Find the maximum area of a rectangle that can be drawn inside a circle's quarter
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How can I find the maximum area of the rectangle here? Given that the circle's radius is 6.
Any hint please?
calculus geometry optimization
asked Jan 16 at 8:39
Just_CauseJust_Cause
1257
$endgroup$
add a comment |
$begingroup$
How can I find the maximum area of the rectangle here? Given that the circle's radius is 6.
Any hint please?
calculus geometry optimization
asked Jan 16 at 8:39
Just_CauseJust_Cause
1257
$endgroup$
$begingroup$
Maybe you can parametrise the rectangle? Maybe set the width of the rectangle to be $x$ and then calculate its area ...
$endgroup$
– Matti P.
Jan 16 at 8:45
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I tried, but I couldn't find any useful relationship between the width and the height.
$endgroup$
– Just_Cause
Jan 16 at 8:49
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Just make a big drawing, draw some auxiliary lines and see what you can come up with.
$endgroup$
– Matti P.
Jan 16 at 9:01
1
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I suspect the square with a vertex at the origin and opposite vertex at $(3sqrt{2},3sqrt{2}$)$ with area $18$ may be bigger
$endgroup$
– Henry
Jan 16 at 9:14
add a comment |
$begingroup$
How can I find the maximum area of the rectangle here? Given that the circle's radius is 6.
Any hint please?
calculus geometry optimization
asked Jan 16 at 8:39
Just_CauseJust_Cause
1257
$endgroup$
How can I find the maximum area of the rectangle here? Given that the circle's radius is 6.
Any hint please?
calculus geometry optimization
calculus geometry optimization
asked Jan 16 at 8:39
Just_CauseJust_Cause
1257
asked Jan 16 at 8:39
Just_CauseJust_Cause
1257
asked Jan 16 at 8:39
Just_CauseJust_Cause
1257
asked Jan 16 at 8:39
Just_CauseJust_Cause
1257
asked Jan 16 at 8:39
Just_CauseJust_Cause
1257
1257
$begingroup$
Maybe you can parametrise the rectangle? Maybe set the width of the rectangle to be $x$ and then calculate its area ...
$endgroup$
– Matti P.
Jan 16 at 8:45
$begingroup$
I tried, but I couldn't find any useful relationship between the width and the height.
$endgroup$
– Just_Cause
Jan 16 at 8:49
$begingroup$
Just make a big drawing, draw some auxiliary lines and see what you can come up with.
$endgroup$
– Matti P.
Jan 16 at 9:01
1
$begingroup$
I suspect the square with a vertex at the origin and opposite vertex at $(3sqrt{2},3sqrt{2}$)$ with area $18$ may be bigger
$endgroup$
– Henry
Jan 16 at 9:14
add a comment |
$begingroup$
Maybe you can parametrise the rectangle? Maybe set the width of the rectangle to be $x$ and then calculate its area ...
$endgroup$
– Matti P.
Jan 16 at 8:45
$begingroup$
I tried, but I couldn't find any useful relationship between the width and the height.
$endgroup$
– Just_Cause
Jan 16 at 8:49
$begingroup$
Just make a big drawing, draw some auxiliary lines and see what you can come up with.
$endgroup$
– Matti P.
Jan 16 at 9:01
1
$begingroup$
I suspect the square with a vertex at the origin and opposite vertex at $(3sqrt{2},3sqrt{2}$)$ with area $18$ may be bigger
$endgroup$
– Henry
Jan 16 at 9:14
$begingroup$
Maybe you can parametrise the rectangle? Maybe set the width of the rectangle to be $x$ and then calculate its area ...
$endgroup$
– Matti P.
Jan 16 at 8:45
$begingroup$
Maybe you can parametrise the rectangle? Maybe set the width of the rectangle to be $x$ and then calculate its area ...
$endgroup$
– Matti P.
Jan 16 at 8:45
$begingroup$
I tried, but I couldn't find any useful relationship between the width and the height.
$endgroup$
– Just_Cause
Jan 16 at 8:49
$begingroup$
I tried, but I couldn't find any useful relationship between the width and the height.
$endgroup$
– Just_Cause
Jan 16 at 8:49
$begingroup$
Just make a big drawing, draw some auxiliary lines and see what you can come up with.
$endgroup$
– Matti P.
Jan 16 at 9:01
$begingroup$
Just make a big drawing, draw some auxiliary lines and see what you can come up with.
$endgroup$
– Matti P.
Jan 16 at 9:01
1
1
$begingroup$
I suspect the square with a vertex at the origin and opposite vertex at $(3sqrt{2},3sqrt{2}$)$ with area $18$ may be bigger
$endgroup$
– Henry
Jan 16 at 9:14
$begingroup$
I suspect the square with a vertex at the origin and opposite vertex at $(3sqrt{2},3sqrt{2}$)$ with area $18$ may be bigger
$endgroup$
– Henry
Jan 16 at 9:14
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Choosing one vertex to be $(r costheta, rsintheta)$ then another is $(r (costheta- sintheta),0)$, a third is $(0,r (costheta- sintheta))$ and the fourth is $(r sintheta, rcostheta)$
The area is then $2r^2 sintheta(costheta- sintheta) = sqrt{8}r^2 sin(theta) sinleft(fracpi4-thetaright) $ which is maximised when $theta= fracpi8$ with the maximum area being $r^2(sqrt{2}-1)$
Added to illustrate comment: If the arrangement is not symmetric, you will not get two vertices on the arc and two on the axes. Instead you will get coomething like this
answered Jan 16 at 9:12
HenryHenry
101k482169
$endgroup$
$begingroup$
Didn't get it how you assigned the coordinates of each vertex, can you explain please?
$endgroup$
– Just_Cause
Jan 16 at 10:49
$begingroup$
@Just_Cause - Are you happy with the $(r costheta, rsintheta)$ as a point on the circle? That lies on the line $y=x + r(cos theta - sin theta)$ and so the next vertex is where that line intersects $y=0$. The other two vertices are reflections in the line $y=x$
$endgroup$
– Henry
Jan 16 at 13:20
$begingroup$
Sure, but how did you assume that the small triangle near the x axis is isosceles? (45 45 90)? Doesn't this lead to the fact that the bigger triangle is isosceles too?
$endgroup$
– Just_Cause
Jan 22 at 11:26
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@Just_Cause If you do not have symmetry then either the vertices are not on the arc and axes or the quadrilateral is not a rectangle. So the triangle is isosceles
$endgroup$
– Henry
Jan 22 at 12:49
$begingroup$
Still don't get what has symmetry got to do here, do you mean the symmetry line of the triangle, you used it to divide the bigger triangle into two halves which would bisect the 90 angle?
$endgroup$
– Just_Cause
Jan 27 at 21:00
|
show 1 more comment
$begingroup$
I would compute the coordinates of the given points:
$$A(x_A,0),B(x_B,sqrt{6^2-x_B^2}),C(x_C,sqrt{6^2-x_C^2}),D(0,y_B)$$ Now it must be
$$vec{AB}=[x_B-x_A,sqrt{6^2-x_B^2}]$$ and $$vec{AD}=[-x_A,y_D]$$ and it must be
$$vec{AB}cdot vec{AD}=0$$
Can you proceed?
answered Jan 16 at 8:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
$begingroup$
Appreciated, but I'm not really required to solve it using vectors.
$endgroup$
– Just_Cause
Jan 16 at 10:44
add a comment |
$begingroup$
Let $ABCD$ be our rectangle, where $A$ be placed on the perpendicular radius and $D$ be placed on the horizontal radius.
Also, let $OE$ be a radius of the circle such that $OEperp AD$, $OEcap BC={F}$, $OEcap AD={G}$ and let $HE$ be a diameter of the circle.
Now, since $G$ is a midpoint of $BC$, we obtain that $G$ is a midpoint of $AD$.
Thus, $OG=AG=DG$ and let $OG=x$, $DC=y$.
Now, since $measuredangle ECH=90^{circ},$ we obtain:
$$CE^2=EGcdot GH$$ or
$$x^2=(6-x-y)(6+x+y)$$ or
$$(x+y)^2=36-x^2$$ or
$$y=sqrt{36-x^2}-x.$$
Id est, $$S_{ABCD}=2xleft(sqrt{36-x^2}-xright).$$
Can you end it now?
answered Jan 16 at 9:15
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
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Sure, but how did you get to $CE^2=EGcdot GH$ ?
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– Just_Cause
Jan 16 at 10:44
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Because the triangle CGH is a right-angled triangle and CE is an altitude of the triangle.
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– Michael Rozenberg
Jan 16 at 11:32
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Oh, i got it. One last thing that confuses me, is it permissible to assume that the radius is perpendicular on AD without an additional information from the question that denotes that? i.e OE bisects AD? As far as i know, i can't put any assumptions unless the question states a key info on some theorem
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– Just_Cause
Jan 16 at 12:34
1
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@Just_Cause If $OEperp AD$ then $OEperp BC$, which gives $BF=FC$ and from here $AG=GD.$
$endgroup$
– Michael Rozenberg
Jan 16 at 13:23
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Choosing one vertex to be $(r costheta, rsintheta)$ then another is $(r (costheta- sintheta),0)$, a third is $(0,r (costheta- sintheta))$ and the fourth is $(r sintheta, rcostheta)$
The area is then $2r^2 sintheta(costheta- sintheta) = sqrt{8}r^2 sin(theta) sinleft(fracpi4-thetaright) $ which is maximised when $theta= fracpi8$ with the maximum area being $r^2(sqrt{2}-1)$
Added to illustrate comment: If the arrangement is not symmetric, you will not get two vertices on the arc and two on the axes. Instead you will get coomething like this
answered Jan 16 at 9:12
HenryHenry
101k482169
$endgroup$
$begingroup$
Didn't get it how you assigned the coordinates of each vertex, can you explain please?
$endgroup$
– Just_Cause
Jan 16 at 10:49
$begingroup$
@Just_Cause - Are you happy with the $(r costheta, rsintheta)$ as a point on the circle? That lies on the line $y=x + r(cos theta - sin theta)$ and so the next vertex is where that line intersects $y=0$. The other two vertices are reflections in the line $y=x$
$endgroup$
– Henry
Jan 16 at 13:20
$begingroup$
Sure, but how did you assume that the small triangle near the x axis is isosceles? (45 45 90)? Doesn't this lead to the fact that the bigger triangle is isosceles too?
$endgroup$
– Just_Cause
Jan 22 at 11:26
$begingroup$
@Just_Cause If you do not have symmetry then either the vertices are not on the arc and axes or the quadrilateral is not a rectangle. So the triangle is isosceles
$endgroup$
– Henry
Jan 22 at 12:49
$begingroup$
Still don't get what has symmetry got to do here, do you mean the symmetry line of the triangle, you used it to divide the bigger triangle into two halves which would bisect the 90 angle?
$endgroup$
– Just_Cause
Jan 27 at 21:00
|
show 1 more comment
$begingroup$
Choosing one vertex to be $(r costheta, rsintheta)$ then another is $(r (costheta- sintheta),0)$, a third is $(0,r (costheta- sintheta))$ and the fourth is $(r sintheta, rcostheta)$
The area is then $2r^2 sintheta(costheta- sintheta) = sqrt{8}r^2 sin(theta) sinleft(fracpi4-thetaright) $ which is maximised when $theta= fracpi8$ with the maximum area being $r^2(sqrt{2}-1)$
Added to illustrate comment: If the arrangement is not symmetric, you will not get two vertices on the arc and two on the axes. Instead you will get coomething like this
answered Jan 16 at 9:12
HenryHenry
101k482169
$endgroup$
$begingroup$
Didn't get it how you assigned the coordinates of each vertex, can you explain please?
$endgroup$
– Just_Cause
Jan 16 at 10:49
$begingroup$
@Just_Cause - Are you happy with the $(r costheta, rsintheta)$ as a point on the circle? That lies on the line $y=x + r(cos theta - sin theta)$ and so the next vertex is where that line intersects $y=0$. The other two vertices are reflections in the line $y=x$
$endgroup$
– Henry
Jan 16 at 13:20
$begingroup$
Sure, but how did you assume that the small triangle near the x axis is isosceles? (45 45 90)? Doesn't this lead to the fact that the bigger triangle is isosceles too?
$endgroup$
– Just_Cause
Jan 22 at 11:26
$begingroup$
@Just_Cause If you do not have symmetry then either the vertices are not on the arc and axes or the quadrilateral is not a rectangle. So the triangle is isosceles
$endgroup$
– Henry
Jan 22 at 12:49
$begingroup$
Still don't get what has symmetry got to do here, do you mean the symmetry line of the triangle, you used it to divide the bigger triangle into two halves which would bisect the 90 angle?
$endgroup$
– Just_Cause
Jan 27 at 21:00
|
show 1 more comment
$begingroup$
Choosing one vertex to be $(r costheta, rsintheta)$ then another is $(r (costheta- sintheta),0)$, a third is $(0,r (costheta- sintheta))$ and the fourth is $(r sintheta, rcostheta)$
The area is then $2r^2 sintheta(costheta- sintheta) = sqrt{8}r^2 sin(theta) sinleft(fracpi4-thetaright) $ which is maximised when $theta= fracpi8$ with the maximum area being $r^2(sqrt{2}-1)$
Added to illustrate comment: If the arrangement is not symmetric, you will not get two vertices on the arc and two on the axes. Instead you will get coomething like this
answered Jan 16 at 9:12
HenryHenry
101k482169
$endgroup$
Choosing one vertex to be $(r costheta, rsintheta)$ then another is $(r (costheta- sintheta),0)$, a third is $(0,r (costheta- sintheta))$ and the fourth is $(r sintheta, rcostheta)$
The area is then $2r^2 sintheta(costheta- sintheta) = sqrt{8}r^2 sin(theta) sinleft(fracpi4-thetaright) $ which is maximised when $theta= fracpi8$ with the maximum area being $r^2(sqrt{2}-1)$
Added to illustrate comment: If the arrangement is not symmetric, you will not get two vertices on the arc and two on the axes. Instead you will get coomething like this
answered Jan 16 at 9:12
HenryHenry
101k482169
edited Jan 28 at 1:15
answered Jan 16 at 9:12
HenryHenry
101k482169
answered Jan 16 at 9:12
HenryHenry
101k482169
answered Jan 16 at 9:12
HenryHenry
101k482169
101k482169
$begingroup$
Didn't get it how you assigned the coordinates of each vertex, can you explain please?
$endgroup$
– Just_Cause
Jan 16 at 10:49
$begingroup$
@Just_Cause - Are you happy with the $(r costheta, rsintheta)$ as a point on the circle? That lies on the line $y=x + r(cos theta - sin theta)$ and so the next vertex is where that line intersects $y=0$. The other two vertices are reflections in the line $y=x$
$endgroup$
– Henry
Jan 16 at 13:20
$begingroup$
Sure, but how did you assume that the small triangle near the x axis is isosceles? (45 45 90)? Doesn't this lead to the fact that the bigger triangle is isosceles too?
$endgroup$
– Just_Cause
Jan 22 at 11:26
$begingroup$
@Just_Cause If you do not have symmetry then either the vertices are not on the arc and axes or the quadrilateral is not a rectangle. So the triangle is isosceles
$endgroup$
– Henry
Jan 22 at 12:49
$begingroup$
Still don't get what has symmetry got to do here, do you mean the symmetry line of the triangle, you used it to divide the bigger triangle into two halves which would bisect the 90 angle?
$endgroup$
– Just_Cause
Jan 27 at 21:00
|
show 1 more comment
$begingroup$
Didn't get it how you assigned the coordinates of each vertex, can you explain please?
$endgroup$
– Just_Cause
Jan 16 at 10:49
$begingroup$
@Just_Cause - Are you happy with the $(r costheta, rsintheta)$ as a point on the circle? That lies on the line $y=x + r(cos theta - sin theta)$ and so the next vertex is where that line intersects $y=0$. The other two vertices are reflections in the line $y=x$
$endgroup$
– Henry
Jan 16 at 13:20
$begingroup$
Sure, but how did you assume that the small triangle near the x axis is isosceles? (45 45 90)? Doesn't this lead to the fact that the bigger triangle is isosceles too?
$endgroup$
– Just_Cause
Jan 22 at 11:26
$begingroup$
@Just_Cause If you do not have symmetry then either the vertices are not on the arc and axes or the quadrilateral is not a rectangle. So the triangle is isosceles
$endgroup$
– Henry
Jan 22 at 12:49
$begingroup$
Still don't get what has symmetry got to do here, do you mean the symmetry line of the triangle, you used it to divide the bigger triangle into two halves which would bisect the 90 angle?
$endgroup$
– Just_Cause
Jan 27 at 21:00
$begingroup$
Didn't get it how you assigned the coordinates of each vertex, can you explain please?
$endgroup$
– Just_Cause
Jan 16 at 10:49
$begingroup$
Didn't get it how you assigned the coordinates of each vertex, can you explain please?
$endgroup$
– Just_Cause
Jan 16 at 10:49
$begingroup$
@Just_Cause - Are you happy with the $(r costheta, rsintheta)$ as a point on the circle? That lies on the line $y=x + r(cos theta - sin theta)$ and so the next vertex is where that line intersects $y=0$. The other two vertices are reflections in the line $y=x$
$endgroup$
– Henry
Jan 16 at 13:20
$begingroup$
@Just_Cause - Are you happy with the $(r costheta, rsintheta)$ as a point on the circle? That lies on the line $y=x + r(cos theta - sin theta)$ and so the next vertex is where that line intersects $y=0$. The other two vertices are reflections in the line $y=x$
$endgroup$
– Henry
Jan 16 at 13:20
$begingroup$
Sure, but how did you assume that the small triangle near the x axis is isosceles? (45 45 90)? Doesn't this lead to the fact that the bigger triangle is isosceles too?
$endgroup$
– Just_Cause
Jan 22 at 11:26
$begingroup$
Sure, but how did you assume that the small triangle near the x axis is isosceles? (45 45 90)? Doesn't this lead to the fact that the bigger triangle is isosceles too?
$endgroup$
– Just_Cause
Jan 22 at 11:26
$begingroup$
@Just_Cause If you do not have symmetry then either the vertices are not on the arc and axes or the quadrilateral is not a rectangle. So the triangle is isosceles
$endgroup$
– Henry
Jan 22 at 12:49
$begingroup$
@Just_Cause If you do not have symmetry then either the vertices are not on the arc and axes or the quadrilateral is not a rectangle. So the triangle is isosceles
$endgroup$
– Henry
Jan 22 at 12:49
$begingroup$
Still don't get what has symmetry got to do here, do you mean the symmetry line of the triangle, you used it to divide the bigger triangle into two halves which would bisect the 90 angle?
$endgroup$
– Just_Cause
Jan 27 at 21:00
$begingroup$
Still don't get what has symmetry got to do here, do you mean the symmetry line of the triangle, you used it to divide the bigger triangle into two halves which would bisect the 90 angle?
$endgroup$
– Just_Cause
Jan 27 at 21:00
|
show 1 more comment
$begingroup$
I would compute the coordinates of the given points:
$$A(x_A,0),B(x_B,sqrt{6^2-x_B^2}),C(x_C,sqrt{6^2-x_C^2}),D(0,y_B)$$ Now it must be
$$vec{AB}=[x_B-x_A,sqrt{6^2-x_B^2}]$$ and $$vec{AD}=[-x_A,y_D]$$ and it must be
$$vec{AB}cdot vec{AD}=0$$
Can you proceed?
answered Jan 16 at 8:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
$begingroup$
Appreciated, but I'm not really required to solve it using vectors.
$endgroup$
– Just_Cause
Jan 16 at 10:44
add a comment |
$begingroup$
I would compute the coordinates of the given points:
$$A(x_A,0),B(x_B,sqrt{6^2-x_B^2}),C(x_C,sqrt{6^2-x_C^2}),D(0,y_B)$$ Now it must be
$$vec{AB}=[x_B-x_A,sqrt{6^2-x_B^2}]$$ and $$vec{AD}=[-x_A,y_D]$$ and it must be
$$vec{AB}cdot vec{AD}=0$$
Can you proceed?
answered Jan 16 at 8:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
$begingroup$
Appreciated, but I'm not really required to solve it using vectors.
$endgroup$
– Just_Cause
Jan 16 at 10:44
add a comment |
$begingroup$
I would compute the coordinates of the given points:
$$A(x_A,0),B(x_B,sqrt{6^2-x_B^2}),C(x_C,sqrt{6^2-x_C^2}),D(0,y_B)$$ Now it must be
$$vec{AB}=[x_B-x_A,sqrt{6^2-x_B^2}]$$ and $$vec{AD}=[-x_A,y_D]$$ and it must be
$$vec{AB}cdot vec{AD}=0$$
Can you proceed?
answered Jan 16 at 8:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
I would compute the coordinates of the given points:
$$A(x_A,0),B(x_B,sqrt{6^2-x_B^2}),C(x_C,sqrt{6^2-x_C^2}),D(0,y_B)$$ Now it must be
$$vec{AB}=[x_B-x_A,sqrt{6^2-x_B^2}]$$ and $$vec{AD}=[-x_A,y_D]$$ and it must be
$$vec{AB}cdot vec{AD}=0$$
Can you proceed?
answered Jan 16 at 8:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 16 at 8:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 16 at 8:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 16 at 8:55
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
78.4k42867
$begingroup$
Appreciated, but I'm not really required to solve it using vectors.
$endgroup$
– Just_Cause
Jan 16 at 10:44
add a comment |
$begingroup$
Appreciated, but I'm not really required to solve it using vectors.
$endgroup$
– Just_Cause
Jan 16 at 10:44
$begingroup$
Appreciated, but I'm not really required to solve it using vectors.
$endgroup$
– Just_Cause
Jan 16 at 10:44
$begingroup$
Appreciated, but I'm not really required to solve it using vectors.
$endgroup$
– Just_Cause
Jan 16 at 10:44
add a comment |
$begingroup$
Let $ABCD$ be our rectangle, where $A$ be placed on the perpendicular radius and $D$ be placed on the horizontal radius.
Also, let $OE$ be a radius of the circle such that $OEperp AD$, $OEcap BC={F}$, $OEcap AD={G}$ and let $HE$ be a diameter of the circle.
Now, since $G$ is a midpoint of $BC$, we obtain that $G$ is a midpoint of $AD$.
Thus, $OG=AG=DG$ and let $OG=x$, $DC=y$.
Now, since $measuredangle ECH=90^{circ},$ we obtain:
$$CE^2=EGcdot GH$$ or
$$x^2=(6-x-y)(6+x+y)$$ or
$$(x+y)^2=36-x^2$$ or
$$y=sqrt{36-x^2}-x.$$
Id est, $$S_{ABCD}=2xleft(sqrt{36-x^2}-xright).$$
Can you end it now?
answered Jan 16 at 9:15
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
$begingroup$
Sure, but how did you get to $CE^2=EGcdot GH$ ?
$endgroup$
– Just_Cause
Jan 16 at 10:44
$begingroup$
Because the triangle CGH is a right-angled triangle and CE is an altitude of the triangle.
$endgroup$
– Michael Rozenberg
Jan 16 at 11:32
$begingroup$
Oh, i got it. One last thing that confuses me, is it permissible to assume that the radius is perpendicular on AD without an additional information from the question that denotes that? i.e OE bisects AD? As far as i know, i can't put any assumptions unless the question states a key info on some theorem
$endgroup$
– Just_Cause
Jan 16 at 12:34
1
$begingroup$
@Just_Cause If $OEperp AD$ then $OEperp BC$, which gives $BF=FC$ and from here $AG=GD.$
$endgroup$
– Michael Rozenberg
Jan 16 at 13:23
add a comment |
$begingroup$
Let $ABCD$ be our rectangle, where $A$ be placed on the perpendicular radius and $D$ be placed on the horizontal radius.
Also, let $OE$ be a radius of the circle such that $OEperp AD$, $OEcap BC={F}$, $OEcap AD={G}$ and let $HE$ be a diameter of the circle.
Now, since $G$ is a midpoint of $BC$, we obtain that $G$ is a midpoint of $AD$.
Thus, $OG=AG=DG$ and let $OG=x$, $DC=y$.
Now, since $measuredangle ECH=90^{circ},$ we obtain:
$$CE^2=EGcdot GH$$ or
$$x^2=(6-x-y)(6+x+y)$$ or
$$(x+y)^2=36-x^2$$ or
$$y=sqrt{36-x^2}-x.$$
Id est, $$S_{ABCD}=2xleft(sqrt{36-x^2}-xright).$$
Can you end it now?
answered Jan 16 at 9:15
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
$begingroup$
Sure, but how did you get to $CE^2=EGcdot GH$ ?
$endgroup$
– Just_Cause
Jan 16 at 10:44
$begingroup$
Because the triangle CGH is a right-angled triangle and CE is an altitude of the triangle.
$endgroup$
– Michael Rozenberg
Jan 16 at 11:32
$begingroup$
Oh, i got it. One last thing that confuses me, is it permissible to assume that the radius is perpendicular on AD without an additional information from the question that denotes that? i.e OE bisects AD? As far as i know, i can't put any assumptions unless the question states a key info on some theorem
$endgroup$
– Just_Cause
Jan 16 at 12:34
1
$begingroup$
@Just_Cause If $OEperp AD$ then $OEperp BC$, which gives $BF=FC$ and from here $AG=GD.$
$endgroup$
– Michael Rozenberg
Jan 16 at 13:23
add a comment |
$begingroup$
Let $ABCD$ be our rectangle, where $A$ be placed on the perpendicular radius and $D$ be placed on the horizontal radius.
Also, let $OE$ be a radius of the circle such that $OEperp AD$, $OEcap BC={F}$, $OEcap AD={G}$ and let $HE$ be a diameter of the circle.
Now, since $G$ is a midpoint of $BC$, we obtain that $G$ is a midpoint of $AD$.
Thus, $OG=AG=DG$ and let $OG=x$, $DC=y$.
Now, since $measuredangle ECH=90^{circ},$ we obtain:
$$CE^2=EGcdot GH$$ or
$$x^2=(6-x-y)(6+x+y)$$ or
$$(x+y)^2=36-x^2$$ or
$$y=sqrt{36-x^2}-x.$$
Id est, $$S_{ABCD}=2xleft(sqrt{36-x^2}-xright).$$
Can you end it now?
answered Jan 16 at 9:15
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
Let $ABCD$ be our rectangle, where $A$ be placed on the perpendicular radius and $D$ be placed on the horizontal radius.
Also, let $OE$ be a radius of the circle such that $OEperp AD$, $OEcap BC={F}$, $OEcap AD={G}$ and let $HE$ be a diameter of the circle.
Now, since $G$ is a midpoint of $BC$, we obtain that $G$ is a midpoint of $AD$.
Thus, $OG=AG=DG$ and let $OG=x$, $DC=y$.
Now, since $measuredangle ECH=90^{circ},$ we obtain:
$$CE^2=EGcdot GH$$ or
$$x^2=(6-x-y)(6+x+y)$$ or
$$(x+y)^2=36-x^2$$ or
$$y=sqrt{36-x^2}-x.$$
Id est, $$S_{ABCD}=2xleft(sqrt{36-x^2}-xright).$$
Can you end it now?
answered Jan 16 at 9:15
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 16 at 9:15
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 16 at 9:15
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 16 at 9:15
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
$begingroup$
Sure, but how did you get to $CE^2=EGcdot GH$ ?
$endgroup$
– Just_Cause
Jan 16 at 10:44
$begingroup$
Because the triangle CGH is a right-angled triangle and CE is an altitude of the triangle.
$endgroup$
– Michael Rozenberg
Jan 16 at 11:32
$begingroup$
Oh, i got it. One last thing that confuses me, is it permissible to assume that the radius is perpendicular on AD without an additional information from the question that denotes that? i.e OE bisects AD? As far as i know, i can't put any assumptions unless the question states a key info on some theorem
$endgroup$
– Just_Cause
Jan 16 at 12:34
1
$begingroup$
@Just_Cause If $OEperp AD$ then $OEperp BC$, which gives $BF=FC$ and from here $AG=GD.$
$endgroup$
– Michael Rozenberg
Jan 16 at 13:23
add a comment |
$begingroup$
Sure, but how did you get to $CE^2=EGcdot GH$ ?
$endgroup$
– Just_Cause
Jan 16 at 10:44
$begingroup$
Because the triangle CGH is a right-angled triangle and CE is an altitude of the triangle.
$endgroup$
– Michael Rozenberg
Jan 16 at 11:32
$begingroup$
Oh, i got it. One last thing that confuses me, is it permissible to assume that the radius is perpendicular on AD without an additional information from the question that denotes that? i.e OE bisects AD? As far as i know, i can't put any assumptions unless the question states a key info on some theorem
$endgroup$
– Just_Cause
Jan 16 at 12:34
1
$begingroup$
@Just_Cause If $OEperp AD$ then $OEperp BC$, which gives $BF=FC$ and from here $AG=GD.$
$endgroup$
– Michael Rozenberg
Jan 16 at 13:23
$begingroup$
Sure, but how did you get to $CE^2=EGcdot GH$ ?
$endgroup$
– Just_Cause
Jan 16 at 10:44
$begingroup$
Sure, but how did you get to $CE^2=EGcdot GH$ ?
$endgroup$
– Just_Cause
Jan 16 at 10:44
$begingroup$
Because the triangle CGH is a right-angled triangle and CE is an altitude of the triangle.
$endgroup$
– Michael Rozenberg
Jan 16 at 11:32
$begingroup$
Because the triangle CGH is a right-angled triangle and CE is an altitude of the triangle.
$endgroup$
– Michael Rozenberg
Jan 16 at 11:32
$begingroup$
Oh, i got it. One last thing that confuses me, is it permissible to assume that the radius is perpendicular on AD without an additional information from the question that denotes that? i.e OE bisects AD? As far as i know, i can't put any assumptions unless the question states a key info on some theorem
$endgroup$
– Just_Cause
Jan 16 at 12:34
$begingroup$
Oh, i got it. One last thing that confuses me, is it permissible to assume that the radius is perpendicular on AD without an additional information from the question that denotes that? i.e OE bisects AD? As far as i know, i can't put any assumptions unless the question states a key info on some theorem
$endgroup$
– Just_Cause
Jan 16 at 12:34
1
1
$begingroup$
@Just_Cause If $OEperp AD$ then $OEperp BC$, which gives $BF=FC$ and from here $AG=GD.$
$endgroup$
– Michael Rozenberg
Jan 16 at 13:23
$begingroup$
@Just_Cause If $OEperp AD$ then $OEperp BC$, which gives $BF=FC$ and from here $AG=GD.$
$endgroup$
– Michael Rozenberg
Jan 16 at 13:23
add a comment |
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$begingroup$
Maybe you can parametrise the rectangle? Maybe set the width of the rectangle to be $x$ and then calculate its area ...
$endgroup$
– Matti P.
Jan 16 at 8:45
$begingroup$
I tried, but I couldn't find any useful relationship between the width and the height.
$endgroup$
– Just_Cause
Jan 16 at 8:49
$begingroup$
Just make a big drawing, draw some auxiliary lines and see what you can come up with.
$endgroup$
– Matti P.
Jan 16 at 9:01
1
$begingroup$
I suspect the square with a vertex at the origin and opposite vertex at $(3sqrt{2},3sqrt{2}$)$ with area $18$ may be bigger
$endgroup$
– Henry
Jan 16 at 9:14