Mixing
Octagon inside a circle
Clash Royale CLAN TAG#URR8PPP
up vote
7
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favorite
An octagon which has side lengths 3, 3, 11, 11, 15, 15, 15 and 15 is inscribed
in a circle. What is the area of the octagon?
I tried using the cosine law on the triangles made when connected with the center but the numbers became really hard to use when I solved for the sine of the angles to easily get the triangle area.
This is a problem from BIMC 2017 individual question 6, where the answer is 567. ;)
geometry contest-math
edited Jul 16 at 10:03
Parcly Taxel
33.6k136588
asked Jul 16 at 8:49
SuperMage1
687210
 |Â
show 1 more comment
up vote
7
down vote
favorite
An octagon which has side lengths 3, 3, 11, 11, 15, 15, 15 and 15 is inscribed
in a circle. What is the area of the octagon?
I tried using the cosine law on the triangles made when connected with the center but the numbers became really hard to use when I solved for the sine of the angles to easily get the triangle area.
This is a problem from BIMC 2017 individual question 6, where the answer is 567. ;)
geometry contest-math
edited Jul 16 at 10:03
Parcly Taxel
33.6k136588
asked Jul 16 at 8:49
SuperMage1
687210
Are those side lengths in order?
– Arthur
Jul 16 at 9:04
3
@Arthur I don't think it matters. For a given radius, the area of the triangles depends only on the sides. But since the central angles have to add up to $2pi,$ you should be able to solve for the radius.
– saulspatz
Jul 16 at 9:07
1
Where does this problem come from? So far, I can't see any way to do it except numerically. Is there reason to believe it has some clever exact solution?
– saulspatz
Jul 16 at 9:09
2
Given that the order does not matter, I'd put them in the order $3, 15, 11, 15, 3, 15, 11, 15$. The symmetry not only makes opposite sides parallel, the sides of length $3$ will be perpendicular to the sides of length $11$. That has to help a lot.
– Jaap Scherphuis
Jul 16 at 9:43
This problem from BIMC 2017 individual question 6 where the answer is 567 ;)
– SuperMage1
Jul 16 at 9:52
 |Â
show 1 more comment
up vote
7
down vote
favorite
up vote
7
down vote
favorite
An octagon which has side lengths 3, 3, 11, 11, 15, 15, 15 and 15 is inscribed
in a circle. What is the area of the octagon?
I tried using the cosine law on the triangles made when connected with the center but the numbers became really hard to use when I solved for the sine of the angles to easily get the triangle area.
This is a problem from BIMC 2017 individual question 6, where the answer is 567. ;)
geometry contest-math
edited Jul 16 at 10:03
Parcly Taxel
33.6k136588
asked Jul 16 at 8:49
SuperMage1
687210
An octagon which has side lengths 3, 3, 11, 11, 15, 15, 15 and 15 is inscribed
in a circle. What is the area of the octagon?
I tried using the cosine law on the triangles made when connected with the center but the numbers became really hard to use when I solved for the sine of the angles to easily get the triangle area.
This is a problem from BIMC 2017 individual question 6, where the answer is 567. ;)
geometry contest-math
edited Jul 16 at 10:03
Parcly Taxel
33.6k136588
asked Jul 16 at 8:49
SuperMage1
687210
edited Jul 16 at 10:03
Parcly Taxel
33.6k136588
edited Jul 16 at 10:03
Parcly Taxel
33.6k136588
edited Jul 16 at 10:03
Parcly Taxel
33.6k136588
33.6k136588
asked Jul 16 at 8:49
SuperMage1
687210
asked Jul 16 at 8:49
SuperMage1
687210
asked Jul 16 at 8:49
SuperMage1
687210
687210
Are those side lengths in order?
– Arthur
Jul 16 at 9:04
3
@Arthur I don't think it matters. For a given radius, the area of the triangles depends only on the sides. But since the central angles have to add up to $2pi,$ you should be able to solve for the radius.
– saulspatz
Jul 16 at 9:07
1
Where does this problem come from? So far, I can't see any way to do it except numerically. Is there reason to believe it has some clever exact solution?
– saulspatz
Jul 16 at 9:09
2
Given that the order does not matter, I'd put them in the order $3, 15, 11, 15, 3, 15, 11, 15$. The symmetry not only makes opposite sides parallel, the sides of length $3$ will be perpendicular to the sides of length $11$. That has to help a lot.
– Jaap Scherphuis
Jul 16 at 9:43
This problem from BIMC 2017 individual question 6 where the answer is 567 ;)
– SuperMage1
Jul 16 at 9:52
 |Â
show 1 more comment
Are those side lengths in order?
– Arthur
Jul 16 at 9:04
3
@Arthur I don't think it matters. For a given radius, the area of the triangles depends only on the sides. But since the central angles have to add up to $2pi,$ you should be able to solve for the radius.
– saulspatz
Jul 16 at 9:07
1
Where does this problem come from? So far, I can't see any way to do it except numerically. Is there reason to believe it has some clever exact solution?
– saulspatz
Jul 16 at 9:09
2
Given that the order does not matter, I'd put them in the order $3, 15, 11, 15, 3, 15, 11, 15$. The symmetry not only makes opposite sides parallel, the sides of length $3$ will be perpendicular to the sides of length $11$. That has to help a lot.
– Jaap Scherphuis
Jul 16 at 9:43
This problem from BIMC 2017 individual question 6 where the answer is 567 ;)
– SuperMage1
Jul 16 at 9:52
Are those side lengths in order?
– Arthur
Jul 16 at 9:04
Are those side lengths in order?
– Arthur
Jul 16 at 9:04
3
3
@Arthur I don't think it matters. For a given radius, the area of the triangles depends only on the sides. But since the central angles have to add up to $2pi,$ you should be able to solve for the radius.
– saulspatz
Jul 16 at 9:07
@Arthur I don't think it matters. For a given radius, the area of the triangles depends only on the sides. But since the central angles have to add up to $2pi,$ you should be able to solve for the radius.
– saulspatz
Jul 16 at 9:07
1
1
Where does this problem come from? So far, I can't see any way to do it except numerically. Is there reason to believe it has some clever exact solution?
– saulspatz
Jul 16 at 9:09
Where does this problem come from? So far, I can't see any way to do it except numerically. Is there reason to believe it has some clever exact solution?
– saulspatz
Jul 16 at 9:09
2
2
Given that the order does not matter, I'd put them in the order $3, 15, 11, 15, 3, 15, 11, 15$. The symmetry not only makes opposite sides parallel, the sides of length $3$ will be perpendicular to the sides of length $11$. That has to help a lot.
– Jaap Scherphuis
Jul 16 at 9:43
Given that the order does not matter, I'd put them in the order $3, 15, 11, 15, 3, 15, 11, 15$. The symmetry not only makes opposite sides parallel, the sides of length $3$ will be perpendicular to the sides of length $11$. That has to help a lot.
– Jaap Scherphuis
Jul 16 at 9:43
This problem from BIMC 2017 individual question 6 where the answer is 567 ;)
– SuperMage1
Jul 16 at 9:52
This problem from BIMC 2017 individual question 6 where the answer is 567 ;)
– SuperMage1
Jul 16 at 9:52
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
13
down vote
Arranging the sides in the order $3,15,11,15,3,15,11,15$, we obtain two overlapping, perpendicular inscribed rectangles of sides $3×h$ and $11×k$ respectively, where $h,k$ are to be determined. The Pythagorean theorem gives
$$3^2+h^2=11^2+k^2$$
$$h^2-k^2=112$$
Furthermore, the rectangles are aligned and have a common centroid. This gives
$$left(frach-112right)^2+left(frack-32right)^2=15^2$$
$$(h-11)^2+(k-3)^2=30^2$$
This immediately suggests finding Pythagorean triples $(a,b,30)$, of which there is only one: $18^2+24^2=30^2$. If we let $h-11=18$ and $k-3=24$, we see that the other equation $h^2-k^2=112$ is "miraculously" satisfied.
Thus $h=29$ and $k=27$. The octagon's area can then be calculated easily by adding the rectangles' areas, subtracting their intersection and adding the triangles' areas:
$$3h+11k-33+2left(frach-112right)left(frack-32right)=87+297-33+2cdot9cdot12=567$$
answered Jul 16 at 10:03
Parcly Taxel
33.6k136588
I've neatened up your figure with Geogebra, if you'd like i.imgur.com/ztpQFxc.png
– Jam
Jul 17 at 12:22
1
@Jam one of the traditions I've had with my answers here is hand-drawn diagrams, particularly because I answer most questions on my mobile phone. But a neat diagram is always good to have.
– Parcly Taxel
Jul 17 at 12:25
add a comment |Â
up vote
2
down vote
Arrange the triangles so there is a $15$ in each quadrant; the $3$ sides are vertical and the $11$ sides are horizontal. Then the distance between $(sqrtr^2-9/4,3/2)$ and $(11/2,sqrtr^2-121/4)$ is 15.
answered Jul 16 at 9:38
Michael
2,577213
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
13
down vote
Arranging the sides in the order $3,15,11,15,3,15,11,15$, we obtain two overlapping, perpendicular inscribed rectangles of sides $3×h$ and $11×k$ respectively, where $h,k$ are to be determined. The Pythagorean theorem gives
$$3^2+h^2=11^2+k^2$$
$$h^2-k^2=112$$
Furthermore, the rectangles are aligned and have a common centroid. This gives
$$left(frach-112right)^2+left(frack-32right)^2=15^2$$
$$(h-11)^2+(k-3)^2=30^2$$
This immediately suggests finding Pythagorean triples $(a,b,30)$, of which there is only one: $18^2+24^2=30^2$. If we let $h-11=18$ and $k-3=24$, we see that the other equation $h^2-k^2=112$ is "miraculously" satisfied.
Thus $h=29$ and $k=27$. The octagon's area can then be calculated easily by adding the rectangles' areas, subtracting their intersection and adding the triangles' areas:
$$3h+11k-33+2left(frach-112right)left(frack-32right)=87+297-33+2cdot9cdot12=567$$
answered Jul 16 at 10:03
Parcly Taxel
33.6k136588
I've neatened up your figure with Geogebra, if you'd like i.imgur.com/ztpQFxc.png
– Jam
Jul 17 at 12:22
1
@Jam one of the traditions I've had with my answers here is hand-drawn diagrams, particularly because I answer most questions on my mobile phone. But a neat diagram is always good to have.
– Parcly Taxel
Jul 17 at 12:25
add a comment |Â
up vote
13
down vote
Arranging the sides in the order $3,15,11,15,3,15,11,15$, we obtain two overlapping, perpendicular inscribed rectangles of sides $3×h$ and $11×k$ respectively, where $h,k$ are to be determined. The Pythagorean theorem gives
$$3^2+h^2=11^2+k^2$$
$$h^2-k^2=112$$
Furthermore, the rectangles are aligned and have a common centroid. This gives
$$left(frach-112right)^2+left(frack-32right)^2=15^2$$
$$(h-11)^2+(k-3)^2=30^2$$
This immediately suggests finding Pythagorean triples $(a,b,30)$, of which there is only one: $18^2+24^2=30^2$. If we let $h-11=18$ and $k-3=24$, we see that the other equation $h^2-k^2=112$ is "miraculously" satisfied.
Thus $h=29$ and $k=27$. The octagon's area can then be calculated easily by adding the rectangles' areas, subtracting their intersection and adding the triangles' areas:
$$3h+11k-33+2left(frach-112right)left(frack-32right)=87+297-33+2cdot9cdot12=567$$
answered Jul 16 at 10:03
Parcly Taxel
33.6k136588
I've neatened up your figure with Geogebra, if you'd like i.imgur.com/ztpQFxc.png
– Jam
Jul 17 at 12:22
1
@Jam one of the traditions I've had with my answers here is hand-drawn diagrams, particularly because I answer most questions on my mobile phone. But a neat diagram is always good to have.
– Parcly Taxel
Jul 17 at 12:25
add a comment |Â
up vote
13
down vote
up vote
13
down vote
Arranging the sides in the order $3,15,11,15,3,15,11,15$, we obtain two overlapping, perpendicular inscribed rectangles of sides $3×h$ and $11×k$ respectively, where $h,k$ are to be determined. The Pythagorean theorem gives
$$3^2+h^2=11^2+k^2$$
$$h^2-k^2=112$$
Furthermore, the rectangles are aligned and have a common centroid. This gives
$$left(frach-112right)^2+left(frack-32right)^2=15^2$$
$$(h-11)^2+(k-3)^2=30^2$$
This immediately suggests finding Pythagorean triples $(a,b,30)$, of which there is only one: $18^2+24^2=30^2$. If we let $h-11=18$ and $k-3=24$, we see that the other equation $h^2-k^2=112$ is "miraculously" satisfied.
Thus $h=29$ and $k=27$. The octagon's area can then be calculated easily by adding the rectangles' areas, subtracting their intersection and adding the triangles' areas:
$$3h+11k-33+2left(frach-112right)left(frack-32right)=87+297-33+2cdot9cdot12=567$$
answered Jul 16 at 10:03
Parcly Taxel
33.6k136588
Arranging the sides in the order $3,15,11,15,3,15,11,15$, we obtain two overlapping, perpendicular inscribed rectangles of sides $3×h$ and $11×k$ respectively, where $h,k$ are to be determined. The Pythagorean theorem gives
$$3^2+h^2=11^2+k^2$$
$$h^2-k^2=112$$
Furthermore, the rectangles are aligned and have a common centroid. This gives
$$left(frach-112right)^2+left(frack-32right)^2=15^2$$
$$(h-11)^2+(k-3)^2=30^2$$
This immediately suggests finding Pythagorean triples $(a,b,30)$, of which there is only one: $18^2+24^2=30^2$. If we let $h-11=18$ and $k-3=24$, we see that the other equation $h^2-k^2=112$ is "miraculously" satisfied.
Thus $h=29$ and $k=27$. The octagon's area can then be calculated easily by adding the rectangles' areas, subtracting their intersection and adding the triangles' areas:
$$3h+11k-33+2left(frach-112right)left(frack-32right)=87+297-33+2cdot9cdot12=567$$
answered Jul 16 at 10:03
Parcly Taxel
33.6k136588
edited Jul 17 at 12:41
answered Jul 16 at 10:03
Parcly Taxel
33.6k136588
answered Jul 16 at 10:03
Parcly Taxel
33.6k136588
answered Jul 16 at 10:03
Parcly Taxel
33.6k136588
33.6k136588
I've neatened up your figure with Geogebra, if you'd like i.imgur.com/ztpQFxc.png
– Jam
Jul 17 at 12:22
1
@Jam one of the traditions I've had with my answers here is hand-drawn diagrams, particularly because I answer most questions on my mobile phone. But a neat diagram is always good to have.
– Parcly Taxel
Jul 17 at 12:25
add a comment |Â
I've neatened up your figure with Geogebra, if you'd like i.imgur.com/ztpQFxc.png
– Jam
Jul 17 at 12:22
1
@Jam one of the traditions I've had with my answers here is hand-drawn diagrams, particularly because I answer most questions on my mobile phone. But a neat diagram is always good to have.
– Parcly Taxel
Jul 17 at 12:25
I've neatened up your figure with Geogebra, if you'd like i.imgur.com/ztpQFxc.png
– Jam
Jul 17 at 12:22
I've neatened up your figure with Geogebra, if you'd like i.imgur.com/ztpQFxc.png
– Jam
Jul 17 at 12:22
1
1
@Jam one of the traditions I've had with my answers here is hand-drawn diagrams, particularly because I answer most questions on my mobile phone. But a neat diagram is always good to have.
– Parcly Taxel
Jul 17 at 12:25
@Jam one of the traditions I've had with my answers here is hand-drawn diagrams, particularly because I answer most questions on my mobile phone. But a neat diagram is always good to have.
– Parcly Taxel
Jul 17 at 12:25
add a comment |Â
up vote
2
down vote
Arrange the triangles so there is a $15$ in each quadrant; the $3$ sides are vertical and the $11$ sides are horizontal. Then the distance between $(sqrtr^2-9/4,3/2)$ and $(11/2,sqrtr^2-121/4)$ is 15.
answered Jul 16 at 9:38
Michael
2,577213
add a comment |Â
up vote
2
down vote
Arrange the triangles so there is a $15$ in each quadrant; the $3$ sides are vertical and the $11$ sides are horizontal. Then the distance between $(sqrtr^2-9/4,3/2)$ and $(11/2,sqrtr^2-121/4)$ is 15.
answered Jul 16 at 9:38
Michael
2,577213
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Arrange the triangles so there is a $15$ in each quadrant; the $3$ sides are vertical and the $11$ sides are horizontal. Then the distance between $(sqrtr^2-9/4,3/2)$ and $(11/2,sqrtr^2-121/4)$ is 15.
answered Jul 16 at 9:38
Michael
2,577213
Arrange the triangles so there is a $15$ in each quadrant; the $3$ sides are vertical and the $11$ sides are horizontal. Then the distance between $(sqrtr^2-9/4,3/2)$ and $(11/2,sqrtr^2-121/4)$ is 15.
answered Jul 16 at 9:38
Michael
2,577213
answered Jul 16 at 9:38
Michael
2,577213
answered Jul 16 at 9:38
Michael
2,577213
answered Jul 16 at 9:38
Michael
2,577213
2,577213
add a comment |Â
add a comment |Â
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Are those side lengths in order?
– Arthur
Jul 16 at 9:04
3
@Arthur I don't think it matters. For a given radius, the area of the triangles depends only on the sides. But since the central angles have to add up to $2pi,$ you should be able to solve for the radius.
– saulspatz
Jul 16 at 9:07
1
Where does this problem come from? So far, I can't see any way to do it except numerically. Is there reason to believe it has some clever exact solution?
– saulspatz
Jul 16 at 9:09
2
Given that the order does not matter, I'd put them in the order $3, 15, 11, 15, 3, 15, 11, 15$. The symmetry not only makes opposite sides parallel, the sides of length $3$ will be perpendicular to the sides of length $11$. That has to help a lot.
– Jaap Scherphuis
Jul 16 at 9:43
This problem from BIMC 2017 individual question 6 where the answer is 567 ;)
– SuperMage1
Jul 16 at 9:52