List of geometric theorems linked by two squares

Question

I'm trying to create a classification for geometric theorems that relate to two squares As a type of organization and classification And the curiosity to explore

I have collected some theorems of this type that I will put in an answer/answers.

I hope you can help me expand my list.

It's important to note that I'm not looking for theorems about squares because it would become too extensive a list, I'm looking for theorems about a number of squares equal to exactly twoTherefore, theorems such as Van Opel's theorem are not accepted in the answers

زكريا حسناوي · Accepted Answer · 2024-05-29 10:30:48Z

For now, I will only post pictures of the theorems. Perhaps later I will add links that talk about the theorems in detail

Potema's theorem: Two squares created on the sides of a triangle $∆ABC$ And we set the point $M$ Which represents the middle $EF$ It will be constant no matter how you move point $C$ As long as $A,B$ are constant

Fensler-Hadwiger theorem We have two squares with a common vertex between them, so they will be the midpoints of the spaces between the opposite vertices, and the centers of the two squares will form the vertices of a third square

Two squares with a common vertex, the previous three properties will be fulfilled

If two squares have a common vertex, the two red lines will be perpendicular and equal

If we have two identical squares with one vertex at the center of the other, the common area between the two squares will be equal to a quarter of the area of one square

In the previous figure there are two properties achieved: The first is that $AM=BM$ The second characteristic is that $PQ=2MN$

Two identical or different squares intersect and are enclosed between them $A,B,C,D$ The aforementioned equality will be achieved

If two identical or different squares have a common vertex, the blue area will be equal to the red area

If two squares have a common vertex, these three lines will meet at one point Moreover, the circles of the two squares pass through the same point, which represents the center of the spiral similarity

If two squares have parallel sides, these lines will meet at one point Furthermore itIf the two squares do not have parallel sides, then these lines will meet at one point. In the previous figure it would be: $\frac{b}{a}=\frac{\sqrt{5}+1}{2}=Φ$

In the previous picture it is: $X=\frac{5(A-B)}{8}$

We have two squares that have a common vertex, the point $F$ moves freely along the straight line $CD$ It will be points $A,C,Q$ Located on one straight line

Two squares in the plane In the general case, these two lines will be perpendicular

Here are some theorems that can be understood from pictures only without words:

@DS There is proof in one of the comments on this post: facebook.com/share/p/TkbGMp1ZfLeodWjB/?mibextid=TrneLp — زكريا حسناوي, Commented Mar 23 at 16:41
@زكرياحسناوي: The list just keeps growing! :) ... Please add names (or at least numbers) to these results, so that they're easier to reference in commentary. Citations would also be helpful, as there may be useful context or related results that help people (like me) avoid spending time investigating generalizations that might already be known. :) — Blue, Commented Apr 22 at 12:27
I have added another answer in which I will include links over time, I have not shared anything on this site for a while, I will share some new intuitions during the coming period, happy that you are interested in this post @blue — زكريا حسناوي, Commented Apr 23 at 8:42

Ethan Bolker · Accepted Answer · 2024-03-23 14:15:52Z

4

Euclid Book II Prop 6 on what we would now describe algebraically as the difference of squares:

answered Mar 23 at 14:15

Ethan Bolker

97.5k7 gold badges114 silver badges204 bronze badges

Add a comment |

Blue · Accepted Answer · 2024-04-01 02:39:59Z

Here are somewhat-natural generalizations of a couple of the two-squares-joined-at-a-vertex results from OP's answer.

Potema's Theorem

Let squares $\square A'B'C'D'$ and $\square A''B''C''D''$ be as shown, with $X$ the midpoint of $X'X''$.

If segment $A'A''$ remains fixed, and segment $D'D''$ keeps a constant length and inclination (in Potema's theorem, $D'=D''$), then midpoint $B$ remains fixed. Moreover, the position of $C$ relative to $D$ matches that of $B$ relative to $A$.

That is, $\square ABCD$ is a parallelogram, with $|AB|$ and $|CD|$ determined by the lengths and relative inclinations of $A'A''$ and $C'C''$.

Specifically, segments $AB$ and $CD$ form congruent triangles with segments of length $|AA'|$ and $|DD'|$ (more-specifically, segments perpendicular to $A'A''$ and $B'B''$). Likewise, segments $BC$ and $AD$ form congruent triangles with segments of length $|BB'|$ and $|CC'|$.

Fensler-Hadwiger

Again, we have squares $\square A'B'C'D'$ and $\square A''B''C''D''$, and midpoints $X$ of $X'X''$.

The square with opposite vertices at the centers of the two squares has its other two vertices equally offset from $B$ and $D$: namely, by distance $|AA'|$ and in directions perpendicular to $A'A''$. (Moreover, they're offset by $|CC'|$ in directions perpendicular to $C'C''$.)

In Fensler-Hedwiger, $A'=A''$, so that the new square's vertices are $B$ and $D$ themselves. Not mentioned in OP's statement of FH is that $BD\perp C'C''$; also, the diameter of the square is equal to $|CC'|$, but this fact is obvious:

There's a bonus square with opposite vertices $A$ and $C$; the remaining vertices are midpoints of $B'D''$ and $B''D'$. This is actually an instance of a deeper theorem: If $P_0P_1\cdots P_n$ and $Q_0Q_1\cdots Q_n$ are similar polygonal paths with opposite orientations in plane (that is, if they're related by a dilation with a negative scale factor), then the midpoints of $P_iQ_i$ form another similar path.

Other sets of midpoints form parallelograms.

$A_1+A_2 = \frac12(M^2+N^2)$

Let squares $\square A'B'C'D'$ and $\square A''B''C''D''$ be as shown, and define $x := |X'X''|$. Writing $S'$ and $S''$ for their areas, and $S$ for the area of the square with opposite vertices at their centers, we can find that

$$a^2+b^2+c^2+d^2 \;=\; 2 \left(\; S' + S'' + 4 S \;\right)$$

For the referenced result from OP's list, we can take $A'=A''$, so that $a=0$. From comments above about the Fensler-Hadwiger theorem, we also have that the diameter of the center square is $\frac12c$. Hence, $S = \frac18 c^2$, and the generalized result specializes back to $b^2+d^2=2(S'+S'')$.

Thank you very much. I know some generalizations that make the aforementioned theorems valid for parallelograms, but I did not imagine that there are generalizations that are valid for squares. Nice work. @blue — زكريا حسناوي, Commented Apr 3 at 3:34