Mixing
Two circles and Four circles inside a regular hexagoan
$begingroup$
Problem Link: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1228
given the side of the regular hexagon. We have to find the radius of the circles in the above four cases in the picture.
can't find the 2nd and 4th case...
1st case: is six equlatiral triangle and the area is sqrt(3)/4*side^2 = 1/2*side*height.
so height =sqrt(3)/2*side
3rd case : is 2*height =sqrt(3)/2*side
so height = (sqrt(3)/2*side)/2
here height is the radius.
now I need to find the radius of circle in the 2nd and fourth case and in all case circle have the same radius.
geometry circles
asked Jan 29 at 11:44
Md. Raihanur RahmanMd. Raihanur Rahman
183
$endgroup$
|
show 1 more comment
$begingroup$
Problem Link: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1228
given the side of the regular hexagon. We have to find the radius of the circles in the above four cases in the picture.
can't find the 2nd and 4th case...
1st case: is six equlatiral triangle and the area is sqrt(3)/4*side^2 = 1/2*side*height.
so height =sqrt(3)/2*side
3rd case : is 2*height =sqrt(3)/2*side
so height = (sqrt(3)/2*side)/2
here height is the radius.
now I need to find the radius of circle in the 2nd and fourth case and in all case circle have the same radius.
geometry circles
asked Jan 29 at 11:44
Md. Raihanur RahmanMd. Raihanur Rahman
183
$endgroup$
$begingroup$
how can I show the picture instead of link
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 11:49
$begingroup$
I don't have enough teputation for that and I've done it for you.
$endgroup$
– José Carlos Santos
Jan 29 at 11:51
$begingroup$
What exactly is the question ?
$endgroup$
– Overmind
Jan 29 at 12:05
$begingroup$
For case 2. Draw radius connecting the center with point in which hexagon is tangent to the circle. Thus find distance of the circle center to the closest hexagon edge...
$endgroup$
– Matteo
Jan 29 at 12:12
$begingroup$
sorry don't understand the case 2.
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 12:28
|
show 1 more comment
$begingroup$
Problem Link: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1228
given the side of the regular hexagon. We have to find the radius of the circles in the above four cases in the picture.
can't find the 2nd and 4th case...
1st case: is six equlatiral triangle and the area is sqrt(3)/4*side^2 = 1/2*side*height.
so height =sqrt(3)/2*side
3rd case : is 2*height =sqrt(3)/2*side
so height = (sqrt(3)/2*side)/2
here height is the radius.
now I need to find the radius of circle in the 2nd and fourth case and in all case circle have the same radius.
geometry circles
asked Jan 29 at 11:44
Md. Raihanur RahmanMd. Raihanur Rahman
183
$endgroup$
Problem Link: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1228
given the side of the regular hexagon. We have to find the radius of the circles in the above four cases in the picture.
can't find the 2nd and 4th case...
1st case: is six equlatiral triangle and the area is sqrt(3)/4*side^2 = 1/2*side*height.
so height =sqrt(3)/2*side
3rd case : is 2*height =sqrt(3)/2*side
so height = (sqrt(3)/2*side)/2
here height is the radius.
now I need to find the radius of circle in the 2nd and fourth case and in all case circle have the same radius.
geometry circles
geometry circles
asked Jan 29 at 11:44
Md. Raihanur RahmanMd. Raihanur Rahman
183
asked Jan 29 at 11:44
Md. Raihanur RahmanMd. Raihanur Rahman
183
edited Feb 26 at 17:07
Md. Raihanur Rahman
asked Jan 29 at 11:44
Md. Raihanur RahmanMd. Raihanur Rahman
183
asked Jan 29 at 11:44
Md. Raihanur RahmanMd. Raihanur Rahman
183
asked Jan 29 at 11:44
Md. Raihanur RahmanMd. Raihanur Rahman
183
183
$begingroup$
how can I show the picture instead of link
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 11:49
$begingroup$
I don't have enough teputation for that and I've done it for you.
$endgroup$
– José Carlos Santos
Jan 29 at 11:51
$begingroup$
What exactly is the question ?
$endgroup$
– Overmind
Jan 29 at 12:05
$begingroup$
For case 2. Draw radius connecting the center with point in which hexagon is tangent to the circle. Thus find distance of the circle center to the closest hexagon edge...
$endgroup$
– Matteo
Jan 29 at 12:12
$begingroup$
sorry don't understand the case 2.
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 12:28
|
show 1 more comment
$begingroup$
how can I show the picture instead of link
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 11:49
$begingroup$
I don't have enough teputation for that and I've done it for you.
$endgroup$
– José Carlos Santos
Jan 29 at 11:51
$begingroup$
What exactly is the question ?
$endgroup$
– Overmind
Jan 29 at 12:05
$begingroup$
For case 2. Draw radius connecting the center with point in which hexagon is tangent to the circle. Thus find distance of the circle center to the closest hexagon edge...
$endgroup$
– Matteo
Jan 29 at 12:12
$begingroup$
sorry don't understand the case 2.
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 12:28
$begingroup$
how can I show the picture instead of link
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 11:49
$begingroup$
how can I show the picture instead of link
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 11:49
$begingroup$
I don't have enough teputation for that and I've done it for you.
$endgroup$
– José Carlos Santos
Jan 29 at 11:51
$begingroup$
I don't have enough teputation for that and I've done it for you.
$endgroup$
– José Carlos Santos
Jan 29 at 11:51
$begingroup$
What exactly is the question ?
$endgroup$
– Overmind
Jan 29 at 12:05
$begingroup$
What exactly is the question ?
$endgroup$
– Overmind
Jan 29 at 12:05
$begingroup$
For case 2. Draw radius connecting the center with point in which hexagon is tangent to the circle. Thus find distance of the circle center to the closest hexagon edge...
$endgroup$
– Matteo
Jan 29 at 12:12
$begingroup$
For case 2. Draw radius connecting the center with point in which hexagon is tangent to the circle. Thus find distance of the circle center to the closest hexagon edge...
$endgroup$
– Matteo
Jan 29 at 12:12
$begingroup$
sorry don't understand the case 2.
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 12:28
$begingroup$
sorry don't understand the case 2.
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 12:28
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Denote the side of the hexagon as $a$:
Case 2:
The distance $H$ between parallel sides of the hexagon is $asqrt3$. On the other side (look at the picture):
$$H=asqrt3=r+(2r)frac{sqrt3}{2}+r=r(2+sqrt3)$$
So the radius of the circle is:
$$r=frac{asqrt3}{2+sqrt3}=a(2sqrt3-3)approx0.464a$$
Case 4:
First equation:
$$PQ=PB+ABcosalpha+ADcosalpha+DQ$$
$$2a=2frac{2rsqrt3}{3}+2(2r)cosalpha$$
$$a=frac{2rsqrt3}{3}+2rcosalpha$$
$$a-frac{2rsqrt3}{3}=2rcosalphatag{1}$$
Second equation:
$$EF=EA+ADsinalpha+DCsinalpha+CF$$
$$asqrt3=2r+2(2r)sinalpha$$
$$frac{asqrt3}{2}=r+2rsinalpha$$
$$frac{asqrt3}{2}-r=2rsinalphatag{2}$$
Square (1) and (2) and add to eliminate $alpha$:
$$(a-frac{2rsqrt3}{3})^2+(frac{asqrt3}{2}-r)^2=4r^2$$
This is a simple quadratic equation with two solutions but only one of them is positive:
$$r=afrac{6sqrt7-7sqrt3}{10}approx0.375a$$
answered Jan 29 at 13:14
OldboyOldboy
9,13111138
$endgroup$
$begingroup$
Lots of thanks I understand the 2nd case ...first I try to solve 4th case my own then I see yours...thanks a lot.
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 13:57
$begingroup$
@Md.RaihanurRahman If you like the answer, consider upvoting and accepting it.
$endgroup$
– Oldboy
Jan 29 at 14:03
$begingroup$
yeah I did that...thank
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 14:05
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Denote the side of the hexagon as $a$:
Case 2:
The distance $H$ between parallel sides of the hexagon is $asqrt3$. On the other side (look at the picture):
$$H=asqrt3=r+(2r)frac{sqrt3}{2}+r=r(2+sqrt3)$$
So the radius of the circle is:
$$r=frac{asqrt3}{2+sqrt3}=a(2sqrt3-3)approx0.464a$$
Case 4:
First equation:
$$PQ=PB+ABcosalpha+ADcosalpha+DQ$$
$$2a=2frac{2rsqrt3}{3}+2(2r)cosalpha$$
$$a=frac{2rsqrt3}{3}+2rcosalpha$$
$$a-frac{2rsqrt3}{3}=2rcosalphatag{1}$$
Second equation:
$$EF=EA+ADsinalpha+DCsinalpha+CF$$
$$asqrt3=2r+2(2r)sinalpha$$
$$frac{asqrt3}{2}=r+2rsinalpha$$
$$frac{asqrt3}{2}-r=2rsinalphatag{2}$$
Square (1) and (2) and add to eliminate $alpha$:
$$(a-frac{2rsqrt3}{3})^2+(frac{asqrt3}{2}-r)^2=4r^2$$
This is a simple quadratic equation with two solutions but only one of them is positive:
$$r=afrac{6sqrt7-7sqrt3}{10}approx0.375a$$
answered Jan 29 at 13:14
OldboyOldboy
9,13111138
$endgroup$
$begingroup$
Lots of thanks I understand the 2nd case ...first I try to solve 4th case my own then I see yours...thanks a lot.
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 13:57
$begingroup$
@Md.RaihanurRahman If you like the answer, consider upvoting and accepting it.
$endgroup$
– Oldboy
Jan 29 at 14:03
$begingroup$
yeah I did that...thank
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 14:05
add a comment |
$begingroup$
Denote the side of the hexagon as $a$:
Case 2:
The distance $H$ between parallel sides of the hexagon is $asqrt3$. On the other side (look at the picture):
$$H=asqrt3=r+(2r)frac{sqrt3}{2}+r=r(2+sqrt3)$$
So the radius of the circle is:
$$r=frac{asqrt3}{2+sqrt3}=a(2sqrt3-3)approx0.464a$$
Case 4:
First equation:
$$PQ=PB+ABcosalpha+ADcosalpha+DQ$$
$$2a=2frac{2rsqrt3}{3}+2(2r)cosalpha$$
$$a=frac{2rsqrt3}{3}+2rcosalpha$$
$$a-frac{2rsqrt3}{3}=2rcosalphatag{1}$$
Second equation:
$$EF=EA+ADsinalpha+DCsinalpha+CF$$
$$asqrt3=2r+2(2r)sinalpha$$
$$frac{asqrt3}{2}=r+2rsinalpha$$
$$frac{asqrt3}{2}-r=2rsinalphatag{2}$$
Square (1) and (2) and add to eliminate $alpha$:
$$(a-frac{2rsqrt3}{3})^2+(frac{asqrt3}{2}-r)^2=4r^2$$
This is a simple quadratic equation with two solutions but only one of them is positive:
$$r=afrac{6sqrt7-7sqrt3}{10}approx0.375a$$
answered Jan 29 at 13:14
OldboyOldboy
9,13111138
$endgroup$
$begingroup$
Lots of thanks I understand the 2nd case ...first I try to solve 4th case my own then I see yours...thanks a lot.
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 13:57
$begingroup$
@Md.RaihanurRahman If you like the answer, consider upvoting and accepting it.
$endgroup$
– Oldboy
Jan 29 at 14:03
$begingroup$
yeah I did that...thank
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 14:05
add a comment |
$begingroup$
Denote the side of the hexagon as $a$:
Case 2:
The distance $H$ between parallel sides of the hexagon is $asqrt3$. On the other side (look at the picture):
$$H=asqrt3=r+(2r)frac{sqrt3}{2}+r=r(2+sqrt3)$$
So the radius of the circle is:
$$r=frac{asqrt3}{2+sqrt3}=a(2sqrt3-3)approx0.464a$$
Case 4:
First equation:
$$PQ=PB+ABcosalpha+ADcosalpha+DQ$$
$$2a=2frac{2rsqrt3}{3}+2(2r)cosalpha$$
$$a=frac{2rsqrt3}{3}+2rcosalpha$$
$$a-frac{2rsqrt3}{3}=2rcosalphatag{1}$$
Second equation:
$$EF=EA+ADsinalpha+DCsinalpha+CF$$
$$asqrt3=2r+2(2r)sinalpha$$
$$frac{asqrt3}{2}=r+2rsinalpha$$
$$frac{asqrt3}{2}-r=2rsinalphatag{2}$$
Square (1) and (2) and add to eliminate $alpha$:
$$(a-frac{2rsqrt3}{3})^2+(frac{asqrt3}{2}-r)^2=4r^2$$
This is a simple quadratic equation with two solutions but only one of them is positive:
$$r=afrac{6sqrt7-7sqrt3}{10}approx0.375a$$
answered Jan 29 at 13:14
OldboyOldboy
9,13111138
$endgroup$
Denote the side of the hexagon as $a$:
Case 2:
The distance $H$ between parallel sides of the hexagon is $asqrt3$. On the other side (look at the picture):
$$H=asqrt3=r+(2r)frac{sqrt3}{2}+r=r(2+sqrt3)$$
So the radius of the circle is:
$$r=frac{asqrt3}{2+sqrt3}=a(2sqrt3-3)approx0.464a$$
Case 4:
First equation:
$$PQ=PB+ABcosalpha+ADcosalpha+DQ$$
$$2a=2frac{2rsqrt3}{3}+2(2r)cosalpha$$
$$a=frac{2rsqrt3}{3}+2rcosalpha$$
$$a-frac{2rsqrt3}{3}=2rcosalphatag{1}$$
Second equation:
$$EF=EA+ADsinalpha+DCsinalpha+CF$$
$$asqrt3=2r+2(2r)sinalpha$$
$$frac{asqrt3}{2}=r+2rsinalpha$$
$$frac{asqrt3}{2}-r=2rsinalphatag{2}$$
Square (1) and (2) and add to eliminate $alpha$:
$$(a-frac{2rsqrt3}{3})^2+(frac{asqrt3}{2}-r)^2=4r^2$$
This is a simple quadratic equation with two solutions but only one of them is positive:
$$r=afrac{6sqrt7-7sqrt3}{10}approx0.375a$$
answered Jan 29 at 13:14
OldboyOldboy
9,13111138
edited Jan 29 at 13:55
answered Jan 29 at 13:14
OldboyOldboy
9,13111138
answered Jan 29 at 13:14
OldboyOldboy
9,13111138
answered Jan 29 at 13:14
OldboyOldboy
9,13111138
9,13111138
$begingroup$
Lots of thanks I understand the 2nd case ...first I try to solve 4th case my own then I see yours...thanks a lot.
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 13:57
$begingroup$
@Md.RaihanurRahman If you like the answer, consider upvoting and accepting it.
$endgroup$
– Oldboy
Jan 29 at 14:03
$begingroup$
yeah I did that...thank
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 14:05
add a comment |
$begingroup$
Lots of thanks I understand the 2nd case ...first I try to solve 4th case my own then I see yours...thanks a lot.
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 13:57
$begingroup$
@Md.RaihanurRahman If you like the answer, consider upvoting and accepting it.
$endgroup$
– Oldboy
Jan 29 at 14:03
$begingroup$
yeah I did that...thank
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 14:05
$begingroup$
Lots of thanks I understand the 2nd case ...first I try to solve 4th case my own then I see yours...thanks a lot.
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 13:57
$begingroup$
Lots of thanks I understand the 2nd case ...first I try to solve 4th case my own then I see yours...thanks a lot.
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 13:57
$begingroup$
@Md.RaihanurRahman If you like the answer, consider upvoting and accepting it.
$endgroup$
– Oldboy
Jan 29 at 14:03
$begingroup$
@Md.RaihanurRahman If you like the answer, consider upvoting and accepting it.
$endgroup$
– Oldboy
Jan 29 at 14:03
$begingroup$
yeah I did that...thank
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 14:05
$begingroup$
yeah I did that...thank
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 14:05
add a comment |
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$begingroup$
how can I show the picture instead of link
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 11:49
$begingroup$
I don't have enough teputation for that and I've done it for you.
$endgroup$
– José Carlos Santos
Jan 29 at 11:51
$begingroup$
What exactly is the question ?
$endgroup$
– Overmind
Jan 29 at 12:05
$begingroup$
For case 2. Draw radius connecting the center with point in which hexagon is tangent to the circle. Thus find distance of the circle center to the closest hexagon edge...
$endgroup$
– Matteo
Jan 29 at 12:12
$begingroup$
sorry don't understand the case 2.
$endgroup$
– Md. Raihanur Rahman
Jan 29 at 12:28