
a, b, c, d = z^2, z, 1, 1/z
u, j = a + I*(b-a), d - I*(b-d)
matrix(3, 3, [[1, c, c.conjugate()], [1, u, u.conjugate()], [1, j, j.conjugate()]]).det()
And the last line delivers the same zero computed above. The colinearity of $\Sigma,N,\Xi$ is also simple, when using complex numbers.
t, k = b - I*(a-b), b + I*(d-b)
sig, xi, n = (t+a)/2, (k+d)/2, (b+c)/2
matrix(3, 3, [[1, sig, sig.conjugate()], [1, n, n.conjugate()], [1, xi, xi.conjugate()]]).det()
Now using the common mid point of the segments $SH$, $\Sigma,\Xi$, we immediately have in $\Delta HSH'$ the parallelism of the side $SH'$ with the corresponding mid line, thus $SH'\|\Sigma N\Xi$. Same mid line argument in $\Delta BUC$ gives $\Sigma N\| UC$. The quadrilateral $A'BDC$ is a parallelogram with $N$ mid point and intersection of the diagonals. The mid line argument in $\Delta DA'K$ gives $DA'\|N\Xi$. Using the parallelogram $ABD'C$ instead, we see that $N$ is the mid point of $AD'$, so the mid line argument in $\Delta ATD'$ gives the final $\Sigma N\| TD'$.
A furter observation is as follows. Let us place the points $A,D$ in the complex plane on the fixed positions with affixes $a=i$, $d=\bar a=-i$. Which is the geometric locus of the points $B$, so that building the isosceles trapezium $ABCD$ (so $c=\bar b$), the square $ABTU$ (so $u=a+i(b-a)$), and the square $DBKJ$ (so $j=d-i(b-d)$) finally the points $U,C,J$ are colinear?
Answer:
sage: var('x,y');
sage: a, b, c, d = i, x+i*y, x-i*y, -i
sage: u, j = a + i*(b-a), d - i*(b-d)
sage: factor(matrix(3, 3, [1, c, c.conjugate(), 1, u, u.conjugate(), 1, j, j.conjugate()]).det())
-4*I*x^2 + 4*I*y^2 + 4*I
So the answer is the hyperbola with equation $x^2=y^2+1$. This is a "complicated conic", possibly making the needed geometric argument not so obvious.
