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












2












$begingroup$


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











$endgroup$












  • $begingroup$
    Do you mean "for all $k$ with $|k|leq3,$" or "for some $k$ with $|k|leq3,$"?
    $endgroup$
    – Christian Blatter
    Dec 9 '18 at 11:01






  • 1




    $begingroup$
    I mean for all $text{k}$ ! Thanks !
    $endgroup$
    – HaiDangel
    Dec 9 '18 at 12:33
















2












$begingroup$


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











$endgroup$












  • $begingroup$
    Do you mean "for all $k$ with $|k|leq3,$" or "for some $k$ with $|k|leq3,$"?
    $endgroup$
    – Christian Blatter
    Dec 9 '18 at 11:01






  • 1




    $begingroup$
    I mean for all $text{k}$ ! Thanks !
    $endgroup$
    – HaiDangel
    Dec 9 '18 at 12:33














2












2








2


3



$begingroup$


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











$endgroup$




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













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 13:19







HaiDangel

















asked Dec 9 '18 at 7:19









HaiDangelHaiDangel

695




695












  • $begingroup$
    Do you mean "for all $k$ with $|k|leq3,$" or "for some $k$ with $|k|leq3,$"?
    $endgroup$
    – Christian Blatter
    Dec 9 '18 at 11:01






  • 1




    $begingroup$
    I mean for all $text{k}$ ! Thanks !
    $endgroup$
    – HaiDangel
    Dec 9 '18 at 12:33


















  • $begingroup$
    Do you mean "for all $k$ with $|k|leq3,$" or "for some $k$ with $|k|leq3,$"?
    $endgroup$
    – Christian Blatter
    Dec 9 '18 at 11:01






  • 1




    $begingroup$
    I mean for all $text{k}$ ! Thanks !
    $endgroup$
    – HaiDangel
    Dec 9 '18 at 12:33
















$begingroup$
Do you mean "for all $k$ with $|k|leq3,$" or "for some $k$ with $|k|leq3,$"?
$endgroup$
– Christian Blatter
Dec 9 '18 at 11:01




$begingroup$
Do you mean "for all $k$ with $|k|leq3,$" or "for some $k$ with $|k|leq3,$"?
$endgroup$
– Christian Blatter
Dec 9 '18 at 11:01




1




1




$begingroup$
I mean for all $text{k}$ ! Thanks !
$endgroup$
– HaiDangel
Dec 9 '18 at 12:33




$begingroup$
I mean for all $text{k}$ ! Thanks !
$endgroup$
– HaiDangel
Dec 9 '18 at 12:33










1 Answer
1






active

oldest

votes


















1












$begingroup$

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









$endgroup$









  • 1




    $begingroup$
    Can you try to solve the rest on your own ? I think your solution is the neatest !
    $endgroup$
    – HaiDangel
    Jan 31 at 3:14











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1 Answer
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1 Answer
1






active

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active

oldest

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active

oldest

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1












$begingroup$

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









$endgroup$









  • 1




    $begingroup$
    Can you try to solve the rest on your own ? I think your solution is the neatest !
    $endgroup$
    – HaiDangel
    Jan 31 at 3:14
















1












$begingroup$

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









$endgroup$









  • 1




    $begingroup$
    Can you try to solve the rest on your own ? I think your solution is the neatest !
    $endgroup$
    – HaiDangel
    Jan 31 at 3:14














1












1








1





$begingroup$

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









$endgroup$



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 '18 at 10:12









Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866




77.8k42866








  • 1




    $begingroup$
    Can you try to solve the rest on your own ? I think your solution is the neatest !
    $endgroup$
    – HaiDangel
    Jan 31 at 3:14














  • 1




    $begingroup$
    Can you try to solve the rest on your own ? I think your solution is the neatest !
    $endgroup$
    – HaiDangel
    Jan 31 at 3:14








1




1




$begingroup$
Can you try to solve the rest on your own ? I think your solution is the neatest !
$endgroup$
– HaiDangel
Jan 31 at 3:14




$begingroup$
Can you try to solve the rest on your own ? I think your solution is the neatest !
$endgroup$
– HaiDangel
Jan 31 at 3:14


















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