Mixing
Geometry challenge question
Clash Royale CLAN TAG#URR8PPP
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The cross section of a tunnel
is a circular arc $CEF$, as shown in the diagram.
Note that the horizontal line $FC$ is a chord,
not a diameter.
The maximum height $|AE|$
of the tunnel is $10$ feet.
A vertical strut ($|BD|=9$ feet)
supports the roof of the tunnel
from the point $B$, which is located at
$27$ feet along the ground from the side
($|BC|=27$).
Calculate the width $|FC|$ of the tunnel at ground level.
My first thought was Pythagoras
to create simultaneous equations
for two right angled triangles
$ABE$ and $BCD$
(in the right hand half of the diagram),
aiming to find the distance $|AB|$
between the two verticals.
However, I kept going round in circles
and could not get away from the "solution"
$|AB|+27=|AB|+27$. Specifically, I can't form a second equation,
perhaps involving radius of the circle,
in order to obtain a unique value
for the distance $|AB|$ between the two vertical lines $AE$ and $BD$.
I also had a look using similar triangles, which also proved fruitless.
geometry
edited Jul 31 at 20:24
g.kov
5,5171716
asked Jul 31 at 9:26
PR59
112
 |Â
show 3 more comments
up vote
0
down vote
favorite
The cross section of a tunnel
is a circular arc $CEF$, as shown in the diagram.
Note that the horizontal line $FC$ is a chord,
not a diameter.
The maximum height $|AE|$
of the tunnel is $10$ feet.
A vertical strut ($|BD|=9$ feet)
supports the roof of the tunnel
from the point $B$, which is located at
$27$ feet along the ground from the side
($|BC|=27$).
Calculate the width $|FC|$ of the tunnel at ground level.
My first thought was Pythagoras
to create simultaneous equations
for two right angled triangles
$ABE$ and $BCD$
(in the right hand half of the diagram),
aiming to find the distance $|AB|$
between the two verticals.
However, I kept going round in circles
and could not get away from the "solution"
$|AB|+27=|AB|+27$. Specifically, I can't form a second equation,
perhaps involving radius of the circle,
in order to obtain a unique value
for the distance $|AB|$ between the two vertical lines $AE$ and $BD$.
I also had a look using similar triangles, which also proved fruitless.
geometry
edited Jul 31 at 20:24
g.kov
5,5171716
asked Jul 31 at 9:26
PR59
112
Note that the drawing is not to scale.
– Joel Reyes Noche
Jul 31 at 9:54
Thanks for pointing out that diagram is not to scale. Thus, horizontal line is a chord, not a diameter. Specifically, I can't form a second equation, perhaps involving radius of the circle, in order to obtain a unique value for the distance between the two vertical lines.
– PR59
Jul 31 at 10:36
Thanks, Yves and jtm....very helpful and much appreciated.
– PR59
Jul 31 at 11:12
The answer is: "the width of the tunnel at ground level" is 80; the radius of the circle is 85.
– g.kov
Jul 31 at 18:57
@Joel Reyes Noche: it is to scale now.
– g.kov
Jul 31 at 20:26
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The cross section of a tunnel
is a circular arc $CEF$, as shown in the diagram.
Note that the horizontal line $FC$ is a chord,
not a diameter.
The maximum height $|AE|$
of the tunnel is $10$ feet.
A vertical strut ($|BD|=9$ feet)
supports the roof of the tunnel
from the point $B$, which is located at
$27$ feet along the ground from the side
($|BC|=27$).
Calculate the width $|FC|$ of the tunnel at ground level.
My first thought was Pythagoras
to create simultaneous equations
for two right angled triangles
$ABE$ and $BCD$
(in the right hand half of the diagram),
aiming to find the distance $|AB|$
between the two verticals.
However, I kept going round in circles
and could not get away from the "solution"
$|AB|+27=|AB|+27$. Specifically, I can't form a second equation,
perhaps involving radius of the circle,
in order to obtain a unique value
for the distance $|AB|$ between the two vertical lines $AE$ and $BD$.
I also had a look using similar triangles, which also proved fruitless.
geometry
edited Jul 31 at 20:24
g.kov
5,5171716
asked Jul 31 at 9:26
PR59
112
The cross section of a tunnel
is a circular arc $CEF$, as shown in the diagram.
Note that the horizontal line $FC$ is a chord,
not a diameter.
The maximum height $|AE|$
of the tunnel is $10$ feet.
A vertical strut ($|BD|=9$ feet)
supports the roof of the tunnel
from the point $B$, which is located at
$27$ feet along the ground from the side
($|BC|=27$).
Calculate the width $|FC|$ of the tunnel at ground level.
My first thought was Pythagoras
to create simultaneous equations
for two right angled triangles
$ABE$ and $BCD$
(in the right hand half of the diagram),
aiming to find the distance $|AB|$
between the two verticals.
However, I kept going round in circles
and could not get away from the "solution"
$|AB|+27=|AB|+27$. Specifically, I can't form a second equation,
perhaps involving radius of the circle,
in order to obtain a unique value
for the distance $|AB|$ between the two vertical lines $AE$ and $BD$.
I also had a look using similar triangles, which also proved fruitless.
geometry
edited Jul 31 at 20:24
g.kov
5,5171716
asked Jul 31 at 9:26
PR59
112
edited Jul 31 at 20:24
g.kov
5,5171716
edited Jul 31 at 20:24
g.kov
5,5171716
edited Jul 31 at 20:24
g.kov
5,5171716
5,5171716
asked Jul 31 at 9:26
PR59
112
asked Jul 31 at 9:26
PR59
112
asked Jul 31 at 9:26
PR59
112
112
Note that the drawing is not to scale.
– Joel Reyes Noche
Jul 31 at 9:54
Thanks for pointing out that diagram is not to scale. Thus, horizontal line is a chord, not a diameter. Specifically, I can't form a second equation, perhaps involving radius of the circle, in order to obtain a unique value for the distance between the two vertical lines.
– PR59
Jul 31 at 10:36
Thanks, Yves and jtm....very helpful and much appreciated.
– PR59
Jul 31 at 11:12
The answer is: "the width of the tunnel at ground level" is 80; the radius of the circle is 85.
– g.kov
Jul 31 at 18:57
@Joel Reyes Noche: it is to scale now.
– g.kov
Jul 31 at 20:26
 |Â
show 3 more comments
Note that the drawing is not to scale.
– Joel Reyes Noche
Jul 31 at 9:54
Thanks for pointing out that diagram is not to scale. Thus, horizontal line is a chord, not a diameter. Specifically, I can't form a second equation, perhaps involving radius of the circle, in order to obtain a unique value for the distance between the two vertical lines.
– PR59
Jul 31 at 10:36
Thanks, Yves and jtm....very helpful and much appreciated.
– PR59
Jul 31 at 11:12
The answer is: "the width of the tunnel at ground level" is 80; the radius of the circle is 85.
– g.kov
Jul 31 at 18:57
@Joel Reyes Noche: it is to scale now.
– g.kov
Jul 31 at 20:26
Note that the drawing is not to scale.
– Joel Reyes Noche
Jul 31 at 9:54
Note that the drawing is not to scale.
– Joel Reyes Noche
Jul 31 at 9:54
Thanks for pointing out that diagram is not to scale. Thus, horizontal line is a chord, not a diameter. Specifically, I can't form a second equation, perhaps involving radius of the circle, in order to obtain a unique value for the distance between the two vertical lines.
– PR59
Jul 31 at 10:36
Thanks for pointing out that diagram is not to scale. Thus, horizontal line is a chord, not a diameter. Specifically, I can't form a second equation, perhaps involving radius of the circle, in order to obtain a unique value for the distance between the two vertical lines.
– PR59
Jul 31 at 10:36
Thanks, Yves and jtm....very helpful and much appreciated.
– PR59
Jul 31 at 11:12
Thanks, Yves and jtm....very helpful and much appreciated.
– PR59
Jul 31 at 11:12
The answer is: "the width of the tunnel at ground level" is 80; the radius of the circle is 85.
– g.kov
Jul 31 at 18:57
The answer is: "the width of the tunnel at ground level" is 80; the radius of the circle is 85.
– g.kov
Jul 31 at 18:57
@Joel Reyes Noche: it is to scale now.
– g.kov
Jul 31 at 20:26
@Joel Reyes Noche: it is to scale now.
– g.kov
Jul 31 at 20:26
 |Â
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
0
down vote
Let the center be $(-y,0)$ and the rightmost point $(0,x)$.
We express that $(0,10), (x,0)$ and $(x-27,9)$ all belong to the circle:
$$(10+y)^2=x^2+y^2=(x-27)^2+(y+9)^2.$$
Then subtracting the last two terms,
$$-54x+27^2+18y+81=0$$
or
$$y=3x+45.$$
Finally,
$$(10+3x+45)^2=x^2+(3x+45)^2$$ and $$x=10sqrt19+30.$$
answered Jul 31 at 10:15
Yves Daoust
110k665203
add a comment |Â
up vote
0
down vote
Let $D',E'$ be the second intersection points
of the verticals $DB,EA$ with the circle.
Also, let $R$ be the radius of the circle
and $y=|AF|=|AC|$.
By the intersecting chords theorem,
beginalign
|EA|cdot|AE'|&=|FA|cdot|AC|
,\
|DB|cdot|BD'|&=|FB|cdot|BC|
,
endalign
we have a system of the two equations in two
unknowns, $R$ and $y$:
beginalign
a(2R-a)&=y^2
,\
b(2(R-a)+b)&=c(2y-c)
.
endalign
After excluding $R$ from this system, we have
a quadratic equation in $y$:
beginalign
y^2-2cdotfracacbcdot y+aleft(fracc^2b-a+bright)&=0
endalign
with two roots:
beginalign
y_pm&=fracacbpmsqrtleft(fracacbright)^2-aleft(fracc^2b-a+bright)
.
endalign
Substitution of $a=10, b=9, c=27$ gives
beginalign
y_+&=40
,\
y_-&=20
,
endalign
and since $y>27$ must hold,
the only suitable value is $y=y_+=40$.
Hence, the answer is $|CF|=2y=80$.
answered yesterday
g.kov
5,5171716
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let the center be $(-y,0)$ and the rightmost point $(0,x)$.
We express that $(0,10), (x,0)$ and $(x-27,9)$ all belong to the circle:
$$(10+y)^2=x^2+y^2=(x-27)^2+(y+9)^2.$$
Then subtracting the last two terms,
$$-54x+27^2+18y+81=0$$
or
$$y=3x+45.$$
Finally,
$$(10+3x+45)^2=x^2+(3x+45)^2$$ and $$x=10sqrt19+30.$$
answered Jul 31 at 10:15
Yves Daoust
110k665203
add a comment |Â
up vote
0
down vote
Let the center be $(-y,0)$ and the rightmost point $(0,x)$.
We express that $(0,10), (x,0)$ and $(x-27,9)$ all belong to the circle:
$$(10+y)^2=x^2+y^2=(x-27)^2+(y+9)^2.$$
Then subtracting the last two terms,
$$-54x+27^2+18y+81=0$$
or
$$y=3x+45.$$
Finally,
$$(10+3x+45)^2=x^2+(3x+45)^2$$ and $$x=10sqrt19+30.$$
answered Jul 31 at 10:15
Yves Daoust
110k665203
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let the center be $(-y,0)$ and the rightmost point $(0,x)$.
We express that $(0,10), (x,0)$ and $(x-27,9)$ all belong to the circle:
$$(10+y)^2=x^2+y^2=(x-27)^2+(y+9)^2.$$
Then subtracting the last two terms,
$$-54x+27^2+18y+81=0$$
or
$$y=3x+45.$$
Finally,
$$(10+3x+45)^2=x^2+(3x+45)^2$$ and $$x=10sqrt19+30.$$
answered Jul 31 at 10:15
Yves Daoust
110k665203
Let the center be $(-y,0)$ and the rightmost point $(0,x)$.
We express that $(0,10), (x,0)$ and $(x-27,9)$ all belong to the circle:
$$(10+y)^2=x^2+y^2=(x-27)^2+(y+9)^2.$$
Then subtracting the last two terms,
$$-54x+27^2+18y+81=0$$
or
$$y=3x+45.$$
Finally,
$$(10+3x+45)^2=x^2+(3x+45)^2$$ and $$x=10sqrt19+30.$$
answered Jul 31 at 10:15
Yves Daoust
110k665203
answered Jul 31 at 10:15
Yves Daoust
110k665203
answered Jul 31 at 10:15
Yves Daoust
110k665203
answered Jul 31 at 10:15
Yves Daoust
110k665203
110k665203
add a comment |Â
add a comment |Â
up vote
0
down vote
Let $D',E'$ be the second intersection points
of the verticals $DB,EA$ with the circle.
Also, let $R$ be the radius of the circle
and $y=|AF|=|AC|$.
By the intersecting chords theorem,
beginalign
|EA|cdot|AE'|&=|FA|cdot|AC|
,\
|DB|cdot|BD'|&=|FB|cdot|BC|
,
endalign
we have a system of the two equations in two
unknowns, $R$ and $y$:
beginalign
a(2R-a)&=y^2
,\
b(2(R-a)+b)&=c(2y-c)
.
endalign
After excluding $R$ from this system, we have
a quadratic equation in $y$:
beginalign
y^2-2cdotfracacbcdot y+aleft(fracc^2b-a+bright)&=0
endalign
with two roots:
beginalign
y_pm&=fracacbpmsqrtleft(fracacbright)^2-aleft(fracc^2b-a+bright)
.
endalign
Substitution of $a=10, b=9, c=27$ gives
beginalign
y_+&=40
,\
y_-&=20
,
endalign
and since $y>27$ must hold,
the only suitable value is $y=y_+=40$.
Hence, the answer is $|CF|=2y=80$.
answered yesterday
g.kov
5,5171716
add a comment |Â
up vote
0
down vote
Let $D',E'$ be the second intersection points
of the verticals $DB,EA$ with the circle.
Also, let $R$ be the radius of the circle
and $y=|AF|=|AC|$.
By the intersecting chords theorem,
beginalign
|EA|cdot|AE'|&=|FA|cdot|AC|
,\
|DB|cdot|BD'|&=|FB|cdot|BC|
,
endalign
we have a system of the two equations in two
unknowns, $R$ and $y$:
beginalign
a(2R-a)&=y^2
,\
b(2(R-a)+b)&=c(2y-c)
.
endalign
After excluding $R$ from this system, we have
a quadratic equation in $y$:
beginalign
y^2-2cdotfracacbcdot y+aleft(fracc^2b-a+bright)&=0
endalign
with two roots:
beginalign
y_pm&=fracacbpmsqrtleft(fracacbright)^2-aleft(fracc^2b-a+bright)
.
endalign
Substitution of $a=10, b=9, c=27$ gives
beginalign
y_+&=40
,\
y_-&=20
,
endalign
and since $y>27$ must hold,
the only suitable value is $y=y_+=40$.
Hence, the answer is $|CF|=2y=80$.
answered yesterday
g.kov
5,5171716
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $D',E'$ be the second intersection points
of the verticals $DB,EA$ with the circle.
Also, let $R$ be the radius of the circle
and $y=|AF|=|AC|$.
By the intersecting chords theorem,
beginalign
|EA|cdot|AE'|&=|FA|cdot|AC|
,\
|DB|cdot|BD'|&=|FB|cdot|BC|
,
endalign
we have a system of the two equations in two
unknowns, $R$ and $y$:
beginalign
a(2R-a)&=y^2
,\
b(2(R-a)+b)&=c(2y-c)
.
endalign
After excluding $R$ from this system, we have
a quadratic equation in $y$:
beginalign
y^2-2cdotfracacbcdot y+aleft(fracc^2b-a+bright)&=0
endalign
with two roots:
beginalign
y_pm&=fracacbpmsqrtleft(fracacbright)^2-aleft(fracc^2b-a+bright)
.
endalign
Substitution of $a=10, b=9, c=27$ gives
beginalign
y_+&=40
,\
y_-&=20
,
endalign
and since $y>27$ must hold,
the only suitable value is $y=y_+=40$.
Hence, the answer is $|CF|=2y=80$.
answered yesterday
g.kov
5,5171716
Let $D',E'$ be the second intersection points
of the verticals $DB,EA$ with the circle.
Also, let $R$ be the radius of the circle
and $y=|AF|=|AC|$.
By the intersecting chords theorem,
beginalign
|EA|cdot|AE'|&=|FA|cdot|AC|
,\
|DB|cdot|BD'|&=|FB|cdot|BC|
,
endalign
we have a system of the two equations in two
unknowns, $R$ and $y$:
beginalign
a(2R-a)&=y^2
,\
b(2(R-a)+b)&=c(2y-c)
.
endalign
After excluding $R$ from this system, we have
a quadratic equation in $y$:
beginalign
y^2-2cdotfracacbcdot y+aleft(fracc^2b-a+bright)&=0
endalign
with two roots:
beginalign
y_pm&=fracacbpmsqrtleft(fracacbright)^2-aleft(fracc^2b-a+bright)
.
endalign
Substitution of $a=10, b=9, c=27$ gives
beginalign
y_+&=40
,\
y_-&=20
,
endalign
and since $y>27$ must hold,
the only suitable value is $y=y_+=40$.
Hence, the answer is $|CF|=2y=80$.
answered yesterday
g.kov
5,5171716
edited yesterday
answered yesterday
g.kov
5,5171716
answered yesterday
g.kov
5,5171716
answered yesterday
g.kov
5,5171716
5,5171716
add a comment |Â
add a comment |Â
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Note that the drawing is not to scale.
– Joel Reyes Noche
Jul 31 at 9:54
Thanks for pointing out that diagram is not to scale. Thus, horizontal line is a chord, not a diameter. Specifically, I can't form a second equation, perhaps involving radius of the circle, in order to obtain a unique value for the distance between the two vertical lines.
– PR59
Jul 31 at 10:36
Thanks, Yves and jtm....very helpful and much appreciated.
– PR59
Jul 31 at 11:12
The answer is: "the width of the tunnel at ground level" is 80; the radius of the circle is 85.
– g.kov
Jul 31 at 18:57
@Joel Reyes Noche: it is to scale now.
– g.kov
Jul 31 at 20:26