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Find the value of $S$ if $S = xover y + yover z + zover x = yover x + zover y + xover z$ and $x + y + z = 0$

2

0

$begingroup$

Find the value of $S$ if $$S = xover y + yover z + zover x = yover x + zover y + xover z$$ and $x + y + z = 0$.

This is a difficult question in my opinion and I was wondering if I could get some help on it. I tried to multiply by $xyz$ but that didn't help. Any help would be appreciated.

algebra-precalculus contest-math fractions

share|cite|improve this question

edited Mar 26 at 20:40

Maria Mazur

50.1k1361125

asked Mar 26 at 7:46

user587054user587054

59211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please name the source of the problem -- which contest is it from, if any? Even "I heard it make its rounds at IMO training in New Zealand" is better than nothing.
  $endgroup$
  – darij grinberg
  Apr 2 at 15:05
  
  

  
  
  
  

add a comment | 

2

0

$begingroup$

Find the value of $S$ if $$S = xover y + yover z + zover x = yover x + zover y + xover z$$ and $x + y + z = 0$.

This is a difficult question in my opinion and I was wondering if I could get some help on it. I tried to multiply by $xyz$ but that didn't help. Any help would be appreciated.

algebra-precalculus contest-math fractions

share|cite|improve this question

edited Mar 26 at 20:40

Maria Mazur

50.1k1361125

asked Mar 26 at 7:46

user587054user587054

59211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please name the source of the problem -- which contest is it from, if any? Even "I heard it make its rounds at IMO training in New Zealand" is better than nothing.
  $endgroup$
  – darij grinberg
  Apr 2 at 15:05
  
  

  
  
  
  

add a comment | 

$begingroup$

Find the value of $S$ if $$S = xover y + yover z + zover x = yover x + zover y + xover z$$ and $x + y + z = 0$.

This is a difficult question in my opinion and I was wondering if I could get some help on it. I tried to multiply by $xyz$ but that didn't help. Any help would be appreciated.

algebra-precalculus contest-math fractions

share|cite|improve this question

edited Mar 26 at 20:40

Maria Mazur

50.1k1361125

asked Mar 26 at 7:46

user587054user587054

59211

$endgroup$

share|cite|improve this question

edited Mar 26 at 20:40

Maria Mazur

50.1k1361125

asked Mar 26 at 7:46

user587054user587054

59211

share|cite|improve this question

edited Mar 26 at 20:40

Maria Mazur

50.1k1361125

asked Mar 26 at 7:46

user587054user587054

59211

edited Mar 26 at 20:40

Maria Mazur

50.1k1361125

edited Mar 26 at 20:40

Maria Mazur

50.1k1361125

asked Mar 26 at 7:46

user587054user587054

59211

asked Mar 26 at 7:46

user587054user587054

59211

3 Answers
3

active

oldest

votes

10

$begingroup$

Note that the sum of any two is the negative of third.

Now if we add two times S then we have $$2S = xover y + yover z + zover x + yover x + zover y + xover z$$
$$ =x+yover z +y+zover x +x+zover y = $$
$$ = -zover z+-xover x +-yover y = -3$$

So $$ S = -3over 2$$

share|cite|improve this answer

edited Mar 28 at 20:19

answered Mar 26 at 7:54

Maria MazurMaria Mazur

50.1k1361125

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Really really elegant :)
  $endgroup$
  – Eureka
  Mar 27 at 20:04
  

  
  
  
  

add a comment | 

1

$begingroup$

$$0=sum_cycleft(fracxy-fracyxright)=fracsumlimits_cyc(x^2z-x^2y)xyz=frac(x-y)(y-z)(z-x)xyz.$$
Since our conditions does not depend on any cyclic permutation of the variables,

we can assume that $x=y$.

Thus, $z=-2y$ and
$$S=1-frac12-2=-frac32.$$

share|cite|improve this answer

answered Mar 26 at 8:32

Michael RozenbergMichael Rozenberg

111k1896201

$endgroup$

add a comment | 

0

$begingroup$

Denote: $y=ax, z=bx$. Then:
$$x+y+z=0 iff x+ax+bx=0 iff a=-1-b;\
S=xover y + yover z + zover x = yover x + zover y + xover z iff \
S=1over a + aover b + b = a + bover a + 1over b iff \
S=-1over 1+b - 1+bover b + b = -1-b - bover 1+b + 1over b iff \
frac(b+2)(2b+1)(b-1)b(1+b)=0 iff \
b_1,2,3=colorred-2,colorgreen-frac12,colorblue1; a_1,2,3=colorred1,colorgreen-frac12,colorblue-2\
S_1=frac1a+frac ab+b=frac1colorred1+fraccolorred1colorred-2+(colorred-2)=-frac32.\
S_2=frac1a+frac ab+b=frac1colorgreen-frac12+fraccolorgreen-frac12colorgreen-frac12+(colorgreen-frac12)=-frac32.\
S_3=frac1a+frac ab+b=frac1colorblue-2+fraccolorblue-2colorblue1+colorblue1=-frac32.\
$$

share|cite|improve this answer

answered Mar 27 at 16:26

farruhotafarruhota

22.2k2942

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  there is one upvote and one downvote so far. wondering why downvote - long?
  $endgroup$
  – farruhota
  Apr 5 at 5:53
  

  
  
  
  

add a comment | 

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10

$begingroup$

Note that the sum of any two is the negative of third.

Now if we add two times S then we have $$2S = xover y + yover z + zover x + yover x + zover y + xover z$$
$$ =x+yover z +y+zover x +x+zover y = $$
$$ = -zover z+-xover x +-yover y = -3$$

So $$ S = -3over 2$$

share|cite|improve this answer

edited Mar 28 at 20:19

answered Mar 26 at 7:54

Maria MazurMaria Mazur

50.1k1361125

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Really really elegant :)
  $endgroup$
  – Eureka
  Mar 27 at 20:04
  

  
  
  
  

add a comment | 

10

$begingroup$

Note that the sum of any two is the negative of third.

Now if we add two times S then we have $$2S = xover y + yover z + zover x + yover x + zover y + xover z$$
$$ =x+yover z +y+zover x +x+zover y = $$
$$ = -zover z+-xover x +-yover y = -3$$

So $$ S = -3over 2$$

share|cite|improve this answer

edited Mar 28 at 20:19

answered Mar 26 at 7:54

Maria MazurMaria Mazur

50.1k1361125

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Really really elegant :)
  $endgroup$
  – Eureka
  Mar 27 at 20:04
  

  
  
  
  

add a comment | 

$begingroup$

Note that the sum of any two is the negative of third.

Now if we add two times S then we have $$2S = xover y + yover z + zover x + yover x + zover y + xover z$$
$$ =x+yover z +y+zover x +x+zover y = $$
$$ = -zover z+-xover x +-yover y = -3$$

So $$ S = -3over 2$$

share|cite|improve this answer

edited Mar 28 at 20:19

answered Mar 26 at 7:54

Maria MazurMaria Mazur

50.1k1361125

$endgroup$

share|cite|improve this answer

edited Mar 28 at 20:19

answered Mar 26 at 7:54

Maria MazurMaria Mazur

50.1k1361125

answered Mar 26 at 7:54

Maria MazurMaria Mazur

50.1k1361125

answered Mar 26 at 7:54

Maria MazurMaria Mazur

50.1k1361125

1

$begingroup$

$$0=sum_cycleft(fracxy-fracyxright)=fracsumlimits_cyc(x^2z-x^2y)xyz=frac(x-y)(y-z)(z-x)xyz.$$
Since our conditions does not depend on any cyclic permutation of the variables,

we can assume that $x=y$.

Thus, $z=-2y$ and
$$S=1-frac12-2=-frac32.$$

share|cite|improve this answer

answered Mar 26 at 8:32

Michael RozenbergMichael Rozenberg

111k1896201

$endgroup$

add a comment | 

1

$begingroup$

$$0=sum_cycleft(fracxy-fracyxright)=fracsumlimits_cyc(x^2z-x^2y)xyz=frac(x-y)(y-z)(z-x)xyz.$$
Since our conditions does not depend on any cyclic permutation of the variables,

we can assume that $x=y$.

Thus, $z=-2y$ and
$$S=1-frac12-2=-frac32.$$

share|cite|improve this answer

answered Mar 26 at 8:32

Michael RozenbergMichael Rozenberg

111k1896201

$endgroup$

add a comment | 

$begingroup$

$$0=sum_cycleft(fracxy-fracyxright)=fracsumlimits_cyc(x^2z-x^2y)xyz=frac(x-y)(y-z)(z-x)xyz.$$
Since our conditions does not depend on any cyclic permutation of the variables,

we can assume that $x=y$.

Thus, $z=-2y$ and
$$S=1-frac12-2=-frac32.$$

share|cite|improve this answer

answered Mar 26 at 8:32

Michael RozenbergMichael Rozenberg

111k1896201

$endgroup$

share|cite|improve this answer

answered Mar 26 at 8:32

Michael RozenbergMichael Rozenberg

111k1896201

answered Mar 26 at 8:32

Michael RozenbergMichael Rozenberg

111k1896201

answered Mar 26 at 8:32

Michael RozenbergMichael Rozenberg

111k1896201

0

$begingroup$

Denote: $y=ax, z=bx$. Then:
$$x+y+z=0 iff x+ax+bx=0 iff a=-1-b;\
S=xover y + yover z + zover x = yover x + zover y + xover z iff \
S=1over a + aover b + b = a + bover a + 1over b iff \
S=-1over 1+b - 1+bover b + b = -1-b - bover 1+b + 1over b iff \
frac(b+2)(2b+1)(b-1)b(1+b)=0 iff \
b_1,2,3=colorred-2,colorgreen-frac12,colorblue1; a_1,2,3=colorred1,colorgreen-frac12,colorblue-2\
S_1=frac1a+frac ab+b=frac1colorred1+fraccolorred1colorred-2+(colorred-2)=-frac32.\
S_2=frac1a+frac ab+b=frac1colorgreen-frac12+fraccolorgreen-frac12colorgreen-frac12+(colorgreen-frac12)=-frac32.\
S_3=frac1a+frac ab+b=frac1colorblue-2+fraccolorblue-2colorblue1+colorblue1=-frac32.\
$$

share|cite|improve this answer

answered Mar 27 at 16:26

farruhotafarruhota

22.2k2942

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  there is one upvote and one downvote so far. wondering why downvote - long?
  $endgroup$
  – farruhota
  Apr 5 at 5:53
  

  
  
  
  

add a comment | 

0

$begingroup$

Denote: $y=ax, z=bx$. Then:
$$x+y+z=0 iff x+ax+bx=0 iff a=-1-b;\
S=xover y + yover z + zover x = yover x + zover y + xover z iff \
S=1over a + aover b + b = a + bover a + 1over b iff \
S=-1over 1+b - 1+bover b + b = -1-b - bover 1+b + 1over b iff \
frac(b+2)(2b+1)(b-1)b(1+b)=0 iff \
b_1,2,3=colorred-2,colorgreen-frac12,colorblue1; a_1,2,3=colorred1,colorgreen-frac12,colorblue-2\
S_1=frac1a+frac ab+b=frac1colorred1+fraccolorred1colorred-2+(colorred-2)=-frac32.\
S_2=frac1a+frac ab+b=frac1colorgreen-frac12+fraccolorgreen-frac12colorgreen-frac12+(colorgreen-frac12)=-frac32.\
S_3=frac1a+frac ab+b=frac1colorblue-2+fraccolorblue-2colorblue1+colorblue1=-frac32.\
$$

share|cite|improve this answer

answered Mar 27 at 16:26

farruhotafarruhota

22.2k2942

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  there is one upvote and one downvote so far. wondering why downvote - long?
  $endgroup$
  – farruhota
  Apr 5 at 5:53
  

  
  
  
  

add a comment | 

$begingroup$

Denote: $y=ax, z=bx$. Then:
$$x+y+z=0 iff x+ax+bx=0 iff a=-1-b;\
S=xover y + yover z + zover x = yover x + zover y + xover z iff \
S=1over a + aover b + b = a + bover a + 1over b iff \
S=-1over 1+b - 1+bover b + b = -1-b - bover 1+b + 1over b iff \
frac(b+2)(2b+1)(b-1)b(1+b)=0 iff \
b_1,2,3=colorred-2,colorgreen-frac12,colorblue1; a_1,2,3=colorred1,colorgreen-frac12,colorblue-2\
S_1=frac1a+frac ab+b=frac1colorred1+fraccolorred1colorred-2+(colorred-2)=-frac32.\
S_2=frac1a+frac ab+b=frac1colorgreen-frac12+fraccolorgreen-frac12colorgreen-frac12+(colorgreen-frac12)=-frac32.\
S_3=frac1a+frac ab+b=frac1colorblue-2+fraccolorblue-2colorblue1+colorblue1=-frac32.\
$$

share|cite|improve this answer

answered Mar 27 at 16:26

farruhotafarruhota

22.2k2942

$endgroup$

share|cite|improve this answer

answered Mar 27 at 16:26

farruhotafarruhota

22.2k2942

answered Mar 27 at 16:26

farruhotafarruhota

22.2k2942

answered Mar 27 at 16:26

farruhotafarruhota

22.2k2942