Mixing
Cyclic pentagon $ABCDE$ has radius three. If $AB=BC=2$ and $CD=DE=4$, find AE. [closed]
$begingroup$
Given cyclic pentagon $ABCDE$ with the radius $3$. If $AB=BC=2$ and $CD=DE=4$, find $AE$.
I think trigonometry tricks are really useful on this problem, but I still can't get the final answer.
trigonometry contest-math polygons
edited Jan 2 at 9:37
Blue
47.7k870151
asked Jan 2 at 8:14
Pray Oren SimanjuntakPray Oren Simanjuntak
61
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closed as off-topic by Paul Frost, user91500, José Carlos Santos, Cesareo, Lord_Farin Jan 2 at 12:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Paul Frost, user91500, José Carlos Santos, Cesareo, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Given cyclic pentagon $ABCDE$ with the radius $3$. If $AB=BC=2$ and $CD=DE=4$, find $AE$.
I think trigonometry tricks are really useful on this problem, but I still can't get the final answer.
trigonometry contest-math polygons
edited Jan 2 at 9:37
Blue
47.7k870151
asked Jan 2 at 8:14
Pray Oren SimanjuntakPray Oren Simanjuntak
61
$endgroup$
closed as off-topic by Paul Frost, user91500, José Carlos Santos, Cesareo, Lord_Farin Jan 2 at 12:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Paul Frost, user91500, José Carlos Santos, Cesareo, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
You should really add what trigonometry tricks you have used to try and solve it
$endgroup$
– Prakhar Nagpal
Jan 2 at 8:37
$begingroup$
I thought the cosine law could be applied on the ABCO, but it's not a cyclic quadrilateral
$endgroup$
– Pray Oren Simanjuntak
Jan 2 at 8:51
$begingroup$
It's $6sinleft(2arcsinfrac{1}{3}+2arcsinfrac{2}{3}right).$
$endgroup$
– Michael Rozenberg
Jan 2 at 10:32
add a comment |
$begingroup$
Given cyclic pentagon $ABCDE$ with the radius $3$. If $AB=BC=2$ and $CD=DE=4$, find $AE$.
I think trigonometry tricks are really useful on this problem, but I still can't get the final answer.
trigonometry contest-math polygons
edited Jan 2 at 9:37
Blue
47.7k870151
asked Jan 2 at 8:14
Pray Oren SimanjuntakPray Oren Simanjuntak
61
$endgroup$
Given cyclic pentagon $ABCDE$ with the radius $3$. If $AB=BC=2$ and $CD=DE=4$, find $AE$.
I think trigonometry tricks are really useful on this problem, but I still can't get the final answer.
trigonometry contest-math polygons
trigonometry contest-math polygons
edited Jan 2 at 9:37
Blue
47.7k870151
asked Jan 2 at 8:14
Pray Oren SimanjuntakPray Oren Simanjuntak
61
edited Jan 2 at 9:37
Blue
47.7k870151
asked Jan 2 at 8:14
Pray Oren SimanjuntakPray Oren Simanjuntak
61
edited Jan 2 at 9:37
Blue
47.7k870151
edited Jan 2 at 9:37
Blue
47.7k870151
edited Jan 2 at 9:37
Blue
47.7k870151
47.7k870151
asked Jan 2 at 8:14
Pray Oren SimanjuntakPray Oren Simanjuntak
61
asked Jan 2 at 8:14
Pray Oren SimanjuntakPray Oren Simanjuntak
61
asked Jan 2 at 8:14
Pray Oren SimanjuntakPray Oren Simanjuntak
61
61
closed as off-topic by Paul Frost, user91500, José Carlos Santos, Cesareo, Lord_Farin Jan 2 at 12:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Paul Frost, user91500, José Carlos Santos, Cesareo, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Paul Frost, user91500, José Carlos Santos, Cesareo, Lord_Farin Jan 2 at 12:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Paul Frost, user91500, José Carlos Santos, Cesareo, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
You should really add what trigonometry tricks you have used to try and solve it
$endgroup$
– Prakhar Nagpal
Jan 2 at 8:37
$begingroup$
I thought the cosine law could be applied on the ABCO, but it's not a cyclic quadrilateral
$endgroup$
– Pray Oren Simanjuntak
Jan 2 at 8:51
$begingroup$
It's $6sinleft(2arcsinfrac{1}{3}+2arcsinfrac{2}{3}right).$
$endgroup$
– Michael Rozenberg
Jan 2 at 10:32
add a comment |
2
$begingroup$
You should really add what trigonometry tricks you have used to try and solve it
$endgroup$
– Prakhar Nagpal
Jan 2 at 8:37
$begingroup$
I thought the cosine law could be applied on the ABCO, but it's not a cyclic quadrilateral
$endgroup$
– Pray Oren Simanjuntak
Jan 2 at 8:51
$begingroup$
It's $6sinleft(2arcsinfrac{1}{3}+2arcsinfrac{2}{3}right).$
$endgroup$
– Michael Rozenberg
Jan 2 at 10:32
2
2
$begingroup$
You should really add what trigonometry tricks you have used to try and solve it
$endgroup$
– Prakhar Nagpal
Jan 2 at 8:37
$begingroup$
You should really add what trigonometry tricks you have used to try and solve it
$endgroup$
– Prakhar Nagpal
Jan 2 at 8:37
$begingroup$
I thought the cosine law could be applied on the ABCO, but it's not a cyclic quadrilateral
$endgroup$
– Pray Oren Simanjuntak
Jan 2 at 8:51
$begingroup$
I thought the cosine law could be applied on the ABCO, but it's not a cyclic quadrilateral
$endgroup$
– Pray Oren Simanjuntak
Jan 2 at 8:51
$begingroup$
It's $6sinleft(2arcsinfrac{1}{3}+2arcsinfrac{2}{3}right).$
$endgroup$
– Michael Rozenberg
Jan 2 at 10:32
$begingroup$
It's $6sinleft(2arcsinfrac{1}{3}+2arcsinfrac{2}{3}right).$
$endgroup$
– Michael Rozenberg
Jan 2 at 10:32
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Calculate the angles at the center:
$$angle AOB = 2 sin^{-1}frac 13 = angle BOC qquad angle COD = 2sin^{-1} frac 23 = angle DOE$$
These angles sum to $4sin^{-1}frac 13 + 4 sin^{-1} frac 23$, and we can simplify this with
begin{align}
& 4sin^{-1}frac 13 + 4 sin^{-1} frac 23 \
= & 4sin^{-1}sin bigg(sin^{-1}frac 13 + sin^{-1} frac 23 bigg) \
= & 4sin^{-1}bigg[sinbigg(sin^{-1}frac 13bigg)cosbigg(sin^{-1}frac 23bigg)+sinbigg(sin^{-1}frac 23bigg)cosbigg(sin^{-1}frac 13bigg)bigg] \
= & 4sin^{-1}bigg[frac 13 cdot frac{sqrt 5}{3} + frac 23 cdot frac{sqrt 8}{3}bigg] \
= & 4sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}
end{align}
Thus, the last angle $angle EOA$ is given by
$$angle EOA = 2pi - 4sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}$$
and $EA$ can be calculated as
begin{align}
EA & = 2cdot OE cdot sin frac{angle EOA}{2} \
& = 2 cdot 3 cdot sinbigg(pi - 2sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}bigg) \
& = 6sinbigg(2sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}bigg) \
& = 12sin bigg(sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}bigg) cos bigg(sin^{-1}frac{sqrt 5 + 4sqrt 2}{9} bigg) \
& = 12 cdot frac{sqrt 5 + 4sqrt 2}{9} cdot frac{sqrt{9^2 - (sqrt 5 + 4sqrt 2)^2}}{9} \
& = 12 cdot frac{sqrt 5 + 4sqrt 2}{9} cdot frac{sqrt{44 - 8sqrt{10}}}{9} \
& = 24 cdot frac{sqrt 5 + 4sqrt 2}{9} cdot frac{sqrt{10}-1}{9} \
& = frac{8sqrt 2 + 56sqrt{5}}{27}
end{align}
Indeed, this is numerically equal to $5.05dots$
answered Jan 2 at 9:17
glowstonetreesglowstonetrees
2,322418
$endgroup$
add a comment |
$begingroup$
I don't think this is the intended solution, as a calculator is necessary and it is absolutely non-elegant. I hope somebody has the time to find a nicer solution.
If you have the radius of the circle and the length of a chord, it's easy to find the angle which defines the chord:
$$sinleft(frac{varphi}{2}right)=frac{c/2}{r}=frac{c}{2r}$$
Consequently you get the twice the angle
$$sinleft(frac{varphi_1}{2}right)=frac{|AB|}{6}=frac23implies varphi_1
cong 83.62^circ$$
(this can be the only solution, as the solution in the second quadrant would result in an angle of size $276^circ$ which makes no sense)
and twice the angle
$$sinleft(frac{varphi_2}{2}right)=frac{|BC|}{6}=frac13implies varphi_2
cong 38.94^circ$$
Now the length of the missing chors is just
$$|AE|=2rcdotsinleft(frac{theta}{2}right)=2rsinleft(frac{2pi-2varphi_1-2varphi_2}{2}right)=2rsinleft(pi-varphi_1-varphi_2right)=2rsinleft(varphi_1+varphi_2right),$$
which can be easily calculated using the above. The result is $|AE|=5.05$
answered Jan 2 at 9:03
b00n heTb00n heT
10.2k12234
$endgroup$
add a comment |
$begingroup$
The most useful trick we can use here? We can permute the order of the sides in a cyclic polygon and still inscribe it in the same circle. After all, the central angle subtended by a side depends only on the side's length and the radius. So the, let's shuffle the sides in a way that produces some symmetry: $A'B'=2$, $B'C'=4$, $E'D'=2$, $D'C'=4$. Then $A'E'$ will be the same as $AE$.
What does this do? We will have $A'C'=E'C'$, so that $A'C'E'$ is an isosceles triangle. Now we need the length $A'C'$, and that's trigonometry. Angles $angle B'C'A'=gamma$ and $angle C'A'B'=alpha$ are acute, with sines $frac{A'B'}{2r}=frac13$ and $frac{B'C'}{2r}=frac23$ respectively. The third angle $angle A'B'C'=beta$ is the complement of $alpha+gamma$, so $sin beta = sin alphacosgamma+cosalphasingamma = frac23cdotfrac{sqrt{8}}{3}+frac13cdotfrac{sqrt{5}}{3}=frac{4sqrt{2}+sqrt{5}}{9}$. Note also that $beta$ is obtuse.
Now, we deal with the isosceles triangle $A'C'E'$. Its angles are $180^circ-beta$, $180^circ-beta$, and $2beta-180^{circ}$. The length $A'E'$ we care about, of course, is opposite the one that's different. Its length is
begin{align*}A'E' &= 2rsin(2beta-180^circ)\
&= -2rsin(2beta) = -4rsinbetacosbeta\
&= -4cdot 3cdot frac{4sqrt{2}+sqrt{5}}{9}cdot frac{-sqrt{81-37-8sqrt{10}}}{9}\
&= frac{4(4sqrt{2}+sqrt{5})sqrt{44-8sqrt{10}}}{27}=frac{4(4sqrt{2}+sqrt{5})(2sqrt{10}-2)}{27}\
&= frac{8(4sqrt{20}+sqrt{50}-4sqrt{2}-sqrt{5})}{27}=frac{8}{27}(7sqrt{5}+sqrt{2})end{align*}
Another verification of the same answer. I hope my somewhat different path is useful here.
answered Jan 2 at 9:45
jmerryjmerry
3,427413
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Calculate the angles at the center:
$$angle AOB = 2 sin^{-1}frac 13 = angle BOC qquad angle COD = 2sin^{-1} frac 23 = angle DOE$$
These angles sum to $4sin^{-1}frac 13 + 4 sin^{-1} frac 23$, and we can simplify this with
begin{align}
& 4sin^{-1}frac 13 + 4 sin^{-1} frac 23 \
= & 4sin^{-1}sin bigg(sin^{-1}frac 13 + sin^{-1} frac 23 bigg) \
= & 4sin^{-1}bigg[sinbigg(sin^{-1}frac 13bigg)cosbigg(sin^{-1}frac 23bigg)+sinbigg(sin^{-1}frac 23bigg)cosbigg(sin^{-1}frac 13bigg)bigg] \
= & 4sin^{-1}bigg[frac 13 cdot frac{sqrt 5}{3} + frac 23 cdot frac{sqrt 8}{3}bigg] \
= & 4sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}
end{align}
Thus, the last angle $angle EOA$ is given by
$$angle EOA = 2pi - 4sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}$$
and $EA$ can be calculated as
begin{align}
EA & = 2cdot OE cdot sin frac{angle EOA}{2} \
& = 2 cdot 3 cdot sinbigg(pi - 2sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}bigg) \
& = 6sinbigg(2sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}bigg) \
& = 12sin bigg(sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}bigg) cos bigg(sin^{-1}frac{sqrt 5 + 4sqrt 2}{9} bigg) \
& = 12 cdot frac{sqrt 5 + 4sqrt 2}{9} cdot frac{sqrt{9^2 - (sqrt 5 + 4sqrt 2)^2}}{9} \
& = 12 cdot frac{sqrt 5 + 4sqrt 2}{9} cdot frac{sqrt{44 - 8sqrt{10}}}{9} \
& = 24 cdot frac{sqrt 5 + 4sqrt 2}{9} cdot frac{sqrt{10}-1}{9} \
& = frac{8sqrt 2 + 56sqrt{5}}{27}
end{align}
Indeed, this is numerically equal to $5.05dots$
answered Jan 2 at 9:17
glowstonetreesglowstonetrees
2,322418
$endgroup$
add a comment |
$begingroup$
Calculate the angles at the center:
$$angle AOB = 2 sin^{-1}frac 13 = angle BOC qquad angle COD = 2sin^{-1} frac 23 = angle DOE$$
These angles sum to $4sin^{-1}frac 13 + 4 sin^{-1} frac 23$, and we can simplify this with
begin{align}
& 4sin^{-1}frac 13 + 4 sin^{-1} frac 23 \
= & 4sin^{-1}sin bigg(sin^{-1}frac 13 + sin^{-1} frac 23 bigg) \
= & 4sin^{-1}bigg[sinbigg(sin^{-1}frac 13bigg)cosbigg(sin^{-1}frac 23bigg)+sinbigg(sin^{-1}frac 23bigg)cosbigg(sin^{-1}frac 13bigg)bigg] \
= & 4sin^{-1}bigg[frac 13 cdot frac{sqrt 5}{3} + frac 23 cdot frac{sqrt 8}{3}bigg] \
= & 4sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}
end{align}
Thus, the last angle $angle EOA$ is given by
$$angle EOA = 2pi - 4sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}$$
and $EA$ can be calculated as
begin{align}
EA & = 2cdot OE cdot sin frac{angle EOA}{2} \
& = 2 cdot 3 cdot sinbigg(pi - 2sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}bigg) \
& = 6sinbigg(2sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}bigg) \
& = 12sin bigg(sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}bigg) cos bigg(sin^{-1}frac{sqrt 5 + 4sqrt 2}{9} bigg) \
& = 12 cdot frac{sqrt 5 + 4sqrt 2}{9} cdot frac{sqrt{9^2 - (sqrt 5 + 4sqrt 2)^2}}{9} \
& = 12 cdot frac{sqrt 5 + 4sqrt 2}{9} cdot frac{sqrt{44 - 8sqrt{10}}}{9} \
& = 24 cdot frac{sqrt 5 + 4sqrt 2}{9} cdot frac{sqrt{10}-1}{9} \
& = frac{8sqrt 2 + 56sqrt{5}}{27}
end{align}
Indeed, this is numerically equal to $5.05dots$
answered Jan 2 at 9:17
glowstonetreesglowstonetrees
2,322418
$endgroup$
add a comment |
$begingroup$
Calculate the angles at the center:
$$angle AOB = 2 sin^{-1}frac 13 = angle BOC qquad angle COD = 2sin^{-1} frac 23 = angle DOE$$
These angles sum to $4sin^{-1}frac 13 + 4 sin^{-1} frac 23$, and we can simplify this with
begin{align}
& 4sin^{-1}frac 13 + 4 sin^{-1} frac 23 \
= & 4sin^{-1}sin bigg(sin^{-1}frac 13 + sin^{-1} frac 23 bigg) \
= & 4sin^{-1}bigg[sinbigg(sin^{-1}frac 13bigg)cosbigg(sin^{-1}frac 23bigg)+sinbigg(sin^{-1}frac 23bigg)cosbigg(sin^{-1}frac 13bigg)bigg] \
= & 4sin^{-1}bigg[frac 13 cdot frac{sqrt 5}{3} + frac 23 cdot frac{sqrt 8}{3}bigg] \
= & 4sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}
end{align}
Thus, the last angle $angle EOA$ is given by
$$angle EOA = 2pi - 4sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}$$
and $EA$ can be calculated as
begin{align}
EA & = 2cdot OE cdot sin frac{angle EOA}{2} \
& = 2 cdot 3 cdot sinbigg(pi - 2sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}bigg) \
& = 6sinbigg(2sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}bigg) \
& = 12sin bigg(sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}bigg) cos bigg(sin^{-1}frac{sqrt 5 + 4sqrt 2}{9} bigg) \
& = 12 cdot frac{sqrt 5 + 4sqrt 2}{9} cdot frac{sqrt{9^2 - (sqrt 5 + 4sqrt 2)^2}}{9} \
& = 12 cdot frac{sqrt 5 + 4sqrt 2}{9} cdot frac{sqrt{44 - 8sqrt{10}}}{9} \
& = 24 cdot frac{sqrt 5 + 4sqrt 2}{9} cdot frac{sqrt{10}-1}{9} \
& = frac{8sqrt 2 + 56sqrt{5}}{27}
end{align}
Indeed, this is numerically equal to $5.05dots$
answered Jan 2 at 9:17
glowstonetreesglowstonetrees
2,322418
$endgroup$
Calculate the angles at the center:
$$angle AOB = 2 sin^{-1}frac 13 = angle BOC qquad angle COD = 2sin^{-1} frac 23 = angle DOE$$
These angles sum to $4sin^{-1}frac 13 + 4 sin^{-1} frac 23$, and we can simplify this with
begin{align}
& 4sin^{-1}frac 13 + 4 sin^{-1} frac 23 \
= & 4sin^{-1}sin bigg(sin^{-1}frac 13 + sin^{-1} frac 23 bigg) \
= & 4sin^{-1}bigg[sinbigg(sin^{-1}frac 13bigg)cosbigg(sin^{-1}frac 23bigg)+sinbigg(sin^{-1}frac 23bigg)cosbigg(sin^{-1}frac 13bigg)bigg] \
= & 4sin^{-1}bigg[frac 13 cdot frac{sqrt 5}{3} + frac 23 cdot frac{sqrt 8}{3}bigg] \
= & 4sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}
end{align}
Thus, the last angle $angle EOA$ is given by
$$angle EOA = 2pi - 4sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}$$
and $EA$ can be calculated as
begin{align}
EA & = 2cdot OE cdot sin frac{angle EOA}{2} \
& = 2 cdot 3 cdot sinbigg(pi - 2sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}bigg) \
& = 6sinbigg(2sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}bigg) \
& = 12sin bigg(sin^{-1}frac{sqrt 5 + 4sqrt 2}{9}bigg) cos bigg(sin^{-1}frac{sqrt 5 + 4sqrt 2}{9} bigg) \
& = 12 cdot frac{sqrt 5 + 4sqrt 2}{9} cdot frac{sqrt{9^2 - (sqrt 5 + 4sqrt 2)^2}}{9} \
& = 12 cdot frac{sqrt 5 + 4sqrt 2}{9} cdot frac{sqrt{44 - 8sqrt{10}}}{9} \
& = 24 cdot frac{sqrt 5 + 4sqrt 2}{9} cdot frac{sqrt{10}-1}{9} \
& = frac{8sqrt 2 + 56sqrt{5}}{27}
end{align}
Indeed, this is numerically equal to $5.05dots$
answered Jan 2 at 9:17
glowstonetreesglowstonetrees
2,322418
answered Jan 2 at 9:17
glowstonetreesglowstonetrees
2,322418
answered Jan 2 at 9:17
glowstonetreesglowstonetrees
2,322418
answered Jan 2 at 9:17
glowstonetreesglowstonetrees
2,322418
2,322418
add a comment |
add a comment |
$begingroup$
I don't think this is the intended solution, as a calculator is necessary and it is absolutely non-elegant. I hope somebody has the time to find a nicer solution.
If you have the radius of the circle and the length of a chord, it's easy to find the angle which defines the chord:
$$sinleft(frac{varphi}{2}right)=frac{c/2}{r}=frac{c}{2r}$$
Consequently you get the twice the angle
$$sinleft(frac{varphi_1}{2}right)=frac{|AB|}{6}=frac23implies varphi_1
cong 83.62^circ$$
(this can be the only solution, as the solution in the second quadrant would result in an angle of size $276^circ$ which makes no sense)
and twice the angle
$$sinleft(frac{varphi_2}{2}right)=frac{|BC|}{6}=frac13implies varphi_2
cong 38.94^circ$$
Now the length of the missing chors is just
$$|AE|=2rcdotsinleft(frac{theta}{2}right)=2rsinleft(frac{2pi-2varphi_1-2varphi_2}{2}right)=2rsinleft(pi-varphi_1-varphi_2right)=2rsinleft(varphi_1+varphi_2right),$$
which can be easily calculated using the above. The result is $|AE|=5.05$
answered Jan 2 at 9:03
b00n heTb00n heT
10.2k12234
$endgroup$
add a comment |
$begingroup$
I don't think this is the intended solution, as a calculator is necessary and it is absolutely non-elegant. I hope somebody has the time to find a nicer solution.
If you have the radius of the circle and the length of a chord, it's easy to find the angle which defines the chord:
$$sinleft(frac{varphi}{2}right)=frac{c/2}{r}=frac{c}{2r}$$
Consequently you get the twice the angle
$$sinleft(frac{varphi_1}{2}right)=frac{|AB|}{6}=frac23implies varphi_1
cong 83.62^circ$$
(this can be the only solution, as the solution in the second quadrant would result in an angle of size $276^circ$ which makes no sense)
and twice the angle
$$sinleft(frac{varphi_2}{2}right)=frac{|BC|}{6}=frac13implies varphi_2
cong 38.94^circ$$
Now the length of the missing chors is just
$$|AE|=2rcdotsinleft(frac{theta}{2}right)=2rsinleft(frac{2pi-2varphi_1-2varphi_2}{2}right)=2rsinleft(pi-varphi_1-varphi_2right)=2rsinleft(varphi_1+varphi_2right),$$
which can be easily calculated using the above. The result is $|AE|=5.05$
answered Jan 2 at 9:03
b00n heTb00n heT
10.2k12234
$endgroup$
add a comment |
$begingroup$
I don't think this is the intended solution, as a calculator is necessary and it is absolutely non-elegant. I hope somebody has the time to find a nicer solution.
If you have the radius of the circle and the length of a chord, it's easy to find the angle which defines the chord:
$$sinleft(frac{varphi}{2}right)=frac{c/2}{r}=frac{c}{2r}$$
Consequently you get the twice the angle
$$sinleft(frac{varphi_1}{2}right)=frac{|AB|}{6}=frac23implies varphi_1
cong 83.62^circ$$
(this can be the only solution, as the solution in the second quadrant would result in an angle of size $276^circ$ which makes no sense)
and twice the angle
$$sinleft(frac{varphi_2}{2}right)=frac{|BC|}{6}=frac13implies varphi_2
cong 38.94^circ$$
Now the length of the missing chors is just
$$|AE|=2rcdotsinleft(frac{theta}{2}right)=2rsinleft(frac{2pi-2varphi_1-2varphi_2}{2}right)=2rsinleft(pi-varphi_1-varphi_2right)=2rsinleft(varphi_1+varphi_2right),$$
which can be easily calculated using the above. The result is $|AE|=5.05$
answered Jan 2 at 9:03
b00n heTb00n heT
10.2k12234
$endgroup$
I don't think this is the intended solution, as a calculator is necessary and it is absolutely non-elegant. I hope somebody has the time to find a nicer solution.
If you have the radius of the circle and the length of a chord, it's easy to find the angle which defines the chord:
$$sinleft(frac{varphi}{2}right)=frac{c/2}{r}=frac{c}{2r}$$
Consequently you get the twice the angle
$$sinleft(frac{varphi_1}{2}right)=frac{|AB|}{6}=frac23implies varphi_1
cong 83.62^circ$$
(this can be the only solution, as the solution in the second quadrant would result in an angle of size $276^circ$ which makes no sense)
and twice the angle
$$sinleft(frac{varphi_2}{2}right)=frac{|BC|}{6}=frac13implies varphi_2
cong 38.94^circ$$
Now the length of the missing chors is just
$$|AE|=2rcdotsinleft(frac{theta}{2}right)=2rsinleft(frac{2pi-2varphi_1-2varphi_2}{2}right)=2rsinleft(pi-varphi_1-varphi_2right)=2rsinleft(varphi_1+varphi_2right),$$
which can be easily calculated using the above. The result is $|AE|=5.05$
answered Jan 2 at 9:03
b00n heTb00n heT
10.2k12234
edited Jan 2 at 9:08
answered Jan 2 at 9:03
b00n heTb00n heT
10.2k12234
answered Jan 2 at 9:03
b00n heTb00n heT
10.2k12234
answered Jan 2 at 9:03
b00n heTb00n heT
10.2k12234
10.2k12234
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The most useful trick we can use here? We can permute the order of the sides in a cyclic polygon and still inscribe it in the same circle. After all, the central angle subtended by a side depends only on the side's length and the radius. So the, let's shuffle the sides in a way that produces some symmetry: $A'B'=2$, $B'C'=4$, $E'D'=2$, $D'C'=4$. Then $A'E'$ will be the same as $AE$.
What does this do? We will have $A'C'=E'C'$, so that $A'C'E'$ is an isosceles triangle. Now we need the length $A'C'$, and that's trigonometry. Angles $angle B'C'A'=gamma$ and $angle C'A'B'=alpha$ are acute, with sines $frac{A'B'}{2r}=frac13$ and $frac{B'C'}{2r}=frac23$ respectively. The third angle $angle A'B'C'=beta$ is the complement of $alpha+gamma$, so $sin beta = sin alphacosgamma+cosalphasingamma = frac23cdotfrac{sqrt{8}}{3}+frac13cdotfrac{sqrt{5}}{3}=frac{4sqrt{2}+sqrt{5}}{9}$. Note also that $beta$ is obtuse.
Now, we deal with the isosceles triangle $A'C'E'$. Its angles are $180^circ-beta$, $180^circ-beta$, and $2beta-180^{circ}$. The length $A'E'$ we care about, of course, is opposite the one that's different. Its length is
begin{align*}A'E' &= 2rsin(2beta-180^circ)\
&= -2rsin(2beta) = -4rsinbetacosbeta\
&= -4cdot 3cdot frac{4sqrt{2}+sqrt{5}}{9}cdot frac{-sqrt{81-37-8sqrt{10}}}{9}\
&= frac{4(4sqrt{2}+sqrt{5})sqrt{44-8sqrt{10}}}{27}=frac{4(4sqrt{2}+sqrt{5})(2sqrt{10}-2)}{27}\
&= frac{8(4sqrt{20}+sqrt{50}-4sqrt{2}-sqrt{5})}{27}=frac{8}{27}(7sqrt{5}+sqrt{2})end{align*}
Another verification of the same answer. I hope my somewhat different path is useful here.
answered Jan 2 at 9:45
jmerryjmerry
3,427413
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add a comment |
$begingroup$
The most useful trick we can use here? We can permute the order of the sides in a cyclic polygon and still inscribe it in the same circle. After all, the central angle subtended by a side depends only on the side's length and the radius. So the, let's shuffle the sides in a way that produces some symmetry: $A'B'=2$, $B'C'=4$, $E'D'=2$, $D'C'=4$. Then $A'E'$ will be the same as $AE$.
What does this do? We will have $A'C'=E'C'$, so that $A'C'E'$ is an isosceles triangle. Now we need the length $A'C'$, and that's trigonometry. Angles $angle B'C'A'=gamma$ and $angle C'A'B'=alpha$ are acute, with sines $frac{A'B'}{2r}=frac13$ and $frac{B'C'}{2r}=frac23$ respectively. The third angle $angle A'B'C'=beta$ is the complement of $alpha+gamma$, so $sin beta = sin alphacosgamma+cosalphasingamma = frac23cdotfrac{sqrt{8}}{3}+frac13cdotfrac{sqrt{5}}{3}=frac{4sqrt{2}+sqrt{5}}{9}$. Note also that $beta$ is obtuse.
Now, we deal with the isosceles triangle $A'C'E'$. Its angles are $180^circ-beta$, $180^circ-beta$, and $2beta-180^{circ}$. The length $A'E'$ we care about, of course, is opposite the one that's different. Its length is
begin{align*}A'E' &= 2rsin(2beta-180^circ)\
&= -2rsin(2beta) = -4rsinbetacosbeta\
&= -4cdot 3cdot frac{4sqrt{2}+sqrt{5}}{9}cdot frac{-sqrt{81-37-8sqrt{10}}}{9}\
&= frac{4(4sqrt{2}+sqrt{5})sqrt{44-8sqrt{10}}}{27}=frac{4(4sqrt{2}+sqrt{5})(2sqrt{10}-2)}{27}\
&= frac{8(4sqrt{20}+sqrt{50}-4sqrt{2}-sqrt{5})}{27}=frac{8}{27}(7sqrt{5}+sqrt{2})end{align*}
Another verification of the same answer. I hope my somewhat different path is useful here.
answered Jan 2 at 9:45
jmerryjmerry
3,427413
$endgroup$
add a comment |
$begingroup$
The most useful trick we can use here? We can permute the order of the sides in a cyclic polygon and still inscribe it in the same circle. After all, the central angle subtended by a side depends only on the side's length and the radius. So the, let's shuffle the sides in a way that produces some symmetry: $A'B'=2$, $B'C'=4$, $E'D'=2$, $D'C'=4$. Then $A'E'$ will be the same as $AE$.
What does this do? We will have $A'C'=E'C'$, so that $A'C'E'$ is an isosceles triangle. Now we need the length $A'C'$, and that's trigonometry. Angles $angle B'C'A'=gamma$ and $angle C'A'B'=alpha$ are acute, with sines $frac{A'B'}{2r}=frac13$ and $frac{B'C'}{2r}=frac23$ respectively. The third angle $angle A'B'C'=beta$ is the complement of $alpha+gamma$, so $sin beta = sin alphacosgamma+cosalphasingamma = frac23cdotfrac{sqrt{8}}{3}+frac13cdotfrac{sqrt{5}}{3}=frac{4sqrt{2}+sqrt{5}}{9}$. Note also that $beta$ is obtuse.
Now, we deal with the isosceles triangle $A'C'E'$. Its angles are $180^circ-beta$, $180^circ-beta$, and $2beta-180^{circ}$. The length $A'E'$ we care about, of course, is opposite the one that's different. Its length is
begin{align*}A'E' &= 2rsin(2beta-180^circ)\
&= -2rsin(2beta) = -4rsinbetacosbeta\
&= -4cdot 3cdot frac{4sqrt{2}+sqrt{5}}{9}cdot frac{-sqrt{81-37-8sqrt{10}}}{9}\
&= frac{4(4sqrt{2}+sqrt{5})sqrt{44-8sqrt{10}}}{27}=frac{4(4sqrt{2}+sqrt{5})(2sqrt{10}-2)}{27}\
&= frac{8(4sqrt{20}+sqrt{50}-4sqrt{2}-sqrt{5})}{27}=frac{8}{27}(7sqrt{5}+sqrt{2})end{align*}
Another verification of the same answer. I hope my somewhat different path is useful here.
answered Jan 2 at 9:45
jmerryjmerry
3,427413
$endgroup$
The most useful trick we can use here? We can permute the order of the sides in a cyclic polygon and still inscribe it in the same circle. After all, the central angle subtended by a side depends only on the side's length and the radius. So the, let's shuffle the sides in a way that produces some symmetry: $A'B'=2$, $B'C'=4$, $E'D'=2$, $D'C'=4$. Then $A'E'$ will be the same as $AE$.
What does this do? We will have $A'C'=E'C'$, so that $A'C'E'$ is an isosceles triangle. Now we need the length $A'C'$, and that's trigonometry. Angles $angle B'C'A'=gamma$ and $angle C'A'B'=alpha$ are acute, with sines $frac{A'B'}{2r}=frac13$ and $frac{B'C'}{2r}=frac23$ respectively. The third angle $angle A'B'C'=beta$ is the complement of $alpha+gamma$, so $sin beta = sin alphacosgamma+cosalphasingamma = frac23cdotfrac{sqrt{8}}{3}+frac13cdotfrac{sqrt{5}}{3}=frac{4sqrt{2}+sqrt{5}}{9}$. Note also that $beta$ is obtuse.
Now, we deal with the isosceles triangle $A'C'E'$. Its angles are $180^circ-beta$, $180^circ-beta$, and $2beta-180^{circ}$. The length $A'E'$ we care about, of course, is opposite the one that's different. Its length is
begin{align*}A'E' &= 2rsin(2beta-180^circ)\
&= -2rsin(2beta) = -4rsinbetacosbeta\
&= -4cdot 3cdot frac{4sqrt{2}+sqrt{5}}{9}cdot frac{-sqrt{81-37-8sqrt{10}}}{9}\
&= frac{4(4sqrt{2}+sqrt{5})sqrt{44-8sqrt{10}}}{27}=frac{4(4sqrt{2}+sqrt{5})(2sqrt{10}-2)}{27}\
&= frac{8(4sqrt{20}+sqrt{50}-4sqrt{2}-sqrt{5})}{27}=frac{8}{27}(7sqrt{5}+sqrt{2})end{align*}
Another verification of the same answer. I hope my somewhat different path is useful here.
answered Jan 2 at 9:45
jmerryjmerry
3,427413
answered Jan 2 at 9:45
jmerryjmerry
3,427413
answered Jan 2 at 9:45
jmerryjmerry
3,427413
answered Jan 2 at 9:45
jmerryjmerry
3,427413
3,427413
add a comment |
add a comment |
2
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You should really add what trigonometry tricks you have used to try and solve it
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– Prakhar Nagpal
Jan 2 at 8:37
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I thought the cosine law could be applied on the ABCO, but it's not a cyclic quadrilateral
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– Pray Oren Simanjuntak
Jan 2 at 8:51
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It's $6sinleft(2arcsinfrac{1}{3}+2arcsinfrac{2}{3}right).$
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– Michael Rozenberg
Jan 2 at 10:32