Mixing
Compass and straightedge construction of a square of an arbitrary line segment
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If I have some arbitrary length line $AB$ and a unit length line $CD$, how can I construct a line whose length is equal to the square of the length of line $AB$ using a compass and a straightedge?
geometry geometric-construction
asked Jun 24 '18 at 11:37
JanekmuricJanekmuric
92211
$endgroup$
add a comment |
$begingroup$
If I have some arbitrary length line $AB$ and a unit length line $CD$, how can I construct a line whose length is equal to the square of the length of line $AB$ using a compass and a straightedge?
geometry geometric-construction
asked Jun 24 '18 at 11:37
JanekmuricJanekmuric
92211
$endgroup$
3
$begingroup$
Hint: $frac{a}{1}=frac{x}{a}$. Also, as a matter of terminology, $AB$ and $CD$ are line segments.
$endgroup$
– Lozenges
Jun 24 '18 at 11:52
add a comment |
$begingroup$
If I have some arbitrary length line $AB$ and a unit length line $CD$, how can I construct a line whose length is equal to the square of the length of line $AB$ using a compass and a straightedge?
geometry geometric-construction
asked Jun 24 '18 at 11:37
JanekmuricJanekmuric
92211
$endgroup$
If I have some arbitrary length line $AB$ and a unit length line $CD$, how can I construct a line whose length is equal to the square of the length of line $AB$ using a compass and a straightedge?
geometry geometric-construction
geometry geometric-construction
asked Jun 24 '18 at 11:37
JanekmuricJanekmuric
92211
asked Jun 24 '18 at 11:37
JanekmuricJanekmuric
92211
asked Jun 24 '18 at 11:37
JanekmuricJanekmuric
92211
asked Jun 24 '18 at 11:37
JanekmuricJanekmuric
92211
asked Jun 24 '18 at 11:37
JanekmuricJanekmuric
92211
92211
3
$begingroup$
Hint: $frac{a}{1}=frac{x}{a}$. Also, as a matter of terminology, $AB$ and $CD$ are line segments.
$endgroup$
– Lozenges
Jun 24 '18 at 11:52
add a comment |
3
$begingroup$
Hint: $frac{a}{1}=frac{x}{a}$. Also, as a matter of terminology, $AB$ and $CD$ are line segments.
$endgroup$
– Lozenges
Jun 24 '18 at 11:52
3
3
$begingroup$
Hint: $frac{a}{1}=frac{x}{a}$. Also, as a matter of terminology, $AB$ and $CD$ are line segments.
$endgroup$
– Lozenges
Jun 24 '18 at 11:52
$begingroup$
Hint: $frac{a}{1}=frac{x}{a}$. Also, as a matter of terminology, $AB$ and $CD$ are line segments.
$endgroup$
– Lozenges
Jun 24 '18 at 11:52
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It is assumed that the simple way to erect a normal at the end of a line segment by Ruler & Compass is already known.
Construct or place the line segments $CD,AP$ as perpendicular straight lines making points $D,A$ to coincide.
Join $CP$ and construct its perpendicular line $PB$ cutting extended line $CD$ at $B$.
Construct a circle $PCQB$ with $CB$ as diameter, mark the the center point of $CB$ as $O,$ center of circle for reference.
By constant product of line segments in a Circle $$ CDcdot AB = AP^2.$$ From this theorem we obtain the straight line segment required as $AB$
since
$$AP=AQ=c, ,CD=1,,$$
and so it follows
$$ AB=c^2.$$
answered Jun 24 '18 at 20:22
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
$begingroup$
Here's the more general construction for the product of two given lengths. On the sides of an angle of vertex $O$ construct points $A$ and $B$ such that $OA$ and $OB$ are the segments you want to multiply. On side $OB$ construct point $U$ such that $OU=1$. Let then $C$ be the intersection of side $OA$ with the line through $B$, parallel to $AU$. The similarity of triangles $OUA$ and $OBC$ gives then
$$
OA:OU=OC:OB,
quadtext{that is:}quad
OC=OAcdot OB.
$$
answered Jun 24 '18 at 21:01
AretinoAretino
23.7k21443
$endgroup$
add a comment |
$begingroup$
Here's yet another construction, in the spirit of Narasimham's answer but with the generality of Aretino's, in that it allows you to construct a segment whose length is the product of any two segments:
Suppose you have two segments of length $a$ and $b$, and a unit-length segment. Choose any point $P$ and draw segments $AP$, $BP$, and $CP$ so that $A,P,B$ are collinear with $P$ between $A$ and $B$, and such that $|AP|=a$, $|BP|=b$, and $|CP|=1$. (See figure.)
Now, construct the circumcircle of triangle $ABC$. This can be done by constructing the perpendicular bisectors of any two of the triangle's sides; they meet at the circumcenter. (See figure.)
Now extend $CP$ until it reaches the other side of the circle at point $D$. Then $CPcdot DP = AP cdot BP$ implies that $DP = ab$. (See figure.)
Narasimham's answer is just a special case of this in which $a=b$ and in which the initial segments $AB$ and $PC$ are constructed to be perpendicular, neither of which is really necessary to make the argument work.
answered Jun 24 '18 at 21:23
mweissmweiss
17.6k23370
$endgroup$
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
$begingroup$
It is assumed that the simple way to erect a normal at the end of a line segment by Ruler & Compass is already known.
Construct or place the line segments $CD,AP$ as perpendicular straight lines making points $D,A$ to coincide.
Join $CP$ and construct its perpendicular line $PB$ cutting extended line $CD$ at $B$.
Construct a circle $PCQB$ with $CB$ as diameter, mark the the center point of $CB$ as $O,$ center of circle for reference.
By constant product of line segments in a Circle $$ CDcdot AB = AP^2.$$ From this theorem we obtain the straight line segment required as $AB$
since
$$AP=AQ=c, ,CD=1,,$$
and so it follows
$$ AB=c^2.$$
answered Jun 24 '18 at 20:22
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
$begingroup$
It is assumed that the simple way to erect a normal at the end of a line segment by Ruler & Compass is already known.
Construct or place the line segments $CD,AP$ as perpendicular straight lines making points $D,A$ to coincide.
Join $CP$ and construct its perpendicular line $PB$ cutting extended line $CD$ at $B$.
Construct a circle $PCQB$ with $CB$ as diameter, mark the the center point of $CB$ as $O,$ center of circle for reference.
By constant product of line segments in a Circle $$ CDcdot AB = AP^2.$$ From this theorem we obtain the straight line segment required as $AB$
since
$$AP=AQ=c, ,CD=1,,$$
and so it follows
$$ AB=c^2.$$
answered Jun 24 '18 at 20:22
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
$begingroup$
It is assumed that the simple way to erect a normal at the end of a line segment by Ruler & Compass is already known.
Construct or place the line segments $CD,AP$ as perpendicular straight lines making points $D,A$ to coincide.
Join $CP$ and construct its perpendicular line $PB$ cutting extended line $CD$ at $B$.
Construct a circle $PCQB$ with $CB$ as diameter, mark the the center point of $CB$ as $O,$ center of circle for reference.
By constant product of line segments in a Circle $$ CDcdot AB = AP^2.$$ From this theorem we obtain the straight line segment required as $AB$
since
$$AP=AQ=c, ,CD=1,,$$
and so it follows
$$ AB=c^2.$$
answered Jun 24 '18 at 20:22
NarasimhamNarasimham
20.8k52158
$endgroup$
It is assumed that the simple way to erect a normal at the end of a line segment by Ruler & Compass is already known.
Construct or place the line segments $CD,AP$ as perpendicular straight lines making points $D,A$ to coincide.
Join $CP$ and construct its perpendicular line $PB$ cutting extended line $CD$ at $B$.
Construct a circle $PCQB$ with $CB$ as diameter, mark the the center point of $CB$ as $O,$ center of circle for reference.
By constant product of line segments in a Circle $$ CDcdot AB = AP^2.$$ From this theorem we obtain the straight line segment required as $AB$
since
$$AP=AQ=c, ,CD=1,,$$
and so it follows
$$ AB=c^2.$$
answered Jun 24 '18 at 20:22
NarasimhamNarasimham
20.8k52158
edited Jan 15 at 9:29
answered Jun 24 '18 at 20:22
NarasimhamNarasimham
20.8k52158
answered Jun 24 '18 at 20:22
NarasimhamNarasimham
20.8k52158
answered Jun 24 '18 at 20:22
NarasimhamNarasimham
20.8k52158
20.8k52158
add a comment |
add a comment |
$begingroup$
Here's the more general construction for the product of two given lengths. On the sides of an angle of vertex $O$ construct points $A$ and $B$ such that $OA$ and $OB$ are the segments you want to multiply. On side $OB$ construct point $U$ such that $OU=1$. Let then $C$ be the intersection of side $OA$ with the line through $B$, parallel to $AU$. The similarity of triangles $OUA$ and $OBC$ gives then
$$
OA:OU=OC:OB,
quadtext{that is:}quad
OC=OAcdot OB.
$$
answered Jun 24 '18 at 21:01
AretinoAretino
23.7k21443
$endgroup$
add a comment |
$begingroup$
Here's the more general construction for the product of two given lengths. On the sides of an angle of vertex $O$ construct points $A$ and $B$ such that $OA$ and $OB$ are the segments you want to multiply. On side $OB$ construct point $U$ such that $OU=1$. Let then $C$ be the intersection of side $OA$ with the line through $B$, parallel to $AU$. The similarity of triangles $OUA$ and $OBC$ gives then
$$
OA:OU=OC:OB,
quadtext{that is:}quad
OC=OAcdot OB.
$$
answered Jun 24 '18 at 21:01
AretinoAretino
23.7k21443
$endgroup$
add a comment |
$begingroup$
Here's the more general construction for the product of two given lengths. On the sides of an angle of vertex $O$ construct points $A$ and $B$ such that $OA$ and $OB$ are the segments you want to multiply. On side $OB$ construct point $U$ such that $OU=1$. Let then $C$ be the intersection of side $OA$ with the line through $B$, parallel to $AU$. The similarity of triangles $OUA$ and $OBC$ gives then
$$
OA:OU=OC:OB,
quadtext{that is:}quad
OC=OAcdot OB.
$$
answered Jun 24 '18 at 21:01
AretinoAretino
23.7k21443
$endgroup$
Here's the more general construction for the product of two given lengths. On the sides of an angle of vertex $O$ construct points $A$ and $B$ such that $OA$ and $OB$ are the segments you want to multiply. On side $OB$ construct point $U$ such that $OU=1$. Let then $C$ be the intersection of side $OA$ with the line through $B$, parallel to $AU$. The similarity of triangles $OUA$ and $OBC$ gives then
$$
OA:OU=OC:OB,
quadtext{that is:}quad
OC=OAcdot OB.
$$
answered Jun 24 '18 at 21:01
AretinoAretino
23.7k21443
answered Jun 24 '18 at 21:01
AretinoAretino
23.7k21443
answered Jun 24 '18 at 21:01
AretinoAretino
23.7k21443
answered Jun 24 '18 at 21:01
AretinoAretino
23.7k21443
23.7k21443
add a comment |
add a comment |
$begingroup$
Here's yet another construction, in the spirit of Narasimham's answer but with the generality of Aretino's, in that it allows you to construct a segment whose length is the product of any two segments:
Suppose you have two segments of length $a$ and $b$, and a unit-length segment. Choose any point $P$ and draw segments $AP$, $BP$, and $CP$ so that $A,P,B$ are collinear with $P$ between $A$ and $B$, and such that $|AP|=a$, $|BP|=b$, and $|CP|=1$. (See figure.)
Now, construct the circumcircle of triangle $ABC$. This can be done by constructing the perpendicular bisectors of any two of the triangle's sides; they meet at the circumcenter. (See figure.)
Now extend $CP$ until it reaches the other side of the circle at point $D$. Then $CPcdot DP = AP cdot BP$ implies that $DP = ab$. (See figure.)
Narasimham's answer is just a special case of this in which $a=b$ and in which the initial segments $AB$ and $PC$ are constructed to be perpendicular, neither of which is really necessary to make the argument work.
answered Jun 24 '18 at 21:23
mweissmweiss
17.6k23370
$endgroup$
add a comment |
$begingroup$
Here's yet another construction, in the spirit of Narasimham's answer but with the generality of Aretino's, in that it allows you to construct a segment whose length is the product of any two segments:
Suppose you have two segments of length $a$ and $b$, and a unit-length segment. Choose any point $P$ and draw segments $AP$, $BP$, and $CP$ so that $A,P,B$ are collinear with $P$ between $A$ and $B$, and such that $|AP|=a$, $|BP|=b$, and $|CP|=1$. (See figure.)
Now, construct the circumcircle of triangle $ABC$. This can be done by constructing the perpendicular bisectors of any two of the triangle's sides; they meet at the circumcenter. (See figure.)
Now extend $CP$ until it reaches the other side of the circle at point $D$. Then $CPcdot DP = AP cdot BP$ implies that $DP = ab$. (See figure.)
Narasimham's answer is just a special case of this in which $a=b$ and in which the initial segments $AB$ and $PC$ are constructed to be perpendicular, neither of which is really necessary to make the argument work.
answered Jun 24 '18 at 21:23
mweissmweiss
17.6k23370
$endgroup$
add a comment |
$begingroup$
Here's yet another construction, in the spirit of Narasimham's answer but with the generality of Aretino's, in that it allows you to construct a segment whose length is the product of any two segments:
Suppose you have two segments of length $a$ and $b$, and a unit-length segment. Choose any point $P$ and draw segments $AP$, $BP$, and $CP$ so that $A,P,B$ are collinear with $P$ between $A$ and $B$, and such that $|AP|=a$, $|BP|=b$, and $|CP|=1$. (See figure.)
Now, construct the circumcircle of triangle $ABC$. This can be done by constructing the perpendicular bisectors of any two of the triangle's sides; they meet at the circumcenter. (See figure.)
Now extend $CP$ until it reaches the other side of the circle at point $D$. Then $CPcdot DP = AP cdot BP$ implies that $DP = ab$. (See figure.)
Narasimham's answer is just a special case of this in which $a=b$ and in which the initial segments $AB$ and $PC$ are constructed to be perpendicular, neither of which is really necessary to make the argument work.
answered Jun 24 '18 at 21:23
mweissmweiss
17.6k23370
$endgroup$
Here's yet another construction, in the spirit of Narasimham's answer but with the generality of Aretino's, in that it allows you to construct a segment whose length is the product of any two segments:
Suppose you have two segments of length $a$ and $b$, and a unit-length segment. Choose any point $P$ and draw segments $AP$, $BP$, and $CP$ so that $A,P,B$ are collinear with $P$ between $A$ and $B$, and such that $|AP|=a$, $|BP|=b$, and $|CP|=1$. (See figure.)
Now, construct the circumcircle of triangle $ABC$. This can be done by constructing the perpendicular bisectors of any two of the triangle's sides; they meet at the circumcenter. (See figure.)
Now extend $CP$ until it reaches the other side of the circle at point $D$. Then $CPcdot DP = AP cdot BP$ implies that $DP = ab$. (See figure.)
Narasimham's answer is just a special case of this in which $a=b$ and in which the initial segments $AB$ and $PC$ are constructed to be perpendicular, neither of which is really necessary to make the argument work.
answered Jun 24 '18 at 21:23
mweissmweiss
17.6k23370
answered Jun 24 '18 at 21:23
mweissmweiss
17.6k23370
answered Jun 24 '18 at 21:23
mweissmweiss
17.6k23370
answered Jun 24 '18 at 21:23
mweissmweiss
17.6k23370
17.6k23370
add a comment |
add a comment |
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3
$begingroup$
Hint: $frac{a}{1}=frac{x}{a}$. Also, as a matter of terminology, $AB$ and $CD$ are line segments.
$endgroup$
– Lozenges
Jun 24 '18 at 11:52