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Find the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$
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If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $fracasqrt b+cd$.Find $a+b+c+d$
The polynomial is $fracx^8-1x-1$ has roots $operatornamecis(2pi k/8)$ for $k in 1, ldots, 7$.
Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.
The area of an octagon (by splitting into triangles) with radius $1$ is $8 cdot frac12 cdot fracsqrt22 = 2sqrt2$.
I am stuck here.The answer for $a+b+c+d=10$
geometry complex-numbers
asked Jul 27 at 10:17
learner_avid
687313
 |Â
show 3 more comments
up vote
1
down vote
favorite
If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $fracasqrt b+cd$.Find $a+b+c+d$
The polynomial is $fracx^8-1x-1$ has roots $operatornamecis(2pi k/8)$ for $k in 1, ldots, 7$.
Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.
The area of an octagon (by splitting into triangles) with radius $1$ is $8 cdot frac12 cdot fracsqrt22 = 2sqrt2$.
I am stuck here.The answer for $a+b+c+d=10$
geometry complex-numbers
asked Jul 27 at 10:17
learner_avid
687313
Are there any restrictions for $a,b,c,d$ such as they have to be integer or they have to be not zero?
– mrtaurho
Jul 27 at 10:20
No there are no such conditions.
– learner_avid
Jul 27 at 10:23
There's something I don't understand: why do you subtract the area of the triangle formed by two adjacent sides?
– Bernard
Jul 27 at 10:29
1
I would guess that $a, b, c, d$ do have to be integers. Why is it hard to compute the area of the triangle? It has vertices $(sqrt2/2, pm sqrt2/2), (1, 0)$. Then you subtract that area and then rewrite it into the form that they want you to write it (presumably in lowest terms, with the radical as simplified as possible, etc).
– Mees de Vries
Jul 27 at 10:29
2
I believe that that you need to add the critical extra words "can be expressed $textbfin its simplest form$ as $fracasqrtb+cd$. Thus $a = 2, b=2, c=0,d=1 implies a+b+c+d=5.$ Are you doing this problem sheet: math.ucla.edu/~radko/circles/lib/data/Handout-556-674.pdf
– Martin Roberts
Jul 27 at 10:35
 |Â
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $fracasqrt b+cd$.Find $a+b+c+d$
The polynomial is $fracx^8-1x-1$ has roots $operatornamecis(2pi k/8)$ for $k in 1, ldots, 7$.
Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.
The area of an octagon (by splitting into triangles) with radius $1$ is $8 cdot frac12 cdot fracsqrt22 = 2sqrt2$.
I am stuck here.The answer for $a+b+c+d=10$
geometry complex-numbers
asked Jul 27 at 10:17
learner_avid
687313
If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $fracasqrt b+cd$.Find $a+b+c+d$
The polynomial is $fracx^8-1x-1$ has roots $operatornamecis(2pi k/8)$ for $k in 1, ldots, 7$.
Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.
The area of an octagon (by splitting into triangles) with radius $1$ is $8 cdot frac12 cdot fracsqrt22 = 2sqrt2$.
I am stuck here.The answer for $a+b+c+d=10$
geometry complex-numbers
asked Jul 27 at 10:17
learner_avid
687313
asked Jul 27 at 10:17
learner_avid
687313
asked Jul 27 at 10:17
learner_avid
687313
asked Jul 27 at 10:17
learner_avid
687313
687313
Are there any restrictions for $a,b,c,d$ such as they have to be integer or they have to be not zero?
– mrtaurho
Jul 27 at 10:20
No there are no such conditions.
– learner_avid
Jul 27 at 10:23
There's something I don't understand: why do you subtract the area of the triangle formed by two adjacent sides?
– Bernard
Jul 27 at 10:29
1
I would guess that $a, b, c, d$ do have to be integers. Why is it hard to compute the area of the triangle? It has vertices $(sqrt2/2, pm sqrt2/2), (1, 0)$. Then you subtract that area and then rewrite it into the form that they want you to write it (presumably in lowest terms, with the radical as simplified as possible, etc).
– Mees de Vries
Jul 27 at 10:29
2
I believe that that you need to add the critical extra words "can be expressed $textbfin its simplest form$ as $fracasqrtb+cd$. Thus $a = 2, b=2, c=0,d=1 implies a+b+c+d=5.$ Are you doing this problem sheet: math.ucla.edu/~radko/circles/lib/data/Handout-556-674.pdf
– Martin Roberts
Jul 27 at 10:35
 |Â
show 3 more comments
Are there any restrictions for $a,b,c,d$ such as they have to be integer or they have to be not zero?
– mrtaurho
Jul 27 at 10:20
No there are no such conditions.
– learner_avid
Jul 27 at 10:23
There's something I don't understand: why do you subtract the area of the triangle formed by two adjacent sides?
– Bernard
Jul 27 at 10:29
1
I would guess that $a, b, c, d$ do have to be integers. Why is it hard to compute the area of the triangle? It has vertices $(sqrt2/2, pm sqrt2/2), (1, 0)$. Then you subtract that area and then rewrite it into the form that they want you to write it (presumably in lowest terms, with the radical as simplified as possible, etc).
– Mees de Vries
Jul 27 at 10:29
2
I believe that that you need to add the critical extra words "can be expressed $textbfin its simplest form$ as $fracasqrtb+cd$. Thus $a = 2, b=2, c=0,d=1 implies a+b+c+d=5.$ Are you doing this problem sheet: math.ucla.edu/~radko/circles/lib/data/Handout-556-674.pdf
– Martin Roberts
Jul 27 at 10:35
Are there any restrictions for $a,b,c,d$ such as they have to be integer or they have to be not zero?
– mrtaurho
Jul 27 at 10:20
Are there any restrictions for $a,b,c,d$ such as they have to be integer or they have to be not zero?
– mrtaurho
Jul 27 at 10:20
No there are no such conditions.
– learner_avid
Jul 27 at 10:23
No there are no such conditions.
– learner_avid
Jul 27 at 10:23
There's something I don't understand: why do you subtract the area of the triangle formed by two adjacent sides?
– Bernard
Jul 27 at 10:29
There's something I don't understand: why do you subtract the area of the triangle formed by two adjacent sides?
– Bernard
Jul 27 at 10:29
1
1
I would guess that $a, b, c, d$ do have to be integers. Why is it hard to compute the area of the triangle? It has vertices $(sqrt2/2, pm sqrt2/2), (1, 0)$. Then you subtract that area and then rewrite it into the form that they want you to write it (presumably in lowest terms, with the radical as simplified as possible, etc).
– Mees de Vries
Jul 27 at 10:29
I would guess that $a, b, c, d$ do have to be integers. Why is it hard to compute the area of the triangle? It has vertices $(sqrt2/2, pm sqrt2/2), (1, 0)$. Then you subtract that area and then rewrite it into the form that they want you to write it (presumably in lowest terms, with the radical as simplified as possible, etc).
– Mees de Vries
Jul 27 at 10:29
2
2
I believe that that you need to add the critical extra words "can be expressed $textbfin its simplest form$ as $fracasqrtb+cd$. Thus $a = 2, b=2, c=0,d=1 implies a+b+c+d=5.$ Are you doing this problem sheet: math.ucla.edu/~radko/circles/lib/data/Handout-556-674.pdf
– Martin Roberts
Jul 27 at 10:35
I believe that that you need to add the critical extra words "can be expressed $textbfin its simplest form$ as $fracasqrtb+cd$. Thus $a = 2, b=2, c=0,d=1 implies a+b+c+d=5.$ Are you doing this problem sheet: math.ucla.edu/~radko/circles/lib/data/Handout-556-674.pdf
– Martin Roberts
Jul 27 at 10:35
 |Â
show 3 more comments
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
The triangle has area
$$
frac12 times sqrt2 times (1- sqrt2/2) = fracsqrt2 - 12,
$$
so the total polygon has area
$$
2sqrt2 - fracsqrt2 - 12 = frac32sqrt2 + frac12 = frac3sqrt2 + 12,
$$
at least in lowest terms. The result is $a + b + c + d = 8$, contrary to the 10 you claimed.
answered Jul 27 at 10:36
Mees de Vries
13.7k12345
add a comment |Â
up vote
1
down vote
beginalign
p_1&=(tfracsqrt22,tfracsqrt22)
,\
p_2&=(0,1)
,\
p_3&=(-tfracsqrt22,tfracsqrt22)
,\ dots
endalign
beginalign
S&=
(p_1x-p_3x)^2+tfrac32(p_1x-p_3x)(p_2y-p_1y)
=frac3sqrt2+12
,\
&=
frac3ksqrt2+k2k
,\
a&=3k,quad b=2,quad c=k,quad d=2k
,\
a+b+c+d&=6k+2,
quad kinmathbbR
.
endalign
In order to get $a+b+c+d=10$,
we need to have $k=tfrac43$.
answered Jul 27 at 11:50
g.kov
5,5321717
add a comment |Â
up vote
0
down vote
$$fracasqrtb+cd~=~2sqrt2~~~~textwith~~~~a+b+c+d~=~10$$
The numerator of the $abcd$-fraction contains one square root plus a number. Hence the sum of the four variable equals an integer the easiest way to get rid of $c$ is by setting it $0$. From thereone we not got $a+b+d=10$. Now $b$ has to be $2$ so that we can arrive at the square root of $2$. Now we got
$$a+d=8~~textand~~fracasqrt2d=2sqrt2$$
So $a$ has to be a multiply of $d$, to be exact $a=2d$. Plugging this into $a+d=8$ leads us to $a=frac163$ and $d=frac83$. Verifying by setting this values in the first equations:
$$beginalign
fracfrac163sqrt2+0frac83~=~2sqrt2~~~~&textand~~~~frac163+2+0+frac83~=~10\
frac16sqrt28~=~2sqrt2~~~~&textand~~~~frac16+8+63~=~10\
2sqrt2~=~2sqrt2~~~~&textand~~~~10~=~10
endalign$$
Therefore $a=frac163,b=2,c=0,frac83$ but this only holds for $a,b,c,d in mathbbQ$.
answered Jul 27 at 10:32
mrtaurho
726219
The area of the octagon is $2sqrt2$ but the area of the polygon is smaller than that becasue you have to subtract the area of the triangle with vertices at $1$ and $frac1pm isqrt2$.
– gandalf61
Jul 27 at 10:46
add a comment |Â
up vote
0
down vote
You got stuck at a very odd point. You basically solved the hard part of the problem. So compute the are of the triangle, subtract it from the area of the octagon, and express the result in the desired form.
Area of triangle: $frac12sqrt2(1-fracsqrt22) = fracsqrt2-12$.
So the area of the polygon is $2sqrt2- fracsqrt2-12= frac3sqrt2+12$.
The result is not unique though. Given any number $t$, $a=3t, b=2, c=t, d=2t$ is a solution (and there are many others). So $a+b+c+d= 6t+2$ can be any number.
Some condition must be missing, check the problem again.
answered Jul 27 at 10:34
A. Pongrácz
1,459115
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The triangle has area
$$
frac12 times sqrt2 times (1- sqrt2/2) = fracsqrt2 - 12,
$$
so the total polygon has area
$$
2sqrt2 - fracsqrt2 - 12 = frac32sqrt2 + frac12 = frac3sqrt2 + 12,
$$
at least in lowest terms. The result is $a + b + c + d = 8$, contrary to the 10 you claimed.
answered Jul 27 at 10:36
Mees de Vries
13.7k12345
add a comment |Â
up vote
1
down vote
accepted
The triangle has area
$$
frac12 times sqrt2 times (1- sqrt2/2) = fracsqrt2 - 12,
$$
so the total polygon has area
$$
2sqrt2 - fracsqrt2 - 12 = frac32sqrt2 + frac12 = frac3sqrt2 + 12,
$$
at least in lowest terms. The result is $a + b + c + d = 8$, contrary to the 10 you claimed.
answered Jul 27 at 10:36
Mees de Vries
13.7k12345
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The triangle has area
$$
frac12 times sqrt2 times (1- sqrt2/2) = fracsqrt2 - 12,
$$
so the total polygon has area
$$
2sqrt2 - fracsqrt2 - 12 = frac32sqrt2 + frac12 = frac3sqrt2 + 12,
$$
at least in lowest terms. The result is $a + b + c + d = 8$, contrary to the 10 you claimed.
answered Jul 27 at 10:36
Mees de Vries
13.7k12345
The triangle has area
$$
frac12 times sqrt2 times (1- sqrt2/2) = fracsqrt2 - 12,
$$
so the total polygon has area
$$
2sqrt2 - fracsqrt2 - 12 = frac32sqrt2 + frac12 = frac3sqrt2 + 12,
$$
at least in lowest terms. The result is $a + b + c + d = 8$, contrary to the 10 you claimed.
answered Jul 27 at 10:36
Mees de Vries
13.7k12345
answered Jul 27 at 10:36
Mees de Vries
13.7k12345
answered Jul 27 at 10:36
Mees de Vries
13.7k12345
answered Jul 27 at 10:36
Mees de Vries
13.7k12345
13.7k12345
add a comment |Â
add a comment |Â
up vote
1
down vote
beginalign
p_1&=(tfracsqrt22,tfracsqrt22)
,\
p_2&=(0,1)
,\
p_3&=(-tfracsqrt22,tfracsqrt22)
,\ dots
endalign
beginalign
S&=
(p_1x-p_3x)^2+tfrac32(p_1x-p_3x)(p_2y-p_1y)
=frac3sqrt2+12
,\
&=
frac3ksqrt2+k2k
,\
a&=3k,quad b=2,quad c=k,quad d=2k
,\
a+b+c+d&=6k+2,
quad kinmathbbR
.
endalign
In order to get $a+b+c+d=10$,
we need to have $k=tfrac43$.
answered Jul 27 at 11:50
g.kov
5,5321717
add a comment |Â
up vote
1
down vote
beginalign
p_1&=(tfracsqrt22,tfracsqrt22)
,\
p_2&=(0,1)
,\
p_3&=(-tfracsqrt22,tfracsqrt22)
,\ dots
endalign
beginalign
S&=
(p_1x-p_3x)^2+tfrac32(p_1x-p_3x)(p_2y-p_1y)
=frac3sqrt2+12
,\
&=
frac3ksqrt2+k2k
,\
a&=3k,quad b=2,quad c=k,quad d=2k
,\
a+b+c+d&=6k+2,
quad kinmathbbR
.
endalign
In order to get $a+b+c+d=10$,
we need to have $k=tfrac43$.
answered Jul 27 at 11:50
g.kov
5,5321717
add a comment |Â
up vote
1
down vote
up vote
1
down vote
beginalign
p_1&=(tfracsqrt22,tfracsqrt22)
,\
p_2&=(0,1)
,\
p_3&=(-tfracsqrt22,tfracsqrt22)
,\ dots
endalign
beginalign
S&=
(p_1x-p_3x)^2+tfrac32(p_1x-p_3x)(p_2y-p_1y)
=frac3sqrt2+12
,\
&=
frac3ksqrt2+k2k
,\
a&=3k,quad b=2,quad c=k,quad d=2k
,\
a+b+c+d&=6k+2,
quad kinmathbbR
.
endalign
In order to get $a+b+c+d=10$,
we need to have $k=tfrac43$.
answered Jul 27 at 11:50
g.kov
5,5321717
beginalign
p_1&=(tfracsqrt22,tfracsqrt22)
,\
p_2&=(0,1)
,\
p_3&=(-tfracsqrt22,tfracsqrt22)
,\ dots
endalign
beginalign
S&=
(p_1x-p_3x)^2+tfrac32(p_1x-p_3x)(p_2y-p_1y)
=frac3sqrt2+12
,\
&=
frac3ksqrt2+k2k
,\
a&=3k,quad b=2,quad c=k,quad d=2k
,\
a+b+c+d&=6k+2,
quad kinmathbbR
.
endalign
In order to get $a+b+c+d=10$,
we need to have $k=tfrac43$.
answered Jul 27 at 11:50
g.kov
5,5321717
answered Jul 27 at 11:50
g.kov
5,5321717
answered Jul 27 at 11:50
g.kov
5,5321717
answered Jul 27 at 11:50
g.kov
5,5321717
5,5321717
add a comment |Â
add a comment |Â
up vote
0
down vote
$$fracasqrtb+cd~=~2sqrt2~~~~textwith~~~~a+b+c+d~=~10$$
The numerator of the $abcd$-fraction contains one square root plus a number. Hence the sum of the four variable equals an integer the easiest way to get rid of $c$ is by setting it $0$. From thereone we not got $a+b+d=10$. Now $b$ has to be $2$ so that we can arrive at the square root of $2$. Now we got
$$a+d=8~~textand~~fracasqrt2d=2sqrt2$$
So $a$ has to be a multiply of $d$, to be exact $a=2d$. Plugging this into $a+d=8$ leads us to $a=frac163$ and $d=frac83$. Verifying by setting this values in the first equations:
$$beginalign
fracfrac163sqrt2+0frac83~=~2sqrt2~~~~&textand~~~~frac163+2+0+frac83~=~10\
frac16sqrt28~=~2sqrt2~~~~&textand~~~~frac16+8+63~=~10\
2sqrt2~=~2sqrt2~~~~&textand~~~~10~=~10
endalign$$
Therefore $a=frac163,b=2,c=0,frac83$ but this only holds for $a,b,c,d in mathbbQ$.
answered Jul 27 at 10:32
mrtaurho
726219
The area of the octagon is $2sqrt2$ but the area of the polygon is smaller than that becasue you have to subtract the area of the triangle with vertices at $1$ and $frac1pm isqrt2$.
– gandalf61
Jul 27 at 10:46
add a comment |Â
up vote
0
down vote
$$fracasqrtb+cd~=~2sqrt2~~~~textwith~~~~a+b+c+d~=~10$$
The numerator of the $abcd$-fraction contains one square root plus a number. Hence the sum of the four variable equals an integer the easiest way to get rid of $c$ is by setting it $0$. From thereone we not got $a+b+d=10$. Now $b$ has to be $2$ so that we can arrive at the square root of $2$. Now we got
$$a+d=8~~textand~~fracasqrt2d=2sqrt2$$
So $a$ has to be a multiply of $d$, to be exact $a=2d$. Plugging this into $a+d=8$ leads us to $a=frac163$ and $d=frac83$. Verifying by setting this values in the first equations:
$$beginalign
fracfrac163sqrt2+0frac83~=~2sqrt2~~~~&textand~~~~frac163+2+0+frac83~=~10\
frac16sqrt28~=~2sqrt2~~~~&textand~~~~frac16+8+63~=~10\
2sqrt2~=~2sqrt2~~~~&textand~~~~10~=~10
endalign$$
Therefore $a=frac163,b=2,c=0,frac83$ but this only holds for $a,b,c,d in mathbbQ$.
answered Jul 27 at 10:32
mrtaurho
726219
The area of the octagon is $2sqrt2$ but the area of the polygon is smaller than that becasue you have to subtract the area of the triangle with vertices at $1$ and $frac1pm isqrt2$.
– gandalf61
Jul 27 at 10:46
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$fracasqrtb+cd~=~2sqrt2~~~~textwith~~~~a+b+c+d~=~10$$
The numerator of the $abcd$-fraction contains one square root plus a number. Hence the sum of the four variable equals an integer the easiest way to get rid of $c$ is by setting it $0$. From thereone we not got $a+b+d=10$. Now $b$ has to be $2$ so that we can arrive at the square root of $2$. Now we got
$$a+d=8~~textand~~fracasqrt2d=2sqrt2$$
So $a$ has to be a multiply of $d$, to be exact $a=2d$. Plugging this into $a+d=8$ leads us to $a=frac163$ and $d=frac83$. Verifying by setting this values in the first equations:
$$beginalign
fracfrac163sqrt2+0frac83~=~2sqrt2~~~~&textand~~~~frac163+2+0+frac83~=~10\
frac16sqrt28~=~2sqrt2~~~~&textand~~~~frac16+8+63~=~10\
2sqrt2~=~2sqrt2~~~~&textand~~~~10~=~10
endalign$$
Therefore $a=frac163,b=2,c=0,frac83$ but this only holds for $a,b,c,d in mathbbQ$.
answered Jul 27 at 10:32
mrtaurho
726219
$$fracasqrtb+cd~=~2sqrt2~~~~textwith~~~~a+b+c+d~=~10$$
The numerator of the $abcd$-fraction contains one square root plus a number. Hence the sum of the four variable equals an integer the easiest way to get rid of $c$ is by setting it $0$. From thereone we not got $a+b+d=10$. Now $b$ has to be $2$ so that we can arrive at the square root of $2$. Now we got
$$a+d=8~~textand~~fracasqrt2d=2sqrt2$$
So $a$ has to be a multiply of $d$, to be exact $a=2d$. Plugging this into $a+d=8$ leads us to $a=frac163$ and $d=frac83$. Verifying by setting this values in the first equations:
$$beginalign
fracfrac163sqrt2+0frac83~=~2sqrt2~~~~&textand~~~~frac163+2+0+frac83~=~10\
frac16sqrt28~=~2sqrt2~~~~&textand~~~~frac16+8+63~=~10\
2sqrt2~=~2sqrt2~~~~&textand~~~~10~=~10
endalign$$
Therefore $a=frac163,b=2,c=0,frac83$ but this only holds for $a,b,c,d in mathbbQ$.
answered Jul 27 at 10:32
mrtaurho
726219
answered Jul 27 at 10:32
mrtaurho
726219
answered Jul 27 at 10:32
mrtaurho
726219
answered Jul 27 at 10:32
mrtaurho
726219
726219
The area of the octagon is $2sqrt2$ but the area of the polygon is smaller than that becasue you have to subtract the area of the triangle with vertices at $1$ and $frac1pm isqrt2$.
– gandalf61
Jul 27 at 10:46
add a comment |Â
The area of the octagon is $2sqrt2$ but the area of the polygon is smaller than that becasue you have to subtract the area of the triangle with vertices at $1$ and $frac1pm isqrt2$.
– gandalf61
Jul 27 at 10:46
The area of the octagon is $2sqrt2$ but the area of the polygon is smaller than that becasue you have to subtract the area of the triangle with vertices at $1$ and $frac1pm isqrt2$.
– gandalf61
Jul 27 at 10:46
The area of the octagon is $2sqrt2$ but the area of the polygon is smaller than that becasue you have to subtract the area of the triangle with vertices at $1$ and $frac1pm isqrt2$.
– gandalf61
Jul 27 at 10:46
add a comment |Â
up vote
0
down vote
You got stuck at a very odd point. You basically solved the hard part of the problem. So compute the are of the triangle, subtract it from the area of the octagon, and express the result in the desired form.
Area of triangle: $frac12sqrt2(1-fracsqrt22) = fracsqrt2-12$.
So the area of the polygon is $2sqrt2- fracsqrt2-12= frac3sqrt2+12$.
The result is not unique though. Given any number $t$, $a=3t, b=2, c=t, d=2t$ is a solution (and there are many others). So $a+b+c+d= 6t+2$ can be any number.
Some condition must be missing, check the problem again.
answered Jul 27 at 10:34
A. Pongrácz
1,459115
add a comment |Â
up vote
0
down vote
You got stuck at a very odd point. You basically solved the hard part of the problem. So compute the are of the triangle, subtract it from the area of the octagon, and express the result in the desired form.
Area of triangle: $frac12sqrt2(1-fracsqrt22) = fracsqrt2-12$.
So the area of the polygon is $2sqrt2- fracsqrt2-12= frac3sqrt2+12$.
The result is not unique though. Given any number $t$, $a=3t, b=2, c=t, d=2t$ is a solution (and there are many others). So $a+b+c+d= 6t+2$ can be any number.
Some condition must be missing, check the problem again.
answered Jul 27 at 10:34
A. Pongrácz
1,459115
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You got stuck at a very odd point. You basically solved the hard part of the problem. So compute the are of the triangle, subtract it from the area of the octagon, and express the result in the desired form.
Area of triangle: $frac12sqrt2(1-fracsqrt22) = fracsqrt2-12$.
So the area of the polygon is $2sqrt2- fracsqrt2-12= frac3sqrt2+12$.
The result is not unique though. Given any number $t$, $a=3t, b=2, c=t, d=2t$ is a solution (and there are many others). So $a+b+c+d= 6t+2$ can be any number.
Some condition must be missing, check the problem again.
answered Jul 27 at 10:34
A. Pongrácz
1,459115
You got stuck at a very odd point. You basically solved the hard part of the problem. So compute the are of the triangle, subtract it from the area of the octagon, and express the result in the desired form.
Area of triangle: $frac12sqrt2(1-fracsqrt22) = fracsqrt2-12$.
So the area of the polygon is $2sqrt2- fracsqrt2-12= frac3sqrt2+12$.
The result is not unique though. Given any number $t$, $a=3t, b=2, c=t, d=2t$ is a solution (and there are many others). So $a+b+c+d= 6t+2$ can be any number.
Some condition must be missing, check the problem again.
answered Jul 27 at 10:34
A. Pongrácz
1,459115
edited Jul 29 at 12:32
answered Jul 27 at 10:34
A. Pongrácz
1,459115
answered Jul 27 at 10:34
A. Pongrácz
1,459115
answered Jul 27 at 10:34
A. Pongrácz
1,459115
1,459115
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Are there any restrictions for $a,b,c,d$ such as they have to be integer or they have to be not zero?
– mrtaurho
Jul 27 at 10:20
No there are no such conditions.
– learner_avid
Jul 27 at 10:23
There's something I don't understand: why do you subtract the area of the triangle formed by two adjacent sides?
– Bernard
Jul 27 at 10:29
1
I would guess that $a, b, c, d$ do have to be integers. Why is it hard to compute the area of the triangle? It has vertices $(sqrt2/2, pm sqrt2/2), (1, 0)$. Then you subtract that area and then rewrite it into the form that they want you to write it (presumably in lowest terms, with the radical as simplified as possible, etc).
– Mees de Vries
Jul 27 at 10:29
2
I believe that that you need to add the critical extra words "can be expressed $textbfin its simplest form$ as $fracasqrtb+cd$. Thus $a = 2, b=2, c=0,d=1 implies a+b+c+d=5.$ Are you doing this problem sheet: math.ucla.edu/~radko/circles/lib/data/Handout-556-674.pdf
– Martin Roberts
Jul 27 at 10:35