Prove $Delta text{ABC}$ is a right triangle?












3














Prove $Delta text{ABC}$ is a right triangle if:
$$left{begin{matrix} sin^{2},text{A}+ sin^{2},text{B}= sin text{C}\ maxleft { measuredangle text{A}- text{k},measuredangle text{B},,measuredangle text{B}- text{k},measuredangle text{A} right }leqq frac{pi }{2}\ left | text{k} right |leqq 3 end{matrix}right.$$
I have a proof for my problem with $text{k}= 0$. See here:



$lceil$ https://diendantoanhoc.net/topic/185093-measuredangle-textc-pi/#entry716757 $rfloor$



I also have another solution but ugly, I try to use similar method with $left | text{k} right |leqq 3$ but without success !






geometry trigonometry inequality







share|cite|improve this question
























  • Do you mean "for all $k$ with $|k|leq3,$" or "for some $k$ with $|k|leq3,$"?
    – Christian Blatter
    Dec 9 at 11:01






  • 1




    I mean for all $text{k}$ ! Thanks !
    – HaiDangel
    Dec 9 at 12:33
















3














Prove $Delta text{ABC}$ is a right triangle if:
$$left{begin{matrix} sin^{2},text{A}+ sin^{2},text{B}= sin text{C}\ maxleft { measuredangle text{A}- text{k},measuredangle text{B},,measuredangle text{B}- text{k},measuredangle text{A} right }leqq frac{pi }{2}\ left | text{k} right |leqq 3 end{matrix}right.$$
I have a proof for my problem with $text{k}= 0$. See here:



$lceil$ https://diendantoanhoc.net/topic/185093-measuredangle-textc-pi/#entry716757 $rfloor$



I also have another solution but ugly, I try to use similar method with $left | text{k} right |leqq 3$ but without success !






geometry trigonometry inequality







share|cite|improve this question
























  • Do you mean "for all $k$ with $|k|leq3,$" or "for some $k$ with $|k|leq3,$"?
    – Christian Blatter
    Dec 9 at 11:01






  • 1




    I mean for all $text{k}$ ! Thanks !
    – HaiDangel
    Dec 9 at 12:33














3












3








3


4





Prove $Delta text{ABC}$ is a right triangle if:
$$left{begin{matrix} sin^{2},text{A}+ sin^{2},text{B}= sin text{C}\ maxleft { measuredangle text{A}- text{k},measuredangle text{B},,measuredangle text{B}- text{k},measuredangle text{A} right }leqq frac{pi }{2}\ left | text{k} right |leqq 3 end{matrix}right.$$
I have a proof for my problem with $text{k}= 0$. See here:



$lceil$ https://diendantoanhoc.net/topic/185093-measuredangle-textc-pi/#entry716757 $rfloor$



I also have another solution but ugly, I try to use similar method with $left | text{k} right |leqq 3$ but without success !






geometry trigonometry inequality







share|cite|improve this question















Prove $Delta text{ABC}$ is a right triangle if:
$$left{begin{matrix} sin^{2},text{A}+ sin^{2},text{B}= sin text{C}\ maxleft { measuredangle text{A}- text{k},measuredangle text{B},,measuredangle text{B}- text{k},measuredangle text{A} right }leqq frac{pi }{2}\ left | text{k} right |leqq 3 end{matrix}right.$$
I have a proof for my problem with $text{k}= 0$. See here:



$lceil$ https://diendantoanhoc.net/topic/185093-measuredangle-textc-pi/#entry716757 $rfloor$



I also have another solution but ugly, I try to use similar method with $left | text{k} right |leqq 3$ but without success !






geometry trigonometry inequality




geometry trigonometry inequality






share|cite|improve this question















share|cite|improve this question













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edited Dec 9 at 13:19

























asked Dec 9 at 7:19









HaiDangel

213




213












  • Do you mean "for all $k$ with $|k|leq3,$" or "for some $k$ with $|k|leq3,$"?
    – Christian Blatter
    Dec 9 at 11:01






  • 1




    I mean for all $text{k}$ ! Thanks !
    – HaiDangel
    Dec 9 at 12:33


















  • Do you mean "for all $k$ with $|k|leq3,$" or "for some $k$ with $|k|leq3,$"?
    – Christian Blatter
    Dec 9 at 11:01






  • 1




    I mean for all $text{k}$ ! Thanks !
    – HaiDangel
    Dec 9 at 12:33
















Do you mean "for all $k$ with $|k|leq3,$" or "for some $k$ with $|k|leq3,$"?
– Christian Blatter
Dec 9 at 11:01




Do you mean "for all $k$ with $|k|leq3,$" or "for some $k$ with $|k|leq3,$"?
– Christian Blatter
Dec 9 at 11:01




1




1




I mean for all $text{k}$ ! Thanks !
– HaiDangel
Dec 9 at 12:33




I mean for all $text{k}$ ! Thanks !
– HaiDangel
Dec 9 at 12:33










2 Answers
2






active

oldest

votes


















0














One idea: Using the formulas $$A=frac{1}{2}absin(gamma)$$ etc we get



$$frac{4A^2}{b^2c^2}+frac{4A^2}{a^2c^2}=frac{2A}{ab}$$ and using $$A=sqrt{s(s-a)(s-b)(s-c)}$$ where $$s=frac{a+b+c}{2}$$ plugging this in our formula we get after squaring and simplifying



$$- left( {a}^{6}-{a}^{4}{b}^{2}-{a}^{4}{c}^{2}-{a}^{2}{b}^{4}-6,{a}^{
2}{b}^{2}{c}^{2}+{b}^{6}-{b}^{4}{c}^{2} right) left( {a}^{2}+{b}^{2
}-{c}^{2} right)
=0$$
it remaines to show that the first factor can not be zero.






share|cite|improve this answer





























    0














    If I read the second condition as
    $$alpha-kbetaleq{piover2},quadbeta-kalphaleq{piover2}qquad forall >kin[-3,3] ,$$
    and add these two inequalities in the case $k:=-3$ I obtain
    $$4(alpha+beta)leq pi ,$$
    or $alpha+betaleq{piover4}$. I don't see how a right triangle can be formed in this way, even without looking at the first condition. Note that in the linked source they are only talking about the case $k=3$.






    share|cite|improve this answer

















    • 1




      $left | text{k} right |leqq 3$ is acceptable ! With $left | text{k} right |leqq 3$, for example: $text{A}= 4,text{B}+ frac{5,pi}{11},,text{C}= frac{6,pi}{11}- 5,text{B},,left | text{k} right |= 4Rightarrow text{B}neq frac{pi}{110} Rightarrow text{C}neq pi$ ! I understand your comment & I think $left | text{k} right |leqq 3$ is neccessary for my problem ! $text{k}= 3$ is my original problem, then I generalised it to $left | text{k} right |leqq 3$ , those are all of mine on Vietnamese Math Forum ( Diendantoanhoc.net ) ! Thanks for your interest !
      – HaiDangel
      Dec 10 at 1:19






    • 1




      There are some misunderstandings about my problem here: $lceil$ artofproblemsolving.com/community/c6h1725878p11175691 $rfloor$
      – HaiDangel
      Dec 10 at 1:29













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

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    0














    One idea: Using the formulas $$A=frac{1}{2}absin(gamma)$$ etc we get



    $$frac{4A^2}{b^2c^2}+frac{4A^2}{a^2c^2}=frac{2A}{ab}$$ and using $$A=sqrt{s(s-a)(s-b)(s-c)}$$ where $$s=frac{a+b+c}{2}$$ plugging this in our formula we get after squaring and simplifying



    $$- left( {a}^{6}-{a}^{4}{b}^{2}-{a}^{4}{c}^{2}-{a}^{2}{b}^{4}-6,{a}^{
    2}{b}^{2}{c}^{2}+{b}^{6}-{b}^{4}{c}^{2} right) left( {a}^{2}+{b}^{2
    }-{c}^{2} right)
    =0$$
    it remaines to show that the first factor can not be zero.






    share|cite|improve this answer


























      0














      One idea: Using the formulas $$A=frac{1}{2}absin(gamma)$$ etc we get



      $$frac{4A^2}{b^2c^2}+frac{4A^2}{a^2c^2}=frac{2A}{ab}$$ and using $$A=sqrt{s(s-a)(s-b)(s-c)}$$ where $$s=frac{a+b+c}{2}$$ plugging this in our formula we get after squaring and simplifying



      $$- left( {a}^{6}-{a}^{4}{b}^{2}-{a}^{4}{c}^{2}-{a}^{2}{b}^{4}-6,{a}^{
      2}{b}^{2}{c}^{2}+{b}^{6}-{b}^{4}{c}^{2} right) left( {a}^{2}+{b}^{2
      }-{c}^{2} right)
      =0$$
      it remaines to show that the first factor can not be zero.






      share|cite|improve this answer
























        0












        0








        0






        One idea: Using the formulas $$A=frac{1}{2}absin(gamma)$$ etc we get



        $$frac{4A^2}{b^2c^2}+frac{4A^2}{a^2c^2}=frac{2A}{ab}$$ and using $$A=sqrt{s(s-a)(s-b)(s-c)}$$ where $$s=frac{a+b+c}{2}$$ plugging this in our formula we get after squaring and simplifying



        $$- left( {a}^{6}-{a}^{4}{b}^{2}-{a}^{4}{c}^{2}-{a}^{2}{b}^{4}-6,{a}^{
        2}{b}^{2}{c}^{2}+{b}^{6}-{b}^{4}{c}^{2} right) left( {a}^{2}+{b}^{2
        }-{c}^{2} right)
        =0$$
        it remaines to show that the first factor can not be zero.






        share|cite|improve this answer












        One idea: Using the formulas $$A=frac{1}{2}absin(gamma)$$ etc we get



        $$frac{4A^2}{b^2c^2}+frac{4A^2}{a^2c^2}=frac{2A}{ab}$$ and using $$A=sqrt{s(s-a)(s-b)(s-c)}$$ where $$s=frac{a+b+c}{2}$$ plugging this in our formula we get after squaring and simplifying



        $$- left( {a}^{6}-{a}^{4}{b}^{2}-{a}^{4}{c}^{2}-{a}^{2}{b}^{4}-6,{a}^{
        2}{b}^{2}{c}^{2}+{b}^{6}-{b}^{4}{c}^{2} right) left( {a}^{2}+{b}^{2
        }-{c}^{2} right)
        =0$$
        it remaines to show that the first factor can not be zero.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 at 10:12









        Dr. Sonnhard Graubner

        73k42865




        73k42865























            0














            If I read the second condition as
            $$alpha-kbetaleq{piover2},quadbeta-kalphaleq{piover2}qquad forall >kin[-3,3] ,$$
            and add these two inequalities in the case $k:=-3$ I obtain
            $$4(alpha+beta)leq pi ,$$
            or $alpha+betaleq{piover4}$. I don't see how a right triangle can be formed in this way, even without looking at the first condition. Note that in the linked source they are only talking about the case $k=3$.






            share|cite|improve this answer

















            • 1




              $left | text{k} right |leqq 3$ is acceptable ! With $left | text{k} right |leqq 3$, for example: $text{A}= 4,text{B}+ frac{5,pi}{11},,text{C}= frac{6,pi}{11}- 5,text{B},,left | text{k} right |= 4Rightarrow text{B}neq frac{pi}{110} Rightarrow text{C}neq pi$ ! I understand your comment & I think $left | text{k} right |leqq 3$ is neccessary for my problem ! $text{k}= 3$ is my original problem, then I generalised it to $left | text{k} right |leqq 3$ , those are all of mine on Vietnamese Math Forum ( Diendantoanhoc.net ) ! Thanks for your interest !
              – HaiDangel
              Dec 10 at 1:19






            • 1




              There are some misunderstandings about my problem here: $lceil$ artofproblemsolving.com/community/c6h1725878p11175691 $rfloor$
              – HaiDangel
              Dec 10 at 1:29


















            0














            If I read the second condition as
            $$alpha-kbetaleq{piover2},quadbeta-kalphaleq{piover2}qquad forall >kin[-3,3] ,$$
            and add these two inequalities in the case $k:=-3$ I obtain
            $$4(alpha+beta)leq pi ,$$
            or $alpha+betaleq{piover4}$. I don't see how a right triangle can be formed in this way, even without looking at the first condition. Note that in the linked source they are only talking about the case $k=3$.






            share|cite|improve this answer

















            • 1




              $left | text{k} right |leqq 3$ is acceptable ! With $left | text{k} right |leqq 3$, for example: $text{A}= 4,text{B}+ frac{5,pi}{11},,text{C}= frac{6,pi}{11}- 5,text{B},,left | text{k} right |= 4Rightarrow text{B}neq frac{pi}{110} Rightarrow text{C}neq pi$ ! I understand your comment & I think $left | text{k} right |leqq 3$ is neccessary for my problem ! $text{k}= 3$ is my original problem, then I generalised it to $left | text{k} right |leqq 3$ , those are all of mine on Vietnamese Math Forum ( Diendantoanhoc.net ) ! Thanks for your interest !
              – HaiDangel
              Dec 10 at 1:19






            • 1




              There are some misunderstandings about my problem here: $lceil$ artofproblemsolving.com/community/c6h1725878p11175691 $rfloor$
              – HaiDangel
              Dec 10 at 1:29
















            0












            0








            0






            If I read the second condition as
            $$alpha-kbetaleq{piover2},quadbeta-kalphaleq{piover2}qquad forall >kin[-3,3] ,$$
            and add these two inequalities in the case $k:=-3$ I obtain
            $$4(alpha+beta)leq pi ,$$
            or $alpha+betaleq{piover4}$. I don't see how a right triangle can be formed in this way, even without looking at the first condition. Note that in the linked source they are only talking about the case $k=3$.






            share|cite|improve this answer












            If I read the second condition as
            $$alpha-kbetaleq{piover2},quadbeta-kalphaleq{piover2}qquad forall >kin[-3,3] ,$$
            and add these two inequalities in the case $k:=-3$ I obtain
            $$4(alpha+beta)leq pi ,$$
            or $alpha+betaleq{piover4}$. I don't see how a right triangle can be formed in this way, even without looking at the first condition. Note that in the linked source they are only talking about the case $k=3$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 9 at 12:00









            Christian Blatter

            172k7112325




            172k7112325








            • 1




              $left | text{k} right |leqq 3$ is acceptable ! With $left | text{k} right |leqq 3$, for example: $text{A}= 4,text{B}+ frac{5,pi}{11},,text{C}= frac{6,pi}{11}- 5,text{B},,left | text{k} right |= 4Rightarrow text{B}neq frac{pi}{110} Rightarrow text{C}neq pi$ ! I understand your comment & I think $left | text{k} right |leqq 3$ is neccessary for my problem ! $text{k}= 3$ is my original problem, then I generalised it to $left | text{k} right |leqq 3$ , those are all of mine on Vietnamese Math Forum ( Diendantoanhoc.net ) ! Thanks for your interest !
              – HaiDangel
              Dec 10 at 1:19






            • 1




              There are some misunderstandings about my problem here: $lceil$ artofproblemsolving.com/community/c6h1725878p11175691 $rfloor$
              – HaiDangel
              Dec 10 at 1:29
















            • 1




              $left | text{k} right |leqq 3$ is acceptable ! With $left | text{k} right |leqq 3$, for example: $text{A}= 4,text{B}+ frac{5,pi}{11},,text{C}= frac{6,pi}{11}- 5,text{B},,left | text{k} right |= 4Rightarrow text{B}neq frac{pi}{110} Rightarrow text{C}neq pi$ ! I understand your comment & I think $left | text{k} right |leqq 3$ is neccessary for my problem ! $text{k}= 3$ is my original problem, then I generalised it to $left | text{k} right |leqq 3$ , those are all of mine on Vietnamese Math Forum ( Diendantoanhoc.net ) ! Thanks for your interest !
              – HaiDangel
              Dec 10 at 1:19






            • 1




              There are some misunderstandings about my problem here: $lceil$ artofproblemsolving.com/community/c6h1725878p11175691 $rfloor$
              – HaiDangel
              Dec 10 at 1:29










            1




            1




            $left | text{k} right |leqq 3$ is acceptable ! With $left | text{k} right |leqq 3$, for example: $text{A}= 4,text{B}+ frac{5,pi}{11},,text{C}= frac{6,pi}{11}- 5,text{B},,left | text{k} right |= 4Rightarrow text{B}neq frac{pi}{110} Rightarrow text{C}neq pi$ ! I understand your comment & I think $left | text{k} right |leqq 3$ is neccessary for my problem ! $text{k}= 3$ is my original problem, then I generalised it to $left | text{k} right |leqq 3$ , those are all of mine on Vietnamese Math Forum ( Diendantoanhoc.net ) ! Thanks for your interest !
            – HaiDangel
            Dec 10 at 1:19




            $left | text{k} right |leqq 3$ is acceptable ! With $left | text{k} right |leqq 3$, for example: $text{A}= 4,text{B}+ frac{5,pi}{11},,text{C}= frac{6,pi}{11}- 5,text{B},,left | text{k} right |= 4Rightarrow text{B}neq frac{pi}{110} Rightarrow text{C}neq pi$ ! I understand your comment & I think $left | text{k} right |leqq 3$ is neccessary for my problem ! $text{k}= 3$ is my original problem, then I generalised it to $left | text{k} right |leqq 3$ , those are all of mine on Vietnamese Math Forum ( Diendantoanhoc.net ) ! Thanks for your interest !
            – HaiDangel
            Dec 10 at 1:19




            1




            1




            There are some misunderstandings about my problem here: $lceil$ artofproblemsolving.com/community/c6h1725878p11175691 $rfloor$
            – HaiDangel
            Dec 10 at 1:29






            There are some misunderstandings about my problem here: $lceil$ artofproblemsolving.com/community/c6h1725878p11175691 $rfloor$
            – HaiDangel
            Dec 10 at 1:29




















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