Mixing
Length of a side of a triangle with angle relations
In the image, $AB=CD=5$, $BC=2$. Then $AD=?$
My try
I extended $BC$ until intersecting $AD$, and i extended $CD$ until intersecting $AB$. After that i tried angle chasing inside the new triangles but i got nothing. Any hints?
This problem is meant to be resolved without trigonometry.
geometry euclidean-geometry triangle
asked Nov 21 '18 at 3:47
Rodrigo Pizarro
834217
add a comment |
In the image, $AB=CD=5$, $BC=2$. Then $AD=?$
My try
I extended $BC$ until intersecting $AD$, and i extended $CD$ until intersecting $AB$. After that i tried angle chasing inside the new triangles but i got nothing. Any hints?
This problem is meant to be resolved without trigonometry.
geometry euclidean-geometry triangle
asked Nov 21 '18 at 3:47
Rodrigo Pizarro
834217
add a comment |
In the image, $AB=CD=5$, $BC=2$. Then $AD=?$
My try
I extended $BC$ until intersecting $AD$, and i extended $CD$ until intersecting $AB$. After that i tried angle chasing inside the new triangles but i got nothing. Any hints?
This problem is meant to be resolved without trigonometry.
geometry euclidean-geometry triangle
asked Nov 21 '18 at 3:47
Rodrigo Pizarro
834217
In the image, $AB=CD=5$, $BC=2$. Then $AD=?$
My try
I extended $BC$ until intersecting $AD$, and i extended $CD$ until intersecting $AB$. After that i tried angle chasing inside the new triangles but i got nothing. Any hints?
This problem is meant to be resolved without trigonometry.
geometry euclidean-geometry triangle
geometry euclidean-geometry triangle
asked Nov 21 '18 at 3:47
Rodrigo Pizarro
834217
asked Nov 21 '18 at 3:47
Rodrigo Pizarro
834217
asked Nov 21 '18 at 3:47
Rodrigo Pizarro
834217
asked Nov 21 '18 at 3:47
Rodrigo Pizarro
834217
asked Nov 21 '18 at 3:47
Rodrigo Pizarro
834217
834217
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
This looks like a trick question:
the construction with all declared
properties does not exist.
Let
$angle BAC=angle ADC=alpha$,
$angle CBA=beta$,
$E=BCcap AD$
and
let's ignore for the moment
that we have a condition
$angle CBA=2angle BAC$.
Then (as it was already noted in other answers)
we must have
begin{align}
angle ECA&=angle EAC=alpha+beta
,\
|ED|&=|AC|=|AE|
,\
|CE|&=|BC|=2
.
end{align}
Let $F$ be the median (and the altitude)
of the isosceles $triangle AEC$.
Then we must
recognize $triangle AFB$
as the famous $3-4-5$
right-angled triangle,
with $|AF|=4$.
Now we can easily find $|AE|$ and $|AD|$:
begin{align}
triangle EFA:quad
|AE|&=sqrt{|AF|^2+|FE|^2}=sqrt{17}
,\
end{align}
and we are tricked to state that
we have the answer:
begin{align}
|AD|&=2|AE|=
2sqrt{17}
.
end{align}
Ant it would be indeed the correct answer,
in case when
we do not have any relations
between the angles $alpha$ and $beta$ .
Otherwise we must check that
the other declared conditions hold,
that is,
begin{align}
angle BAC=alpha&=x
,\
angle CBA=beta&=2x
,
end{align}
in other words
begin{align}
angle CBA&=2angle BAC
.
end{align}
Let's check if the following is true:
begin{align}
triangle AFC:quadalpha+beta&
=x+2x=3x=arctan4
,\
triangle AFB:phantom{quadalpha+}beta
&=2x
=
arctantfrac43
.
end{align}
So, the following must be true:
begin{align}
tfrac13arctan4&=
tfrac12arctantfrac43
,
end{align}
but this is false, since
begin{align}
tfrac13arctan4
&approx 0.4419392213
,\
tfrac12arctantfrac43
&approx 0.4636476090
,
end{align}
thus the original question
does not have a valid solution.
Edit
Another illustration that given geometric construction
is invalid.
Triangle $ABC$ with given constraints on the angles and side lengths
uniquely defines $x$:
begin{align}
triangle ABC:quad
frac{|BC|}{sin x}&=
frac{|AB|}{sin 3x}
,\
frac{2}{sin x}&=
frac{5}{sin x ,(3-4sin^2x)}
,\
sin^2x&=tfrac18
,\
sin x&=tfrac{sqrt2}4
,\
x&=arcsintfrac{sqrt2}4approx 20.7^circ
,
end{align}
Given $x$, we can construct $triangle ABC$
and the point $D$, but the ray $DX:angle XDC=x$
will miss the point $A$,
and the real picture should in fact look like this:
answered Nov 21 '18 at 9:35
g.kov
6,0971718
add a comment |
The idea of this answer is the same as the one in Lee's answer. One can get $AD$ without using trigonometry.
Let $E$ be the intersection point of $AD$ with $BC$.
Then, since $angle{ACE}=3x$, we get $angle{DCE}=5x-3x=2x$.
It follows that $triangle{CBA}equivtriangle{ECD}$. So, $CE=2$ and $ED=CA$.
Since $angle{CEA}=x+2x=3x$, we see that $triangle{ACE}$ is an isosceles triangle. It follows that $AE=AC$.
Let $AC=p$. Then, using
$$CA^2+CD^2=2(CE^2+AE^2)tag1$$
(see here) one gets
$$p^2+5^2=2(2^2+p^2)implies p=sqrt{17}$$
from which
$$AD=2p=2sqrt{17}$$
follows.
The wiki page uses trigonometry for proving $(1)$.
One can prove $(1)$ without using trigonometry.
Let $vec{EC}=vec c,vec{EA}=vec a$. Then, $vec{ED}=-vec a$ and
$$begin{align}CA^2+CD^2&=|vec{a}-vec{c}|^2+|-vec{a}-vec{c}|^2
\\&=2|vec a|^2+2|vec c|^2
\\&=2EA^2+2EC^2qquadsquareend{align}$$
answered Nov 21 '18 at 5:16
mathlove
91.7k881214
add a comment |
Lets say $BC$ intersects $AD$ at point $E$, then $ABC$ is similar to $DCE$, those $AC=ED$ and $CE=2$. Also notice that $ACE$ is isosceles, thus $AC=ED=AE$, i.e. $AD=2AC$.
I don't know, but maybe it helps to solve your problem without trigonometry
answered Nov 21 '18 at 4:37
Lee
1409
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
This looks like a trick question:
the construction with all declared
properties does not exist.
Let
$angle BAC=angle ADC=alpha$,
$angle CBA=beta$,
$E=BCcap AD$
and
let's ignore for the moment
that we have a condition
$angle CBA=2angle BAC$.
Then (as it was already noted in other answers)
we must have
begin{align}
angle ECA&=angle EAC=alpha+beta
,\
|ED|&=|AC|=|AE|
,\
|CE|&=|BC|=2
.
end{align}
Let $F$ be the median (and the altitude)
of the isosceles $triangle AEC$.
Then we must
recognize $triangle AFB$
as the famous $3-4-5$
right-angled triangle,
with $|AF|=4$.
Now we can easily find $|AE|$ and $|AD|$:
begin{align}
triangle EFA:quad
|AE|&=sqrt{|AF|^2+|FE|^2}=sqrt{17}
,\
end{align}
and we are tricked to state that
we have the answer:
begin{align}
|AD|&=2|AE|=
2sqrt{17}
.
end{align}
Ant it would be indeed the correct answer,
in case when
we do not have any relations
between the angles $alpha$ and $beta$ .
Otherwise we must check that
the other declared conditions hold,
that is,
begin{align}
angle BAC=alpha&=x
,\
angle CBA=beta&=2x
,
end{align}
in other words
begin{align}
angle CBA&=2angle BAC
.
end{align}
Let's check if the following is true:
begin{align}
triangle AFC:quadalpha+beta&
=x+2x=3x=arctan4
,\
triangle AFB:phantom{quadalpha+}beta
&=2x
=
arctantfrac43
.
end{align}
So, the following must be true:
begin{align}
tfrac13arctan4&=
tfrac12arctantfrac43
,
end{align}
but this is false, since
begin{align}
tfrac13arctan4
&approx 0.4419392213
,\
tfrac12arctantfrac43
&approx 0.4636476090
,
end{align}
thus the original question
does not have a valid solution.
Edit
Another illustration that given geometric construction
is invalid.
Triangle $ABC$ with given constraints on the angles and side lengths
uniquely defines $x$:
begin{align}
triangle ABC:quad
frac{|BC|}{sin x}&=
frac{|AB|}{sin 3x}
,\
frac{2}{sin x}&=
frac{5}{sin x ,(3-4sin^2x)}
,\
sin^2x&=tfrac18
,\
sin x&=tfrac{sqrt2}4
,\
x&=arcsintfrac{sqrt2}4approx 20.7^circ
,
end{align}
Given $x$, we can construct $triangle ABC$
and the point $D$, but the ray $DX:angle XDC=x$
will miss the point $A$,
and the real picture should in fact look like this:
answered Nov 21 '18 at 9:35
g.kov
6,0971718
add a comment |
This looks like a trick question:
the construction with all declared
properties does not exist.
Let
$angle BAC=angle ADC=alpha$,
$angle CBA=beta$,
$E=BCcap AD$
and
let's ignore for the moment
that we have a condition
$angle CBA=2angle BAC$.
Then (as it was already noted in other answers)
we must have
begin{align}
angle ECA&=angle EAC=alpha+beta
,\
|ED|&=|AC|=|AE|
,\
|CE|&=|BC|=2
.
end{align}
Let $F$ be the median (and the altitude)
of the isosceles $triangle AEC$.
Then we must
recognize $triangle AFB$
as the famous $3-4-5$
right-angled triangle,
with $|AF|=4$.
Now we can easily find $|AE|$ and $|AD|$:
begin{align}
triangle EFA:quad
|AE|&=sqrt{|AF|^2+|FE|^2}=sqrt{17}
,\
end{align}
and we are tricked to state that
we have the answer:
begin{align}
|AD|&=2|AE|=
2sqrt{17}
.
end{align}
Ant it would be indeed the correct answer,
in case when
we do not have any relations
between the angles $alpha$ and $beta$ .
Otherwise we must check that
the other declared conditions hold,
that is,
begin{align}
angle BAC=alpha&=x
,\
angle CBA=beta&=2x
,
end{align}
in other words
begin{align}
angle CBA&=2angle BAC
.
end{align}
Let's check if the following is true:
begin{align}
triangle AFC:quadalpha+beta&
=x+2x=3x=arctan4
,\
triangle AFB:phantom{quadalpha+}beta
&=2x
=
arctantfrac43
.
end{align}
So, the following must be true:
begin{align}
tfrac13arctan4&=
tfrac12arctantfrac43
,
end{align}
but this is false, since
begin{align}
tfrac13arctan4
&approx 0.4419392213
,\
tfrac12arctantfrac43
&approx 0.4636476090
,
end{align}
thus the original question
does not have a valid solution.
Edit
Another illustration that given geometric construction
is invalid.
Triangle $ABC$ with given constraints on the angles and side lengths
uniquely defines $x$:
begin{align}
triangle ABC:quad
frac{|BC|}{sin x}&=
frac{|AB|}{sin 3x}
,\
frac{2}{sin x}&=
frac{5}{sin x ,(3-4sin^2x)}
,\
sin^2x&=tfrac18
,\
sin x&=tfrac{sqrt2}4
,\
x&=arcsintfrac{sqrt2}4approx 20.7^circ
,
end{align}
Given $x$, we can construct $triangle ABC$
and the point $D$, but the ray $DX:angle XDC=x$
will miss the point $A$,
and the real picture should in fact look like this:
answered Nov 21 '18 at 9:35
g.kov
6,0971718
add a comment |
This looks like a trick question:
the construction with all declared
properties does not exist.
Let
$angle BAC=angle ADC=alpha$,
$angle CBA=beta$,
$E=BCcap AD$
and
let's ignore for the moment
that we have a condition
$angle CBA=2angle BAC$.
Then (as it was already noted in other answers)
we must have
begin{align}
angle ECA&=angle EAC=alpha+beta
,\
|ED|&=|AC|=|AE|
,\
|CE|&=|BC|=2
.
end{align}
Let $F$ be the median (and the altitude)
of the isosceles $triangle AEC$.
Then we must
recognize $triangle AFB$
as the famous $3-4-5$
right-angled triangle,
with $|AF|=4$.
Now we can easily find $|AE|$ and $|AD|$:
begin{align}
triangle EFA:quad
|AE|&=sqrt{|AF|^2+|FE|^2}=sqrt{17}
,\
end{align}
and we are tricked to state that
we have the answer:
begin{align}
|AD|&=2|AE|=
2sqrt{17}
.
end{align}
Ant it would be indeed the correct answer,
in case when
we do not have any relations
between the angles $alpha$ and $beta$ .
Otherwise we must check that
the other declared conditions hold,
that is,
begin{align}
angle BAC=alpha&=x
,\
angle CBA=beta&=2x
,
end{align}
in other words
begin{align}
angle CBA&=2angle BAC
.
end{align}
Let's check if the following is true:
begin{align}
triangle AFC:quadalpha+beta&
=x+2x=3x=arctan4
,\
triangle AFB:phantom{quadalpha+}beta
&=2x
=
arctantfrac43
.
end{align}
So, the following must be true:
begin{align}
tfrac13arctan4&=
tfrac12arctantfrac43
,
end{align}
but this is false, since
begin{align}
tfrac13arctan4
&approx 0.4419392213
,\
tfrac12arctantfrac43
&approx 0.4636476090
,
end{align}
thus the original question
does not have a valid solution.
Edit
Another illustration that given geometric construction
is invalid.
Triangle $ABC$ with given constraints on the angles and side lengths
uniquely defines $x$:
begin{align}
triangle ABC:quad
frac{|BC|}{sin x}&=
frac{|AB|}{sin 3x}
,\
frac{2}{sin x}&=
frac{5}{sin x ,(3-4sin^2x)}
,\
sin^2x&=tfrac18
,\
sin x&=tfrac{sqrt2}4
,\
x&=arcsintfrac{sqrt2}4approx 20.7^circ
,
end{align}
Given $x$, we can construct $triangle ABC$
and the point $D$, but the ray $DX:angle XDC=x$
will miss the point $A$,
and the real picture should in fact look like this:
answered Nov 21 '18 at 9:35
g.kov
6,0971718
This looks like a trick question:
the construction with all declared
properties does not exist.
Let
$angle BAC=angle ADC=alpha$,
$angle CBA=beta$,
$E=BCcap AD$
and
let's ignore for the moment
that we have a condition
$angle CBA=2angle BAC$.
Then (as it was already noted in other answers)
we must have
begin{align}
angle ECA&=angle EAC=alpha+beta
,\
|ED|&=|AC|=|AE|
,\
|CE|&=|BC|=2
.
end{align}
Let $F$ be the median (and the altitude)
of the isosceles $triangle AEC$.
Then we must
recognize $triangle AFB$
as the famous $3-4-5$
right-angled triangle,
with $|AF|=4$.
Now we can easily find $|AE|$ and $|AD|$:
begin{align}
triangle EFA:quad
|AE|&=sqrt{|AF|^2+|FE|^2}=sqrt{17}
,\
end{align}
and we are tricked to state that
we have the answer:
begin{align}
|AD|&=2|AE|=
2sqrt{17}
.
end{align}
Ant it would be indeed the correct answer,
in case when
we do not have any relations
between the angles $alpha$ and $beta$ .
Otherwise we must check that
the other declared conditions hold,
that is,
begin{align}
angle BAC=alpha&=x
,\
angle CBA=beta&=2x
,
end{align}
in other words
begin{align}
angle CBA&=2angle BAC
.
end{align}
Let's check if the following is true:
begin{align}
triangle AFC:quadalpha+beta&
=x+2x=3x=arctan4
,\
triangle AFB:phantom{quadalpha+}beta
&=2x
=
arctantfrac43
.
end{align}
So, the following must be true:
begin{align}
tfrac13arctan4&=
tfrac12arctantfrac43
,
end{align}
but this is false, since
begin{align}
tfrac13arctan4
&approx 0.4419392213
,\
tfrac12arctantfrac43
&approx 0.4636476090
,
end{align}
thus the original question
does not have a valid solution.
Edit
Another illustration that given geometric construction
is invalid.
Triangle $ABC$ with given constraints on the angles and side lengths
uniquely defines $x$:
begin{align}
triangle ABC:quad
frac{|BC|}{sin x}&=
frac{|AB|}{sin 3x}
,\
frac{2}{sin x}&=
frac{5}{sin x ,(3-4sin^2x)}
,\
sin^2x&=tfrac18
,\
sin x&=tfrac{sqrt2}4
,\
x&=arcsintfrac{sqrt2}4approx 20.7^circ
,
end{align}
Given $x$, we can construct $triangle ABC$
and the point $D$, but the ray $DX:angle XDC=x$
will miss the point $A$,
and the real picture should in fact look like this:
answered Nov 21 '18 at 9:35
g.kov
6,0971718
edited Nov 21 '18 at 11:11
answered Nov 21 '18 at 9:35
g.kov
6,0971718
answered Nov 21 '18 at 9:35
g.kov
6,0971718
answered Nov 21 '18 at 9:35
g.kov
6,0971718
6,0971718
add a comment |
add a comment |
The idea of this answer is the same as the one in Lee's answer. One can get $AD$ without using trigonometry.
Let $E$ be the intersection point of $AD$ with $BC$.
Then, since $angle{ACE}=3x$, we get $angle{DCE}=5x-3x=2x$.
It follows that $triangle{CBA}equivtriangle{ECD}$. So, $CE=2$ and $ED=CA$.
Since $angle{CEA}=x+2x=3x$, we see that $triangle{ACE}$ is an isosceles triangle. It follows that $AE=AC$.
Let $AC=p$. Then, using
$$CA^2+CD^2=2(CE^2+AE^2)tag1$$
(see here) one gets
$$p^2+5^2=2(2^2+p^2)implies p=sqrt{17}$$
from which
$$AD=2p=2sqrt{17}$$
follows.
The wiki page uses trigonometry for proving $(1)$.
One can prove $(1)$ without using trigonometry.
Let $vec{EC}=vec c,vec{EA}=vec a$. Then, $vec{ED}=-vec a$ and
$$begin{align}CA^2+CD^2&=|vec{a}-vec{c}|^2+|-vec{a}-vec{c}|^2
\\&=2|vec a|^2+2|vec c|^2
\\&=2EA^2+2EC^2qquadsquareend{align}$$
answered Nov 21 '18 at 5:16
mathlove
91.7k881214
add a comment |
The idea of this answer is the same as the one in Lee's answer. One can get $AD$ without using trigonometry.
Let $E$ be the intersection point of $AD$ with $BC$.
Then, since $angle{ACE}=3x$, we get $angle{DCE}=5x-3x=2x$.
It follows that $triangle{CBA}equivtriangle{ECD}$. So, $CE=2$ and $ED=CA$.
Since $angle{CEA}=x+2x=3x$, we see that $triangle{ACE}$ is an isosceles triangle. It follows that $AE=AC$.
Let $AC=p$. Then, using
$$CA^2+CD^2=2(CE^2+AE^2)tag1$$
(see here) one gets
$$p^2+5^2=2(2^2+p^2)implies p=sqrt{17}$$
from which
$$AD=2p=2sqrt{17}$$
follows.
The wiki page uses trigonometry for proving $(1)$.
One can prove $(1)$ without using trigonometry.
Let $vec{EC}=vec c,vec{EA}=vec a$. Then, $vec{ED}=-vec a$ and
$$begin{align}CA^2+CD^2&=|vec{a}-vec{c}|^2+|-vec{a}-vec{c}|^2
\\&=2|vec a|^2+2|vec c|^2
\\&=2EA^2+2EC^2qquadsquareend{align}$$
answered Nov 21 '18 at 5:16
mathlove
91.7k881214
add a comment |
The idea of this answer is the same as the one in Lee's answer. One can get $AD$ without using trigonometry.
Let $E$ be the intersection point of $AD$ with $BC$.
Then, since $angle{ACE}=3x$, we get $angle{DCE}=5x-3x=2x$.
It follows that $triangle{CBA}equivtriangle{ECD}$. So, $CE=2$ and $ED=CA$.
Since $angle{CEA}=x+2x=3x$, we see that $triangle{ACE}$ is an isosceles triangle. It follows that $AE=AC$.
Let $AC=p$. Then, using
$$CA^2+CD^2=2(CE^2+AE^2)tag1$$
(see here) one gets
$$p^2+5^2=2(2^2+p^2)implies p=sqrt{17}$$
from which
$$AD=2p=2sqrt{17}$$
follows.
The wiki page uses trigonometry for proving $(1)$.
One can prove $(1)$ without using trigonometry.
Let $vec{EC}=vec c,vec{EA}=vec a$. Then, $vec{ED}=-vec a$ and
$$begin{align}CA^2+CD^2&=|vec{a}-vec{c}|^2+|-vec{a}-vec{c}|^2
\\&=2|vec a|^2+2|vec c|^2
\\&=2EA^2+2EC^2qquadsquareend{align}$$
answered Nov 21 '18 at 5:16
mathlove
91.7k881214
The idea of this answer is the same as the one in Lee's answer. One can get $AD$ without using trigonometry.
Let $E$ be the intersection point of $AD$ with $BC$.
Then, since $angle{ACE}=3x$, we get $angle{DCE}=5x-3x=2x$.
It follows that $triangle{CBA}equivtriangle{ECD}$. So, $CE=2$ and $ED=CA$.
Since $angle{CEA}=x+2x=3x$, we see that $triangle{ACE}$ is an isosceles triangle. It follows that $AE=AC$.
Let $AC=p$. Then, using
$$CA^2+CD^2=2(CE^2+AE^2)tag1$$
(see here) one gets
$$p^2+5^2=2(2^2+p^2)implies p=sqrt{17}$$
from which
$$AD=2p=2sqrt{17}$$
follows.
The wiki page uses trigonometry for proving $(1)$.
One can prove $(1)$ without using trigonometry.
Let $vec{EC}=vec c,vec{EA}=vec a$. Then, $vec{ED}=-vec a$ and
$$begin{align}CA^2+CD^2&=|vec{a}-vec{c}|^2+|-vec{a}-vec{c}|^2
\\&=2|vec a|^2+2|vec c|^2
\\&=2EA^2+2EC^2qquadsquareend{align}$$
answered Nov 21 '18 at 5:16
mathlove
91.7k881214
answered Nov 21 '18 at 5:16
mathlove
91.7k881214
answered Nov 21 '18 at 5:16
mathlove
91.7k881214
answered Nov 21 '18 at 5:16
mathlove
91.7k881214
91.7k881214
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Lets say $BC$ intersects $AD$ at point $E$, then $ABC$ is similar to $DCE$, those $AC=ED$ and $CE=2$. Also notice that $ACE$ is isosceles, thus $AC=ED=AE$, i.e. $AD=2AC$.
I don't know, but maybe it helps to solve your problem without trigonometry
answered Nov 21 '18 at 4:37
Lee
1409
add a comment |
Lets say $BC$ intersects $AD$ at point $E$, then $ABC$ is similar to $DCE$, those $AC=ED$ and $CE=2$. Also notice that $ACE$ is isosceles, thus $AC=ED=AE$, i.e. $AD=2AC$.
I don't know, but maybe it helps to solve your problem without trigonometry
answered Nov 21 '18 at 4:37
Lee
1409
add a comment |
Lets say $BC$ intersects $AD$ at point $E$, then $ABC$ is similar to $DCE$, those $AC=ED$ and $CE=2$. Also notice that $ACE$ is isosceles, thus $AC=ED=AE$, i.e. $AD=2AC$.
I don't know, but maybe it helps to solve your problem without trigonometry
answered Nov 21 '18 at 4:37
Lee
1409
Lets say $BC$ intersects $AD$ at point $E$, then $ABC$ is similar to $DCE$, those $AC=ED$ and $CE=2$. Also notice that $ACE$ is isosceles, thus $AC=ED=AE$, i.e. $AD=2AC$.
I don't know, but maybe it helps to solve your problem without trigonometry
answered Nov 21 '18 at 4:37
Lee
1409
answered Nov 21 '18 at 4:37
Lee
1409
answered Nov 21 '18 at 4:37
Lee
1409
answered Nov 21 '18 at 4:37
Lee
1409
1409
add a comment |
add a comment |
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