$dfrac{cos alpha}{cos (beta-gamma)}$ [on hold]
I have a minimum and maximum problem connected to trigonometry. Please help, it is from an American olympiad from 2013.
Suppose that the angles of a triangle are $alpha, beta$ and $gamma$. Let $H$ be the set $left{dfrac{cos alpha}{cos (beta-gamma)}, dfrac{cos beta}{cos (gamma-alpha)}, dfrac{cos gamma}{cos (alpha-beta)}right}$.
a) Find the minimal possible value of the largest item of $H$.
b) Find the maximal possible value of the smallest item of $H$.
geometry trigonometry contest-math maxima-minima
edited 2 days ago
user376343
2,7882822
asked 2 days ago
Peter Boss
623
put on hold as off-topic by amWhy, Eevee Trainer, Paul Frost, Carl Schildkraut, KReiser 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Eevee Trainer, Paul Frost, Carl Schildkraut, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I have a minimum and maximum problem connected to trigonometry. Please help, it is from an American olympiad from 2013.
Suppose that the angles of a triangle are $alpha, beta$ and $gamma$. Let $H$ be the set $left{dfrac{cos alpha}{cos (beta-gamma)}, dfrac{cos beta}{cos (gamma-alpha)}, dfrac{cos gamma}{cos (alpha-beta)}right}$.
a) Find the minimal possible value of the largest item of $H$.
b) Find the maximal possible value of the smallest item of $H$.
geometry trigonometry contest-math maxima-minima
edited 2 days ago
user376343
2,7882822
asked 2 days ago
Peter Boss
623
put on hold as off-topic by amWhy, Eevee Trainer, Paul Frost, Carl Schildkraut, KReiser 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Eevee Trainer, Paul Frost, Carl Schildkraut, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
2
Without bothering to do any math, relying purely by instinct, my guess is that the equilateral triangle is relevant to both questions. Therefore, my first approach would be to attempt to prove this.
– user2661923
2 days ago
1
Welcome to MSE. To help us provide the most useful & educational answer for you, please let us know what you've tried already, any particular difficulties you encountered, etc. Thanks.
– John Omielan
2 days ago
add a comment |
I have a minimum and maximum problem connected to trigonometry. Please help, it is from an American olympiad from 2013.
Suppose that the angles of a triangle are $alpha, beta$ and $gamma$. Let $H$ be the set $left{dfrac{cos alpha}{cos (beta-gamma)}, dfrac{cos beta}{cos (gamma-alpha)}, dfrac{cos gamma}{cos (alpha-beta)}right}$.
a) Find the minimal possible value of the largest item of $H$.
b) Find the maximal possible value of the smallest item of $H$.
geometry trigonometry contest-math maxima-minima
edited 2 days ago
user376343
2,7882822
asked 2 days ago
Peter Boss
623
I have a minimum and maximum problem connected to trigonometry. Please help, it is from an American olympiad from 2013.
Suppose that the angles of a triangle are $alpha, beta$ and $gamma$. Let $H$ be the set $left{dfrac{cos alpha}{cos (beta-gamma)}, dfrac{cos beta}{cos (gamma-alpha)}, dfrac{cos gamma}{cos (alpha-beta)}right}$.
a) Find the minimal possible value of the largest item of $H$.
b) Find the maximal possible value of the smallest item of $H$.
geometry trigonometry contest-math maxima-minima
geometry trigonometry contest-math maxima-minima
edited 2 days ago
user376343
2,7882822
asked 2 days ago
Peter Boss
623
edited 2 days ago
user376343
2,7882822
asked 2 days ago
Peter Boss
623
edited 2 days ago
user376343
2,7882822
edited 2 days ago
user376343
2,7882822
edited 2 days ago
user376343
2,7882822
2,7882822
asked 2 days ago
Peter Boss
623
asked 2 days ago
Peter Boss
623
asked 2 days ago
Peter Boss
623
623
put on hold as off-topic by amWhy, Eevee Trainer, Paul Frost, Carl Schildkraut, KReiser 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Eevee Trainer, Paul Frost, Carl Schildkraut, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by amWhy, Eevee Trainer, Paul Frost, Carl Schildkraut, KReiser 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Eevee Trainer, Paul Frost, Carl Schildkraut, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
2
Without bothering to do any math, relying purely by instinct, my guess is that the equilateral triangle is relevant to both questions. Therefore, my first approach would be to attempt to prove this.
– user2661923
2 days ago
1
Welcome to MSE. To help us provide the most useful & educational answer for you, please let us know what you've tried already, any particular difficulties you encountered, etc. Thanks.
– John Omielan
2 days ago
add a comment |
2
Without bothering to do any math, relying purely by instinct, my guess is that the equilateral triangle is relevant to both questions. Therefore, my first approach would be to attempt to prove this.
– user2661923
2 days ago
1
Welcome to MSE. To help us provide the most useful & educational answer for you, please let us know what you've tried already, any particular difficulties you encountered, etc. Thanks.
– John Omielan
2 days ago
2
2
Without bothering to do any math, relying purely by instinct, my guess is that the equilateral triangle is relevant to both questions. Therefore, my first approach would be to attempt to prove this.
– user2661923
2 days ago
Without bothering to do any math, relying purely by instinct, my guess is that the equilateral triangle is relevant to both questions. Therefore, my first approach would be to attempt to prove this.
– user2661923
2 days ago
1
1
Welcome to MSE. To help us provide the most useful & educational answer for you, please let us know what you've tried already, any particular difficulties you encountered, etc. Thanks.
– John Omielan
2 days ago
Welcome to MSE. To help us provide the most useful & educational answer for you, please let us know what you've tried already, any particular difficulties you encountered, etc. Thanks.
– John Omielan
2 days ago
add a comment |
1 Answer
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WLOG, let $A=alphage B=betage C=gamma$
$cfrac {cos A}{cos (B-C)}=-cfrac{cos(B+C)}{cos(B-C)}=1-cfrac 2{1+tan Btan C}tag 1$
If $A=pi/2$, $cfrac {cos A}{cos (B-C)}=0, cfrac {cos B}{cos (C-A)}=cfrac {cos B}{cos B}=1=cfrac {cos C}{cos (A-B)}$. Hence minMax=$1$, maxMin=$0$;
If $A>pi/2$, from $(1)$, it's easy to see the maximum in $H$ is $1-cfrac 2{1+tan A tan B}$, and the minimum is $1-cfrac 2{1+tan B tan C}$. When $Ato pi/2$, minMax $to 1$; when $B=C to pi/4$, maxMin $to 0$;
If $A<pi/2$, maximum in H is $1-cfrac 2{1+tan A tan B}$, and the minimum is $1-cfrac 2{1+tan B tan C}$. It can be easily shown mininum of maximum is $1/2$ when $A=B=pi/3$ and maximum of minimum is $1/2$ when $B=C=pi/3$.
Combine all cases, we have min max=max min = $1/2$ when $A=B=C=pi/3$
PS: $f(x)=1-cfrac 2{1+x}, x>0$, as $x$ increases, $cfrac 2{1+x}$ decreases, and $f(x)$ increases. the minimum of $tan A tan B (pi/2>Age Bge Cge pi/3)$ occurs when $A=B=pi/3$. etc, etc.
answered 2 days ago
Lance
61229
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
WLOG, let $A=alphage B=betage C=gamma$
$cfrac {cos A}{cos (B-C)}=-cfrac{cos(B+C)}{cos(B-C)}=1-cfrac 2{1+tan Btan C}tag 1$
If $A=pi/2$, $cfrac {cos A}{cos (B-C)}=0, cfrac {cos B}{cos (C-A)}=cfrac {cos B}{cos B}=1=cfrac {cos C}{cos (A-B)}$. Hence minMax=$1$, maxMin=$0$;
If $A>pi/2$, from $(1)$, it's easy to see the maximum in $H$ is $1-cfrac 2{1+tan A tan B}$, and the minimum is $1-cfrac 2{1+tan B tan C}$. When $Ato pi/2$, minMax $to 1$; when $B=C to pi/4$, maxMin $to 0$;
If $A<pi/2$, maximum in H is $1-cfrac 2{1+tan A tan B}$, and the minimum is $1-cfrac 2{1+tan B tan C}$. It can be easily shown mininum of maximum is $1/2$ when $A=B=pi/3$ and maximum of minimum is $1/2$ when $B=C=pi/3$.
Combine all cases, we have min max=max min = $1/2$ when $A=B=C=pi/3$
PS: $f(x)=1-cfrac 2{1+x}, x>0$, as $x$ increases, $cfrac 2{1+x}$ decreases, and $f(x)$ increases. the minimum of $tan A tan B (pi/2>Age Bge Cge pi/3)$ occurs when $A=B=pi/3$. etc, etc.
answered 2 days ago
Lance
61229
add a comment |
WLOG, let $A=alphage B=betage C=gamma$
$cfrac {cos A}{cos (B-C)}=-cfrac{cos(B+C)}{cos(B-C)}=1-cfrac 2{1+tan Btan C}tag 1$
If $A=pi/2$, $cfrac {cos A}{cos (B-C)}=0, cfrac {cos B}{cos (C-A)}=cfrac {cos B}{cos B}=1=cfrac {cos C}{cos (A-B)}$. Hence minMax=$1$, maxMin=$0$;
If $A>pi/2$, from $(1)$, it's easy to see the maximum in $H$ is $1-cfrac 2{1+tan A tan B}$, and the minimum is $1-cfrac 2{1+tan B tan C}$. When $Ato pi/2$, minMax $to 1$; when $B=C to pi/4$, maxMin $to 0$;
If $A<pi/2$, maximum in H is $1-cfrac 2{1+tan A tan B}$, and the minimum is $1-cfrac 2{1+tan B tan C}$. It can be easily shown mininum of maximum is $1/2$ when $A=B=pi/3$ and maximum of minimum is $1/2$ when $B=C=pi/3$.
Combine all cases, we have min max=max min = $1/2$ when $A=B=C=pi/3$
PS: $f(x)=1-cfrac 2{1+x}, x>0$, as $x$ increases, $cfrac 2{1+x}$ decreases, and $f(x)$ increases. the minimum of $tan A tan B (pi/2>Age Bge Cge pi/3)$ occurs when $A=B=pi/3$. etc, etc.
answered 2 days ago
Lance
61229
add a comment |
WLOG, let $A=alphage B=betage C=gamma$
$cfrac {cos A}{cos (B-C)}=-cfrac{cos(B+C)}{cos(B-C)}=1-cfrac 2{1+tan Btan C}tag 1$
If $A=pi/2$, $cfrac {cos A}{cos (B-C)}=0, cfrac {cos B}{cos (C-A)}=cfrac {cos B}{cos B}=1=cfrac {cos C}{cos (A-B)}$. Hence minMax=$1$, maxMin=$0$;
If $A>pi/2$, from $(1)$, it's easy to see the maximum in $H$ is $1-cfrac 2{1+tan A tan B}$, and the minimum is $1-cfrac 2{1+tan B tan C}$. When $Ato pi/2$, minMax $to 1$; when $B=C to pi/4$, maxMin $to 0$;
If $A<pi/2$, maximum in H is $1-cfrac 2{1+tan A tan B}$, and the minimum is $1-cfrac 2{1+tan B tan C}$. It can be easily shown mininum of maximum is $1/2$ when $A=B=pi/3$ and maximum of minimum is $1/2$ when $B=C=pi/3$.
Combine all cases, we have min max=max min = $1/2$ when $A=B=C=pi/3$
PS: $f(x)=1-cfrac 2{1+x}, x>0$, as $x$ increases, $cfrac 2{1+x}$ decreases, and $f(x)$ increases. the minimum of $tan A tan B (pi/2>Age Bge Cge pi/3)$ occurs when $A=B=pi/3$. etc, etc.
answered 2 days ago
Lance
61229
WLOG, let $A=alphage B=betage C=gamma$
$cfrac {cos A}{cos (B-C)}=-cfrac{cos(B+C)}{cos(B-C)}=1-cfrac 2{1+tan Btan C}tag 1$
If $A=pi/2$, $cfrac {cos A}{cos (B-C)}=0, cfrac {cos B}{cos (C-A)}=cfrac {cos B}{cos B}=1=cfrac {cos C}{cos (A-B)}$. Hence minMax=$1$, maxMin=$0$;
If $A>pi/2$, from $(1)$, it's easy to see the maximum in $H$ is $1-cfrac 2{1+tan A tan B}$, and the minimum is $1-cfrac 2{1+tan B tan C}$. When $Ato pi/2$, minMax $to 1$; when $B=C to pi/4$, maxMin $to 0$;
If $A<pi/2$, maximum in H is $1-cfrac 2{1+tan A tan B}$, and the minimum is $1-cfrac 2{1+tan B tan C}$. It can be easily shown mininum of maximum is $1/2$ when $A=B=pi/3$ and maximum of minimum is $1/2$ when $B=C=pi/3$.
Combine all cases, we have min max=max min = $1/2$ when $A=B=C=pi/3$
PS: $f(x)=1-cfrac 2{1+x}, x>0$, as $x$ increases, $cfrac 2{1+x}$ decreases, and $f(x)$ increases. the minimum of $tan A tan B (pi/2>Age Bge Cge pi/3)$ occurs when $A=B=pi/3$. etc, etc.
answered 2 days ago
Lance
61229
edited 2 days ago
answered 2 days ago
Lance
61229
answered 2 days ago
Lance
61229
answered 2 days ago
Lance
61229
61229
add a comment |
add a comment |
2
Without bothering to do any math, relying purely by instinct, my guess is that the equilateral triangle is relevant to both questions. Therefore, my first approach would be to attempt to prove this.
– user2661923
2 days ago
1
Welcome to MSE. To help us provide the most useful & educational answer for you, please let us know what you've tried already, any particular difficulties you encountered, etc. Thanks.
– John Omielan
2 days ago