nyuuzyou/wikis huggingface dataset:
Construction of Point in Cartesian Plane with Rational Coordinates

== Theorem ==
Let $\CC$ be a Cartesian plane.
Let $P = \tuple {x, y}$ be a rational point in $\CC$.
Then $P$ is constructible using a compass and straightedge construction.
== Proof ==
Let $x = \dfrac m n$ where $m, n \in \Z_{\ne 0}$ are non-zero integers.
Let $O$ denote the point $\tuple {0, 0}$.
Let $A$ denote the point $\tuple {1, 0}$.
Let $M$ denote the point $\tuple {0, m}$.
Let $N$ denote the point $\tuple {0, n}$.
The $x$-axis is identified with the straight line through $O$ and $A$.
The $y$-axis is constructed as the line perpendicular to $OA$ through $O$.
Using Construction of Lattice Point in Cartesian Plane, the points $M$ and $\N$ are constructed.
The line $NA$ is drawn.
Using Construction of Parallel Line, $MQ$ is drawn, where $Q$ is the point of intersection of $MQ$ with the $x$-axis.
We have that $\dfrac {OM} {ON} = \dfrac m n$
As $\triangle ONA$ is similar to $\triangle OMQ$, it follows that $\dfrac {OM} {ON} = \dfrac {OQ} {OA}$
Therefore $Q$ is the point $\tuple {\dfrac m n, 0}$.
:
Let $y = \dfrac r s$ where $r, s \in \Z_{\ne 0}$ are non-zero integers.
Using Construction of Lattice Point in Cartesian Plane, the points $\tuple {r, 0}$ and $\tuple {s, 0}$ are constructed.
Using the same technique as above, mutatis mutandis, the point $\tuple {0, \dfrac r s}$ is constructed.
Using Construction of Parallel Line, a straight line is drawn through $\tuple {\dfrac m n, 0}$ parallel to the $y$-axis.
Using Construction of Parallel Line, a straight line is drawn through $\tuple {0, \dfrac r s}$ parallel to the $x$-axis.
By definition of Cartesian plane, their intersection is at $\tuple {\dfrac m n, \dfrac r s}$, which is the required point $P = \tuple {x, y}$.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 40$. Construction with Ruler and Compasses
Category:Analytic Geometry
Category:Compass and Straightedge Constructions
Construction of Polyhedron in Outer of Concentric Spheres/Porism

== Theorem ==
== Proof ==
:
We have that the polyhedron as constructed in is divided into pyramids which are similar in number and arrangement.
But from :
:similar pyramids are to one another in the triplicate ratio of their corresponding sides.
Therefore the pyramid whose base is $KBPS$ and whose apex is $A$ has to the similarly arranged pyramid in the other sphere the triplicate ratio of their corresponding sides: that is, $AB$ to the radius of the other sphere.
The same applies to all the other pyramids.
It follows from :
:the sum total of all the pyramids that form the whole polyhedron in the one sphere has to the sum total of all the pyramids that form the whole polyhedron in the other sphere the triplicate ratio of the radii of the spheres.
That is, the triplicate ratio of the diameters of the spheres.
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Spheres
Category:Polyhedra
Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater

== Theorem ==
=== Lemma 1 ===
=== Lemma 2 ===
== Proof ==
Let $\rho$ be a rational straight line.
Let $m$ and $n$ be natural numbers such that $m^2 - n^2$ is not square.
Let $x$ be a straight line such that:
:$(1): \quad m^2 : \paren {m^2 - n^2} = \rho^2 : x^2$
Thus:
:$x^2 = \dfrac {m^2 - n^2} {m^2} \rho^2$
and so:
:$x = \rho \sqrt {1 - k^2}$
where $k = \dfrac n m$.
From $(1)$:
:$x^2 \frown \rho^2$
where $\frown$ denotes commensurability in length.
Thus $x$ is a rational straight line, but:
:$x \smile \rho$
where $\smile$ denotes incommensurability in length.
From $(1)$:
:$ m^2 : n^2 = \rho^2 : \rho^2 - x^2$
so that:
:$\sqrt {\rho^2 - x^2} \frown \rho$
and in fact:
:$\sqrt {\rho^2 - x^2} = k \rho$
where $k$ is a rational number.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater/Lemma 1

== Lemma to Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater ==
== Proof ==
Let $a$ and $b$ be two natural numbers such that $a > b$.
Let $a$ and $b$ be either both even or both odd.
From:
:Even Number minus Even Number is Even
and:
:Odd Number minus Odd Number is Even
their difference $c = a - b$ is even.
Let $d = \dfrac c 2$.
Let $a$ and $b$ be similar plane numbers.
By definition of similar plane numbers it is noted that $a$ and $b$ may both be square.
Then by Square of Sum less Square:
:$a b + d^2 = \paren {a - d}^2$
From Product of Similar Plane Numbers is Square, $a b$ is a square number.
So two square numbers $a b$ and $d^2$ have been found whose sum is square.
Similarly:
:$a b = \paren {a - d}^2 - d^2$
from which it is noted that two square numbers $\paren {a - d}^2$ and $d^2$ have been found whose difference is square.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater/Lemma 2

== Lemma to Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater ==
== Proof ==
Let $a$ and $b$ be square numbers such that $a > b$.
Let $c = a - b$ be even.
Let $d = \dfrac c 2$
From Lemma 1:
:$a b + d^2 = \paren {a - d}^2$
Then:
:$a b + \paren {d - 1}^2 < \paren {a - d}^2$
Suppose $a b + \paren {d - 1}^2$ were square.
Then either:
:$a b + \paren {d - 1}^2 = \paren {a - d - 1}^2$
or:
:$a b + \paren {d - 1}^2 < \paren {a - d - 1}^2$
It cannot be greater, without being equal to \paren {a - d}^2 because they are consecutive numbers.
Suppose:
:$a b + \paren {d - 1}^2 = \paren {a - d - 1}^2$
Since $c = 2 d$ it follows that $c - 2 = 2 \paren {d - 1}$
Therefore from Square of Sum less Square:
:$\paren {a - 2} b + \paren {d - 1}^2 = \paren {a - d - 1}^2$
But :
:$a b + \paren {d - 1}^2 = \paren {a - d - 1}^2$
Therefore:
:$\paren {a - 2} b + \paren {d - 1}^2 = a b + \paren {d - 1}^2$
Thus:
:$\paren {a - 2} b = a b$
which is absurd.
Therefore $a b + \paren {d - 1}^2 \ne \paren {a - d - 1}^2$.
Suppose $a b + \paren {d - 1}^2 = f^2$ for some natural number $f$.
Let $h = 2 \paren {a - d - f}$.
It follows that:
:$\paren {a - b - h} = 2 \paren {f - b}$
So from Square of Sum less Square:
:$\paren {a - h} b + \paren {f - b}^2 = f^2$
But :
:$a b + \paren {d - 1}^2 = f^2$
Thus:
:$\paren {a - h} b + \paren {f - b}^2 = a b + \paren {d - 1}^2$
which is absurd.
Therefore:
:$a b + \paren {d - 1}^2 \not < \paren {a - d - 1}^2$
Both possibilities have been shown not to be possible.
Hence $a b + \paren {d - 1}^2$ is not square.
Thus we have two square numbers whose sum is not square, as required.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Incommensurable with Greater

== Theorem ==
== Proof ==
Let $\rho$ be a rational straight line.
Let $m$ and $n$ be natural numbers such that $m^2 + n^2$ is not square.
Let $x$ be such that:
:$m^2 + n^2 : m^2 = \rho^2 + x^2$
and so:
:$x^2 = \dfrac {m^2} {m^2 + n^2} \rho^2$
or:
:$x = \dfrac {\rho} {\sqrt {1 + k^2} }$
for some rational $k$.
Then $\rho$ and $\dfrac {\rho} {\sqrt {1 + k^2} }$ fit the condition.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Regular 257-Gon

== Theorem ==
It is possible to construct a regular polygon with $257$ sides) using a compass and straightedge construction.
== Proof ==
From Construction of Regular Prime $p$-Gon Exists iff $p$ is Fermat Prime it is known that this construction is possible.
== Historical Note ==
== Sources ==
* : $257$
* : $257$
Category:Compass and Straightedge Constructions
Category:Regular Polygons
Category:257
Construction of Regular 65,537-Gon

== Theorem ==
It is possible to construct a regular polygon with $65 \, 537$ sides) using a compass and straightedge construction.
== Proof ==
From Construction of Regular Prime $p$-Gon Exists iff $p$ is Fermat Prime it is known that this construction is possible.
== Historical Note ==
== Sources ==
* : $65,537$
* : $65,537$
Category:Compass and Straightedge Constructions
Category:Regular Polygons
Category:65,537
Construction of Regular 65,537-Gon/Historical Note

== Historical Note on Construction of Regular $65 \, 537$-Gon ==
It was proved by in $1801$ that the construction is possible.
The first actual construction of a regular $65,537$-gon was attempted by , who calculated the $384$ required quadratic equations in $1894$ after a decade of work. However, it has been suggested that there are mistakes in his work.
== Sources ==
* : $65,537$
* : $65,537$
Category:Historical Notes
Construction of Regular Heptadecagon

== Theorem ==
It is possible to construct a regular hepadecagon (that is, a regular polygon with $17$ sides) using a compass and straightedge construction.
== Construction ==
:
The construction will inscribe a regular hepadecagon inside any arbitrary circle.
By :
:construct a circle with center $O$ and radius $OA$.
By :
:produce $OA$ to $B$, hence making $AB$ a diameter of this circle.
By :
:construct $OC$ perpendicular to $OA$.
By twice:
:construct $OD$ whose length is $\dfrac 1 4$ the length of $OC$.
By :
:join $DA$.
By twice:
:construct $\angle ODE$ to be $\dfrac 1 4$ the angle $\angle ODA$.
By and :
:construct $\angle EDF$ to be half a right angle.
Using and :
:construct a semicircle on $AF$ intersecting $OC$ at $G$.
By :
:construct a semicircle with center $E$ and radius $EG$, intersecting $AB$ at $H$ and $K$.
By :
:construct $HL$ and $KM$ perpendicular to $OA$, intersecting the circle $ACB$ at $L$ and $M$.
By :
:bisect $\angle LOM$ to obtain angle $\angle NOM$.
By :
:join $NM$.
$NM$ is one of the sides of a regular hepadecagon which has been inscribed inside circle $ACB$.
== Proof ==
It remains to be demonstrated that the line segment $NM$ is the side of a regular hepadecagon inscribed in circle $ACB$.
This will be done by demonstrating that $\angle NOM$ is equal to $\dfrac {2 \pi} {17}$ radians, that is, $\dfrac 1 {17}$ of the full circle $ACB$.
For convenience, let the radius $OA$ be equal to $4 a$.
By Pythagoras's Theorem, $AD = a \sqrt {17}$.
By definition of tangent, $OE = a \map \arctan {\dfrac {\angle ODA} 4}$.
By construction, $\angle EDF = \dfrac \pi 4$ radians.
Thus:
== Also see ==
* Construction of Regular Prime $p$-Gon Exists iff $p$ is Fermat Prime
== Historical Note ==
== Sources ==
* : $17$
* : $17$
Category:Compass and Straightedge Constructions
Category:Regular Polygons
Category:17
Construction of Regular Heptadecagon/Historical Note

== Historical Note on Construction of Regular Heptadecagon ==
The existence of the construction of the regular heptadecagon was first demonstrated by on $30$th March $1796$, at the age of $19$.
Some sources suggest that it was this discovery that led him to consider mathematics as a career option.
The construction given here is the one given by in $1893$.
== Sources ==
* : Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)
* : Chapter $\text {A}.25$: Gauss ()
* : Chapter $7$: Patterns in Numbers: Gauss
Category:Historical Notes
Construction of Regular Octahedron within Given Sphere

== Theorem ==
== Proof ==
:
Let $AB$ be the diameter of the given sphere.
Let $AB$ be bisected at $C$.
Let $ADB$ be a semicircle on the diameter $AB$.
Let $CD$ be drawn from $C$ perpendicular to $AB$.
Let $DB$ be joined.
Let the square $EFGH$ be set out whose sides are all equal to $DB$.
Let $HF$ and $EG$ be joined.
Let $K$ be the point where $HF$ and $EG$ meet.
From :
:let $KL$ be drawn from $K$ perpendicular to the plane of the square $EFGH$.
Let $KM$ be produced through to the other side of the plane of the square $EFGH$.
Let $KL$ and $KM$ be equal to any one of the straight lines $EK, FK, GK, HK$.
Let $LE, LF, LG, LH, ME, MF, MG, MH$ be joined.
We have that:
:$KE = KH$
and:
:$\angle EKH$ is a right angle.
Therefore from :
:$HE^2 = 2 \cdot EK^2$
We have that:
:$LK = KE$
and:
:$\angle LKE$ is a right angle.
Therefore from :
:$EL^2 = 2 \cdot EK^2$
But it has already been proved that:
:$HE^2 = 2 \cdot EK^2$
Therefore:
:$LE^2 = EH^2$
and so:
:$LE = EH$
For the same reason:
:$LH = HE$
Therefore $\triangle LEH$ is an equilateral triangle.
Similarly we can prove that each of the remaining triangles of which the sides of $EFGH$ are the bases are equilateral.
Therefore an octahedron has been constructed which is contained by eight equilateral triangles.
Next it is to be demonstrated that the octahedron $EFGHLM$ can be inscribed in the given sphere.
We have that:
:$LM = KM = KE$
Therefore the semicircle described on $LM$ also passes through $E$.
Let $LM$ remain fixed, and let that semicircle be carried around and restored to the same position from which it began.
For the same reason as for $E$ it follows that the semicircle passes through the points $F, G, H$.
Thus the octahedron $EFGHLM$ has been inscribed in a sphere.
We have that:
:$LK = KM$
while $KE$ is common.
These also contain right angles.
Therefore from :
:$LE = EM$
From :
:$\angle LEM$ is a right angle.
Therefore from :
:$LM^2 = 2 \cdot LE^2$
We also have that:
:$AC = CB$
and so:
:$AB = 2 \cdot BC$
But:
:$AB : BC = AB^2 : BD^2$
Therefore:
:$AB^2 = 2 \cdot BD^2$
But it has already been proved that:
:$LM^2 = 2 \cdot LE^2$
and because $EH = DB$:
:$DB^2 = LE^2$
Therefore:
:$AB^2 = LM^2$
and so:
:$AB = LM$
But $AB$ is the diameter of the given sphere.
Therefore $LM$ equals the diameter of the given sphere.
Therefore the (regular) octahedron $EFGHLM$ has been inscribed in the given sphere.
At the same time it has been demonstrated that the square on the diameter of the given sphere is double the square on the edge of the (regular) octahedron $EFGHLM$ inscribed within it.
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Regular Octahedra
Category:Spheres
Construction of Regular Pentagon using Rusty Compass

== Theorem ==
Using a straightedge and rusty compass, it is possible to inscribe a regular pentagon inside a circle.
== Construction ==
:
Let the rusty compass be set to the radius $AD$ of the circle $\CC$ whose center is at $D$.
Let $ADO$ be a diameter of $\CC$.
Using Construction of Perpendicular using Rusty Compass, construct a straight line at right angles to $AD$ from the endpoint $A$.
Mark off $AE$ on this perpendicular so that $AE = AD$.
Bisect $AD$ at $Z$ and draw $ZE$.
On $ZE$, mark off $ZH = DA$ and bisect $ZH$ at $T$.
Construct a straight line at right angles to $EZ$ at $T$.
Let this perpendicular meet $DA$ which has been produced to $I$.
Construct a circle whose center is at $I$ with radius $AD$.
Let this circle meet circle $\CC$ at $L$ and $M$.
Bisect $MO$ and $LO$ and construct perpendiculars to $MO$ and $NO$ respectively at these points of bisection.
Let these perpendiculars meet $\CC$ at $N$ and $P$.
The vertices of the required regular pentagon are $L$, $M$, $N$, $O$ and $P$.
== Proof ==
== Historical Note ==
== Sources ==
* 
* : Abul Wafa (): $45$
Category:Pentagons
Category:Rusty Compass Constructions
Construction of Regular Prime p-Gon Exists iff p is Fermat Prime

== Theorem ==
Let $p$ be a prime number.
Then there exists a compass and straightedge construction for a regular $p$-gon $p$ is a Fermat prime.
== Proof ==
== Also see ==
* 
* 
* Construction of Regular Heptadecagon
== Historical Note ==
== Sources ==
* : Chapter $\text{IV}$: The Prince of Amateurs
* : $17$
* : $257$
* : $17$
* : $257$
* 
* 
Category:Compass and Straightedge Constructions
Category:Regular Polygons
Category:Fermat Primes
Construction of Regular Prime p-Gon Exists iff p is Fermat Prime/Historical Note

== Historical Note on Construction of Regular Prime $p$-Gon Exists iff $p$ is Fermat Prime ==
The result Construction of Regular Prime $p$-Gon Exists iff $p$ is Fermat Prime was stated, but not proved, by , who demonstrated the result for $n = 17$ in $1796$, when he was $18$.
The case $p = 257$ was demonstrated by in $1822$, and again by in $1832$.
The case $p = 65 \, 537$ was attempted by , who offered a construction in $1894$ after a decade of work. However, it has been suggested that there are mistakes in his work.
The cases where $p = 3$ and $p = 5$ were known to the ancient Greeks and are given in .
== Sources ==
* : Chapter $\text{IV}$: The Prince of Amateurs
* : $17$
* : Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)
* : Chapter $\text {A}.25$: Gauss ()
* : $17$
* 
* 
* : Chapter $7$: Patterns in Numbers: Gauss
Category:Historical Notes
Construction of Regular Tetrahedron within Given Sphere

== Theorem ==
=== Lemma ===
== Proof ==
:
Let $AB$ be the diameter of the given sphere.
Let $AB$ be cut at $C$ where $AC = 2 \cdot CB$.
Let $ADB$ be a semicircle on the diameter $AB$.
Let $CD$ be drawn from $C$ perpendicular to $AB$.
Let $DA$ be joined.
Let the circle $EFG$ be constructed whose radius equals $DC$.
From :
:let the equilateral triangle $EFG$ be inscribed within the circle $EFG$.
From :
:let the center of the circle $EFG$ be $H$.
Let $EH, HF, HG$ be joined.
From :
:let $HK$ be drawn from $H$ perpendicular to the plane of the circle $EFG$.
Let $HK$ equal to $AC$ be cut off from $HK$.
Let $KE, KF, KG$ be joined.
We have that $KH$ is perpendicular to the plane of the circle $EFG$.
So from :
:$KH$ is perpendicular to each of the straight lines which meet it and are in the plane of the circle $EFG$.
But each of $HE, HF, HG$ join $HK$ and are in the plane of the circle $EFG$.
Therefore $HK$ is perpendicular to each of $HE, HF, HG$.
We have that:
:$AC = HK$
and:
:$CD = HE$
Both $AC, CD$ and $HK, HE$ contain right angles.
Therefore from :
:$DA = KE$
For the same reason:
:$KF = DA$
:$KG = DA$
Therefore the three straight lines $KE, KF$ and $KG$ are equal to one another.
We have that:
:$AC = 2 \cdot CB$
Therefore:
:$AB = 3 \cdot CB$
But from :
:$AB : BC = AD^2 : DC^2$
Therefore:
:$AD^2 = 3 \cdot DC^2$
But from :
:$FE^2 = 3 \cdot EH^2$
We have that:
:$DC = EH$
Therefore:
:$DA = EF$
But we have that:
:$DA = KE = KF = KG$
Therefore:
:$EF = FG = GE = KE = KF = KG$
Therefore each of $\triangle EFG, \triangle KEF, \triangle KFG, \triangle KEG$ are equilateral.
Therefore a regular tetrahedron $EFGK$ has been constructed out of four equilateral triangles.
In this construction, $\triangle EFG$ is its base and $K$ is its apex.
It is next to be demonstrated that $EFGK$ is inscribed in the given sphere.
Let $KH$ be produced to $L$ such that $HL = CB$.
From :
:$AC : CD = CD : CB$
while:
:$AC = KH$
:$CD = HE$
:$CB = HL$
Therefore:
:$KH : HE = EH : HL$
Therefore from :
:$KH \cdot HL = EH^2$
We have that $\angle KHE$ and $\angle EHL$ are both right angles.
Therefore from:
:
and:
:
the semicircle described on $KL$ also passes through $E$.
Let $KL$ remain fixed, and let that semicircle be carried around and restored to the same position from which it began.
Then it will pass through the points $F$ and $G$.
We have that:
:$KH = AC$
and:
:$HL = AB$
Thus:
:$KL = AB$
Therefore $KL$ is the diameter of the given sphere.
Thus the regular tetrahedron $EFGK$ has been inscribed in the given sphere.
It remains to be shown that the square on the diameter of the given sphere is $1 \frac 1 2$ the square on the side of the regular tetrahedron $EFGK$.
We have that:
:$AC = 2 \cdot CB$
Therefore:
:$AB = 3 \cdot CB$
and so:
:$2 \cdot AB = 3 \cdot AC$
But:
:$BA : AC = BA^2 : AD^2$
Therefore:
:$BA^2 = \dfrac 3 2 AD^2$
But we have that:
:$BA$ equals the diameter of the given sphere
and:
:$AD$ equals the side of the regular tetrahedron $EFGK$.
Hence the result.
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Regular Tetrahedra
Category:Spheres
Construction of Regular Tetrahedron within Given Sphere/Lemma

== Lemma to Construction of Regular Tetrahedron within Given Sphere ==
== Proof ==
:
Let the figure of the semicircle be set out.
Let $DB$ be joined.
Let the square $EC$ be described on $AC$.
Let the parallelogram $FB$ be completed.
From:
:
and:
:
it follows that:
:$BA : AD = DA : AC$
Therefore from :
:$BA \cdot AC = AD^2$
From :
:$AB : BC = EB : EF$
But $EA = AC$ and so:
:$EB = BA \cdot AC$
and:
:$BF = AC \cdot CB$
Therefore:
:$AB : BC = BA \cdot AC : AC \cdot CB$
We also have:
:$BA \cdot AC = AD^2$
We have that $\angle ADB$ is a right angle.
So from :
:$DC$ is a mean proportional between $AC$ and $CB$
and so:
:$AC \cdot CB = DC^2$
Therefore:
:$AB : BC = AD^2 : DC^2$
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Regular Tetrahedra
Construction of Rhombic Dodecahedron

== Theorem ==
The rhombic dodecahedron can be constructed as follows:
Take a cube $K$ embedded in $3$-dimensional space.
Place $6$ more cubes, each congruent with $K$, so that one face of each coincides with a different face of $K$.
Join the vertices of $K$ to the centers of the adjacent cubes to describe square pyramids whose apices are the centers of the adjacent cubes and whose bases are the faces of $K$.
The polyhedron formed by the $6$ square pyramids so formed, together with $K$, is a rhombic dodecahedron.
== Proof ==
== Sources ==
* : $12$
* : $12$
Category:Rhombic Dodecahedra
Construction of Right Angle by Stretched Rope

== Theorem ==
Let a rope be knotted at regular intervals so that there are $12$ equal segments separated by knots.
Tie the rope in a loop consisting of those $12$ segments.
Fix one of the knots to the ground at the point you want the right angle to be placed.
Stretch the rope tightly in the direction of one of the legs of the right angle you want to create and fix the fourth knot to the ground.
Stretch the rope tightly in the direction of the other leg of the right angle you want to create and fix the third knot in the ground, at the same time making sure the remaining segment of $5$ knots is also tight.
The point where the $3$-knot section and the $4$-knot section meet, the rope will be bent at a right angle.
:
== Proof ==
The rope will be stretched in the form of a triangle with sides $3$, $4$ and $5$.
We have that:
:$3^2 + 4^2 = 9 + 16 = 25 = 5^2$
and so by Pythagoras's Theorem the triangle is right angled.
== Historical Note ==
== Sources ==
* : Chapter $1$: Hippocrates' Quadrature of the Lune ( )
* : Chapter $\text {B}.1$: The Pythagorean Theorem
Category:Pythagoras's Theorem
Category:Classic Problems
Construction of Right Angle by Stretched Rope/Historical Note

== Historical Note on Construction of Right Angle by Stretched Rope ==
It was been suggested by that the technique for Construction of Right Angle by Stretched Rope may have been used by the ancient Egyptians to survey their fields.
It is believed that they were aware of the properties of the $3-4-5$ triangle, and so this is a feasible hypothesis, but no evidence exists that they ever used this specific technique.
Despite the fact that there exists no evidence for this, it is sometimes seen reported as though it were a 'well-known fact', whereas it is nothing of the sort.
 hedges his bets in his of $1990$, and introduces this idea with the disclaimer: There is a tradition ...
== Sources ==
* : Chapter $1$: Hippocrates' Quadrature of the Lune ( )
* : Chapter $\text {B}.1$: The Pythagorean Theorem
* : Squares Without Pythagoras
Category:Historical Notes