Japanese Temple Problem From 1844











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I recently learnt a Japanese geometry temple problem.



The problem is the following:



Five squares are arranged as the image shows. Prove that the area of triangle T and the area of square S are equal.





This is problem 6 in this article.
I am thinking about law of cosines, but I have not been able to prove the theorem. Any hints would be appreciated.










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  • 1




    If you like this kind of problems I cannot recommend enough the book "Sacred Mathematics - Japanese Temple Geometry" by Fukagawa Hidetoshi and Tony Rothman. It also includes a couple of "relatives" to this problem.
    – mickep
    Nov 24 at 14:03










  • I checked it out. A great book indeed, thank you.
    – Larry
    Nov 24 at 14:22










  • The shapes $S,T$ share a vertex $A$ where $6$ angles meet, $3$ of them right angles. Thus the other $3$ angles sum to $pi/2$. Let $theta$ be the angle in $T$, so the two other acute angles and the right angle between them from $S$ sum to $pi-theta$. As $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$ where $BA,AC$ are sides of the upper squares. The challenge then is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
    – TheSimpliFire
    Nov 24 at 14:24






  • 1




    ^ From J.G.'s answer.
    – TheSimpliFire
    Nov 24 at 14:26















up vote
60
down vote

favorite
43












I recently learnt a Japanese geometry temple problem.



The problem is the following:



Five squares are arranged as the image shows. Prove that the area of triangle T and the area of square S are equal.





This is problem 6 in this article.
I am thinking about law of cosines, but I have not been able to prove the theorem. Any hints would be appreciated.










share|cite|improve this question




















  • 1




    If you like this kind of problems I cannot recommend enough the book "Sacred Mathematics - Japanese Temple Geometry" by Fukagawa Hidetoshi and Tony Rothman. It also includes a couple of "relatives" to this problem.
    – mickep
    Nov 24 at 14:03










  • I checked it out. A great book indeed, thank you.
    – Larry
    Nov 24 at 14:22










  • The shapes $S,T$ share a vertex $A$ where $6$ angles meet, $3$ of them right angles. Thus the other $3$ angles sum to $pi/2$. Let $theta$ be the angle in $T$, so the two other acute angles and the right angle between them from $S$ sum to $pi-theta$. As $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$ where $BA,AC$ are sides of the upper squares. The challenge then is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
    – TheSimpliFire
    Nov 24 at 14:24






  • 1




    ^ From J.G.'s answer.
    – TheSimpliFire
    Nov 24 at 14:26













up vote
60
down vote

favorite
43









up vote
60
down vote

favorite
43






43





I recently learnt a Japanese geometry temple problem.



The problem is the following:



Five squares are arranged as the image shows. Prove that the area of triangle T and the area of square S are equal.





This is problem 6 in this article.
I am thinking about law of cosines, but I have not been able to prove the theorem. Any hints would be appreciated.










share|cite|improve this question















I recently learnt a Japanese geometry temple problem.



The problem is the following:



Five squares are arranged as the image shows. Prove that the area of triangle T and the area of square S are equal.





This is problem 6 in this article.
I am thinking about law of cosines, but I have not been able to prove the theorem. Any hints would be appreciated.







geometry sangaku






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edited Nov 22 at 20:49









Jean-Claude Arbaut

14.7k63363




14.7k63363










asked Nov 22 at 20:40









Larry

1,3202722




1,3202722








  • 1




    If you like this kind of problems I cannot recommend enough the book "Sacred Mathematics - Japanese Temple Geometry" by Fukagawa Hidetoshi and Tony Rothman. It also includes a couple of "relatives" to this problem.
    – mickep
    Nov 24 at 14:03










  • I checked it out. A great book indeed, thank you.
    – Larry
    Nov 24 at 14:22










  • The shapes $S,T$ share a vertex $A$ where $6$ angles meet, $3$ of them right angles. Thus the other $3$ angles sum to $pi/2$. Let $theta$ be the angle in $T$, so the two other acute angles and the right angle between them from $S$ sum to $pi-theta$. As $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$ where $BA,AC$ are sides of the upper squares. The challenge then is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
    – TheSimpliFire
    Nov 24 at 14:24






  • 1




    ^ From J.G.'s answer.
    – TheSimpliFire
    Nov 24 at 14:26














  • 1




    If you like this kind of problems I cannot recommend enough the book "Sacred Mathematics - Japanese Temple Geometry" by Fukagawa Hidetoshi and Tony Rothman. It also includes a couple of "relatives" to this problem.
    – mickep
    Nov 24 at 14:03










  • I checked it out. A great book indeed, thank you.
    – Larry
    Nov 24 at 14:22










  • The shapes $S,T$ share a vertex $A$ where $6$ angles meet, $3$ of them right angles. Thus the other $3$ angles sum to $pi/2$. Let $theta$ be the angle in $T$, so the two other acute angles and the right angle between them from $S$ sum to $pi-theta$. As $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$ where $BA,AC$ are sides of the upper squares. The challenge then is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
    – TheSimpliFire
    Nov 24 at 14:24






  • 1




    ^ From J.G.'s answer.
    – TheSimpliFire
    Nov 24 at 14:26








1




1




If you like this kind of problems I cannot recommend enough the book "Sacred Mathematics - Japanese Temple Geometry" by Fukagawa Hidetoshi and Tony Rothman. It also includes a couple of "relatives" to this problem.
– mickep
Nov 24 at 14:03




If you like this kind of problems I cannot recommend enough the book "Sacred Mathematics - Japanese Temple Geometry" by Fukagawa Hidetoshi and Tony Rothman. It also includes a couple of "relatives" to this problem.
– mickep
Nov 24 at 14:03












I checked it out. A great book indeed, thank you.
– Larry
Nov 24 at 14:22




I checked it out. A great book indeed, thank you.
– Larry
Nov 24 at 14:22












The shapes $S,T$ share a vertex $A$ where $6$ angles meet, $3$ of them right angles. Thus the other $3$ angles sum to $pi/2$. Let $theta$ be the angle in $T$, so the two other acute angles and the right angle between them from $S$ sum to $pi-theta$. As $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$ where $BA,AC$ are sides of the upper squares. The challenge then is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
– TheSimpliFire
Nov 24 at 14:24




The shapes $S,T$ share a vertex $A$ where $6$ angles meet, $3$ of them right angles. Thus the other $3$ angles sum to $pi/2$. Let $theta$ be the angle in $T$, so the two other acute angles and the right angle between them from $S$ sum to $pi-theta$. As $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$ where $BA,AC$ are sides of the upper squares. The challenge then is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.
– TheSimpliFire
Nov 24 at 14:24




1




1




^ From J.G.'s answer.
– TheSimpliFire
Nov 24 at 14:26




^ From J.G.'s answer.
– TheSimpliFire
Nov 24 at 14:26










6 Answers
6






active

oldest

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up vote
40
down vote



accepted










We will, first of all, prove a very interesting property




$mathbf{Lemma;1}$



Given two squares PQRS and PTUV (as shown on the picture), the triangles $Delta STP$ and $Delta PVQ$ have equal area.




$mathbf {Proof}$



enter image description here



Denote by $alpha$ the angle SPT and by $[...]$ the area of the polygon "...". Hence
$$[Delta STP]=frac{overline {PS}cdotoverline {PT}cdot sin(alpha)}{2}$$ $$[Delta PVQ]=frac{overline {QP}*overline {PV}cdotsinBigl(360°-(90°+90+alpha)Bigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsinBigl(180°-alphaBigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsin(alpha)}{2}$$



Since $overline {PS}=overline {PQ}$ and $overline {PT}=overline {PV}$ $$[Delta STP]=[Delta PVQ]$$



Now, back to the problem



enter image description here
Let $overline {AB}=a$ and $overline {IJ}=b$. Note first of all that $$Delta BEC cong Delta EIF$$
See why? $mathbf {Hint:}$




It is obvious that $overline {CE}=overline {EF}$. Use the properties of right triangles in order to show that all angles are equal.




Thus $${(overline{CE})^2}={a^2}+{b^2}=S$$



Note furthermore that $$[Delta BEC]=[Delta EIF]=frac{ab}{2}$$
By Lemma 1:
$$[Delta DCG]=[Delta BEC]=frac{ab}{2}=[Delta EIF]=[Delta GFK]$$
The area of the polygon AJKGD is thus
$$[AJKGD]=[ABCD]+[CEFG]+[FIJK]+4[Delta DCG]=2Bigl({a^2}+{b^2}Bigr)+2ab$$



The area of the trapezoid AJKD is moreover
$$[AJKD]=frac{(a+b)(2a+2b)}{2}={a^2}+2ab+{b^2}$$



Finally
$$T=[Delta DKG]=[AJKGD]-[AJKD]={a^2}+{b^2}=S Rightarrow S=T$$






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  • Comments are not for extended discussion; this conversation has been moved to chat.
    – Aloizio Macedo
    Nov 24 at 13:16


















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enter image description here



$$|square P_1 P_2 P_3 P_4| = (a+b)^2 = frac12(a+b)(2a+2b) = |square Q_1 Q_2 Q_3 Q_4|quad=:R$$




$$S ;=; R - 4cdotfrac12ab ;=; T$$




(This space intentionally left blank.)






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  • 3




    I wonder what tools you use to create awesome graphs like this
    – Larry
    Nov 23 at 14:24








  • 5




    @Larry: I use GeoGebra.
    – Blue
    Nov 23 at 19:47






  • 1




    I see, thank you.
    – Larry
    Nov 23 at 19:49






  • 1




    @crskhr: $a$ and $b$ are the lengths labeled at the bottom of the figure; they match the sides of the bottom squares. And "$square$" in this context means merely "quadrilateral" of any shape, despite the symbol itself being more structured; compare this to "$triangle$" (in, say, "$triangle ABC$"), which means "triangle" of any shape, despite the symbol itself resembling an equilateral figure. (See, for instance, Wikipedia's entry for mathematical symbols.)
    – Blue
    Nov 25 at 11:40








  • 1




    @Blue Thanks a lot for clarifying :)
    – crskhr
    Nov 25 at 14:34


















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17
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Because there are so many squares, coordinates are easy to compute.



enter image description here



The area of the shaded square is clearly $u^2+v^2$.



The area of the shaded triangle is one-half of the absolute value of the determinant of the array



$$left[ begin{array}{c}
1 & 1 & 1 \
2u-v & 3u & 2u \
3u+v & u+3v & u+v
end{array} right]$$



which is also $u^2+v^2$.






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    up vote
    12
    down vote













    The four triangles adjacent to $S$ (two of them right, two of them obtuse) all have the same area. (Each has the same base and height as the one on the opposite side of the square, while the two right triangles are congruent.). Now rotate each of the obtuse triangles by $90^circ$ so that they are adjacent to $T$, as shown.



    enter image description here



    What now needs to be proved is that the two shaded pentagons have equal area. This can be done by observing that each pentagon decomposes into an isosceles right triangle and a trapezoid. The isosceles right triangles are congruent; the trapezoids have equal area as their two bases are the same and their heights are the same.






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      up vote
      10
      down vote













      While the other solutions are obviously correct, they are also unnecessarily complicated.

      Since the angle of the squares is not specified, it must be true for all angles, so* why not pick one which is simple to work with and results in a degenerate case.



      enter image description here



      *) The assumption of truth is not required since we first do show that S=T is, in fact, true (in one simple case) and skirt the rules from there on by leaving the extrapolation to the reader.

      While this approach is best suited for puzzles, looking at the edge cases first is a quick way to disprove things by example, or at least check that your calculations are correct






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      • 7




        That's a beautiful drawing, but I'm not sure if "Since the angle of the squares is not specified, it must be true for all angles" is logically sound here. It seems to be assuming the truth of the answer.
        – Owen
        Nov 23 at 23:19






      • 5




        Tongue in cheek: Isn't this almost equivalent to "P is true because the question asks me to prove it"?
        – IanF1
        Nov 24 at 8:01






      • 1




        I'm a bit puzzled why an off-topic aside ("testing degenerate cases is useful" and "regularly check your tire pressure" are equally relevant remarks to the posed problem) with a drawing and nothing else gets so many upvotes. "To disprove by example" is useless here, as we already know it's true; that's the puzzle: show it. This differs from the other proofs NOT by removing unneccessary complications but by removing the PROOF part; without a hint of a path to a proof, or any insight? Very weird.
        – user3445853
        Nov 24 at 20:38






      • 2




        If the absurdity of this isn't clear enough: Continue the logic and let the surface of the yellow square go to zero. Now both triangles are equal sized (as both are zero). So then drawing a seagreen square on its own is someway somehow corroboration that the original question is true, and the "extrapolation" of the method of proof is left to the reader?! Neat. I don't know why I ever worked for math.
        – user3445853
        Nov 24 at 20:40










      • The value of this view is in showing that, if there is a constant difference between the areas of the triangle and square, then that difference is zero. (Or, if there's a constant ratio, then the ratio is one; etc.) So, had the task been to find that constant difference (or ratio, or whatever), then consulting a particularly-convenient configuration would've made for a proper solution. In any case, this serves as a sanity check on the problem.
        – Blue
        16 hours ago


















      up vote
      1
      down vote













      This is a long comment.



      The shapes $S,,T$ share a vertex $A$ where six angles meet, three of them right angles. The other three angles therefore sum to a right angle. Let $theta$ be the angle in $T$, so the two other acute angles, and the right angle between them from $S$, sum to $pi-theta$. Since $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$, where $BA,,AC$ are sides of the upper squares.



      The challenge, then, is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.






      share|cite|improve this answer





















      • I don't think that $BACD$ has twice the area of $ABC$. If you try to move $ABC$ by shearing to get a triangle with side $BC$ and something on the bottom line, the third point isn't $D$.
        – D. Thomine
        Nov 22 at 21:35










      • $BACD$ is a rhombus because its opposite sides are parallel. So it does have twice the area of $ABC$.
        – I like Serena
        Nov 22 at 21:41










      • $BACD$ is not a rhombus.
        – D. Thomine
        Nov 22 at 21:42






      • 2




        @Servaes It's too long. "Long comment" is something people often say when their contribution has to be an "answer" because of software limitations, but is far from a complete answer.
        – J.G.
        Nov 23 at 10:22






      • 1




        I have converted your answer into a comment.
        – TheSimpliFire
        Nov 24 at 14:28











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      6 Answers
      6






      active

      oldest

      votes








      6 Answers
      6






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

      votes








      up vote
      40
      down vote



      accepted










      We will, first of all, prove a very interesting property




      $mathbf{Lemma;1}$



      Given two squares PQRS and PTUV (as shown on the picture), the triangles $Delta STP$ and $Delta PVQ$ have equal area.




      $mathbf {Proof}$



      enter image description here



      Denote by $alpha$ the angle SPT and by $[...]$ the area of the polygon "...". Hence
      $$[Delta STP]=frac{overline {PS}cdotoverline {PT}cdot sin(alpha)}{2}$$ $$[Delta PVQ]=frac{overline {QP}*overline {PV}cdotsinBigl(360°-(90°+90+alpha)Bigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsinBigl(180°-alphaBigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsin(alpha)}{2}$$



      Since $overline {PS}=overline {PQ}$ and $overline {PT}=overline {PV}$ $$[Delta STP]=[Delta PVQ]$$



      Now, back to the problem



      enter image description here
      Let $overline {AB}=a$ and $overline {IJ}=b$. Note first of all that $$Delta BEC cong Delta EIF$$
      See why? $mathbf {Hint:}$




      It is obvious that $overline {CE}=overline {EF}$. Use the properties of right triangles in order to show that all angles are equal.




      Thus $${(overline{CE})^2}={a^2}+{b^2}=S$$



      Note furthermore that $$[Delta BEC]=[Delta EIF]=frac{ab}{2}$$
      By Lemma 1:
      $$[Delta DCG]=[Delta BEC]=frac{ab}{2}=[Delta EIF]=[Delta GFK]$$
      The area of the polygon AJKGD is thus
      $$[AJKGD]=[ABCD]+[CEFG]+[FIJK]+4[Delta DCG]=2Bigl({a^2}+{b^2}Bigr)+2ab$$



      The area of the trapezoid AJKD is moreover
      $$[AJKD]=frac{(a+b)(2a+2b)}{2}={a^2}+2ab+{b^2}$$



      Finally
      $$T=[Delta DKG]=[AJKGD]-[AJKD]={a^2}+{b^2}=S Rightarrow S=T$$






      share|cite|improve this answer























      • Comments are not for extended discussion; this conversation has been moved to chat.
        – Aloizio Macedo
        Nov 24 at 13:16















      up vote
      40
      down vote



      accepted










      We will, first of all, prove a very interesting property




      $mathbf{Lemma;1}$



      Given two squares PQRS and PTUV (as shown on the picture), the triangles $Delta STP$ and $Delta PVQ$ have equal area.




      $mathbf {Proof}$



      enter image description here



      Denote by $alpha$ the angle SPT and by $[...]$ the area of the polygon "...". Hence
      $$[Delta STP]=frac{overline {PS}cdotoverline {PT}cdot sin(alpha)}{2}$$ $$[Delta PVQ]=frac{overline {QP}*overline {PV}cdotsinBigl(360°-(90°+90+alpha)Bigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsinBigl(180°-alphaBigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsin(alpha)}{2}$$



      Since $overline {PS}=overline {PQ}$ and $overline {PT}=overline {PV}$ $$[Delta STP]=[Delta PVQ]$$



      Now, back to the problem



      enter image description here
      Let $overline {AB}=a$ and $overline {IJ}=b$. Note first of all that $$Delta BEC cong Delta EIF$$
      See why? $mathbf {Hint:}$




      It is obvious that $overline {CE}=overline {EF}$. Use the properties of right triangles in order to show that all angles are equal.




      Thus $${(overline{CE})^2}={a^2}+{b^2}=S$$



      Note furthermore that $$[Delta BEC]=[Delta EIF]=frac{ab}{2}$$
      By Lemma 1:
      $$[Delta DCG]=[Delta BEC]=frac{ab}{2}=[Delta EIF]=[Delta GFK]$$
      The area of the polygon AJKGD is thus
      $$[AJKGD]=[ABCD]+[CEFG]+[FIJK]+4[Delta DCG]=2Bigl({a^2}+{b^2}Bigr)+2ab$$



      The area of the trapezoid AJKD is moreover
      $$[AJKD]=frac{(a+b)(2a+2b)}{2}={a^2}+2ab+{b^2}$$



      Finally
      $$T=[Delta DKG]=[AJKGD]-[AJKD]={a^2}+{b^2}=S Rightarrow S=T$$






      share|cite|improve this answer























      • Comments are not for extended discussion; this conversation has been moved to chat.
        – Aloizio Macedo
        Nov 24 at 13:16













      up vote
      40
      down vote



      accepted







      up vote
      40
      down vote



      accepted






      We will, first of all, prove a very interesting property




      $mathbf{Lemma;1}$



      Given two squares PQRS and PTUV (as shown on the picture), the triangles $Delta STP$ and $Delta PVQ$ have equal area.




      $mathbf {Proof}$



      enter image description here



      Denote by $alpha$ the angle SPT and by $[...]$ the area of the polygon "...". Hence
      $$[Delta STP]=frac{overline {PS}cdotoverline {PT}cdot sin(alpha)}{2}$$ $$[Delta PVQ]=frac{overline {QP}*overline {PV}cdotsinBigl(360°-(90°+90+alpha)Bigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsinBigl(180°-alphaBigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsin(alpha)}{2}$$



      Since $overline {PS}=overline {PQ}$ and $overline {PT}=overline {PV}$ $$[Delta STP]=[Delta PVQ]$$



      Now, back to the problem



      enter image description here
      Let $overline {AB}=a$ and $overline {IJ}=b$. Note first of all that $$Delta BEC cong Delta EIF$$
      See why? $mathbf {Hint:}$




      It is obvious that $overline {CE}=overline {EF}$. Use the properties of right triangles in order to show that all angles are equal.




      Thus $${(overline{CE})^2}={a^2}+{b^2}=S$$



      Note furthermore that $$[Delta BEC]=[Delta EIF]=frac{ab}{2}$$
      By Lemma 1:
      $$[Delta DCG]=[Delta BEC]=frac{ab}{2}=[Delta EIF]=[Delta GFK]$$
      The area of the polygon AJKGD is thus
      $$[AJKGD]=[ABCD]+[CEFG]+[FIJK]+4[Delta DCG]=2Bigl({a^2}+{b^2}Bigr)+2ab$$



      The area of the trapezoid AJKD is moreover
      $$[AJKD]=frac{(a+b)(2a+2b)}{2}={a^2}+2ab+{b^2}$$



      Finally
      $$T=[Delta DKG]=[AJKGD]-[AJKD]={a^2}+{b^2}=S Rightarrow S=T$$






      share|cite|improve this answer














      We will, first of all, prove a very interesting property




      $mathbf{Lemma;1}$



      Given two squares PQRS and PTUV (as shown on the picture), the triangles $Delta STP$ and $Delta PVQ$ have equal area.




      $mathbf {Proof}$



      enter image description here



      Denote by $alpha$ the angle SPT and by $[...]$ the area of the polygon "...". Hence
      $$[Delta STP]=frac{overline {PS}cdotoverline {PT}cdot sin(alpha)}{2}$$ $$[Delta PVQ]=frac{overline {QP}*overline {PV}cdotsinBigl(360°-(90°+90+alpha)Bigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsinBigl(180°-alphaBigr)}{2}=frac{overline {QP}cdotoverline {PV}cdotsin(alpha)}{2}$$



      Since $overline {PS}=overline {PQ}$ and $overline {PT}=overline {PV}$ $$[Delta STP]=[Delta PVQ]$$



      Now, back to the problem



      enter image description here
      Let $overline {AB}=a$ and $overline {IJ}=b$. Note first of all that $$Delta BEC cong Delta EIF$$
      See why? $mathbf {Hint:}$




      It is obvious that $overline {CE}=overline {EF}$. Use the properties of right triangles in order to show that all angles are equal.




      Thus $${(overline{CE})^2}={a^2}+{b^2}=S$$



      Note furthermore that $$[Delta BEC]=[Delta EIF]=frac{ab}{2}$$
      By Lemma 1:
      $$[Delta DCG]=[Delta BEC]=frac{ab}{2}=[Delta EIF]=[Delta GFK]$$
      The area of the polygon AJKGD is thus
      $$[AJKGD]=[ABCD]+[CEFG]+[FIJK]+4[Delta DCG]=2Bigl({a^2}+{b^2}Bigr)+2ab$$



      The area of the trapezoid AJKD is moreover
      $$[AJKD]=frac{(a+b)(2a+2b)}{2}={a^2}+2ab+{b^2}$$



      Finally
      $$T=[Delta DKG]=[AJKGD]-[AJKD]={a^2}+{b^2}=S Rightarrow S=T$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 23 at 7:43









      Mutantoe

      558411




      558411










      answered Nov 22 at 21:39









      Dr. Mathva

      906315




      906315












      • Comments are not for extended discussion; this conversation has been moved to chat.
        – Aloizio Macedo
        Nov 24 at 13:16


















      • Comments are not for extended discussion; this conversation has been moved to chat.
        – Aloizio Macedo
        Nov 24 at 13:16
















      Comments are not for extended discussion; this conversation has been moved to chat.
      – Aloizio Macedo
      Nov 24 at 13:16




      Comments are not for extended discussion; this conversation has been moved to chat.
      – Aloizio Macedo
      Nov 24 at 13:16










      up vote
      30
      down vote













      enter image description here



      $$|square P_1 P_2 P_3 P_4| = (a+b)^2 = frac12(a+b)(2a+2b) = |square Q_1 Q_2 Q_3 Q_4|quad=:R$$




      $$S ;=; R - 4cdotfrac12ab ;=; T$$




      (This space intentionally left blank.)






      share|cite|improve this answer



















      • 3




        I wonder what tools you use to create awesome graphs like this
        – Larry
        Nov 23 at 14:24








      • 5




        @Larry: I use GeoGebra.
        – Blue
        Nov 23 at 19:47






      • 1




        I see, thank you.
        – Larry
        Nov 23 at 19:49






      • 1




        @crskhr: $a$ and $b$ are the lengths labeled at the bottom of the figure; they match the sides of the bottom squares. And "$square$" in this context means merely "quadrilateral" of any shape, despite the symbol itself being more structured; compare this to "$triangle$" (in, say, "$triangle ABC$"), which means "triangle" of any shape, despite the symbol itself resembling an equilateral figure. (See, for instance, Wikipedia's entry for mathematical symbols.)
        – Blue
        Nov 25 at 11:40








      • 1




        @Blue Thanks a lot for clarifying :)
        – crskhr
        Nov 25 at 14:34















      up vote
      30
      down vote













      enter image description here



      $$|square P_1 P_2 P_3 P_4| = (a+b)^2 = frac12(a+b)(2a+2b) = |square Q_1 Q_2 Q_3 Q_4|quad=:R$$




      $$S ;=; R - 4cdotfrac12ab ;=; T$$




      (This space intentionally left blank.)






      share|cite|improve this answer



















      • 3




        I wonder what tools you use to create awesome graphs like this
        – Larry
        Nov 23 at 14:24








      • 5




        @Larry: I use GeoGebra.
        – Blue
        Nov 23 at 19:47






      • 1




        I see, thank you.
        – Larry
        Nov 23 at 19:49






      • 1




        @crskhr: $a$ and $b$ are the lengths labeled at the bottom of the figure; they match the sides of the bottom squares. And "$square$" in this context means merely "quadrilateral" of any shape, despite the symbol itself being more structured; compare this to "$triangle$" (in, say, "$triangle ABC$"), which means "triangle" of any shape, despite the symbol itself resembling an equilateral figure. (See, for instance, Wikipedia's entry for mathematical symbols.)
        – Blue
        Nov 25 at 11:40








      • 1




        @Blue Thanks a lot for clarifying :)
        – crskhr
        Nov 25 at 14:34













      up vote
      30
      down vote










      up vote
      30
      down vote









      enter image description here



      $$|square P_1 P_2 P_3 P_4| = (a+b)^2 = frac12(a+b)(2a+2b) = |square Q_1 Q_2 Q_3 Q_4|quad=:R$$




      $$S ;=; R - 4cdotfrac12ab ;=; T$$




      (This space intentionally left blank.)






      share|cite|improve this answer














      enter image description here



      $$|square P_1 P_2 P_3 P_4| = (a+b)^2 = frac12(a+b)(2a+2b) = |square Q_1 Q_2 Q_3 Q_4|quad=:R$$




      $$S ;=; R - 4cdotfrac12ab ;=; T$$




      (This space intentionally left blank.)







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 23 at 6:02

























      answered Nov 23 at 3:39









      Blue

      47.3k870149




      47.3k870149








      • 3




        I wonder what tools you use to create awesome graphs like this
        – Larry
        Nov 23 at 14:24








      • 5




        @Larry: I use GeoGebra.
        – Blue
        Nov 23 at 19:47






      • 1




        I see, thank you.
        – Larry
        Nov 23 at 19:49






      • 1




        @crskhr: $a$ and $b$ are the lengths labeled at the bottom of the figure; they match the sides of the bottom squares. And "$square$" in this context means merely "quadrilateral" of any shape, despite the symbol itself being more structured; compare this to "$triangle$" (in, say, "$triangle ABC$"), which means "triangle" of any shape, despite the symbol itself resembling an equilateral figure. (See, for instance, Wikipedia's entry for mathematical symbols.)
        – Blue
        Nov 25 at 11:40








      • 1




        @Blue Thanks a lot for clarifying :)
        – crskhr
        Nov 25 at 14:34














      • 3




        I wonder what tools you use to create awesome graphs like this
        – Larry
        Nov 23 at 14:24








      • 5




        @Larry: I use GeoGebra.
        – Blue
        Nov 23 at 19:47






      • 1




        I see, thank you.
        – Larry
        Nov 23 at 19:49






      • 1




        @crskhr: $a$ and $b$ are the lengths labeled at the bottom of the figure; they match the sides of the bottom squares. And "$square$" in this context means merely "quadrilateral" of any shape, despite the symbol itself being more structured; compare this to "$triangle$" (in, say, "$triangle ABC$"), which means "triangle" of any shape, despite the symbol itself resembling an equilateral figure. (See, for instance, Wikipedia's entry for mathematical symbols.)
        – Blue
        Nov 25 at 11:40








      • 1




        @Blue Thanks a lot for clarifying :)
        – crskhr
        Nov 25 at 14:34








      3




      3




      I wonder what tools you use to create awesome graphs like this
      – Larry
      Nov 23 at 14:24






      I wonder what tools you use to create awesome graphs like this
      – Larry
      Nov 23 at 14:24






      5




      5




      @Larry: I use GeoGebra.
      – Blue
      Nov 23 at 19:47




      @Larry: I use GeoGebra.
      – Blue
      Nov 23 at 19:47




      1




      1




      I see, thank you.
      – Larry
      Nov 23 at 19:49




      I see, thank you.
      – Larry
      Nov 23 at 19:49




      1




      1




      @crskhr: $a$ and $b$ are the lengths labeled at the bottom of the figure; they match the sides of the bottom squares. And "$square$" in this context means merely "quadrilateral" of any shape, despite the symbol itself being more structured; compare this to "$triangle$" (in, say, "$triangle ABC$"), which means "triangle" of any shape, despite the symbol itself resembling an equilateral figure. (See, for instance, Wikipedia's entry for mathematical symbols.)
      – Blue
      Nov 25 at 11:40






      @crskhr: $a$ and $b$ are the lengths labeled at the bottom of the figure; they match the sides of the bottom squares. And "$square$" in this context means merely "quadrilateral" of any shape, despite the symbol itself being more structured; compare this to "$triangle$" (in, say, "$triangle ABC$"), which means "triangle" of any shape, despite the symbol itself resembling an equilateral figure. (See, for instance, Wikipedia's entry for mathematical symbols.)
      – Blue
      Nov 25 at 11:40






      1




      1




      @Blue Thanks a lot for clarifying :)
      – crskhr
      Nov 25 at 14:34




      @Blue Thanks a lot for clarifying :)
      – crskhr
      Nov 25 at 14:34










      up vote
      17
      down vote













      Because there are so many squares, coordinates are easy to compute.



      enter image description here



      The area of the shaded square is clearly $u^2+v^2$.



      The area of the shaded triangle is one-half of the absolute value of the determinant of the array



      $$left[ begin{array}{c}
      1 & 1 & 1 \
      2u-v & 3u & 2u \
      3u+v & u+3v & u+v
      end{array} right]$$



      which is also $u^2+v^2$.






      share|cite|improve this answer

























        up vote
        17
        down vote













        Because there are so many squares, coordinates are easy to compute.



        enter image description here



        The area of the shaded square is clearly $u^2+v^2$.



        The area of the shaded triangle is one-half of the absolute value of the determinant of the array



        $$left[ begin{array}{c}
        1 & 1 & 1 \
        2u-v & 3u & 2u \
        3u+v & u+3v & u+v
        end{array} right]$$



        which is also $u^2+v^2$.






        share|cite|improve this answer























          up vote
          17
          down vote










          up vote
          17
          down vote









          Because there are so many squares, coordinates are easy to compute.



          enter image description here



          The area of the shaded square is clearly $u^2+v^2$.



          The area of the shaded triangle is one-half of the absolute value of the determinant of the array



          $$left[ begin{array}{c}
          1 & 1 & 1 \
          2u-v & 3u & 2u \
          3u+v & u+3v & u+v
          end{array} right]$$



          which is also $u^2+v^2$.






          share|cite|improve this answer












          Because there are so many squares, coordinates are easy to compute.



          enter image description here



          The area of the shaded square is clearly $u^2+v^2$.



          The area of the shaded triangle is one-half of the absolute value of the determinant of the array



          $$left[ begin{array}{c}
          1 & 1 & 1 \
          2u-v & 3u & 2u \
          3u+v & u+3v & u+v
          end{array} right]$$



          which is also $u^2+v^2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 at 14:50









          steven gregory

          17.6k32257




          17.6k32257






















              up vote
              12
              down vote













              The four triangles adjacent to $S$ (two of them right, two of them obtuse) all have the same area. (Each has the same base and height as the one on the opposite side of the square, while the two right triangles are congruent.). Now rotate each of the obtuse triangles by $90^circ$ so that they are adjacent to $T$, as shown.



              enter image description here



              What now needs to be proved is that the two shaded pentagons have equal area. This can be done by observing that each pentagon decomposes into an isosceles right triangle and a trapezoid. The isosceles right triangles are congruent; the trapezoids have equal area as their two bases are the same and their heights are the same.






              share|cite|improve this answer



























                up vote
                12
                down vote













                The four triangles adjacent to $S$ (two of them right, two of them obtuse) all have the same area. (Each has the same base and height as the one on the opposite side of the square, while the two right triangles are congruent.). Now rotate each of the obtuse triangles by $90^circ$ so that they are adjacent to $T$, as shown.



                enter image description here



                What now needs to be proved is that the two shaded pentagons have equal area. This can be done by observing that each pentagon decomposes into an isosceles right triangle and a trapezoid. The isosceles right triangles are congruent; the trapezoids have equal area as their two bases are the same and their heights are the same.






                share|cite|improve this answer

























                  up vote
                  12
                  down vote










                  up vote
                  12
                  down vote









                  The four triangles adjacent to $S$ (two of them right, two of them obtuse) all have the same area. (Each has the same base and height as the one on the opposite side of the square, while the two right triangles are congruent.). Now rotate each of the obtuse triangles by $90^circ$ so that they are adjacent to $T$, as shown.



                  enter image description here



                  What now needs to be proved is that the two shaded pentagons have equal area. This can be done by observing that each pentagon decomposes into an isosceles right triangle and a trapezoid. The isosceles right triangles are congruent; the trapezoids have equal area as their two bases are the same and their heights are the same.






                  share|cite|improve this answer














                  The four triangles adjacent to $S$ (two of them right, two of them obtuse) all have the same area. (Each has the same base and height as the one on the opposite side of the square, while the two right triangles are congruent.). Now rotate each of the obtuse triangles by $90^circ$ so that they are adjacent to $T$, as shown.



                  enter image description here



                  What now needs to be proved is that the two shaded pentagons have equal area. This can be done by observing that each pentagon decomposes into an isosceles right triangle and a trapezoid. The isosceles right triangles are congruent; the trapezoids have equal area as their two bases are the same and their heights are the same.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 24 at 15:01

























                  answered Nov 24 at 9:25









                  Will Orrick

                  13.4k13359




                  13.4k13359






















                      up vote
                      10
                      down vote













                      While the other solutions are obviously correct, they are also unnecessarily complicated.

                      Since the angle of the squares is not specified, it must be true for all angles, so* why not pick one which is simple to work with and results in a degenerate case.



                      enter image description here



                      *) The assumption of truth is not required since we first do show that S=T is, in fact, true (in one simple case) and skirt the rules from there on by leaving the extrapolation to the reader.

                      While this approach is best suited for puzzles, looking at the edge cases first is a quick way to disprove things by example, or at least check that your calculations are correct






                      share|cite|improve this answer



















                      • 7




                        That's a beautiful drawing, but I'm not sure if "Since the angle of the squares is not specified, it must be true for all angles" is logically sound here. It seems to be assuming the truth of the answer.
                        – Owen
                        Nov 23 at 23:19






                      • 5




                        Tongue in cheek: Isn't this almost equivalent to "P is true because the question asks me to prove it"?
                        – IanF1
                        Nov 24 at 8:01






                      • 1




                        I'm a bit puzzled why an off-topic aside ("testing degenerate cases is useful" and "regularly check your tire pressure" are equally relevant remarks to the posed problem) with a drawing and nothing else gets so many upvotes. "To disprove by example" is useless here, as we already know it's true; that's the puzzle: show it. This differs from the other proofs NOT by removing unneccessary complications but by removing the PROOF part; without a hint of a path to a proof, or any insight? Very weird.
                        – user3445853
                        Nov 24 at 20:38






                      • 2




                        If the absurdity of this isn't clear enough: Continue the logic and let the surface of the yellow square go to zero. Now both triangles are equal sized (as both are zero). So then drawing a seagreen square on its own is someway somehow corroboration that the original question is true, and the "extrapolation" of the method of proof is left to the reader?! Neat. I don't know why I ever worked for math.
                        – user3445853
                        Nov 24 at 20:40










                      • The value of this view is in showing that, if there is a constant difference between the areas of the triangle and square, then that difference is zero. (Or, if there's a constant ratio, then the ratio is one; etc.) So, had the task been to find that constant difference (or ratio, or whatever), then consulting a particularly-convenient configuration would've made for a proper solution. In any case, this serves as a sanity check on the problem.
                        – Blue
                        16 hours ago















                      up vote
                      10
                      down vote













                      While the other solutions are obviously correct, they are also unnecessarily complicated.

                      Since the angle of the squares is not specified, it must be true for all angles, so* why not pick one which is simple to work with and results in a degenerate case.



                      enter image description here



                      *) The assumption of truth is not required since we first do show that S=T is, in fact, true (in one simple case) and skirt the rules from there on by leaving the extrapolation to the reader.

                      While this approach is best suited for puzzles, looking at the edge cases first is a quick way to disprove things by example, or at least check that your calculations are correct






                      share|cite|improve this answer



















                      • 7




                        That's a beautiful drawing, but I'm not sure if "Since the angle of the squares is not specified, it must be true for all angles" is logically sound here. It seems to be assuming the truth of the answer.
                        – Owen
                        Nov 23 at 23:19






                      • 5




                        Tongue in cheek: Isn't this almost equivalent to "P is true because the question asks me to prove it"?
                        – IanF1
                        Nov 24 at 8:01






                      • 1




                        I'm a bit puzzled why an off-topic aside ("testing degenerate cases is useful" and "regularly check your tire pressure" are equally relevant remarks to the posed problem) with a drawing and nothing else gets so many upvotes. "To disprove by example" is useless here, as we already know it's true; that's the puzzle: show it. This differs from the other proofs NOT by removing unneccessary complications but by removing the PROOF part; without a hint of a path to a proof, or any insight? Very weird.
                        – user3445853
                        Nov 24 at 20:38






                      • 2




                        If the absurdity of this isn't clear enough: Continue the logic and let the surface of the yellow square go to zero. Now both triangles are equal sized (as both are zero). So then drawing a seagreen square on its own is someway somehow corroboration that the original question is true, and the "extrapolation" of the method of proof is left to the reader?! Neat. I don't know why I ever worked for math.
                        – user3445853
                        Nov 24 at 20:40










                      • The value of this view is in showing that, if there is a constant difference between the areas of the triangle and square, then that difference is zero. (Or, if there's a constant ratio, then the ratio is one; etc.) So, had the task been to find that constant difference (or ratio, or whatever), then consulting a particularly-convenient configuration would've made for a proper solution. In any case, this serves as a sanity check on the problem.
                        – Blue
                        16 hours ago













                      up vote
                      10
                      down vote










                      up vote
                      10
                      down vote









                      While the other solutions are obviously correct, they are also unnecessarily complicated.

                      Since the angle of the squares is not specified, it must be true for all angles, so* why not pick one which is simple to work with and results in a degenerate case.



                      enter image description here



                      *) The assumption of truth is not required since we first do show that S=T is, in fact, true (in one simple case) and skirt the rules from there on by leaving the extrapolation to the reader.

                      While this approach is best suited for puzzles, looking at the edge cases first is a quick way to disprove things by example, or at least check that your calculations are correct






                      share|cite|improve this answer














                      While the other solutions are obviously correct, they are also unnecessarily complicated.

                      Since the angle of the squares is not specified, it must be true for all angles, so* why not pick one which is simple to work with and results in a degenerate case.



                      enter image description here



                      *) The assumption of truth is not required since we first do show that S=T is, in fact, true (in one simple case) and skirt the rules from there on by leaving the extrapolation to the reader.

                      While this approach is best suited for puzzles, looking at the edge cases first is a quick way to disprove things by example, or at least check that your calculations are correct







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Nov 24 at 9:49

























                      answered Nov 23 at 19:06









                      DenDenDo

                      57948




                      57948








                      • 7




                        That's a beautiful drawing, but I'm not sure if "Since the angle of the squares is not specified, it must be true for all angles" is logically sound here. It seems to be assuming the truth of the answer.
                        – Owen
                        Nov 23 at 23:19






                      • 5




                        Tongue in cheek: Isn't this almost equivalent to "P is true because the question asks me to prove it"?
                        – IanF1
                        Nov 24 at 8:01






                      • 1




                        I'm a bit puzzled why an off-topic aside ("testing degenerate cases is useful" and "regularly check your tire pressure" are equally relevant remarks to the posed problem) with a drawing and nothing else gets so many upvotes. "To disprove by example" is useless here, as we already know it's true; that's the puzzle: show it. This differs from the other proofs NOT by removing unneccessary complications but by removing the PROOF part; without a hint of a path to a proof, or any insight? Very weird.
                        – user3445853
                        Nov 24 at 20:38






                      • 2




                        If the absurdity of this isn't clear enough: Continue the logic and let the surface of the yellow square go to zero. Now both triangles are equal sized (as both are zero). So then drawing a seagreen square on its own is someway somehow corroboration that the original question is true, and the "extrapolation" of the method of proof is left to the reader?! Neat. I don't know why I ever worked for math.
                        – user3445853
                        Nov 24 at 20:40










                      • The value of this view is in showing that, if there is a constant difference between the areas of the triangle and square, then that difference is zero. (Or, if there's a constant ratio, then the ratio is one; etc.) So, had the task been to find that constant difference (or ratio, or whatever), then consulting a particularly-convenient configuration would've made for a proper solution. In any case, this serves as a sanity check on the problem.
                        – Blue
                        16 hours ago














                      • 7




                        That's a beautiful drawing, but I'm not sure if "Since the angle of the squares is not specified, it must be true for all angles" is logically sound here. It seems to be assuming the truth of the answer.
                        – Owen
                        Nov 23 at 23:19






                      • 5




                        Tongue in cheek: Isn't this almost equivalent to "P is true because the question asks me to prove it"?
                        – IanF1
                        Nov 24 at 8:01






                      • 1




                        I'm a bit puzzled why an off-topic aside ("testing degenerate cases is useful" and "regularly check your tire pressure" are equally relevant remarks to the posed problem) with a drawing and nothing else gets so many upvotes. "To disprove by example" is useless here, as we already know it's true; that's the puzzle: show it. This differs from the other proofs NOT by removing unneccessary complications but by removing the PROOF part; without a hint of a path to a proof, or any insight? Very weird.
                        – user3445853
                        Nov 24 at 20:38






                      • 2




                        If the absurdity of this isn't clear enough: Continue the logic and let the surface of the yellow square go to zero. Now both triangles are equal sized (as both are zero). So then drawing a seagreen square on its own is someway somehow corroboration that the original question is true, and the "extrapolation" of the method of proof is left to the reader?! Neat. I don't know why I ever worked for math.
                        – user3445853
                        Nov 24 at 20:40










                      • The value of this view is in showing that, if there is a constant difference between the areas of the triangle and square, then that difference is zero. (Or, if there's a constant ratio, then the ratio is one; etc.) So, had the task been to find that constant difference (or ratio, or whatever), then consulting a particularly-convenient configuration would've made for a proper solution. In any case, this serves as a sanity check on the problem.
                        – Blue
                        16 hours ago








                      7




                      7




                      That's a beautiful drawing, but I'm not sure if "Since the angle of the squares is not specified, it must be true for all angles" is logically sound here. It seems to be assuming the truth of the answer.
                      – Owen
                      Nov 23 at 23:19




                      That's a beautiful drawing, but I'm not sure if "Since the angle of the squares is not specified, it must be true for all angles" is logically sound here. It seems to be assuming the truth of the answer.
                      – Owen
                      Nov 23 at 23:19




                      5




                      5




                      Tongue in cheek: Isn't this almost equivalent to "P is true because the question asks me to prove it"?
                      – IanF1
                      Nov 24 at 8:01




                      Tongue in cheek: Isn't this almost equivalent to "P is true because the question asks me to prove it"?
                      – IanF1
                      Nov 24 at 8:01




                      1




                      1




                      I'm a bit puzzled why an off-topic aside ("testing degenerate cases is useful" and "regularly check your tire pressure" are equally relevant remarks to the posed problem) with a drawing and nothing else gets so many upvotes. "To disprove by example" is useless here, as we already know it's true; that's the puzzle: show it. This differs from the other proofs NOT by removing unneccessary complications but by removing the PROOF part; without a hint of a path to a proof, or any insight? Very weird.
                      – user3445853
                      Nov 24 at 20:38




                      I'm a bit puzzled why an off-topic aside ("testing degenerate cases is useful" and "regularly check your tire pressure" are equally relevant remarks to the posed problem) with a drawing and nothing else gets so many upvotes. "To disprove by example" is useless here, as we already know it's true; that's the puzzle: show it. This differs from the other proofs NOT by removing unneccessary complications but by removing the PROOF part; without a hint of a path to a proof, or any insight? Very weird.
                      – user3445853
                      Nov 24 at 20:38




                      2




                      2




                      If the absurdity of this isn't clear enough: Continue the logic and let the surface of the yellow square go to zero. Now both triangles are equal sized (as both are zero). So then drawing a seagreen square on its own is someway somehow corroboration that the original question is true, and the "extrapolation" of the method of proof is left to the reader?! Neat. I don't know why I ever worked for math.
                      – user3445853
                      Nov 24 at 20:40




                      If the absurdity of this isn't clear enough: Continue the logic and let the surface of the yellow square go to zero. Now both triangles are equal sized (as both are zero). So then drawing a seagreen square on its own is someway somehow corroboration that the original question is true, and the "extrapolation" of the method of proof is left to the reader?! Neat. I don't know why I ever worked for math.
                      – user3445853
                      Nov 24 at 20:40












                      The value of this view is in showing that, if there is a constant difference between the areas of the triangle and square, then that difference is zero. (Or, if there's a constant ratio, then the ratio is one; etc.) So, had the task been to find that constant difference (or ratio, or whatever), then consulting a particularly-convenient configuration would've made for a proper solution. In any case, this serves as a sanity check on the problem.
                      – Blue
                      16 hours ago




                      The value of this view is in showing that, if there is a constant difference between the areas of the triangle and square, then that difference is zero. (Or, if there's a constant ratio, then the ratio is one; etc.) So, had the task been to find that constant difference (or ratio, or whatever), then consulting a particularly-convenient configuration would've made for a proper solution. In any case, this serves as a sanity check on the problem.
                      – Blue
                      16 hours ago










                      up vote
                      1
                      down vote













                      This is a long comment.



                      The shapes $S,,T$ share a vertex $A$ where six angles meet, three of them right angles. The other three angles therefore sum to a right angle. Let $theta$ be the angle in $T$, so the two other acute angles, and the right angle between them from $S$, sum to $pi-theta$. Since $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$, where $BA,,AC$ are sides of the upper squares.



                      The challenge, then, is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.






                      share|cite|improve this answer





















                      • I don't think that $BACD$ has twice the area of $ABC$. If you try to move $ABC$ by shearing to get a triangle with side $BC$ and something on the bottom line, the third point isn't $D$.
                        – D. Thomine
                        Nov 22 at 21:35










                      • $BACD$ is a rhombus because its opposite sides are parallel. So it does have twice the area of $ABC$.
                        – I like Serena
                        Nov 22 at 21:41










                      • $BACD$ is not a rhombus.
                        – D. Thomine
                        Nov 22 at 21:42






                      • 2




                        @Servaes It's too long. "Long comment" is something people often say when their contribution has to be an "answer" because of software limitations, but is far from a complete answer.
                        – J.G.
                        Nov 23 at 10:22






                      • 1




                        I have converted your answer into a comment.
                        – TheSimpliFire
                        Nov 24 at 14:28















                      up vote
                      1
                      down vote













                      This is a long comment.



                      The shapes $S,,T$ share a vertex $A$ where six angles meet, three of them right angles. The other three angles therefore sum to a right angle. Let $theta$ be the angle in $T$, so the two other acute angles, and the right angle between them from $S$, sum to $pi-theta$. Since $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$, where $BA,,AC$ are sides of the upper squares.



                      The challenge, then, is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.






                      share|cite|improve this answer





















                      • I don't think that $BACD$ has twice the area of $ABC$. If you try to move $ABC$ by shearing to get a triangle with side $BC$ and something on the bottom line, the third point isn't $D$.
                        – D. Thomine
                        Nov 22 at 21:35










                      • $BACD$ is a rhombus because its opposite sides are parallel. So it does have twice the area of $ABC$.
                        – I like Serena
                        Nov 22 at 21:41










                      • $BACD$ is not a rhombus.
                        – D. Thomine
                        Nov 22 at 21:42






                      • 2




                        @Servaes It's too long. "Long comment" is something people often say when their contribution has to be an "answer" because of software limitations, but is far from a complete answer.
                        – J.G.
                        Nov 23 at 10:22






                      • 1




                        I have converted your answer into a comment.
                        – TheSimpliFire
                        Nov 24 at 14:28













                      up vote
                      1
                      down vote










                      up vote
                      1
                      down vote









                      This is a long comment.



                      The shapes $S,,T$ share a vertex $A$ where six angles meet, three of them right angles. The other three angles therefore sum to a right angle. Let $theta$ be the angle in $T$, so the two other acute angles, and the right angle between them from $S$, sum to $pi-theta$. Since $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$, where $BA,,AC$ are sides of the upper squares.



                      The challenge, then, is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.






                      share|cite|improve this answer












                      This is a long comment.



                      The shapes $S,,T$ share a vertex $A$ where six angles meet, three of them right angles. The other three angles therefore sum to a right angle. Let $theta$ be the angle in $T$, so the two other acute angles, and the right angle between them from $S$, sum to $pi-theta$. Since $sintheta=sin(pi-theta)$, $T$ has the same area as $triangle ABC$, where $BA,,AC$ are sides of the upper squares.



                      The challenge, then, is to show $S$ has that much area too. Let $D$ denote the vertex of $S$ opposite $A$. I suspect we can show the quadrilateral $BACD$ has twice the area of $S$, and also twice the area of $triangle ABC$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 22 at 21:11









                      J.G.

                      21.2k21933




                      21.2k21933












                      • I don't think that $BACD$ has twice the area of $ABC$. If you try to move $ABC$ by shearing to get a triangle with side $BC$ and something on the bottom line, the third point isn't $D$.
                        – D. Thomine
                        Nov 22 at 21:35










                      • $BACD$ is a rhombus because its opposite sides are parallel. So it does have twice the area of $ABC$.
                        – I like Serena
                        Nov 22 at 21:41










                      • $BACD$ is not a rhombus.
                        – D. Thomine
                        Nov 22 at 21:42






                      • 2




                        @Servaes It's too long. "Long comment" is something people often say when their contribution has to be an "answer" because of software limitations, but is far from a complete answer.
                        – J.G.
                        Nov 23 at 10:22






                      • 1




                        I have converted your answer into a comment.
                        – TheSimpliFire
                        Nov 24 at 14:28


















                      • I don't think that $BACD$ has twice the area of $ABC$. If you try to move $ABC$ by shearing to get a triangle with side $BC$ and something on the bottom line, the third point isn't $D$.
                        – D. Thomine
                        Nov 22 at 21:35










                      • $BACD$ is a rhombus because its opposite sides are parallel. So it does have twice the area of $ABC$.
                        – I like Serena
                        Nov 22 at 21:41










                      • $BACD$ is not a rhombus.
                        – D. Thomine
                        Nov 22 at 21:42






                      • 2




                        @Servaes It's too long. "Long comment" is something people often say when their contribution has to be an "answer" because of software limitations, but is far from a complete answer.
                        – J.G.
                        Nov 23 at 10:22






                      • 1




                        I have converted your answer into a comment.
                        – TheSimpliFire
                        Nov 24 at 14:28
















                      I don't think that $BACD$ has twice the area of $ABC$. If you try to move $ABC$ by shearing to get a triangle with side $BC$ and something on the bottom line, the third point isn't $D$.
                      – D. Thomine
                      Nov 22 at 21:35




                      I don't think that $BACD$ has twice the area of $ABC$. If you try to move $ABC$ by shearing to get a triangle with side $BC$ and something on the bottom line, the third point isn't $D$.
                      – D. Thomine
                      Nov 22 at 21:35












                      $BACD$ is a rhombus because its opposite sides are parallel. So it does have twice the area of $ABC$.
                      – I like Serena
                      Nov 22 at 21:41




                      $BACD$ is a rhombus because its opposite sides are parallel. So it does have twice the area of $ABC$.
                      – I like Serena
                      Nov 22 at 21:41












                      $BACD$ is not a rhombus.
                      – D. Thomine
                      Nov 22 at 21:42




                      $BACD$ is not a rhombus.
                      – D. Thomine
                      Nov 22 at 21:42




                      2




                      2




                      @Servaes It's too long. "Long comment" is something people often say when their contribution has to be an "answer" because of software limitations, but is far from a complete answer.
                      – J.G.
                      Nov 23 at 10:22




                      @Servaes It's too long. "Long comment" is something people often say when their contribution has to be an "answer" because of software limitations, but is far from a complete answer.
                      – J.G.
                      Nov 23 at 10:22




                      1




                      1




                      I have converted your answer into a comment.
                      – TheSimpliFire
                      Nov 24 at 14:28




                      I have converted your answer into a comment.
                      – TheSimpliFire
                      Nov 24 at 14:28


















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