Prove $Delta text{ABC}$ is a right triangle?
$begingroup$
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
asked Dec 9 '18 at 7:19
HaiDangelHaiDangel
695
$endgroup$
add a comment |
$begingroup$
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
asked Dec 9 '18 at 7:19
HaiDangelHaiDangel
695
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 9 '18 at 7:19
HaiDangelHaiDangel
695
$endgroup$
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
asked Dec 9 '18 at 7:19
HaiDangelHaiDangel
695
asked Dec 9 '18 at 7:19
HaiDangelHaiDangel
695
edited Dec 9 '18 at 13:19
HaiDangel
asked Dec 9 '18 at 7:19
HaiDangelHaiDangel
695
asked Dec 9 '18 at 7:19
HaiDangelHaiDangel
695
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 9 '18 at 10:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
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
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Dec 9 '18 at 10:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
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
add a comment |
$begingroup$
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.
answered Dec 9 '18 at 10:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
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
add a comment |
$begingroup$
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.
answered Dec 9 '18 at 10:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
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.
answered Dec 9 '18 at 10:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Dec 9 '18 at 10:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Dec 9 '18 at 10:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
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
add a comment |
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
add a comment |
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$begingroup$
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