Mixing
In acute $triangle ABC$, show $DE+DF leq BC$, where $D$, $E$, $F$ are the feet of the altitudes from $A$,...
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This question already has an answer here:
BM01 2011/12 Question 6 Geometry Problem
1 answer
Let $triangle ABC$ be an acute angled triangle. The feet of the altitudes from $A$, $B$, and $C$ are $D$, $E$, and $F$, respectively. Prove that
$$DE+DF leq BC$$
and determine the triangles for which equality holds.
(Note: The altitude from A is the line through A which is perpendicular to BC. The foot of this altitude is the point D where it meets BC. The other altitudes are similarly defined. )
geometry euclidean-geometry geometric-transformation
edited Jan 19 at 18:59
greedoid
45.2k1158113
asked Jan 19 at 18:13
John AburiJohn Aburi
142
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marked as duplicate by Misha Lavrov, Did, Shaun, Yanko, hunter Jan 27 at 18:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
BM01 2011/12 Question 6 Geometry Problem
1 answer
Let $triangle ABC$ be an acute angled triangle. The feet of the altitudes from $A$, $B$, and $C$ are $D$, $E$, and $F$, respectively. Prove that
$$DE+DF leq BC$$
and determine the triangles for which equality holds.
(Note: The altitude from A is the line through A which is perpendicular to BC. The foot of this altitude is the point D where it meets BC. The other altitudes are similarly defined. )
geometry euclidean-geometry geometric-transformation
edited Jan 19 at 18:59
greedoid
45.2k1158113
asked Jan 19 at 18:13
John AburiJohn Aburi
142
$endgroup$
marked as duplicate by Misha Lavrov, Did, Shaun, Yanko, hunter Jan 27 at 18:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Welcome to Math.SE! The community here typically frowns upon questions that are nothing more than isolated problem statements. Help us help you. Tell us what you know about the problem, and/or where you got stuck. (Put such information in the question itself, not in comments that may not get read.) This information helps answerers target their responses to your skill level and specific difficulty, without wasting anyone's time telling you things you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.)
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– Blue
Jan 19 at 18:21
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This question already has an answer here:
BM01 2011/12 Question 6 Geometry Problem
1 answer
Let $triangle ABC$ be an acute angled triangle. The feet of the altitudes from $A$, $B$, and $C$ are $D$, $E$, and $F$, respectively. Prove that
$$DE+DF leq BC$$
and determine the triangles for which equality holds.
(Note: The altitude from A is the line through A which is perpendicular to BC. The foot of this altitude is the point D where it meets BC. The other altitudes are similarly defined. )
geometry euclidean-geometry geometric-transformation
edited Jan 19 at 18:59
greedoid
45.2k1158113
asked Jan 19 at 18:13
John AburiJohn Aburi
142
$endgroup$
This question already has an answer here:
BM01 2011/12 Question 6 Geometry Problem
1 answer
Let $triangle ABC$ be an acute angled triangle. The feet of the altitudes from $A$, $B$, and $C$ are $D$, $E$, and $F$, respectively. Prove that
$$DE+DF leq BC$$
and determine the triangles for which equality holds.
(Note: The altitude from A is the line through A which is perpendicular to BC. The foot of this altitude is the point D where it meets BC. The other altitudes are similarly defined. )
This question already has an answer here:
BM01 2011/12 Question 6 Geometry Problem
1 answer
geometry euclidean-geometry geometric-transformation
geometry euclidean-geometry geometric-transformation
edited Jan 19 at 18:59
greedoid
45.2k1158113
asked Jan 19 at 18:13
John AburiJohn Aburi
142
edited Jan 19 at 18:59
greedoid
45.2k1158113
asked Jan 19 at 18:13
John AburiJohn Aburi
142
edited Jan 19 at 18:59
greedoid
45.2k1158113
edited Jan 19 at 18:59
greedoid
45.2k1158113
edited Jan 19 at 18:59
greedoid
45.2k1158113
45.2k1158113
asked Jan 19 at 18:13
John AburiJohn Aburi
142
asked Jan 19 at 18:13
John AburiJohn Aburi
142
asked Jan 19 at 18:13
John AburiJohn Aburi
142
142
marked as duplicate by Misha Lavrov, Did, Shaun, Yanko, hunter Jan 27 at 18:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Misha Lavrov, Did, Shaun, Yanko, hunter Jan 27 at 18:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Welcome to Math.SE! The community here typically frowns upon questions that are nothing more than isolated problem statements. Help us help you. Tell us what you know about the problem, and/or where you got stuck. (Put such information in the question itself, not in comments that may not get read.) This information helps answerers target their responses to your skill level and specific difficulty, without wasting anyone's time telling you things you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.)
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– Blue
Jan 19 at 18:21
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Welcome to Math.SE! The community here typically frowns upon questions that are nothing more than isolated problem statements. Help us help you. Tell us what you know about the problem, and/or where you got stuck. (Put such information in the question itself, not in comments that may not get read.) This information helps answerers target their responses to your skill level and specific difficulty, without wasting anyone's time telling you things you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.)
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– Blue
Jan 19 at 18:21
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Welcome to Math.SE! The community here typically frowns upon questions that are nothing more than isolated problem statements. Help us help you. Tell us what you know about the problem, and/or where you got stuck. (Put such information in the question itself, not in comments that may not get read.) This information helps answerers target their responses to your skill level and specific difficulty, without wasting anyone's time telling you things you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.)
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– Blue
Jan 19 at 18:21
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Welcome to Math.SE! The community here typically frowns upon questions that are nothing more than isolated problem statements. Help us help you. Tell us what you know about the problem, and/or where you got stuck. (Put such information in the question itself, not in comments that may not get read.) This information helps answerers target their responses to your skill level and specific difficulty, without wasting anyone's time telling you things you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.)
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– Blue
Jan 19 at 18:21
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2 Answers
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Due to the existence of the 9-point circle, the circumradius of the orthic triangle $DEF$ equals $frac{R}{2}$.
Since the angles of the orthic triangle equal $pi-2A,pi-2B,pi-2C$, your inequality is equivalent to
$$ sin(2B)+sin(2C)leq 2sin(A) $$
or to
$$ 2sin(B)cos(B)+2sin(C)cos(C)leq 2sin(B)cos(C)+2sin(C)cos(B) $$
or to
$$ (cos B-cos C)(sin B-sin C) leq 0 $$
which is trivial since over the interval $left(0,frac{pi}{2}right)$ the sine function is increasing and the cosine function is decreasing. Equality holds only for isosceles triangles.
answered Jan 19 at 18:29
Jack D'AurizioJack D'Aurizio
290k33282664
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Let $F'$ be a reflection of $F$ across $BC$, so $FD = F'D$. Since $BC$ is diameter of circle around cyclic quadrilateral $BCEF$ and $angle FDA = angle EDA$ we see that $F'$ is on this circle and that $F',D,E$ are colinear. Since $BC$ is diameter we have $$BCgeq F'E = F'D+DE = FD+DE$$
answered Jan 19 at 18:36
greedoidgreedoid
45.2k1158113
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1
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(+1) very elegant.
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– Jack D'Aurizio
Jan 19 at 18:56
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What is the reason for ∠FDA=∠EDA ?
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– John Aburi
Jan 20 at 17:26
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Is it necessary to give a reason for F′,D,E are collinear?
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– John Aburi
Jan 20 at 17:27
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For the first question, look at cyclic quadrilateral $BCEF$ and for the second, yes it is necesery, this means F'D,E are on one chord and it is less then diameter.
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– greedoid
Jan 21 at 16:16
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Due to the existence of the 9-point circle, the circumradius of the orthic triangle $DEF$ equals $frac{R}{2}$.
Since the angles of the orthic triangle equal $pi-2A,pi-2B,pi-2C$, your inequality is equivalent to
$$ sin(2B)+sin(2C)leq 2sin(A) $$
or to
$$ 2sin(B)cos(B)+2sin(C)cos(C)leq 2sin(B)cos(C)+2sin(C)cos(B) $$
or to
$$ (cos B-cos C)(sin B-sin C) leq 0 $$
which is trivial since over the interval $left(0,frac{pi}{2}right)$ the sine function is increasing and the cosine function is decreasing. Equality holds only for isosceles triangles.
answered Jan 19 at 18:29
Jack D'AurizioJack D'Aurizio
290k33282664
$endgroup$
add a comment |
$begingroup$
Due to the existence of the 9-point circle, the circumradius of the orthic triangle $DEF$ equals $frac{R}{2}$.
Since the angles of the orthic triangle equal $pi-2A,pi-2B,pi-2C$, your inequality is equivalent to
$$ sin(2B)+sin(2C)leq 2sin(A) $$
or to
$$ 2sin(B)cos(B)+2sin(C)cos(C)leq 2sin(B)cos(C)+2sin(C)cos(B) $$
or to
$$ (cos B-cos C)(sin B-sin C) leq 0 $$
which is trivial since over the interval $left(0,frac{pi}{2}right)$ the sine function is increasing and the cosine function is decreasing. Equality holds only for isosceles triangles.
answered Jan 19 at 18:29
Jack D'AurizioJack D'Aurizio
290k33282664
$endgroup$
add a comment |
$begingroup$
Due to the existence of the 9-point circle, the circumradius of the orthic triangle $DEF$ equals $frac{R}{2}$.
Since the angles of the orthic triangle equal $pi-2A,pi-2B,pi-2C$, your inequality is equivalent to
$$ sin(2B)+sin(2C)leq 2sin(A) $$
or to
$$ 2sin(B)cos(B)+2sin(C)cos(C)leq 2sin(B)cos(C)+2sin(C)cos(B) $$
or to
$$ (cos B-cos C)(sin B-sin C) leq 0 $$
which is trivial since over the interval $left(0,frac{pi}{2}right)$ the sine function is increasing and the cosine function is decreasing. Equality holds only for isosceles triangles.
answered Jan 19 at 18:29
Jack D'AurizioJack D'Aurizio
290k33282664
$endgroup$
Due to the existence of the 9-point circle, the circumradius of the orthic triangle $DEF$ equals $frac{R}{2}$.
Since the angles of the orthic triangle equal $pi-2A,pi-2B,pi-2C$, your inequality is equivalent to
$$ sin(2B)+sin(2C)leq 2sin(A) $$
or to
$$ 2sin(B)cos(B)+2sin(C)cos(C)leq 2sin(B)cos(C)+2sin(C)cos(B) $$
or to
$$ (cos B-cos C)(sin B-sin C) leq 0 $$
which is trivial since over the interval $left(0,frac{pi}{2}right)$ the sine function is increasing and the cosine function is decreasing. Equality holds only for isosceles triangles.
answered Jan 19 at 18:29
Jack D'AurizioJack D'Aurizio
290k33282664
answered Jan 19 at 18:29
Jack D'AurizioJack D'Aurizio
290k33282664
answered Jan 19 at 18:29
Jack D'AurizioJack D'Aurizio
290k33282664
answered Jan 19 at 18:29
Jack D'AurizioJack D'Aurizio
290k33282664
290k33282664
add a comment |
add a comment |
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Let $F'$ be a reflection of $F$ across $BC$, so $FD = F'D$. Since $BC$ is diameter of circle around cyclic quadrilateral $BCEF$ and $angle FDA = angle EDA$ we see that $F'$ is on this circle and that $F',D,E$ are colinear. Since $BC$ is diameter we have $$BCgeq F'E = F'D+DE = FD+DE$$
answered Jan 19 at 18:36
greedoidgreedoid
45.2k1158113
$endgroup$
1
$begingroup$
(+1) very elegant.
$endgroup$
– Jack D'Aurizio
Jan 19 at 18:56
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What is the reason for ∠FDA=∠EDA ?
$endgroup$
– John Aburi
Jan 20 at 17:26
$begingroup$
Is it necessary to give a reason for F′,D,E are collinear?
$endgroup$
– John Aburi
Jan 20 at 17:27
$begingroup$
For the first question, look at cyclic quadrilateral $BCEF$ and for the second, yes it is necesery, this means F'D,E are on one chord and it is less then diameter.
$endgroup$
– greedoid
Jan 21 at 16:16
add a comment |
$begingroup$
Let $F'$ be a reflection of $F$ across $BC$, so $FD = F'D$. Since $BC$ is diameter of circle around cyclic quadrilateral $BCEF$ and $angle FDA = angle EDA$ we see that $F'$ is on this circle and that $F',D,E$ are colinear. Since $BC$ is diameter we have $$BCgeq F'E = F'D+DE = FD+DE$$
answered Jan 19 at 18:36
greedoidgreedoid
45.2k1158113
$endgroup$
1
$begingroup$
(+1) very elegant.
$endgroup$
– Jack D'Aurizio
Jan 19 at 18:56
$begingroup$
What is the reason for ∠FDA=∠EDA ?
$endgroup$
– John Aburi
Jan 20 at 17:26
$begingroup$
Is it necessary to give a reason for F′,D,E are collinear?
$endgroup$
– John Aburi
Jan 20 at 17:27
$begingroup$
For the first question, look at cyclic quadrilateral $BCEF$ and for the second, yes it is necesery, this means F'D,E are on one chord and it is less then diameter.
$endgroup$
– greedoid
Jan 21 at 16:16
add a comment |
$begingroup$
Let $F'$ be a reflection of $F$ across $BC$, so $FD = F'D$. Since $BC$ is diameter of circle around cyclic quadrilateral $BCEF$ and $angle FDA = angle EDA$ we see that $F'$ is on this circle and that $F',D,E$ are colinear. Since $BC$ is diameter we have $$BCgeq F'E = F'D+DE = FD+DE$$
answered Jan 19 at 18:36
greedoidgreedoid
45.2k1158113
$endgroup$
Let $F'$ be a reflection of $F$ across $BC$, so $FD = F'D$. Since $BC$ is diameter of circle around cyclic quadrilateral $BCEF$ and $angle FDA = angle EDA$ we see that $F'$ is on this circle and that $F',D,E$ are colinear. Since $BC$ is diameter we have $$BCgeq F'E = F'D+DE = FD+DE$$
answered Jan 19 at 18:36
greedoidgreedoid
45.2k1158113
edited Jan 19 at 19:01
answered Jan 19 at 18:36
greedoidgreedoid
45.2k1158113
answered Jan 19 at 18:36
greedoidgreedoid
45.2k1158113
answered Jan 19 at 18:36
greedoidgreedoid
45.2k1158113
45.2k1158113
1
$begingroup$
(+1) very elegant.
$endgroup$
– Jack D'Aurizio
Jan 19 at 18:56
$begingroup$
What is the reason for ∠FDA=∠EDA ?
$endgroup$
– John Aburi
Jan 20 at 17:26
$begingroup$
Is it necessary to give a reason for F′,D,E are collinear?
$endgroup$
– John Aburi
Jan 20 at 17:27
$begingroup$
For the first question, look at cyclic quadrilateral $BCEF$ and for the second, yes it is necesery, this means F'D,E are on one chord and it is less then diameter.
$endgroup$
– greedoid
Jan 21 at 16:16
add a comment |
1
$begingroup$
(+1) very elegant.
$endgroup$
– Jack D'Aurizio
Jan 19 at 18:56
$begingroup$
What is the reason for ∠FDA=∠EDA ?
$endgroup$
– John Aburi
Jan 20 at 17:26
$begingroup$
Is it necessary to give a reason for F′,D,E are collinear?
$endgroup$
– John Aburi
Jan 20 at 17:27
$begingroup$
For the first question, look at cyclic quadrilateral $BCEF$ and for the second, yes it is necesery, this means F'D,E are on one chord and it is less then diameter.
$endgroup$
– greedoid
Jan 21 at 16:16
1
1
$begingroup$
(+1) very elegant.
$endgroup$
– Jack D'Aurizio
Jan 19 at 18:56
$begingroup$
(+1) very elegant.
$endgroup$
– Jack D'Aurizio
Jan 19 at 18:56
$begingroup$
What is the reason for ∠FDA=∠EDA ?
$endgroup$
– John Aburi
Jan 20 at 17:26
$begingroup$
What is the reason for ∠FDA=∠EDA ?
$endgroup$
– John Aburi
Jan 20 at 17:26
$begingroup$
Is it necessary to give a reason for F′,D,E are collinear?
$endgroup$
– John Aburi
Jan 20 at 17:27
$begingroup$
Is it necessary to give a reason for F′,D,E are collinear?
$endgroup$
– John Aburi
Jan 20 at 17:27
$begingroup$
For the first question, look at cyclic quadrilateral $BCEF$ and for the second, yes it is necesery, this means F'D,E are on one chord and it is less then diameter.
$endgroup$
– greedoid
Jan 21 at 16:16
$begingroup$
For the first question, look at cyclic quadrilateral $BCEF$ and for the second, yes it is necesery, this means F'D,E are on one chord and it is less then diameter.
$endgroup$
– greedoid
Jan 21 at 16:16
add a comment |
$begingroup$
Welcome to Math.SE! The community here typically frowns upon questions that are nothing more than isolated problem statements. Help us help you. Tell us what you know about the problem, and/or where you got stuck. (Put such information in the question itself, not in comments that may not get read.) This information helps answerers target their responses to your skill level and specific difficulty, without wasting anyone's time telling you things you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.)
$endgroup$
– Blue
Jan 19 at 18:21