geometry inequality
$begingroup$
$M$ is a point in $triangle ABC$. $AM$ intersect with $BC$ at $A_1$. $BM$ intersect with $AC$ at $B_1$. $CM$ intersect with $AB$ at $C_1$.
Proof that: $$AA_1 times BB_1 times CC_1 geq 27 (MA_1 times MB_1 times MC_1)$$
geometry inequality triangle geometric-inequalities
edited Dec 9 '18 at 12:54
Martin Sleziak
44.9k10121274
asked Dec 9 '18 at 9:42
Nguyen Xuan MinhNguyen Xuan Minh
214
$endgroup$
add a comment |
$begingroup$
$M$ is a point in $triangle ABC$. $AM$ intersect with $BC$ at $A_1$. $BM$ intersect with $AC$ at $B_1$. $CM$ intersect with $AB$ at $C_1$.
Proof that: $$AA_1 times BB_1 times CC_1 geq 27 (MA_1 times MB_1 times MC_1)$$
geometry inequality triangle geometric-inequalities
edited Dec 9 '18 at 12:54
Martin Sleziak
44.9k10121274
asked Dec 9 '18 at 9:42
Nguyen Xuan MinhNguyen Xuan Minh
214
$endgroup$
1
$begingroup$
Hint: consider areas and use AM-GM.
$endgroup$
– user10354138
Dec 9 '18 at 9:49
$begingroup$
What are your thoughts? Write your working in the edit. Please don't think it as spoon-feeding, question taking & answer giving website.
$endgroup$
– jayant98
Dec 9 '18 at 9:52
add a comment |
$begingroup$
$M$ is a point in $triangle ABC$. $AM$ intersect with $BC$ at $A_1$. $BM$ intersect with $AC$ at $B_1$. $CM$ intersect with $AB$ at $C_1$.
Proof that: $$AA_1 times BB_1 times CC_1 geq 27 (MA_1 times MB_1 times MC_1)$$
geometry inequality triangle geometric-inequalities
edited Dec 9 '18 at 12:54
Martin Sleziak
44.9k10121274
asked Dec 9 '18 at 9:42
Nguyen Xuan MinhNguyen Xuan Minh
214
$endgroup$
$M$ is a point in $triangle ABC$. $AM$ intersect with $BC$ at $A_1$. $BM$ intersect with $AC$ at $B_1$. $CM$ intersect with $AB$ at $C_1$.
Proof that: $$AA_1 times BB_1 times CC_1 geq 27 (MA_1 times MB_1 times MC_1)$$
geometry inequality triangle geometric-inequalities
geometry inequality triangle geometric-inequalities
edited Dec 9 '18 at 12:54
Martin Sleziak
44.9k10121274
asked Dec 9 '18 at 9:42
Nguyen Xuan MinhNguyen Xuan Minh
214
edited Dec 9 '18 at 12:54
Martin Sleziak
44.9k10121274
asked Dec 9 '18 at 9:42
Nguyen Xuan MinhNguyen Xuan Minh
214
edited Dec 9 '18 at 12:54
Martin Sleziak
44.9k10121274
edited Dec 9 '18 at 12:54
Martin Sleziak
44.9k10121274
edited Dec 9 '18 at 12:54
Martin Sleziak
44.9k10121274
44.9k10121274
asked Dec 9 '18 at 9:42
Nguyen Xuan MinhNguyen Xuan Minh
214
asked Dec 9 '18 at 9:42
Nguyen Xuan MinhNguyen Xuan Minh
214
asked Dec 9 '18 at 9:42
Nguyen Xuan MinhNguyen Xuan Minh
214
214
1
$begingroup$
Hint: consider areas and use AM-GM.
$endgroup$
– user10354138
Dec 9 '18 at 9:49
$begingroup$
What are your thoughts? Write your working in the edit. Please don't think it as spoon-feeding, question taking & answer giving website.
$endgroup$
– jayant98
Dec 9 '18 at 9:52
add a comment |
1
$begingroup$
Hint: consider areas and use AM-GM.
$endgroup$
– user10354138
Dec 9 '18 at 9:49
$begingroup$
What are your thoughts? Write your working in the edit. Please don't think it as spoon-feeding, question taking & answer giving website.
$endgroup$
– jayant98
Dec 9 '18 at 9:52
1
1
$begingroup$
Hint: consider areas and use AM-GM.
$endgroup$
– user10354138
Dec 9 '18 at 9:49
$begingroup$
Hint: consider areas and use AM-GM.
$endgroup$
– user10354138
Dec 9 '18 at 9:49
$begingroup$
What are your thoughts? Write your working in the edit. Please don't think it as spoon-feeding, question taking & answer giving website.
$endgroup$
– jayant98
Dec 9 '18 at 9:52
$begingroup$
What are your thoughts? Write your working in the edit. Please don't think it as spoon-feeding, question taking & answer giving website.
$endgroup$
– jayant98
Dec 9 '18 at 9:52
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$ Leftrightarrow S_{AMB}cdot S_{BMC}cdot S_{CMA}le left(dfrac{ S_{AMB}+S_{BMC}+S_{CMA}}{3} right )^3 = dfrac{1}{27}S^3 $$
answered Dec 9 '18 at 10:07
Sergic PrimazonSergic Primazon
53225
$endgroup$
add a comment |
$begingroup$
Transform the ratio of sides to ratio of the areas of triangles. Then the inequality is equivalent to:
$27abcleq(a+b+c)^3$, where $a$, $b$, and $c$ stand for the areas of three inner triangles.
edited Dec 9 '18 at 10:36
Anubhab Ghosal
1,19619
answered Dec 9 '18 at 9:51
Ziqing XUZiqing XU
11
$endgroup$
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 10:07
add a comment |
$begingroup$
By AM-GM
$$frac{MA_1cdot MB_1cdot MC_1}{AA_1cdot BB_1cdot CC_1}=frac{S_{Delta BMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMB}}{S_{Delta ABC}}leq$$
$$leqleft(frac{frac{S_{Delta BMC}}{S_{Delta ABC}}+frac{S_{Delta AMC}}{S_{Delta ABC}}+frac{S_{Delta AMB}}{S_{Delta ABC}}}{3}right)^3=frac{1}{27}.$$
answered Dec 9 '18 at 10:42
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$ Leftrightarrow S_{AMB}cdot S_{BMC}cdot S_{CMA}le left(dfrac{ S_{AMB}+S_{BMC}+S_{CMA}}{3} right )^3 = dfrac{1}{27}S^3 $$
answered Dec 9 '18 at 10:07
Sergic PrimazonSergic Primazon
53225
$endgroup$
add a comment |
$begingroup$
$$ Leftrightarrow S_{AMB}cdot S_{BMC}cdot S_{CMA}le left(dfrac{ S_{AMB}+S_{BMC}+S_{CMA}}{3} right )^3 = dfrac{1}{27}S^3 $$
answered Dec 9 '18 at 10:07
Sergic PrimazonSergic Primazon
53225
$endgroup$
add a comment |
$begingroup$
$$ Leftrightarrow S_{AMB}cdot S_{BMC}cdot S_{CMA}le left(dfrac{ S_{AMB}+S_{BMC}+S_{CMA}}{3} right )^3 = dfrac{1}{27}S^3 $$
answered Dec 9 '18 at 10:07
Sergic PrimazonSergic Primazon
53225
$endgroup$
$$ Leftrightarrow S_{AMB}cdot S_{BMC}cdot S_{CMA}le left(dfrac{ S_{AMB}+S_{BMC}+S_{CMA}}{3} right )^3 = dfrac{1}{27}S^3 $$
answered Dec 9 '18 at 10:07
Sergic PrimazonSergic Primazon
53225
answered Dec 9 '18 at 10:07
Sergic PrimazonSergic Primazon
53225
answered Dec 9 '18 at 10:07
Sergic PrimazonSergic Primazon
53225
answered Dec 9 '18 at 10:07
Sergic PrimazonSergic Primazon
53225
53225
add a comment |
add a comment |
$begingroup$
Transform the ratio of sides to ratio of the areas of triangles. Then the inequality is equivalent to:
$27abcleq(a+b+c)^3$, where $a$, $b$, and $c$ stand for the areas of three inner triangles.
edited Dec 9 '18 at 10:36
Anubhab Ghosal
1,19619
answered Dec 9 '18 at 9:51
Ziqing XUZiqing XU
11
$endgroup$
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 10:07
add a comment |
$begingroup$
Transform the ratio of sides to ratio of the areas of triangles. Then the inequality is equivalent to:
$27abcleq(a+b+c)^3$, where $a$, $b$, and $c$ stand for the areas of three inner triangles.
edited Dec 9 '18 at 10:36
Anubhab Ghosal
1,19619
answered Dec 9 '18 at 9:51
Ziqing XUZiqing XU
11
$endgroup$
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 10:07
add a comment |
$begingroup$
Transform the ratio of sides to ratio of the areas of triangles. Then the inequality is equivalent to:
$27abcleq(a+b+c)^3$, where $a$, $b$, and $c$ stand for the areas of three inner triangles.
edited Dec 9 '18 at 10:36
Anubhab Ghosal
1,19619
answered Dec 9 '18 at 9:51
Ziqing XUZiqing XU
11
$endgroup$
Transform the ratio of sides to ratio of the areas of triangles. Then the inequality is equivalent to:
$27abcleq(a+b+c)^3$, where $a$, $b$, and $c$ stand for the areas of three inner triangles.
edited Dec 9 '18 at 10:36
Anubhab Ghosal
1,19619
answered Dec 9 '18 at 9:51
Ziqing XUZiqing XU
11
edited Dec 9 '18 at 10:36
Anubhab Ghosal
1,19619
edited Dec 9 '18 at 10:36
Anubhab Ghosal
1,19619
edited Dec 9 '18 at 10:36
Anubhab Ghosal
1,19619
1,19619
answered Dec 9 '18 at 9:51
Ziqing XUZiqing XU
11
answered Dec 9 '18 at 9:51
Ziqing XUZiqing XU
11
answered Dec 9 '18 at 9:51
Ziqing XUZiqing XU
11
11
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 10:07
add a comment |
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 10:07
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 10:07
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 10:07
add a comment |
$begingroup$
By AM-GM
$$frac{MA_1cdot MB_1cdot MC_1}{AA_1cdot BB_1cdot CC_1}=frac{S_{Delta BMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMB}}{S_{Delta ABC}}leq$$
$$leqleft(frac{frac{S_{Delta BMC}}{S_{Delta ABC}}+frac{S_{Delta AMC}}{S_{Delta ABC}}+frac{S_{Delta AMB}}{S_{Delta ABC}}}{3}right)^3=frac{1}{27}.$$
answered Dec 9 '18 at 10:42
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
By AM-GM
$$frac{MA_1cdot MB_1cdot MC_1}{AA_1cdot BB_1cdot CC_1}=frac{S_{Delta BMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMB}}{S_{Delta ABC}}leq$$
$$leqleft(frac{frac{S_{Delta BMC}}{S_{Delta ABC}}+frac{S_{Delta AMC}}{S_{Delta ABC}}+frac{S_{Delta AMB}}{S_{Delta ABC}}}{3}right)^3=frac{1}{27}.$$
answered Dec 9 '18 at 10:42
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
By AM-GM
$$frac{MA_1cdot MB_1cdot MC_1}{AA_1cdot BB_1cdot CC_1}=frac{S_{Delta BMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMB}}{S_{Delta ABC}}leq$$
$$leqleft(frac{frac{S_{Delta BMC}}{S_{Delta ABC}}+frac{S_{Delta AMC}}{S_{Delta ABC}}+frac{S_{Delta AMB}}{S_{Delta ABC}}}{3}right)^3=frac{1}{27}.$$
answered Dec 9 '18 at 10:42
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
By AM-GM
$$frac{MA_1cdot MB_1cdot MC_1}{AA_1cdot BB_1cdot CC_1}=frac{S_{Delta BMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMC}}{S_{Delta ABC}}cdotfrac{S_{Delta AMB}}{S_{Delta ABC}}leq$$
$$leqleft(frac{frac{S_{Delta BMC}}{S_{Delta ABC}}+frac{S_{Delta AMC}}{S_{Delta ABC}}+frac{S_{Delta AMB}}{S_{Delta ABC}}}{3}right)^3=frac{1}{27}.$$
answered Dec 9 '18 at 10:42
Michael RozenbergMichael Rozenberg
108k1895200
answered Dec 9 '18 at 10:42
Michael RozenbergMichael Rozenberg
108k1895200
answered Dec 9 '18 at 10:42
Michael RozenbergMichael Rozenberg
108k1895200
answered Dec 9 '18 at 10:42
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
add a comment |
add a comment |
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$begingroup$
Hint: consider areas and use AM-GM.
$endgroup$
– user10354138
Dec 9 '18 at 9:49
$begingroup$
What are your thoughts? Write your working in the edit. Please don't think it as spoon-feeding, question taking & answer giving website.
$endgroup$
– jayant98
Dec 9 '18 at 9:52