Is here a specific name for the following theorem? (Sides of inscribed squares of a triangle meet at points...












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enter image description hereenter image description here



For any acute and right triangle, two of the intersections(C and B) of the two inscribed squares and the vertex A of the triangle are collinear.



I have a proof for the theorem, but I have not found any specific name for it. I searched "inscribed squares in a triangle," and I still could not find the same shape. Can anyone provide a specific name for this, or a way that I can find the theorem online?



For the similar question concerning the cubes inscribed in a tetrahedron, please visit Extend squares inscribed in a triangle to the cubes inscribed in a tetrahedron
.



enter image description here



Proof:



$tan(angle ACB) = frac{IK}{KC}$



$because LG = IK$



$therefore frac{LG}{KC} = frac{IK}{KC} = tan(angle ACB)$



Similarly,



$frac{LH}{JC} = tan(angle ACB)$



$because angle LHO = angle LGO = angle OLC = angle OKC = 90^{circ}$



$therefore angle JCK = angle GLH = 360^{circ} - angle LHO - angle LGO - angle GOH = 360^{circ} - angle OJC - angle OKC - angle JOK $



$because frac{LH}{JC} = frac{LG}{KC} = tan(angle ACB), angle JCK = angle GLH$



$therefore Delta LHG simeq Delta CJK $ (SAS)



$therefore angle LGH = angle JKC$



$because angle LGO = angle OKC = 90^{circ}$



$therefore angle HGO = angle OKJ = angle LGO - angle LGH = angle OKC -angle JKC$



Similarly, $angle GHO = angle OJK$



$therefore Delta GOH simeq Delta KOJ$ (AA)



$therefore frac{GH}{JK} = frac{LG}{KC}$



$therefore frac{GO}{OK} = frac{LG}{KC} = frac{GH}{JK}$



$because angle LGO = angle OKC$



$therefore Delta LGO simeq Delta CKO$ (SAS)



$therefore angle GOL = angle KOC$



Similarly, $angle HOL = angle JOC$



$because angle IOJ = angle POK$



$therefore angle LOG + angle IOJ + angle JOC = angle COK + angle POK + angle HOL = frac{1}{2}(360^{circ}) = 180^{circ}$



$therefore$ Points C, O, L are collinear



$Q.E.D$










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  • 1




    $begingroup$
    Which angle is 90 degrees? (unclear in diagram)
    $endgroup$
    – coffeemath
    Aug 28 '18 at 1:25






  • 2




    $begingroup$
    I added a new picture. Is that what you want me to make clear about?
    $endgroup$
    – Larry
    Aug 28 '18 at 1:29






  • 3




    $begingroup$
    I've never heard of this before. It may be genuinely new.
    $endgroup$
    – mweiss
    Aug 28 '18 at 2:59






  • 2




    $begingroup$
    I don't know if the very-interesting theorem has a name ("Larry's Theorem"?), but I know it can be generalized a bit: If you broaden the definition of "inscribed square" to "a square with vertices on the extended sides of a triangle" (in particular, with two adjacent vertices on a particular side-line), then your result applies to obtuse triangles, too. Moreover, under this broader notion, there are two "inscribed" squares on each side-line; a "small" and a "large". Your result applies to the large squares, as well. (Mixing large and small doesn't seem obviously fruitful, but who knows?)
    $endgroup$
    – Blue
    Aug 28 '18 at 11:06






  • 3




    $begingroup$
    @EdPegg: Sadly, the three lines aren't concurrent.
    $endgroup$
    – Blue
    Aug 30 '18 at 4:19
















21












$begingroup$


enter image description hereenter image description here



For any acute and right triangle, two of the intersections(C and B) of the two inscribed squares and the vertex A of the triangle are collinear.



I have a proof for the theorem, but I have not found any specific name for it. I searched "inscribed squares in a triangle," and I still could not find the same shape. Can anyone provide a specific name for this, or a way that I can find the theorem online?



For the similar question concerning the cubes inscribed in a tetrahedron, please visit Extend squares inscribed in a triangle to the cubes inscribed in a tetrahedron
.



enter image description here



Proof:



$tan(angle ACB) = frac{IK}{KC}$



$because LG = IK$



$therefore frac{LG}{KC} = frac{IK}{KC} = tan(angle ACB)$



Similarly,



$frac{LH}{JC} = tan(angle ACB)$



$because angle LHO = angle LGO = angle OLC = angle OKC = 90^{circ}$



$therefore angle JCK = angle GLH = 360^{circ} - angle LHO - angle LGO - angle GOH = 360^{circ} - angle OJC - angle OKC - angle JOK $



$because frac{LH}{JC} = frac{LG}{KC} = tan(angle ACB), angle JCK = angle GLH$



$therefore Delta LHG simeq Delta CJK $ (SAS)



$therefore angle LGH = angle JKC$



$because angle LGO = angle OKC = 90^{circ}$



$therefore angle HGO = angle OKJ = angle LGO - angle LGH = angle OKC -angle JKC$



Similarly, $angle GHO = angle OJK$



$therefore Delta GOH simeq Delta KOJ$ (AA)



$therefore frac{GH}{JK} = frac{LG}{KC}$



$therefore frac{GO}{OK} = frac{LG}{KC} = frac{GH}{JK}$



$because angle LGO = angle OKC$



$therefore Delta LGO simeq Delta CKO$ (SAS)



$therefore angle GOL = angle KOC$



Similarly, $angle HOL = angle JOC$



$because angle IOJ = angle POK$



$therefore angle LOG + angle IOJ + angle JOC = angle COK + angle POK + angle HOL = frac{1}{2}(360^{circ}) = 180^{circ}$



$therefore$ Points C, O, L are collinear



$Q.E.D$










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  • 1




    $begingroup$
    Which angle is 90 degrees? (unclear in diagram)
    $endgroup$
    – coffeemath
    Aug 28 '18 at 1:25






  • 2




    $begingroup$
    I added a new picture. Is that what you want me to make clear about?
    $endgroup$
    – Larry
    Aug 28 '18 at 1:29






  • 3




    $begingroup$
    I've never heard of this before. It may be genuinely new.
    $endgroup$
    – mweiss
    Aug 28 '18 at 2:59






  • 2




    $begingroup$
    I don't know if the very-interesting theorem has a name ("Larry's Theorem"?), but I know it can be generalized a bit: If you broaden the definition of "inscribed square" to "a square with vertices on the extended sides of a triangle" (in particular, with two adjacent vertices on a particular side-line), then your result applies to obtuse triangles, too. Moreover, under this broader notion, there are two "inscribed" squares on each side-line; a "small" and a "large". Your result applies to the large squares, as well. (Mixing large and small doesn't seem obviously fruitful, but who knows?)
    $endgroup$
    – Blue
    Aug 28 '18 at 11:06






  • 3




    $begingroup$
    @EdPegg: Sadly, the three lines aren't concurrent.
    $endgroup$
    – Blue
    Aug 30 '18 at 4:19














21












21








21


4



$begingroup$


enter image description hereenter image description here



For any acute and right triangle, two of the intersections(C and B) of the two inscribed squares and the vertex A of the triangle are collinear.



I have a proof for the theorem, but I have not found any specific name for it. I searched "inscribed squares in a triangle," and I still could not find the same shape. Can anyone provide a specific name for this, or a way that I can find the theorem online?



For the similar question concerning the cubes inscribed in a tetrahedron, please visit Extend squares inscribed in a triangle to the cubes inscribed in a tetrahedron
.



enter image description here



Proof:



$tan(angle ACB) = frac{IK}{KC}$



$because LG = IK$



$therefore frac{LG}{KC} = frac{IK}{KC} = tan(angle ACB)$



Similarly,



$frac{LH}{JC} = tan(angle ACB)$



$because angle LHO = angle LGO = angle OLC = angle OKC = 90^{circ}$



$therefore angle JCK = angle GLH = 360^{circ} - angle LHO - angle LGO - angle GOH = 360^{circ} - angle OJC - angle OKC - angle JOK $



$because frac{LH}{JC} = frac{LG}{KC} = tan(angle ACB), angle JCK = angle GLH$



$therefore Delta LHG simeq Delta CJK $ (SAS)



$therefore angle LGH = angle JKC$



$because angle LGO = angle OKC = 90^{circ}$



$therefore angle HGO = angle OKJ = angle LGO - angle LGH = angle OKC -angle JKC$



Similarly, $angle GHO = angle OJK$



$therefore Delta GOH simeq Delta KOJ$ (AA)



$therefore frac{GH}{JK} = frac{LG}{KC}$



$therefore frac{GO}{OK} = frac{LG}{KC} = frac{GH}{JK}$



$because angle LGO = angle OKC$



$therefore Delta LGO simeq Delta CKO$ (SAS)



$therefore angle GOL = angle KOC$



Similarly, $angle HOL = angle JOC$



$because angle IOJ = angle POK$



$therefore angle LOG + angle IOJ + angle JOC = angle COK + angle POK + angle HOL = frac{1}{2}(360^{circ}) = 180^{circ}$



$therefore$ Points C, O, L are collinear



$Q.E.D$










share|cite|improve this question











$endgroup$




enter image description hereenter image description here



For any acute and right triangle, two of the intersections(C and B) of the two inscribed squares and the vertex A of the triangle are collinear.



I have a proof for the theorem, but I have not found any specific name for it. I searched "inscribed squares in a triangle," and I still could not find the same shape. Can anyone provide a specific name for this, or a way that I can find the theorem online?



For the similar question concerning the cubes inscribed in a tetrahedron, please visit Extend squares inscribed in a triangle to the cubes inscribed in a tetrahedron
.



enter image description here



Proof:



$tan(angle ACB) = frac{IK}{KC}$



$because LG = IK$



$therefore frac{LG}{KC} = frac{IK}{KC} = tan(angle ACB)$



Similarly,



$frac{LH}{JC} = tan(angle ACB)$



$because angle LHO = angle LGO = angle OLC = angle OKC = 90^{circ}$



$therefore angle JCK = angle GLH = 360^{circ} - angle LHO - angle LGO - angle GOH = 360^{circ} - angle OJC - angle OKC - angle JOK $



$because frac{LH}{JC} = frac{LG}{KC} = tan(angle ACB), angle JCK = angle GLH$



$therefore Delta LHG simeq Delta CJK $ (SAS)



$therefore angle LGH = angle JKC$



$because angle LGO = angle OKC = 90^{circ}$



$therefore angle HGO = angle OKJ = angle LGO - angle LGH = angle OKC -angle JKC$



Similarly, $angle GHO = angle OJK$



$therefore Delta GOH simeq Delta KOJ$ (AA)



$therefore frac{GH}{JK} = frac{LG}{KC}$



$therefore frac{GO}{OK} = frac{LG}{KC} = frac{GH}{JK}$



$because angle LGO = angle OKC$



$therefore Delta LGO simeq Delta CKO$ (SAS)



$therefore angle GOL = angle KOC$



Similarly, $angle HOL = angle JOC$



$because angle IOJ = angle POK$



$therefore angle LOG + angle IOJ + angle JOC = angle COK + angle POK + angle HOL = frac{1}{2}(360^{circ}) = 180^{circ}$



$therefore$ Points C, O, L are collinear



$Q.E.D$







geometry






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edited Dec 20 '18 at 21:19







Larry

















asked Aug 28 '18 at 1:22









LarryLarry

2,46831130




2,46831130








  • 1




    $begingroup$
    Which angle is 90 degrees? (unclear in diagram)
    $endgroup$
    – coffeemath
    Aug 28 '18 at 1:25






  • 2




    $begingroup$
    I added a new picture. Is that what you want me to make clear about?
    $endgroup$
    – Larry
    Aug 28 '18 at 1:29






  • 3




    $begingroup$
    I've never heard of this before. It may be genuinely new.
    $endgroup$
    – mweiss
    Aug 28 '18 at 2:59






  • 2




    $begingroup$
    I don't know if the very-interesting theorem has a name ("Larry's Theorem"?), but I know it can be generalized a bit: If you broaden the definition of "inscribed square" to "a square with vertices on the extended sides of a triangle" (in particular, with two adjacent vertices on a particular side-line), then your result applies to obtuse triangles, too. Moreover, under this broader notion, there are two "inscribed" squares on each side-line; a "small" and a "large". Your result applies to the large squares, as well. (Mixing large and small doesn't seem obviously fruitful, but who knows?)
    $endgroup$
    – Blue
    Aug 28 '18 at 11:06






  • 3




    $begingroup$
    @EdPegg: Sadly, the three lines aren't concurrent.
    $endgroup$
    – Blue
    Aug 30 '18 at 4:19














  • 1




    $begingroup$
    Which angle is 90 degrees? (unclear in diagram)
    $endgroup$
    – coffeemath
    Aug 28 '18 at 1:25






  • 2




    $begingroup$
    I added a new picture. Is that what you want me to make clear about?
    $endgroup$
    – Larry
    Aug 28 '18 at 1:29






  • 3




    $begingroup$
    I've never heard of this before. It may be genuinely new.
    $endgroup$
    – mweiss
    Aug 28 '18 at 2:59






  • 2




    $begingroup$
    I don't know if the very-interesting theorem has a name ("Larry's Theorem"?), but I know it can be generalized a bit: If you broaden the definition of "inscribed square" to "a square with vertices on the extended sides of a triangle" (in particular, with two adjacent vertices on a particular side-line), then your result applies to obtuse triangles, too. Moreover, under this broader notion, there are two "inscribed" squares on each side-line; a "small" and a "large". Your result applies to the large squares, as well. (Mixing large and small doesn't seem obviously fruitful, but who knows?)
    $endgroup$
    – Blue
    Aug 28 '18 at 11:06






  • 3




    $begingroup$
    @EdPegg: Sadly, the three lines aren't concurrent.
    $endgroup$
    – Blue
    Aug 30 '18 at 4:19








1




1




$begingroup$
Which angle is 90 degrees? (unclear in diagram)
$endgroup$
– coffeemath
Aug 28 '18 at 1:25




$begingroup$
Which angle is 90 degrees? (unclear in diagram)
$endgroup$
– coffeemath
Aug 28 '18 at 1:25




2




2




$begingroup$
I added a new picture. Is that what you want me to make clear about?
$endgroup$
– Larry
Aug 28 '18 at 1:29




$begingroup$
I added a new picture. Is that what you want me to make clear about?
$endgroup$
– Larry
Aug 28 '18 at 1:29




3




3




$begingroup$
I've never heard of this before. It may be genuinely new.
$endgroup$
– mweiss
Aug 28 '18 at 2:59




$begingroup$
I've never heard of this before. It may be genuinely new.
$endgroup$
– mweiss
Aug 28 '18 at 2:59




2




2




$begingroup$
I don't know if the very-interesting theorem has a name ("Larry's Theorem"?), but I know it can be generalized a bit: If you broaden the definition of "inscribed square" to "a square with vertices on the extended sides of a triangle" (in particular, with two adjacent vertices on a particular side-line), then your result applies to obtuse triangles, too. Moreover, under this broader notion, there are two "inscribed" squares on each side-line; a "small" and a "large". Your result applies to the large squares, as well. (Mixing large and small doesn't seem obviously fruitful, but who knows?)
$endgroup$
– Blue
Aug 28 '18 at 11:06




$begingroup$
I don't know if the very-interesting theorem has a name ("Larry's Theorem"?), but I know it can be generalized a bit: If you broaden the definition of "inscribed square" to "a square with vertices on the extended sides of a triangle" (in particular, with two adjacent vertices on a particular side-line), then your result applies to obtuse triangles, too. Moreover, under this broader notion, there are two "inscribed" squares on each side-line; a "small" and a "large". Your result applies to the large squares, as well. (Mixing large and small doesn't seem obviously fruitful, but who knows?)
$endgroup$
– Blue
Aug 28 '18 at 11:06




3




3




$begingroup$
@EdPegg: Sadly, the three lines aren't concurrent.
$endgroup$
– Blue
Aug 30 '18 at 4:19




$begingroup$
@EdPegg: Sadly, the three lines aren't concurrent.
$endgroup$
– Blue
Aug 30 '18 at 4:19










3 Answers
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In light of the Bailey and Detemple (B&T) article mentioned in @Larry's answer, I'm updating my response. See the Edit History for a previous version of this answer.





As B&T describe, the key notion here is a general property of squares inscribed in angles; having a square's vertex lie on the third side of a triangle is an unnecessary restriction. (That restriction would have some relevance if the three resulting lines of intersection for a triangle were concurrent; alas, they are not.) But in fact, the property generalizes beyond squares to similar parallelograms.



Visually, the result is clear:



enter image description here



Articulating it in words is a little tricky. Bear with me.




Definitions. Given an angle with side-lines $a$ and $b$, we call a parallelogram "inscribed with base-line $a$" if two adjacent vertices (the "base vertices") lie on $a$ and a third vertex lies on $b$. We'll call the side-lines not parallel to $a$ the "lateral lines" of the parallelogram.




Let an angle have vertex $O$ and side-lines $a$ and $b$. Inscribe a parallelogram with base vertices $A$ and $A^prime$ on $a$, and with $A^{primeprime}$ (adjacent to $A$) on $b$; let $A^star$ be the fourth vertex. Likewise, inscribe parallelogram $square B^prime B B^{primeprime} B^star$ with base-line $b$.




Theorem. If the parallelograms are similar and "compatibly slanted" (that is, if $square A^prime A A^{primeprime}A^star sim square B^prime B B^{primeprime} B^{star}$, and $angle OAA^{primeprime}congangle OBB^{primeprime}$), then $O$ is collinear with the points where the lateral lines through $A$ and $A^prime$ meet the respective lateral lines through $B$ and $B^prime$.




For proof, let the lateral lines through $A$ and $B$ meet at $P$. Let $overleftrightarrow{OP}$ meet the lateral line through $A^prime$ at $P^prime$. Via the similarity $triangle OAP sim OA^prime P^prime$, we deduce
$$frac{|PP^prime|}{|OP|} = frac{|AA^prime|}{|OA|} = frac{|AA^prime|}{|AA^{primeprime}|}frac{|AA^{primeprime}|}{|OA|} tag{1}$$
The parallelogram similarity allows us to re-write the first factor in $(1)$, while $triangle OAA^{primeprime}simtriangle OBB^{primeprime}$ allows us to re-write the second:
$$frac{|PP^prime|}{|OP|} = frac{|BB^prime|}{|BB^{primeprime}|}frac{|BB^{primeprime}|}{|OB|} tag{2}$$
We may conclude, then, that $P^prime$ lies on the lateral line through $B^prime$, giving the Theorem. $square$





Note that the Theorem holds for arbitrary pairs of parallelograms inscribed in an angle; moreover, it holds whether $overrightarrow{PP^prime}$ points in the same or opposite direction as $overrightarrow{OP}$. @Larry and B&T restrict the parallelograms to squares, and @Larry constrains the "fourth vertices" $A^star$ and $B^star$ to a third line (B&T consider that as a special case). They all also implicitly assume the "same direction" configuration, but here we see that the "opposite direction" case is equally valid:



enter image description here





Here's a related (new?) result, for which I currently have only an ugly coordinate proof.




A triangle admits three more "inscribed" squares; these have opposite vertices on a particular side-line, and one vertex on each other side-line. Given $triangle ABC$, draw the new squares associated with sides $overline{AB}$ and $overline{CA}$. The side-lines of the squares meet in four point-pairs, each of which are collinear with vertex $A$. (The lines also meet in another eight points that seem to have no remarkable collinearity property.)




In the figure, the four green lines show collinearity of top triangle vertex with four pairs of points-of-intersection of two squares' side-lines. When two lines meet at one of these points, the lines parallel to those lines meet at the other point of the pair.



enter image description here



This result, too, generalizes to parallelograms that need not have "fourth vertices" constrained to a third line. Here's an image:



enter image description here






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  • 1




    $begingroup$
    Now I understand what you meant by the "large" square. That is amazing!
    $endgroup$
    – Larry
    Sep 2 '18 at 13:34










  • $begingroup$
    Nice proof, and I think it'd be even easier to understand by removing the length labels from the first diagram.
    $endgroup$
    – Matt F.
    Sep 9 '18 at 17:54










  • $begingroup$
    @Larry: Thanks for the bounty! :) BTW: You should post the tetrahedron variant as a separate question.
    $endgroup$
    – Blue
    Sep 12 '18 at 11:14










  • $begingroup$
    :) You’re welcome. I was thinking about posting it as a separate question, but I thought the question would be easier to understand with the context. Maybe I should post another question.
    $endgroup$
    – Larry
    Sep 12 '18 at 11:57






  • 1




    $begingroup$
    @Larry: Post another question, but add a link back to this one for context. (And add a link from this one to the other for people who might be interested.) Simple! :)
    $endgroup$
    – Blue
    Sep 12 '18 at 12:42



















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Here is an alternative proof, using the diagram below.



As an aside, note that proving $Q'Q parallel R'P$ would allow us to conclude the rest from a special case of Desargues's theorem. We would have:




  • $Q'S parallel R'O$, $ SQ parallel OP$, $ QQ' parallel PR'$, so

  • $Q'S cap R'O$ and $SQ cap OP$ and $QQ' cap PR'$ are all collinear on the line at infinity, so

  • the triangles $Q'SQ$ and $R'OP$ are in perspective axially, so

  • the triangles are in perspective centrally, and

  • $Q'R'$, $SO$, $QP$ are all coincident.


Perhaps someone else will see how to prove that first parallelism geometrically.



diagram



Meanwhile, here's a vectorial proof of that parallelism and of the theorem as a whole, for anyone who wants an algebraic approach.



Define $b,c,d,e,f$ as vectors, and $t,u$ as scalars such that




  • $b$ goes from $O$ to $S$

  • $c$ goes from $P'$ to $Q'$, $d$ goes from $R'$ to $Q'$, $tc$ goes from $O$ to $R'$.

  • $e$ goes from $R$ to $Q$, $ f$ goes from $P$ to $Q$, $ ue$ goes from $O$ to $P$.


Now we record several scalar facts:
begin{align}
PQR text{ are vertices of a square}&: e.f = 0, e.e = f.f & (1)\
P'Q'R' text{ are vertices of a square}&: c.d = 0, c.c = d.d & (2)\
c+f text{ rotated }90^o text{ gives }d+e&: |c+f|=|d+e| &(3)\
P' text{ lies on }PQ&: (tc+d-c).e=ue.e &(4)\
R text{ lies on }Q'R'&: (ue+f-e).c=tc.c &(5)\
S text{ lies on }P'Q'&: b.d=d.d &(6)\
S text{ lies on }QR&: b.f=f.f &(7)\
1+2+3 &: c.f=d.e & (8)\
4+5+8 &: dfrac{t}{c.e+e.e}=dfrac{u}{c.c+c.e} & (9)
end{align}
This is enough to prove that $R'P=ue-tc$ and $Q'Q=ue+f-tc-d$ are both perpendicular to $c+e$, and therefore parallel as for the above use of Desargues's theorem. Meanwhile, proceeding vectorically,
we check that $$(d.e)b=(e.e)c + (c.c)e$$ because both sides have the same dot product with $d$ (by 2,6), and both sides have the same dot product with $f$ (by 1,7,8), and we assume that $d$ and $f$ span the plane. We conclude that
$$frac{c.c}{b.c}(c.e+e.e)=d.e=frac{e.e}{b.e}(c.c+c.e) (10)$$
So multiplying (9) and (10) gives
$$frac{c.c}{b.c} t=frac{e.e}{b.e} u =: x$$
Finally, let the point $X$ be $xb$ away from $O$. Then:




  • $xb.c=tc.c$, so $X$ is on $Q'R'$;

  • $xb.e=ue.e$, so $X$ is on $PQ$;

  • by construction $X$ is on $OS$;

  • so $Q'R'$, $PQ$, and $OS$ are coincident, as desired.






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    2












    $begingroup$

    Although interesting, this theorem is not new. I recently found the article "Squares Inscribed in Angles and Triangles" on JSTOR.



    enter image description here



    Bailey, Herbert, and Duane Detemple. “Squares Inscribed in Angles and Triangles.” Mathematics Magazine, vol. 71, no. 4, 1998, pp. 278–284. JSTOR, JSTOR, www.jstor.org/stable/2690699.



    It would be nice if I am the person who discovered the fact, but since the theorem was published in 1998, the credit does not belong to me.






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    $endgroup$













    • $begingroup$
      That someone discovered the fact before you does not diminish your accomplishment. The late, great Steve Fisk called this kind of thing a mere "accident of time". :)
      $endgroup$
      – Blue
      Oct 19 '18 at 0:01






    • 1




      $begingroup$
      Thank you for your encouragement.
      $endgroup$
      – Larry
      Oct 19 '18 at 0:06






    • 1




      $begingroup$
      The reference you found shows that the collinearity property holds for arbitrary squares inscribed in the two sides of an angle; the third side of the triangle is a distraction (especially since there's no concurrence). I'm a little surprised I didn't see this earlier: My proof never used the third components in my ratios $cot C:pm 1:cot A$ and $cot C:pm 1:cot B$. I'll refine that proof. In fact, I can generalize so that the result applies not merely to inscribed squares, but inscribed similar parallelograms. (My ostensibly-"new" result may generalize in the same way.)
      $endgroup$
      – Blue
      Oct 19 '18 at 2:15













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    3 Answers
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    3 Answers
    3






    active

    oldest

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    active

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    active

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    6





    +50







    $begingroup$

    In light of the Bailey and Detemple (B&T) article mentioned in @Larry's answer, I'm updating my response. See the Edit History for a previous version of this answer.





    As B&T describe, the key notion here is a general property of squares inscribed in angles; having a square's vertex lie on the third side of a triangle is an unnecessary restriction. (That restriction would have some relevance if the three resulting lines of intersection for a triangle were concurrent; alas, they are not.) But in fact, the property generalizes beyond squares to similar parallelograms.



    Visually, the result is clear:



    enter image description here



    Articulating it in words is a little tricky. Bear with me.




    Definitions. Given an angle with side-lines $a$ and $b$, we call a parallelogram "inscribed with base-line $a$" if two adjacent vertices (the "base vertices") lie on $a$ and a third vertex lies on $b$. We'll call the side-lines not parallel to $a$ the "lateral lines" of the parallelogram.




    Let an angle have vertex $O$ and side-lines $a$ and $b$. Inscribe a parallelogram with base vertices $A$ and $A^prime$ on $a$, and with $A^{primeprime}$ (adjacent to $A$) on $b$; let $A^star$ be the fourth vertex. Likewise, inscribe parallelogram $square B^prime B B^{primeprime} B^star$ with base-line $b$.




    Theorem. If the parallelograms are similar and "compatibly slanted" (that is, if $square A^prime A A^{primeprime}A^star sim square B^prime B B^{primeprime} B^{star}$, and $angle OAA^{primeprime}congangle OBB^{primeprime}$), then $O$ is collinear with the points where the lateral lines through $A$ and $A^prime$ meet the respective lateral lines through $B$ and $B^prime$.




    For proof, let the lateral lines through $A$ and $B$ meet at $P$. Let $overleftrightarrow{OP}$ meet the lateral line through $A^prime$ at $P^prime$. Via the similarity $triangle OAP sim OA^prime P^prime$, we deduce
    $$frac{|PP^prime|}{|OP|} = frac{|AA^prime|}{|OA|} = frac{|AA^prime|}{|AA^{primeprime}|}frac{|AA^{primeprime}|}{|OA|} tag{1}$$
    The parallelogram similarity allows us to re-write the first factor in $(1)$, while $triangle OAA^{primeprime}simtriangle OBB^{primeprime}$ allows us to re-write the second:
    $$frac{|PP^prime|}{|OP|} = frac{|BB^prime|}{|BB^{primeprime}|}frac{|BB^{primeprime}|}{|OB|} tag{2}$$
    We may conclude, then, that $P^prime$ lies on the lateral line through $B^prime$, giving the Theorem. $square$





    Note that the Theorem holds for arbitrary pairs of parallelograms inscribed in an angle; moreover, it holds whether $overrightarrow{PP^prime}$ points in the same or opposite direction as $overrightarrow{OP}$. @Larry and B&T restrict the parallelograms to squares, and @Larry constrains the "fourth vertices" $A^star$ and $B^star$ to a third line (B&T consider that as a special case). They all also implicitly assume the "same direction" configuration, but here we see that the "opposite direction" case is equally valid:



    enter image description here





    Here's a related (new?) result, for which I currently have only an ugly coordinate proof.




    A triangle admits three more "inscribed" squares; these have opposite vertices on a particular side-line, and one vertex on each other side-line. Given $triangle ABC$, draw the new squares associated with sides $overline{AB}$ and $overline{CA}$. The side-lines of the squares meet in four point-pairs, each of which are collinear with vertex $A$. (The lines also meet in another eight points that seem to have no remarkable collinearity property.)




    In the figure, the four green lines show collinearity of top triangle vertex with four pairs of points-of-intersection of two squares' side-lines. When two lines meet at one of these points, the lines parallel to those lines meet at the other point of the pair.



    enter image description here



    This result, too, generalizes to parallelograms that need not have "fourth vertices" constrained to a third line. Here's an image:



    enter image description here






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Now I understand what you meant by the "large" square. That is amazing!
      $endgroup$
      – Larry
      Sep 2 '18 at 13:34










    • $begingroup$
      Nice proof, and I think it'd be even easier to understand by removing the length labels from the first diagram.
      $endgroup$
      – Matt F.
      Sep 9 '18 at 17:54










    • $begingroup$
      @Larry: Thanks for the bounty! :) BTW: You should post the tetrahedron variant as a separate question.
      $endgroup$
      – Blue
      Sep 12 '18 at 11:14










    • $begingroup$
      :) You’re welcome. I was thinking about posting it as a separate question, but I thought the question would be easier to understand with the context. Maybe I should post another question.
      $endgroup$
      – Larry
      Sep 12 '18 at 11:57






    • 1




      $begingroup$
      @Larry: Post another question, but add a link back to this one for context. (And add a link from this one to the other for people who might be interested.) Simple! :)
      $endgroup$
      – Blue
      Sep 12 '18 at 12:42
















    6





    +50







    $begingroup$

    In light of the Bailey and Detemple (B&T) article mentioned in @Larry's answer, I'm updating my response. See the Edit History for a previous version of this answer.





    As B&T describe, the key notion here is a general property of squares inscribed in angles; having a square's vertex lie on the third side of a triangle is an unnecessary restriction. (That restriction would have some relevance if the three resulting lines of intersection for a triangle were concurrent; alas, they are not.) But in fact, the property generalizes beyond squares to similar parallelograms.



    Visually, the result is clear:



    enter image description here



    Articulating it in words is a little tricky. Bear with me.




    Definitions. Given an angle with side-lines $a$ and $b$, we call a parallelogram "inscribed with base-line $a$" if two adjacent vertices (the "base vertices") lie on $a$ and a third vertex lies on $b$. We'll call the side-lines not parallel to $a$ the "lateral lines" of the parallelogram.




    Let an angle have vertex $O$ and side-lines $a$ and $b$. Inscribe a parallelogram with base vertices $A$ and $A^prime$ on $a$, and with $A^{primeprime}$ (adjacent to $A$) on $b$; let $A^star$ be the fourth vertex. Likewise, inscribe parallelogram $square B^prime B B^{primeprime} B^star$ with base-line $b$.




    Theorem. If the parallelograms are similar and "compatibly slanted" (that is, if $square A^prime A A^{primeprime}A^star sim square B^prime B B^{primeprime} B^{star}$, and $angle OAA^{primeprime}congangle OBB^{primeprime}$), then $O$ is collinear with the points where the lateral lines through $A$ and $A^prime$ meet the respective lateral lines through $B$ and $B^prime$.




    For proof, let the lateral lines through $A$ and $B$ meet at $P$. Let $overleftrightarrow{OP}$ meet the lateral line through $A^prime$ at $P^prime$. Via the similarity $triangle OAP sim OA^prime P^prime$, we deduce
    $$frac{|PP^prime|}{|OP|} = frac{|AA^prime|}{|OA|} = frac{|AA^prime|}{|AA^{primeprime}|}frac{|AA^{primeprime}|}{|OA|} tag{1}$$
    The parallelogram similarity allows us to re-write the first factor in $(1)$, while $triangle OAA^{primeprime}simtriangle OBB^{primeprime}$ allows us to re-write the second:
    $$frac{|PP^prime|}{|OP|} = frac{|BB^prime|}{|BB^{primeprime}|}frac{|BB^{primeprime}|}{|OB|} tag{2}$$
    We may conclude, then, that $P^prime$ lies on the lateral line through $B^prime$, giving the Theorem. $square$





    Note that the Theorem holds for arbitrary pairs of parallelograms inscribed in an angle; moreover, it holds whether $overrightarrow{PP^prime}$ points in the same or opposite direction as $overrightarrow{OP}$. @Larry and B&T restrict the parallelograms to squares, and @Larry constrains the "fourth vertices" $A^star$ and $B^star$ to a third line (B&T consider that as a special case). They all also implicitly assume the "same direction" configuration, but here we see that the "opposite direction" case is equally valid:



    enter image description here





    Here's a related (new?) result, for which I currently have only an ugly coordinate proof.




    A triangle admits three more "inscribed" squares; these have opposite vertices on a particular side-line, and one vertex on each other side-line. Given $triangle ABC$, draw the new squares associated with sides $overline{AB}$ and $overline{CA}$. The side-lines of the squares meet in four point-pairs, each of which are collinear with vertex $A$. (The lines also meet in another eight points that seem to have no remarkable collinearity property.)




    In the figure, the four green lines show collinearity of top triangle vertex with four pairs of points-of-intersection of two squares' side-lines. When two lines meet at one of these points, the lines parallel to those lines meet at the other point of the pair.



    enter image description here



    This result, too, generalizes to parallelograms that need not have "fourth vertices" constrained to a third line. Here's an image:



    enter image description here






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Now I understand what you meant by the "large" square. That is amazing!
      $endgroup$
      – Larry
      Sep 2 '18 at 13:34










    • $begingroup$
      Nice proof, and I think it'd be even easier to understand by removing the length labels from the first diagram.
      $endgroup$
      – Matt F.
      Sep 9 '18 at 17:54










    • $begingroup$
      @Larry: Thanks for the bounty! :) BTW: You should post the tetrahedron variant as a separate question.
      $endgroup$
      – Blue
      Sep 12 '18 at 11:14










    • $begingroup$
      :) You’re welcome. I was thinking about posting it as a separate question, but I thought the question would be easier to understand with the context. Maybe I should post another question.
      $endgroup$
      – Larry
      Sep 12 '18 at 11:57






    • 1




      $begingroup$
      @Larry: Post another question, but add a link back to this one for context. (And add a link from this one to the other for people who might be interested.) Simple! :)
      $endgroup$
      – Blue
      Sep 12 '18 at 12:42














    6





    +50







    6





    +50



    6




    +50



    $begingroup$

    In light of the Bailey and Detemple (B&T) article mentioned in @Larry's answer, I'm updating my response. See the Edit History for a previous version of this answer.





    As B&T describe, the key notion here is a general property of squares inscribed in angles; having a square's vertex lie on the third side of a triangle is an unnecessary restriction. (That restriction would have some relevance if the three resulting lines of intersection for a triangle were concurrent; alas, they are not.) But in fact, the property generalizes beyond squares to similar parallelograms.



    Visually, the result is clear:



    enter image description here



    Articulating it in words is a little tricky. Bear with me.




    Definitions. Given an angle with side-lines $a$ and $b$, we call a parallelogram "inscribed with base-line $a$" if two adjacent vertices (the "base vertices") lie on $a$ and a third vertex lies on $b$. We'll call the side-lines not parallel to $a$ the "lateral lines" of the parallelogram.




    Let an angle have vertex $O$ and side-lines $a$ and $b$. Inscribe a parallelogram with base vertices $A$ and $A^prime$ on $a$, and with $A^{primeprime}$ (adjacent to $A$) on $b$; let $A^star$ be the fourth vertex. Likewise, inscribe parallelogram $square B^prime B B^{primeprime} B^star$ with base-line $b$.




    Theorem. If the parallelograms are similar and "compatibly slanted" (that is, if $square A^prime A A^{primeprime}A^star sim square B^prime B B^{primeprime} B^{star}$, and $angle OAA^{primeprime}congangle OBB^{primeprime}$), then $O$ is collinear with the points where the lateral lines through $A$ and $A^prime$ meet the respective lateral lines through $B$ and $B^prime$.




    For proof, let the lateral lines through $A$ and $B$ meet at $P$. Let $overleftrightarrow{OP}$ meet the lateral line through $A^prime$ at $P^prime$. Via the similarity $triangle OAP sim OA^prime P^prime$, we deduce
    $$frac{|PP^prime|}{|OP|} = frac{|AA^prime|}{|OA|} = frac{|AA^prime|}{|AA^{primeprime}|}frac{|AA^{primeprime}|}{|OA|} tag{1}$$
    The parallelogram similarity allows us to re-write the first factor in $(1)$, while $triangle OAA^{primeprime}simtriangle OBB^{primeprime}$ allows us to re-write the second:
    $$frac{|PP^prime|}{|OP|} = frac{|BB^prime|}{|BB^{primeprime}|}frac{|BB^{primeprime}|}{|OB|} tag{2}$$
    We may conclude, then, that $P^prime$ lies on the lateral line through $B^prime$, giving the Theorem. $square$





    Note that the Theorem holds for arbitrary pairs of parallelograms inscribed in an angle; moreover, it holds whether $overrightarrow{PP^prime}$ points in the same or opposite direction as $overrightarrow{OP}$. @Larry and B&T restrict the parallelograms to squares, and @Larry constrains the "fourth vertices" $A^star$ and $B^star$ to a third line (B&T consider that as a special case). They all also implicitly assume the "same direction" configuration, but here we see that the "opposite direction" case is equally valid:



    enter image description here





    Here's a related (new?) result, for which I currently have only an ugly coordinate proof.




    A triangle admits three more "inscribed" squares; these have opposite vertices on a particular side-line, and one vertex on each other side-line. Given $triangle ABC$, draw the new squares associated with sides $overline{AB}$ and $overline{CA}$. The side-lines of the squares meet in four point-pairs, each of which are collinear with vertex $A$. (The lines also meet in another eight points that seem to have no remarkable collinearity property.)




    In the figure, the four green lines show collinearity of top triangle vertex with four pairs of points-of-intersection of two squares' side-lines. When two lines meet at one of these points, the lines parallel to those lines meet at the other point of the pair.



    enter image description here



    This result, too, generalizes to parallelograms that need not have "fourth vertices" constrained to a third line. Here's an image:



    enter image description here






    share|cite|improve this answer











    $endgroup$



    In light of the Bailey and Detemple (B&T) article mentioned in @Larry's answer, I'm updating my response. See the Edit History for a previous version of this answer.





    As B&T describe, the key notion here is a general property of squares inscribed in angles; having a square's vertex lie on the third side of a triangle is an unnecessary restriction. (That restriction would have some relevance if the three resulting lines of intersection for a triangle were concurrent; alas, they are not.) But in fact, the property generalizes beyond squares to similar parallelograms.



    Visually, the result is clear:



    enter image description here



    Articulating it in words is a little tricky. Bear with me.




    Definitions. Given an angle with side-lines $a$ and $b$, we call a parallelogram "inscribed with base-line $a$" if two adjacent vertices (the "base vertices") lie on $a$ and a third vertex lies on $b$. We'll call the side-lines not parallel to $a$ the "lateral lines" of the parallelogram.




    Let an angle have vertex $O$ and side-lines $a$ and $b$. Inscribe a parallelogram with base vertices $A$ and $A^prime$ on $a$, and with $A^{primeprime}$ (adjacent to $A$) on $b$; let $A^star$ be the fourth vertex. Likewise, inscribe parallelogram $square B^prime B B^{primeprime} B^star$ with base-line $b$.




    Theorem. If the parallelograms are similar and "compatibly slanted" (that is, if $square A^prime A A^{primeprime}A^star sim square B^prime B B^{primeprime} B^{star}$, and $angle OAA^{primeprime}congangle OBB^{primeprime}$), then $O$ is collinear with the points where the lateral lines through $A$ and $A^prime$ meet the respective lateral lines through $B$ and $B^prime$.




    For proof, let the lateral lines through $A$ and $B$ meet at $P$. Let $overleftrightarrow{OP}$ meet the lateral line through $A^prime$ at $P^prime$. Via the similarity $triangle OAP sim OA^prime P^prime$, we deduce
    $$frac{|PP^prime|}{|OP|} = frac{|AA^prime|}{|OA|} = frac{|AA^prime|}{|AA^{primeprime}|}frac{|AA^{primeprime}|}{|OA|} tag{1}$$
    The parallelogram similarity allows us to re-write the first factor in $(1)$, while $triangle OAA^{primeprime}simtriangle OBB^{primeprime}$ allows us to re-write the second:
    $$frac{|PP^prime|}{|OP|} = frac{|BB^prime|}{|BB^{primeprime}|}frac{|BB^{primeprime}|}{|OB|} tag{2}$$
    We may conclude, then, that $P^prime$ lies on the lateral line through $B^prime$, giving the Theorem. $square$





    Note that the Theorem holds for arbitrary pairs of parallelograms inscribed in an angle; moreover, it holds whether $overrightarrow{PP^prime}$ points in the same or opposite direction as $overrightarrow{OP}$. @Larry and B&T restrict the parallelograms to squares, and @Larry constrains the "fourth vertices" $A^star$ and $B^star$ to a third line (B&T consider that as a special case). They all also implicitly assume the "same direction" configuration, but here we see that the "opposite direction" case is equally valid:



    enter image description here





    Here's a related (new?) result, for which I currently have only an ugly coordinate proof.




    A triangle admits three more "inscribed" squares; these have opposite vertices on a particular side-line, and one vertex on each other side-line. Given $triangle ABC$, draw the new squares associated with sides $overline{AB}$ and $overline{CA}$. The side-lines of the squares meet in four point-pairs, each of which are collinear with vertex $A$. (The lines also meet in another eight points that seem to have no remarkable collinearity property.)




    In the figure, the four green lines show collinearity of top triangle vertex with four pairs of points-of-intersection of two squares' side-lines. When two lines meet at one of these points, the lines parallel to those lines meet at the other point of the pair.



    enter image description here



    This result, too, generalizes to parallelograms that need not have "fourth vertices" constrained to a third line. Here's an image:



    enter image description here







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Oct 19 '18 at 14:31

























    answered Aug 30 '18 at 13:10









    BlueBlue

    49k870156




    49k870156








    • 1




      $begingroup$
      Now I understand what you meant by the "large" square. That is amazing!
      $endgroup$
      – Larry
      Sep 2 '18 at 13:34










    • $begingroup$
      Nice proof, and I think it'd be even easier to understand by removing the length labels from the first diagram.
      $endgroup$
      – Matt F.
      Sep 9 '18 at 17:54










    • $begingroup$
      @Larry: Thanks for the bounty! :) BTW: You should post the tetrahedron variant as a separate question.
      $endgroup$
      – Blue
      Sep 12 '18 at 11:14










    • $begingroup$
      :) You’re welcome. I was thinking about posting it as a separate question, but I thought the question would be easier to understand with the context. Maybe I should post another question.
      $endgroup$
      – Larry
      Sep 12 '18 at 11:57






    • 1




      $begingroup$
      @Larry: Post another question, but add a link back to this one for context. (And add a link from this one to the other for people who might be interested.) Simple! :)
      $endgroup$
      – Blue
      Sep 12 '18 at 12:42














    • 1




      $begingroup$
      Now I understand what you meant by the "large" square. That is amazing!
      $endgroup$
      – Larry
      Sep 2 '18 at 13:34










    • $begingroup$
      Nice proof, and I think it'd be even easier to understand by removing the length labels from the first diagram.
      $endgroup$
      – Matt F.
      Sep 9 '18 at 17:54










    • $begingroup$
      @Larry: Thanks for the bounty! :) BTW: You should post the tetrahedron variant as a separate question.
      $endgroup$
      – Blue
      Sep 12 '18 at 11:14










    • $begingroup$
      :) You’re welcome. I was thinking about posting it as a separate question, but I thought the question would be easier to understand with the context. Maybe I should post another question.
      $endgroup$
      – Larry
      Sep 12 '18 at 11:57






    • 1




      $begingroup$
      @Larry: Post another question, but add a link back to this one for context. (And add a link from this one to the other for people who might be interested.) Simple! :)
      $endgroup$
      – Blue
      Sep 12 '18 at 12:42








    1




    1




    $begingroup$
    Now I understand what you meant by the "large" square. That is amazing!
    $endgroup$
    – Larry
    Sep 2 '18 at 13:34




    $begingroup$
    Now I understand what you meant by the "large" square. That is amazing!
    $endgroup$
    – Larry
    Sep 2 '18 at 13:34












    $begingroup$
    Nice proof, and I think it'd be even easier to understand by removing the length labels from the first diagram.
    $endgroup$
    – Matt F.
    Sep 9 '18 at 17:54




    $begingroup$
    Nice proof, and I think it'd be even easier to understand by removing the length labels from the first diagram.
    $endgroup$
    – Matt F.
    Sep 9 '18 at 17:54












    $begingroup$
    @Larry: Thanks for the bounty! :) BTW: You should post the tetrahedron variant as a separate question.
    $endgroup$
    – Blue
    Sep 12 '18 at 11:14




    $begingroup$
    @Larry: Thanks for the bounty! :) BTW: You should post the tetrahedron variant as a separate question.
    $endgroup$
    – Blue
    Sep 12 '18 at 11:14












    $begingroup$
    :) You’re welcome. I was thinking about posting it as a separate question, but I thought the question would be easier to understand with the context. Maybe I should post another question.
    $endgroup$
    – Larry
    Sep 12 '18 at 11:57




    $begingroup$
    :) You’re welcome. I was thinking about posting it as a separate question, but I thought the question would be easier to understand with the context. Maybe I should post another question.
    $endgroup$
    – Larry
    Sep 12 '18 at 11:57




    1




    1




    $begingroup$
    @Larry: Post another question, but add a link back to this one for context. (And add a link from this one to the other for people who might be interested.) Simple! :)
    $endgroup$
    – Blue
    Sep 12 '18 at 12:42




    $begingroup$
    @Larry: Post another question, but add a link back to this one for context. (And add a link from this one to the other for people who might be interested.) Simple! :)
    $endgroup$
    – Blue
    Sep 12 '18 at 12:42











    2












    $begingroup$

    Here is an alternative proof, using the diagram below.



    As an aside, note that proving $Q'Q parallel R'P$ would allow us to conclude the rest from a special case of Desargues's theorem. We would have:




    • $Q'S parallel R'O$, $ SQ parallel OP$, $ QQ' parallel PR'$, so

    • $Q'S cap R'O$ and $SQ cap OP$ and $QQ' cap PR'$ are all collinear on the line at infinity, so

    • the triangles $Q'SQ$ and $R'OP$ are in perspective axially, so

    • the triangles are in perspective centrally, and

    • $Q'R'$, $SO$, $QP$ are all coincident.


    Perhaps someone else will see how to prove that first parallelism geometrically.



    diagram



    Meanwhile, here's a vectorial proof of that parallelism and of the theorem as a whole, for anyone who wants an algebraic approach.



    Define $b,c,d,e,f$ as vectors, and $t,u$ as scalars such that




    • $b$ goes from $O$ to $S$

    • $c$ goes from $P'$ to $Q'$, $d$ goes from $R'$ to $Q'$, $tc$ goes from $O$ to $R'$.

    • $e$ goes from $R$ to $Q$, $ f$ goes from $P$ to $Q$, $ ue$ goes from $O$ to $P$.


    Now we record several scalar facts:
    begin{align}
    PQR text{ are vertices of a square}&: e.f = 0, e.e = f.f & (1)\
    P'Q'R' text{ are vertices of a square}&: c.d = 0, c.c = d.d & (2)\
    c+f text{ rotated }90^o text{ gives }d+e&: |c+f|=|d+e| &(3)\
    P' text{ lies on }PQ&: (tc+d-c).e=ue.e &(4)\
    R text{ lies on }Q'R'&: (ue+f-e).c=tc.c &(5)\
    S text{ lies on }P'Q'&: b.d=d.d &(6)\
    S text{ lies on }QR&: b.f=f.f &(7)\
    1+2+3 &: c.f=d.e & (8)\
    4+5+8 &: dfrac{t}{c.e+e.e}=dfrac{u}{c.c+c.e} & (9)
    end{align}
    This is enough to prove that $R'P=ue-tc$ and $Q'Q=ue+f-tc-d$ are both perpendicular to $c+e$, and therefore parallel as for the above use of Desargues's theorem. Meanwhile, proceeding vectorically,
    we check that $$(d.e)b=(e.e)c + (c.c)e$$ because both sides have the same dot product with $d$ (by 2,6), and both sides have the same dot product with $f$ (by 1,7,8), and we assume that $d$ and $f$ span the plane. We conclude that
    $$frac{c.c}{b.c}(c.e+e.e)=d.e=frac{e.e}{b.e}(c.c+c.e) (10)$$
    So multiplying (9) and (10) gives
    $$frac{c.c}{b.c} t=frac{e.e}{b.e} u =: x$$
    Finally, let the point $X$ be $xb$ away from $O$. Then:




    • $xb.c=tc.c$, so $X$ is on $Q'R'$;

    • $xb.e=ue.e$, so $X$ is on $PQ$;

    • by construction $X$ is on $OS$;

    • so $Q'R'$, $PQ$, and $OS$ are coincident, as desired.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Here is an alternative proof, using the diagram below.



      As an aside, note that proving $Q'Q parallel R'P$ would allow us to conclude the rest from a special case of Desargues's theorem. We would have:




      • $Q'S parallel R'O$, $ SQ parallel OP$, $ QQ' parallel PR'$, so

      • $Q'S cap R'O$ and $SQ cap OP$ and $QQ' cap PR'$ are all collinear on the line at infinity, so

      • the triangles $Q'SQ$ and $R'OP$ are in perspective axially, so

      • the triangles are in perspective centrally, and

      • $Q'R'$, $SO$, $QP$ are all coincident.


      Perhaps someone else will see how to prove that first parallelism geometrically.



      diagram



      Meanwhile, here's a vectorial proof of that parallelism and of the theorem as a whole, for anyone who wants an algebraic approach.



      Define $b,c,d,e,f$ as vectors, and $t,u$ as scalars such that




      • $b$ goes from $O$ to $S$

      • $c$ goes from $P'$ to $Q'$, $d$ goes from $R'$ to $Q'$, $tc$ goes from $O$ to $R'$.

      • $e$ goes from $R$ to $Q$, $ f$ goes from $P$ to $Q$, $ ue$ goes from $O$ to $P$.


      Now we record several scalar facts:
      begin{align}
      PQR text{ are vertices of a square}&: e.f = 0, e.e = f.f & (1)\
      P'Q'R' text{ are vertices of a square}&: c.d = 0, c.c = d.d & (2)\
      c+f text{ rotated }90^o text{ gives }d+e&: |c+f|=|d+e| &(3)\
      P' text{ lies on }PQ&: (tc+d-c).e=ue.e &(4)\
      R text{ lies on }Q'R'&: (ue+f-e).c=tc.c &(5)\
      S text{ lies on }P'Q'&: b.d=d.d &(6)\
      S text{ lies on }QR&: b.f=f.f &(7)\
      1+2+3 &: c.f=d.e & (8)\
      4+5+8 &: dfrac{t}{c.e+e.e}=dfrac{u}{c.c+c.e} & (9)
      end{align}
      This is enough to prove that $R'P=ue-tc$ and $Q'Q=ue+f-tc-d$ are both perpendicular to $c+e$, and therefore parallel as for the above use of Desargues's theorem. Meanwhile, proceeding vectorically,
      we check that $$(d.e)b=(e.e)c + (c.c)e$$ because both sides have the same dot product with $d$ (by 2,6), and both sides have the same dot product with $f$ (by 1,7,8), and we assume that $d$ and $f$ span the plane. We conclude that
      $$frac{c.c}{b.c}(c.e+e.e)=d.e=frac{e.e}{b.e}(c.c+c.e) (10)$$
      So multiplying (9) and (10) gives
      $$frac{c.c}{b.c} t=frac{e.e}{b.e} u =: x$$
      Finally, let the point $X$ be $xb$ away from $O$. Then:




      • $xb.c=tc.c$, so $X$ is on $Q'R'$;

      • $xb.e=ue.e$, so $X$ is on $PQ$;

      • by construction $X$ is on $OS$;

      • so $Q'R'$, $PQ$, and $OS$ are coincident, as desired.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Here is an alternative proof, using the diagram below.



        As an aside, note that proving $Q'Q parallel R'P$ would allow us to conclude the rest from a special case of Desargues's theorem. We would have:




        • $Q'S parallel R'O$, $ SQ parallel OP$, $ QQ' parallel PR'$, so

        • $Q'S cap R'O$ and $SQ cap OP$ and $QQ' cap PR'$ are all collinear on the line at infinity, so

        • the triangles $Q'SQ$ and $R'OP$ are in perspective axially, so

        • the triangles are in perspective centrally, and

        • $Q'R'$, $SO$, $QP$ are all coincident.


        Perhaps someone else will see how to prove that first parallelism geometrically.



        diagram



        Meanwhile, here's a vectorial proof of that parallelism and of the theorem as a whole, for anyone who wants an algebraic approach.



        Define $b,c,d,e,f$ as vectors, and $t,u$ as scalars such that




        • $b$ goes from $O$ to $S$

        • $c$ goes from $P'$ to $Q'$, $d$ goes from $R'$ to $Q'$, $tc$ goes from $O$ to $R'$.

        • $e$ goes from $R$ to $Q$, $ f$ goes from $P$ to $Q$, $ ue$ goes from $O$ to $P$.


        Now we record several scalar facts:
        begin{align}
        PQR text{ are vertices of a square}&: e.f = 0, e.e = f.f & (1)\
        P'Q'R' text{ are vertices of a square}&: c.d = 0, c.c = d.d & (2)\
        c+f text{ rotated }90^o text{ gives }d+e&: |c+f|=|d+e| &(3)\
        P' text{ lies on }PQ&: (tc+d-c).e=ue.e &(4)\
        R text{ lies on }Q'R'&: (ue+f-e).c=tc.c &(5)\
        S text{ lies on }P'Q'&: b.d=d.d &(6)\
        S text{ lies on }QR&: b.f=f.f &(7)\
        1+2+3 &: c.f=d.e & (8)\
        4+5+8 &: dfrac{t}{c.e+e.e}=dfrac{u}{c.c+c.e} & (9)
        end{align}
        This is enough to prove that $R'P=ue-tc$ and $Q'Q=ue+f-tc-d$ are both perpendicular to $c+e$, and therefore parallel as for the above use of Desargues's theorem. Meanwhile, proceeding vectorically,
        we check that $$(d.e)b=(e.e)c + (c.c)e$$ because both sides have the same dot product with $d$ (by 2,6), and both sides have the same dot product with $f$ (by 1,7,8), and we assume that $d$ and $f$ span the plane. We conclude that
        $$frac{c.c}{b.c}(c.e+e.e)=d.e=frac{e.e}{b.e}(c.c+c.e) (10)$$
        So multiplying (9) and (10) gives
        $$frac{c.c}{b.c} t=frac{e.e}{b.e} u =: x$$
        Finally, let the point $X$ be $xb$ away from $O$. Then:




        • $xb.c=tc.c$, so $X$ is on $Q'R'$;

        • $xb.e=ue.e$, so $X$ is on $PQ$;

        • by construction $X$ is on $OS$;

        • so $Q'R'$, $PQ$, and $OS$ are coincident, as desired.






        share|cite|improve this answer











        $endgroup$



        Here is an alternative proof, using the diagram below.



        As an aside, note that proving $Q'Q parallel R'P$ would allow us to conclude the rest from a special case of Desargues's theorem. We would have:




        • $Q'S parallel R'O$, $ SQ parallel OP$, $ QQ' parallel PR'$, so

        • $Q'S cap R'O$ and $SQ cap OP$ and $QQ' cap PR'$ are all collinear on the line at infinity, so

        • the triangles $Q'SQ$ and $R'OP$ are in perspective axially, so

        • the triangles are in perspective centrally, and

        • $Q'R'$, $SO$, $QP$ are all coincident.


        Perhaps someone else will see how to prove that first parallelism geometrically.



        diagram



        Meanwhile, here's a vectorial proof of that parallelism and of the theorem as a whole, for anyone who wants an algebraic approach.



        Define $b,c,d,e,f$ as vectors, and $t,u$ as scalars such that




        • $b$ goes from $O$ to $S$

        • $c$ goes from $P'$ to $Q'$, $d$ goes from $R'$ to $Q'$, $tc$ goes from $O$ to $R'$.

        • $e$ goes from $R$ to $Q$, $ f$ goes from $P$ to $Q$, $ ue$ goes from $O$ to $P$.


        Now we record several scalar facts:
        begin{align}
        PQR text{ are vertices of a square}&: e.f = 0, e.e = f.f & (1)\
        P'Q'R' text{ are vertices of a square}&: c.d = 0, c.c = d.d & (2)\
        c+f text{ rotated }90^o text{ gives }d+e&: |c+f|=|d+e| &(3)\
        P' text{ lies on }PQ&: (tc+d-c).e=ue.e &(4)\
        R text{ lies on }Q'R'&: (ue+f-e).c=tc.c &(5)\
        S text{ lies on }P'Q'&: b.d=d.d &(6)\
        S text{ lies on }QR&: b.f=f.f &(7)\
        1+2+3 &: c.f=d.e & (8)\
        4+5+8 &: dfrac{t}{c.e+e.e}=dfrac{u}{c.c+c.e} & (9)
        end{align}
        This is enough to prove that $R'P=ue-tc$ and $Q'Q=ue+f-tc-d$ are both perpendicular to $c+e$, and therefore parallel as for the above use of Desargues's theorem. Meanwhile, proceeding vectorically,
        we check that $$(d.e)b=(e.e)c + (c.c)e$$ because both sides have the same dot product with $d$ (by 2,6), and both sides have the same dot product with $f$ (by 1,7,8), and we assume that $d$ and $f$ span the plane. We conclude that
        $$frac{c.c}{b.c}(c.e+e.e)=d.e=frac{e.e}{b.e}(c.c+c.e) (10)$$
        So multiplying (9) and (10) gives
        $$frac{c.c}{b.c} t=frac{e.e}{b.e} u =: x$$
        Finally, let the point $X$ be $xb$ away from $O$. Then:




        • $xb.c=tc.c$, so $X$ is on $Q'R'$;

        • $xb.e=ue.e$, so $X$ is on $PQ$;

        • by construction $X$ is on $OS$;

        • so $Q'R'$, $PQ$, and $OS$ are coincident, as desired.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Sep 9 '18 at 3:47

























        answered Sep 9 '18 at 3:34









        Matt F.Matt F.

        2,180617




        2,180617























            2












            $begingroup$

            Although interesting, this theorem is not new. I recently found the article "Squares Inscribed in Angles and Triangles" on JSTOR.



            enter image description here



            Bailey, Herbert, and Duane Detemple. “Squares Inscribed in Angles and Triangles.” Mathematics Magazine, vol. 71, no. 4, 1998, pp. 278–284. JSTOR, JSTOR, www.jstor.org/stable/2690699.



            It would be nice if I am the person who discovered the fact, but since the theorem was published in 1998, the credit does not belong to me.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              That someone discovered the fact before you does not diminish your accomplishment. The late, great Steve Fisk called this kind of thing a mere "accident of time". :)
              $endgroup$
              – Blue
              Oct 19 '18 at 0:01






            • 1




              $begingroup$
              Thank you for your encouragement.
              $endgroup$
              – Larry
              Oct 19 '18 at 0:06






            • 1




              $begingroup$
              The reference you found shows that the collinearity property holds for arbitrary squares inscribed in the two sides of an angle; the third side of the triangle is a distraction (especially since there's no concurrence). I'm a little surprised I didn't see this earlier: My proof never used the third components in my ratios $cot C:pm 1:cot A$ and $cot C:pm 1:cot B$. I'll refine that proof. In fact, I can generalize so that the result applies not merely to inscribed squares, but inscribed similar parallelograms. (My ostensibly-"new" result may generalize in the same way.)
              $endgroup$
              – Blue
              Oct 19 '18 at 2:15


















            2












            $begingroup$

            Although interesting, this theorem is not new. I recently found the article "Squares Inscribed in Angles and Triangles" on JSTOR.



            enter image description here



            Bailey, Herbert, and Duane Detemple. “Squares Inscribed in Angles and Triangles.” Mathematics Magazine, vol. 71, no. 4, 1998, pp. 278–284. JSTOR, JSTOR, www.jstor.org/stable/2690699.



            It would be nice if I am the person who discovered the fact, but since the theorem was published in 1998, the credit does not belong to me.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              That someone discovered the fact before you does not diminish your accomplishment. The late, great Steve Fisk called this kind of thing a mere "accident of time". :)
              $endgroup$
              – Blue
              Oct 19 '18 at 0:01






            • 1




              $begingroup$
              Thank you for your encouragement.
              $endgroup$
              – Larry
              Oct 19 '18 at 0:06






            • 1




              $begingroup$
              The reference you found shows that the collinearity property holds for arbitrary squares inscribed in the two sides of an angle; the third side of the triangle is a distraction (especially since there's no concurrence). I'm a little surprised I didn't see this earlier: My proof never used the third components in my ratios $cot C:pm 1:cot A$ and $cot C:pm 1:cot B$. I'll refine that proof. In fact, I can generalize so that the result applies not merely to inscribed squares, but inscribed similar parallelograms. (My ostensibly-"new" result may generalize in the same way.)
              $endgroup$
              – Blue
              Oct 19 '18 at 2:15
















            2












            2








            2





            $begingroup$

            Although interesting, this theorem is not new. I recently found the article "Squares Inscribed in Angles and Triangles" on JSTOR.



            enter image description here



            Bailey, Herbert, and Duane Detemple. “Squares Inscribed in Angles and Triangles.” Mathematics Magazine, vol. 71, no. 4, 1998, pp. 278–284. JSTOR, JSTOR, www.jstor.org/stable/2690699.



            It would be nice if I am the person who discovered the fact, but since the theorem was published in 1998, the credit does not belong to me.






            share|cite|improve this answer









            $endgroup$



            Although interesting, this theorem is not new. I recently found the article "Squares Inscribed in Angles and Triangles" on JSTOR.



            enter image description here



            Bailey, Herbert, and Duane Detemple. “Squares Inscribed in Angles and Triangles.” Mathematics Magazine, vol. 71, no. 4, 1998, pp. 278–284. JSTOR, JSTOR, www.jstor.org/stable/2690699.



            It would be nice if I am the person who discovered the fact, but since the theorem was published in 1998, the credit does not belong to me.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 18 '18 at 19:42









            LarryLarry

            2,46831130




            2,46831130












            • $begingroup$
              That someone discovered the fact before you does not diminish your accomplishment. The late, great Steve Fisk called this kind of thing a mere "accident of time". :)
              $endgroup$
              – Blue
              Oct 19 '18 at 0:01






            • 1




              $begingroup$
              Thank you for your encouragement.
              $endgroup$
              – Larry
              Oct 19 '18 at 0:06






            • 1




              $begingroup$
              The reference you found shows that the collinearity property holds for arbitrary squares inscribed in the two sides of an angle; the third side of the triangle is a distraction (especially since there's no concurrence). I'm a little surprised I didn't see this earlier: My proof never used the third components in my ratios $cot C:pm 1:cot A$ and $cot C:pm 1:cot B$. I'll refine that proof. In fact, I can generalize so that the result applies not merely to inscribed squares, but inscribed similar parallelograms. (My ostensibly-"new" result may generalize in the same way.)
              $endgroup$
              – Blue
              Oct 19 '18 at 2:15




















            • $begingroup$
              That someone discovered the fact before you does not diminish your accomplishment. The late, great Steve Fisk called this kind of thing a mere "accident of time". :)
              $endgroup$
              – Blue
              Oct 19 '18 at 0:01






            • 1




              $begingroup$
              Thank you for your encouragement.
              $endgroup$
              – Larry
              Oct 19 '18 at 0:06






            • 1




              $begingroup$
              The reference you found shows that the collinearity property holds for arbitrary squares inscribed in the two sides of an angle; the third side of the triangle is a distraction (especially since there's no concurrence). I'm a little surprised I didn't see this earlier: My proof never used the third components in my ratios $cot C:pm 1:cot A$ and $cot C:pm 1:cot B$. I'll refine that proof. In fact, I can generalize so that the result applies not merely to inscribed squares, but inscribed similar parallelograms. (My ostensibly-"new" result may generalize in the same way.)
              $endgroup$
              – Blue
              Oct 19 '18 at 2:15


















            $begingroup$
            That someone discovered the fact before you does not diminish your accomplishment. The late, great Steve Fisk called this kind of thing a mere "accident of time". :)
            $endgroup$
            – Blue
            Oct 19 '18 at 0:01




            $begingroup$
            That someone discovered the fact before you does not diminish your accomplishment. The late, great Steve Fisk called this kind of thing a mere "accident of time". :)
            $endgroup$
            – Blue
            Oct 19 '18 at 0:01




            1




            1




            $begingroup$
            Thank you for your encouragement.
            $endgroup$
            – Larry
            Oct 19 '18 at 0:06




            $begingroup$
            Thank you for your encouragement.
            $endgroup$
            – Larry
            Oct 19 '18 at 0:06




            1




            1




            $begingroup$
            The reference you found shows that the collinearity property holds for arbitrary squares inscribed in the two sides of an angle; the third side of the triangle is a distraction (especially since there's no concurrence). I'm a little surprised I didn't see this earlier: My proof never used the third components in my ratios $cot C:pm 1:cot A$ and $cot C:pm 1:cot B$. I'll refine that proof. In fact, I can generalize so that the result applies not merely to inscribed squares, but inscribed similar parallelograms. (My ostensibly-"new" result may generalize in the same way.)
            $endgroup$
            – Blue
            Oct 19 '18 at 2:15






            $begingroup$
            The reference you found shows that the collinearity property holds for arbitrary squares inscribed in the two sides of an angle; the third side of the triangle is a distraction (especially since there's no concurrence). I'm a little surprised I didn't see this earlier: My proof never used the third components in my ratios $cot C:pm 1:cot A$ and $cot C:pm 1:cot B$. I'll refine that proof. In fact, I can generalize so that the result applies not merely to inscribed squares, but inscribed similar parallelograms. (My ostensibly-"new" result may generalize in the same way.)
            $endgroup$
            – Blue
            Oct 19 '18 at 2:15




















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