If $a^2+b^{2 }= c^2$ what is the value of $a^2-b^{2 }$ in terms of $c$?
Math Olympiad Question:
If $a^2+b^{2 }= c^2$ what is the value of $a^2-b^{2 }$ in terms of $c$?
I was solving some problems of the Indian National Mathematics Olympiad and I found this problem to be very intriguing. At first look it looks pretty straight forward but this problem requires the solution to be in terms of $c$ only, meaning that if you have the value of c, put that in your solution and you should get the same value as $a^2-b^{2 }$.
I tried simple algebra, trigonometry, geometry and pattern recognition but none of these methods were quite successful and could give a reasonable proof + solution.
My question is how to find the solution and prove this question? Is it possible to use Geometry and solve?
EDIT -
Sorry my bad, I think $a>b$ and $a$ and $b$ are integer values as in the starting of the exam it's written that all values will be integer and non-negative.
FINAL EDIT:
I think this question has been misprinted. Sorry for the inconvenience.
algebra-precalculus
asked Nov 25 '18 at 9:05
strangeindian
455
|
show 3 more comments
Math Olympiad Question:
If $a^2+b^{2 }= c^2$ what is the value of $a^2-b^{2 }$ in terms of $c$?
I was solving some problems of the Indian National Mathematics Olympiad and I found this problem to be very intriguing. At first look it looks pretty straight forward but this problem requires the solution to be in terms of $c$ only, meaning that if you have the value of c, put that in your solution and you should get the same value as $a^2-b^{2 }$.
I tried simple algebra, trigonometry, geometry and pattern recognition but none of these methods were quite successful and could give a reasonable proof + solution.
My question is how to find the solution and prove this question? Is it possible to use Geometry and solve?
EDIT -
Sorry my bad, I think $a>b$ and $a$ and $b$ are integer values as in the starting of the exam it's written that all values will be integer and non-negative.
FINAL EDIT:
I think this question has been misprinted. Sorry for the inconvenience.
algebra-precalculus
asked Nov 25 '18 at 9:05
strangeindian
455
8
Are you sure this is possible? I don't think $a^2 - b^2$ is well-defined because let's say $c = 10$ then we can have $a = 8$ and $b = 6$, so $a^2 - b^2 = 28$ but we can also have $a = 10$ and $b = 0$, so $a^2 - b^2 = 100$.
– sedrick
Nov 25 '18 at 9:09
I have gotten till $a^2-b^2=sqrt {c^2-2ab}cdot sqrt {c^2+2ab}$. I don't know if it possible to simplify this...
– Mohammad Zuhair Khan
Nov 25 '18 at 9:10
2
Why do you think $a^2-b^2$ is a function of $c$? Even if you force $a$, $b$, and $c$ to be positive integers, $gcd(a,b)=1$, and $ageq b$, there are still multiple values of $a^2-b^2$. For example, $(a,b)=(56,33)$ and $(a,b)=(63,16)$ gives $c=65$, but $a^2-b^2$ is not unique.
– Batominovski
Nov 25 '18 at 9:12
It means if you take some three values that work for the formula $a^2+b^{2 }= c^2$ then the same three values should work for $a^2-b^{2 }$ = $x$ where $x$ is your solution in terms of c.
– strangeindian
Nov 25 '18 at 9:12
Another example to consider $7^2+24^2=15^2+20^2=25^2$
– Mark Bennet
Nov 25 '18 at 9:12
|
show 3 more comments
Math Olympiad Question:
If $a^2+b^{2 }= c^2$ what is the value of $a^2-b^{2 }$ in terms of $c$?
I was solving some problems of the Indian National Mathematics Olympiad and I found this problem to be very intriguing. At first look it looks pretty straight forward but this problem requires the solution to be in terms of $c$ only, meaning that if you have the value of c, put that in your solution and you should get the same value as $a^2-b^{2 }$.
I tried simple algebra, trigonometry, geometry and pattern recognition but none of these methods were quite successful and could give a reasonable proof + solution.
My question is how to find the solution and prove this question? Is it possible to use Geometry and solve?
EDIT -
Sorry my bad, I think $a>b$ and $a$ and $b$ are integer values as in the starting of the exam it's written that all values will be integer and non-negative.
FINAL EDIT:
I think this question has been misprinted. Sorry for the inconvenience.
algebra-precalculus
asked Nov 25 '18 at 9:05
strangeindian
455
Math Olympiad Question:
If $a^2+b^{2 }= c^2$ what is the value of $a^2-b^{2 }$ in terms of $c$?
I was solving some problems of the Indian National Mathematics Olympiad and I found this problem to be very intriguing. At first look it looks pretty straight forward but this problem requires the solution to be in terms of $c$ only, meaning that if you have the value of c, put that in your solution and you should get the same value as $a^2-b^{2 }$.
I tried simple algebra, trigonometry, geometry and pattern recognition but none of these methods were quite successful and could give a reasonable proof + solution.
My question is how to find the solution and prove this question? Is it possible to use Geometry and solve?
EDIT -
Sorry my bad, I think $a>b$ and $a$ and $b$ are integer values as in the starting of the exam it's written that all values will be integer and non-negative.
FINAL EDIT:
I think this question has been misprinted. Sorry for the inconvenience.
algebra-precalculus
algebra-precalculus
asked Nov 25 '18 at 9:05
strangeindian
455
asked Nov 25 '18 at 9:05
strangeindian
455
edited Nov 25 '18 at 9:23
asked Nov 25 '18 at 9:05
strangeindian
455
asked Nov 25 '18 at 9:05
strangeindian
455
asked Nov 25 '18 at 9:05
strangeindian
455
455
8
Are you sure this is possible? I don't think $a^2 - b^2$ is well-defined because let's say $c = 10$ then we can have $a = 8$ and $b = 6$, so $a^2 - b^2 = 28$ but we can also have $a = 10$ and $b = 0$, so $a^2 - b^2 = 100$.
– sedrick
Nov 25 '18 at 9:09
I have gotten till $a^2-b^2=sqrt {c^2-2ab}cdot sqrt {c^2+2ab}$. I don't know if it possible to simplify this...
– Mohammad Zuhair Khan
Nov 25 '18 at 9:10
2
Why do you think $a^2-b^2$ is a function of $c$? Even if you force $a$, $b$, and $c$ to be positive integers, $gcd(a,b)=1$, and $ageq b$, there are still multiple values of $a^2-b^2$. For example, $(a,b)=(56,33)$ and $(a,b)=(63,16)$ gives $c=65$, but $a^2-b^2$ is not unique.
– Batominovski
Nov 25 '18 at 9:12
It means if you take some three values that work for the formula $a^2+b^{2 }= c^2$ then the same three values should work for $a^2-b^{2 }$ = $x$ where $x$ is your solution in terms of c.
– strangeindian
Nov 25 '18 at 9:12
Another example to consider $7^2+24^2=15^2+20^2=25^2$
– Mark Bennet
Nov 25 '18 at 9:12
|
show 3 more comments
8
Are you sure this is possible? I don't think $a^2 - b^2$ is well-defined because let's say $c = 10$ then we can have $a = 8$ and $b = 6$, so $a^2 - b^2 = 28$ but we can also have $a = 10$ and $b = 0$, so $a^2 - b^2 = 100$.
– sedrick
Nov 25 '18 at 9:09
I have gotten till $a^2-b^2=sqrt {c^2-2ab}cdot sqrt {c^2+2ab}$. I don't know if it possible to simplify this...
– Mohammad Zuhair Khan
Nov 25 '18 at 9:10
2
Why do you think $a^2-b^2$ is a function of $c$? Even if you force $a$, $b$, and $c$ to be positive integers, $gcd(a,b)=1$, and $ageq b$, there are still multiple values of $a^2-b^2$. For example, $(a,b)=(56,33)$ and $(a,b)=(63,16)$ gives $c=65$, but $a^2-b^2$ is not unique.
– Batominovski
Nov 25 '18 at 9:12
It means if you take some three values that work for the formula $a^2+b^{2 }= c^2$ then the same three values should work for $a^2-b^{2 }$ = $x$ where $x$ is your solution in terms of c.
– strangeindian
Nov 25 '18 at 9:12
Another example to consider $7^2+24^2=15^2+20^2=25^2$
– Mark Bennet
Nov 25 '18 at 9:12
8
8
Are you sure this is possible? I don't think $a^2 - b^2$ is well-defined because let's say $c = 10$ then we can have $a = 8$ and $b = 6$, so $a^2 - b^2 = 28$ but we can also have $a = 10$ and $b = 0$, so $a^2 - b^2 = 100$.
– sedrick
Nov 25 '18 at 9:09
Are you sure this is possible? I don't think $a^2 - b^2$ is well-defined because let's say $c = 10$ then we can have $a = 8$ and $b = 6$, so $a^2 - b^2 = 28$ but we can also have $a = 10$ and $b = 0$, so $a^2 - b^2 = 100$.
– sedrick
Nov 25 '18 at 9:09
I have gotten till $a^2-b^2=sqrt {c^2-2ab}cdot sqrt {c^2+2ab}$. I don't know if it possible to simplify this...
– Mohammad Zuhair Khan
Nov 25 '18 at 9:10
I have gotten till $a^2-b^2=sqrt {c^2-2ab}cdot sqrt {c^2+2ab}$. I don't know if it possible to simplify this...
– Mohammad Zuhair Khan
Nov 25 '18 at 9:10
2
2
Why do you think $a^2-b^2$ is a function of $c$? Even if you force $a$, $b$, and $c$ to be positive integers, $gcd(a,b)=1$, and $ageq b$, there are still multiple values of $a^2-b^2$. For example, $(a,b)=(56,33)$ and $(a,b)=(63,16)$ gives $c=65$, but $a^2-b^2$ is not unique.
– Batominovski
Nov 25 '18 at 9:12
Why do you think $a^2-b^2$ is a function of $c$? Even if you force $a$, $b$, and $c$ to be positive integers, $gcd(a,b)=1$, and $ageq b$, there are still multiple values of $a^2-b^2$. For example, $(a,b)=(56,33)$ and $(a,b)=(63,16)$ gives $c=65$, but $a^2-b^2$ is not unique.
– Batominovski
Nov 25 '18 at 9:12
It means if you take some three values that work for the formula $a^2+b^{2 }= c^2$ then the same three values should work for $a^2-b^{2 }$ = $x$ where $x$ is your solution in terms of c.
– strangeindian
Nov 25 '18 at 9:12
It means if you take some three values that work for the formula $a^2+b^{2 }= c^2$ then the same three values should work for $a^2-b^{2 }$ = $x$ where $x$ is your solution in terms of c.
– strangeindian
Nov 25 '18 at 9:12
Another example to consider $7^2+24^2=15^2+20^2=25^2$
– Mark Bennet
Nov 25 '18 at 9:12
Another example to consider $7^2+24^2=15^2+20^2=25^2$
– Mark Bennet
Nov 25 '18 at 9:12
|
show 3 more comments
2 Answers
2
active
oldest
votes
You are essentially looking at points on a circle of radius $c$.
If $x=arctan left(frac baright)$
then $a=c cos x,text{ } b= csin x$ and $a^2+b^2=c^2$
while $a^2-b^2=c^2 (cos^2 x- sin^2 x)=c^2 cos 2x$.
$cos 2x$ can take any value between $-1$ and $+1$, and there are rational points on the circle arbitrarily close to any point you choose. So for appropriate choices of $a,b,c$ (even if restricted to integers) you will find the whole range of possibilities. Where you are on the circle depends on the angle $x$ and not the radius $c$.
answered Nov 25 '18 at 9:23
Mark Bennet
80.5k981179
I like this solution. Maybe it wasn't a misprint but required exponential thinking like this.
– strangeindian
Nov 25 '18 at 9:25
add a comment |
This is not possible.
Take, for example, the following:
$$
4^2+3^2 = 5^2+0^2 = sqrt{21}^2+ 2^2 = 5^2
$$
but
$$
4^2 - 3^2 = 7\
5^2 - 0^2 =25\
sqrt{21}^2-2^2 = 17
$$
So the value of $a^2-b^2$ does not depend only on $c$.
answered Nov 25 '18 at 9:14
bruderjakob17
1587
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are essentially looking at points on a circle of radius $c$.
If $x=arctan left(frac baright)$
then $a=c cos x,text{ } b= csin x$ and $a^2+b^2=c^2$
while $a^2-b^2=c^2 (cos^2 x- sin^2 x)=c^2 cos 2x$.
$cos 2x$ can take any value between $-1$ and $+1$, and there are rational points on the circle arbitrarily close to any point you choose. So for appropriate choices of $a,b,c$ (even if restricted to integers) you will find the whole range of possibilities. Where you are on the circle depends on the angle $x$ and not the radius $c$.
answered Nov 25 '18 at 9:23
Mark Bennet
80.5k981179
I like this solution. Maybe it wasn't a misprint but required exponential thinking like this.
– strangeindian
Nov 25 '18 at 9:25
add a comment |
You are essentially looking at points on a circle of radius $c$.
If $x=arctan left(frac baright)$
then $a=c cos x,text{ } b= csin x$ and $a^2+b^2=c^2$
while $a^2-b^2=c^2 (cos^2 x- sin^2 x)=c^2 cos 2x$.
$cos 2x$ can take any value between $-1$ and $+1$, and there are rational points on the circle arbitrarily close to any point you choose. So for appropriate choices of $a,b,c$ (even if restricted to integers) you will find the whole range of possibilities. Where you are on the circle depends on the angle $x$ and not the radius $c$.
answered Nov 25 '18 at 9:23
Mark Bennet
80.5k981179
I like this solution. Maybe it wasn't a misprint but required exponential thinking like this.
– strangeindian
Nov 25 '18 at 9:25
add a comment |
You are essentially looking at points on a circle of radius $c$.
If $x=arctan left(frac baright)$
then $a=c cos x,text{ } b= csin x$ and $a^2+b^2=c^2$
while $a^2-b^2=c^2 (cos^2 x- sin^2 x)=c^2 cos 2x$.
$cos 2x$ can take any value between $-1$ and $+1$, and there are rational points on the circle arbitrarily close to any point you choose. So for appropriate choices of $a,b,c$ (even if restricted to integers) you will find the whole range of possibilities. Where you are on the circle depends on the angle $x$ and not the radius $c$.
answered Nov 25 '18 at 9:23
Mark Bennet
80.5k981179
You are essentially looking at points on a circle of radius $c$.
If $x=arctan left(frac baright)$
then $a=c cos x,text{ } b= csin x$ and $a^2+b^2=c^2$
while $a^2-b^2=c^2 (cos^2 x- sin^2 x)=c^2 cos 2x$.
$cos 2x$ can take any value between $-1$ and $+1$, and there are rational points on the circle arbitrarily close to any point you choose. So for appropriate choices of $a,b,c$ (even if restricted to integers) you will find the whole range of possibilities. Where you are on the circle depends on the angle $x$ and not the radius $c$.
answered Nov 25 '18 at 9:23
Mark Bennet
80.5k981179
answered Nov 25 '18 at 9:23
Mark Bennet
80.5k981179
answered Nov 25 '18 at 9:23
Mark Bennet
80.5k981179
answered Nov 25 '18 at 9:23
Mark Bennet
80.5k981179
80.5k981179
I like this solution. Maybe it wasn't a misprint but required exponential thinking like this.
– strangeindian
Nov 25 '18 at 9:25
add a comment |
I like this solution. Maybe it wasn't a misprint but required exponential thinking like this.
– strangeindian
Nov 25 '18 at 9:25
I like this solution. Maybe it wasn't a misprint but required exponential thinking like this.
– strangeindian
Nov 25 '18 at 9:25
I like this solution. Maybe it wasn't a misprint but required exponential thinking like this.
– strangeindian
Nov 25 '18 at 9:25
add a comment |
This is not possible.
Take, for example, the following:
$$
4^2+3^2 = 5^2+0^2 = sqrt{21}^2+ 2^2 = 5^2
$$
but
$$
4^2 - 3^2 = 7\
5^2 - 0^2 =25\
sqrt{21}^2-2^2 = 17
$$
So the value of $a^2-b^2$ does not depend only on $c$.
answered Nov 25 '18 at 9:14
bruderjakob17
1587
add a comment |
This is not possible.
Take, for example, the following:
$$
4^2+3^2 = 5^2+0^2 = sqrt{21}^2+ 2^2 = 5^2
$$
but
$$
4^2 - 3^2 = 7\
5^2 - 0^2 =25\
sqrt{21}^2-2^2 = 17
$$
So the value of $a^2-b^2$ does not depend only on $c$.
answered Nov 25 '18 at 9:14
bruderjakob17
1587
add a comment |
This is not possible.
Take, for example, the following:
$$
4^2+3^2 = 5^2+0^2 = sqrt{21}^2+ 2^2 = 5^2
$$
but
$$
4^2 - 3^2 = 7\
5^2 - 0^2 =25\
sqrt{21}^2-2^2 = 17
$$
So the value of $a^2-b^2$ does not depend only on $c$.
answered Nov 25 '18 at 9:14
bruderjakob17
1587
This is not possible.
Take, for example, the following:
$$
4^2+3^2 = 5^2+0^2 = sqrt{21}^2+ 2^2 = 5^2
$$
but
$$
4^2 - 3^2 = 7\
5^2 - 0^2 =25\
sqrt{21}^2-2^2 = 17
$$
So the value of $a^2-b^2$ does not depend only on $c$.
answered Nov 25 '18 at 9:14
bruderjakob17
1587
answered Nov 25 '18 at 9:14
bruderjakob17
1587
answered Nov 25 '18 at 9:14
bruderjakob17
1587
answered Nov 25 '18 at 9:14
bruderjakob17
1587
1587
add a comment |
add a comment |
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8
Are you sure this is possible? I don't think $a^2 - b^2$ is well-defined because let's say $c = 10$ then we can have $a = 8$ and $b = 6$, so $a^2 - b^2 = 28$ but we can also have $a = 10$ and $b = 0$, so $a^2 - b^2 = 100$.
– sedrick
Nov 25 '18 at 9:09
I have gotten till $a^2-b^2=sqrt {c^2-2ab}cdot sqrt {c^2+2ab}$. I don't know if it possible to simplify this...
– Mohammad Zuhair Khan
Nov 25 '18 at 9:10
2
Why do you think $a^2-b^2$ is a function of $c$? Even if you force $a$, $b$, and $c$ to be positive integers, $gcd(a,b)=1$, and $ageq b$, there are still multiple values of $a^2-b^2$. For example, $(a,b)=(56,33)$ and $(a,b)=(63,16)$ gives $c=65$, but $a^2-b^2$ is not unique.
– Batominovski
Nov 25 '18 at 9:12
It means if you take some three values that work for the formula $a^2+b^{2 }= c^2$ then the same three values should work for $a^2-b^{2 }$ = $x$ where $x$ is your solution in terms of c.
– strangeindian
Nov 25 '18 at 9:12
Another example to consider $7^2+24^2=15^2+20^2=25^2$
– Mark Bennet
Nov 25 '18 at 9:12