Difference between revisions of "Arbelos"

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|ImageName=Arbelos
 
|ImageName=Arbelos
 
|Image=shoemakers Knife.jpg
 
|Image=shoemakers Knife.jpg
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[[Image:Arbelos(1).jpg|350px]]
 
[[Image:Arbelos(1).jpg|350px]]
  
In geometry, an '''arbelos''' is a figure bounded by three semicircles, tangent in pairs, and with diameters lying on the same line. Graphically, an arbelos is the green region in the picture above. The position of central point B is arbitrary and can be anywhere along the diameter. Here we assume that the three diameters are <math>r, (1- r), and 1,</math> as the image shows.  
+
In geometry, an '''arbelos''' is a figure bounded by three semicircles, tangent in pairs, and with diameters lying on the same line. Graphically, an arbelos is the green region in the picture above. The position of central point B is arbitrary and can be anywhere along the diameter. Here we assume that the three diameters are <math>r, (1- r), </math>and <math>1,</math> as the image shows.  
  
 
From now on, let's think in a semicircle way!
 
From now on, let's think in a semicircle way!
 +
 +
Try it yourself!
 +
{{#iframe:https://www.cs.drexel.edu/~jcm357/MathImagesDU12/arbelos/arbelos.html|600|350}}
  
 
|ImageDesc===Properties==  
 
|ImageDesc===Properties==  
[[Image:Arbelos(2).jpg|Figure 1-Properties of Arbelos|thumb|450px|right]]
 
  
 
Arbelos has lots of unexpected but interesting properties.
 
Arbelos has lots of unexpected but interesting properties.
  
 +
[[Image:Arbelos(2).jpg|Figure 1-Properties of Arbelos|thumb|450px|right]]
 
===Arc Length===
 
===Arc Length===
{{hide|1=
 
 
The arc length along the bottom of the arbelos is the same as the arc length along the enclosing semicircle.
 
The arc length along the bottom of the arbelos is the same as the arc length along the enclosing semicircle.
  
 
Proof:
 
Proof:
  
The circumference of a circle is proportional to its diameter. Therefore,
+
The circumference of a circle is proportional to its diameter. See Figure 1 and we can find that
 +
 
 +
:<math>C_{AB} = \frac{1}{2}\pi r </math>
 +
:<math>C_{BC} = \frac{1}{2}\pi (1 - r) </math>
 +
 
 +
Add them together and we get
  
 
:<math>C_{AB} + C_{BC} = \frac{1}{2} \pi r + \frac{1}{2} \pi (1-r) = \frac{1}{2} \pi (r + 1 -r ) = \frac{1}{2} \pi </math>
 
:<math>C_{AB} + C_{BC} = \frac{1}{2} \pi r + \frac{1}{2} \pi (1-r) = \frac{1}{2} \pi (r + 1 -r ) = \frac{1}{2} \pi </math>
:<math>C_{AC} = \frac{1}{2} \pi </math>
+
 
:<math>C_{AB} + C_{BC} = C_{AC} </math>
+
whereas
}}
+
 
 +
:<math>C_{AC} = \frac{1}{2} \pi </math>.
 +
 
 +
Therefore, they have the same length: <math>C_{AB} + C_{BC} = C_{AC} </math>.
  
 
===Area===
 
===Area===
{{hide|1=
 
 
Starting from point <math>B</math>, draw a line that is perpendicular to <math>AC</math>. It meets the enclosing semicircle at point <math>D</math>. Then the area of the arbelos equals the area of a circle with diameter <math>BD</math>.  
 
Starting from point <math>B</math>, draw a line that is perpendicular to <math>AC</math>. It meets the enclosing semicircle at point <math>D</math>. Then the area of the arbelos equals the area of a circle with diameter <math>BD</math>.  
  
 
{{SwitchPreview|ShowMessage=Click to show proof|HideMessage=Click to hide proof|PreviewText=If you don't know how to prove it, start with what you already know.|FullText=
 
{{SwitchPreview|ShowMessage=Click to show proof|HideMessage=Click to hide proof|PreviewText=If you don't know how to prove it, start with what you already know.|FullText=
  
If you don't know how to prove it, start with what you already know: the area of the circle is the same as the area of the arbelos. First, try to use line segments to represent those two areas.
+
First, try to use line segments to represent those two areas.
The area of a circle is <math>\pi \frac{d^2}{4}</math>,
+
The area of a circle is <math>\pi \frac{d^2}{4}</math> and therefore the area of a semicircle is <math>\frac{1}{2}\pi \frac{d^2}{4}</math>.
:{{EquationRef2|Eq. 1}}<math>\therefore A_{circle} = \pi\left( \frac{(BD)^2}{4}\right) </math>
 
  
( Here, the name of the line segment in parentheses indicates the length of the segment.)  
+
:{{EquationRef2|Eq. 1}}<math> A_{circle} = \pi\left( \frac{(BD)^2}{4}\right) </math> (''The name of the line segment in parentheses indicates the length of the segment.'')  
  
 
We can tell from the image that <math> A_{arbelos} = A_{semicircleAC} - A_{semicircleAB} - A_{semicircleBC} </math>,
 
We can tell from the image that <math> A_{arbelos} = A_{semicircleAC} - A_{semicircleAB} - A_{semicircleBC} </math>,
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:{{EquationRef2|Eq. 2}} <math> A_{arbelos}= \pi \frac{ (AB) \cdot (BC)}{4}</math>  
 
:{{EquationRef2|Eq. 2}} <math> A_{arbelos}= \pi \frac{ (AB) \cdot (BC)}{4}</math>  
  
According to {{EquationNote|Eq. 1}} and {{EquationNote|Eq. 2}},  we translate our goal, <math>A_{cirlce}= A_{arbelos}</math>, into <math>\pi\left(\frac{(BD)^2}{4}\right)= \pi \frac{(AB) \cdot (BC)}{4}</math>. Simplify this "big" equation, and you will find that you only need to prove that <math>(BD)^2 = (AB)(BC)</math>.  
+
According to {{EquationNote|Eq. 1}} and {{EquationNote|Eq. 2}},  we translate our goal, <math>A_{cirlce}= A_{arbelos}</math>, into <math>\pi\left(\frac{(BD)^2}{4}\right)= \pi \frac{(AB) \cdot (BC)}{4}</math>. Simplify this "big" equation, and you will find that you only need to prove that <math>(BD)^2 = (AB)(BC)</math>.
 +
 
 +
 
 +
:<math>\blacktriangleright</math>We have now translated the seemingly sophisticated problem of proving that the area of the circle was the same as the area of the arbelos into something nice :::and simple. Reduction to absurdity is a very important method when dealing with mathematical problems where you don't have a clue.
  
'''We have now translated the seemingly sophisticated problem of proving that the area of the circle was the same as the area of the arbelos into something nice and simple. Reduction to absurdity is a very important method when dealing with mathematical problems where you don't have a clue.'''
 
  
 
Now let's prove that <math>(BD)^2 = (AB)(BC)</math>:  
 
Now let's prove that <math>(BD)^2 = (AB)(BC)</math>:  
  
According to <balloon title="In geometry, if A, D and C are points on a circle where AC is a diameter of the circle, then angle ADC is a right angle.">Thales' Theorem</balloon>,<math>\triangle ADC </math> is a right triangle, thus <math>\angle{ADC} = 90^ \circ</math>.
+
According to <balloon title="In geometry, if A, D and C are points on a circle where AC is a diameter of the circle, then angle ADC is a right angle.">Thales' Theorem</balloon>,<math>\triangle ADC </math> is a right triangle, thus <math>\angle{ADC} = 90^ \circ</math>. Since we drew a line from B that is perpendicular to AC as stated in the problem, it is given that <math>DB \perp AC</math>. Therefore, <math>\triangle ABD </math> is a right triangle with <math>\angle{ABD} = 90^\circ</math>. Thus, <math>\angle{ADC} = \angle{ABD} </math>.  
  
Since we drew a line from B that is perpendicular to AC as stated in the problem, it is given that <math>DB \perp AC</math>.
+
:{{EquationRef2|Eq. 3}}<math>\angle{DAB} = \angle{DAC}</math> because they are the same angle.
 
 
Therefore, <math>\triangle ABD </math> is a right triangle with <math>\angle{ABD} = 90^\circ</math>.
 
:<math>\therefore \angle{ADC} = \angle{ABD} </math>
 
:{{EquationRef2|Eq. 1}}<math>\angle{DAB} = \angle{DAC}</math> because they are the same angle.
 
  
 
In a triangle, the sum of the three angles is <math>180^\circ</math>.  
 
In a triangle, the sum of the three angles is <math>180^\circ</math>.  
  
:{{EquationRef2|Eq. 2}}<math>\angle{ADB} = 180^\circ -\angle{ABD} - \angle{DAB} = 90^\circ - \angle{DAB}</math>
+
:{{EquationRef2|Eq. 4}}<math>\angle{ADB} = 180^\circ -\angle{ABD} - \angle{DAB} = 90^\circ - \angle{DAB}</math>
:{{EquationRef2|Eq. 3}}<math> \angle{ACD} = 180^\circ - \angle{ADC} - \angle{DAC} = 90^\circ - \angle{DAC}</math>
+
:{{EquationRef2|Eq. 5}}<math> \angle{ACD} = 180^\circ - \angle{ADC} - \angle{DAC} = 90^\circ - \angle{DAC}</math>
  
Because of {{EquationNote|Eq. 1}}, therefore {{EquationNote|Eq. 2}} = {{EquationNote|Eq. 3}}, <math>\angle{ADB} = \angle{ACD} </math>
+
Because of {{EquationNote|Eq. 3}}, therefore {{EquationNote|Eq. 4}} = {{EquationNote|Eq. 5}}, <math>\angle{ADB} = \angle{ACD} </math>
  
 
Therefore, <math>\triangle ABD </math> and <math>\triangle BCD</math> are similar triangles since both of their corresponding angles are equal in measure.
 
Therefore, <math>\triangle ABD </math> and <math>\triangle BCD</math> are similar triangles since both of their corresponding angles are equal in measure.
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Note that <math>C_1 </math> and <math>C_2</math> make a whole circle, which is the area of the circle with diameter BD, whereas <math>A </math> represents the area of the arbelos. Thus, this last equation proves that the area of the arbelos equals the area of the circle with diameter BD.
 
Note that <math>C_1 </math> and <math>C_2</math> make a whole circle, which is the area of the circle with diameter BD, whereas <math>A </math> represents the area of the arbelos. Thus, this last equation proves that the area of the arbelos equals the area of the circle with diameter BD.
  
}}}}
+
}}
  
 
===Rectangles===
 
===Rectangles===
{{hide|1=
+
 
 +
[[image:Arbelos(2).jpg|400px]]
 +
 
 
Assume line segment <math>AD</math> intersects semicircle <math>AB</math> at point <math>E</math>. Line segment <math>CD</math> intersects semicircle <math>BC</math> at point <math>F</math>. Then the quadrilateral, <math>BEDF</math>, is a rectangle.
 
Assume line segment <math>AD</math> intersects semicircle <math>AB</math> at point <math>E</math>. Line segment <math>CD</math> intersects semicircle <math>BC</math> at point <math>F</math>. Then the quadrilateral, <math>BEDF</math>, is a rectangle.
  
 
{{SwitchPreview|ShowMessage=Click to show proof|HideMessage=Click to hide proof|PreviewText=According to Thales' Theorem,|FullText=
 
{{SwitchPreview|ShowMessage=Click to show proof|HideMessage=Click to hide proof|PreviewText=According to Thales' Theorem,|FullText=
 
According to <balloon title="In geometry, if A, D and C are points on a circle where AC is a diameter of the circle, then angle ADC is a right angle.">Thales' Theorem</balloon>,  
 
According to <balloon title="In geometry, if A, D and C are points on a circle where AC is a diameter of the circle, then angle ADC is a right angle.">Thales' Theorem</balloon>,  
:<math>\angle {ADC} = \angle{AEB} = \angle{BFC} = 90^\circ</math>
+
<math>\angle{ADC}</math>, <math>\angle{AEB}</math>, and <math>\angle{BFC}</math> are all right angles. So <math>\angle {ADC} = \angle{AEB} = \angle{BFC} = 90^\circ</math>.
 +
 
 +
Thus <math>BEDF</math> has three right angles now: <math>\angle{BED}, \angle{BFD}</math> and <math>\angle{EDF}</math>.
  
Thus <math>BEDF</math> has three right angles now: <math>\angle{BED}, \angle{BFD}</math> and <math>\angle{EDF}</math>. Its last angle <math>\angle{EBF} </math> is also a right angle because the sum of the four angles of a quadrilateral is <math>360^\circ</math> and <math>\angle{EBF} = 360^\circ - 3 \cdot 90^\circ = 90^\circ</math>. Therefore, <math>BEDF</math> is a rectangle.
+
Its last angle <math>\angle{EBF} </math> is also a right angle because the sum of the four angles of a quadrilateral is <math>360^\circ</math> and  
  
Note: Since <math>BEDF</math> is a rectangle, <math>EF</math> and <math>BD</math> are equal and  bisect each other. <math>O</math> is the midpoint.
+
:<math>\angle{EBF} = 360^\circ - 3 \cdot 90^\circ = 90^\circ</math>.
}}}}
+
 
 +
Therefore, <math>BEDF</math> is a rectangle. Note that since <math>BEDF</math> is a rectangle, <math>EF</math> and <math>BD</math> are equal and  bisect each other. <math>O</math> is the midpoint.
 +
}}
  
 
===Tangents===
 
===Tangents===
{{hide|1=
 
 
A line that connects the two points <math>E, F</math> is tangent to the two smaller semicircles at <math>E</math> and <math>F</math>.   
 
A line that connects the two points <math>E, F</math> is tangent to the two smaller semicircles at <math>E</math> and <math>F</math>.   
 
[[Image:Arbelos(7)-.jpg|Figure 2-Tangent|thumb|400px|left]]
 
[[Image:Arbelos(7)-.jpg|Figure 2-Tangent|thumb|400px|left]]
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One more thing: it is not a coincidence that <math>B, E, D, F</math> are on a common circle. In a triangle, three perpendicular bisectors meet in a point. In a right triangle, that point is the midpoint of the hypotenuse. The distance from this point to each vertex of the triangle is the same. So a circle centered at the midpoint of the hypotenuse passes through all three vertices of a right triangle. If we treat the rectangle <math>BEDF</math> as two right triangles, <math>\triangle DEF</math> and <math>\triangle BEF</math>, a circle centered at <math>O</math> passes through all the vertices of these two triangles. Therefore the four vertices <math>B, E, D</math> and <math>F</math> always lie along a common circle.
 
One more thing: it is not a coincidence that <math>B, E, D, F</math> are on a common circle. In a triangle, three perpendicular bisectors meet in a point. In a right triangle, that point is the midpoint of the hypotenuse. The distance from this point to each vertex of the triangle is the same. So a circle centered at the midpoint of the hypotenuse passes through all three vertices of a right triangle. If we treat the rectangle <math>BEDF</math> as two right triangles, <math>\triangle DEF</math> and <math>\triangle BEF</math>, a circle centered at <math>O</math> passes through all the vertices of these two triangles. Therefore the four vertices <math>B, E, D</math> and <math>F</math> always lie along a common circle.
}}
 
  
 
Those are the four main properties of the arbelos. But Archimedes did not stop exploring this amazing figure; he found more fascinating things inside the arbelos.
 
Those are the four main properties of the arbelos. But Archimedes did not stop exploring this amazing figure; he found more fascinating things inside the arbelos.
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===Proving The Twin Circles Congruent===
 
===Proving The Twin Circles Congruent===
{{hide|1=
 
 
Why do the twin circles have the same diameter? Try it by yourself first.  
 
Why do the twin circles have the same diameter? Try it by yourself first.  
  
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Let’s start with the left twin circle and find its diameter in three steps.
 
Let’s start with the left twin circle and find its diameter in three steps.
  
First of all, assume the radius of the left circle is <math>R</math>. (Again, the name of the line segment indicates the length of the segment.)
+
First of all, assume the radius of the left circle is <math>R</math>. (''Again, the name of the line segment indicates the length of the segment.'')
  
 
1. Use the <balloon title="In brief, Pythagorean Theorem states that the square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of the other two sides.">Pythagorean Theorem</balloon> to express <math>EF</math> in terms of <math>EM, MF, EN, NF.</math>
 
1. Use the <balloon title="In brief, Pythagorean Theorem states that the square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of the other two sides.">Pythagorean Theorem</balloon> to express <math>EF</math> in terms of <math>EM, MF, EN, NF.</math>
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Therefore, the diameter of the left twin circle is <math>r(1- r)</math>.
 
Therefore, the diameter of the left twin circle is <math>r(1- r)</math>.
  
Same solution for the right twin circle. Do you get the same result?
+
Because of the symmetry of the arbelos, we can go through this same solution for the right circle. Do you get the same result?
  
 
:Do you want to know how Archimedes proved it 2000 years ago?
 
:Do you want to know how Archimedes proved it 2000 years ago?
:Learn more in [http://www.cut-the-knot.org/Curriculum/Geometry/BookOfLemmas/BOL5.shtml Proposition 5] of ‘‘ Book of Lemmas ’’. }}
+
:Learn more in [http://www.cut-the-knot.org/Curriculum/Geometry/BookOfLemmas/BOL5.shtml Proposition 5] of ‘‘ Book of Lemmas ’’.  
  
 
===Archimedes’ Circles and the Problem of Apollonius===
 
===Archimedes’ Circles and the Problem of Apollonius===
{{hide|1=
 
  
 
[[Image:Arbelos(17)-.jpg|250px]]
 
[[Image:Arbelos(17)-.jpg|250px]]
Constructing the twin circles based on three tangencies is a special case of the '''[[Problem of Apollonius]]''' (e.g. the white circle is tangent to three other circles: orange, pink, green. The construction of this white circle is the Problem of Apollonius.). }}
+
Constructing the twin circles based on three tangencies is a special case of the '''[[Problem of Apollonius]]''' (e.g. the white circle is tangent to three other circles: orange, pink, green. The construction of this white circle is the Problem of Apollonius.).  
  
 
==Bankoff Circle==
 
==Bankoff Circle==
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:{{EquationRef2|Eq. 1}} <math> Area = \frac{R}{2}</math>
 
:{{EquationRef2|Eq. 1}} <math> Area = \frac{R}{2}</math>
  
According to '''Heron's Formula''', it states that the area of a triangle with sides a, b, and c is <math>Area = \sqrt{s(s-a)(s-b)(s-c)} </math> where s is the semiperimeter of the triangle.  
+
According to '''Heron's Formula''', it states that the area of a triangle with sides a, b, and c is <math>Area = \sqrt{s(s-a)(s-b)(s-c)} </math> where s is half of the perimeter of the triangle.  
 
:Let  
 
:Let  
 
:<math>a = BC = \frac{1-r}{2} + R</math>
 
:<math>a = BC = \frac{1-r}{2} + R</math>
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:{{EquationRef2|Eq. 3}}<math> Area = \frac{x}{2} \left( 1 + 2R\right) </math>
 
:{{EquationRef2|Eq. 3}}<math> Area = \frac{x}{2} \left( 1 + 2R\right) </math>
  
We have three equations all representing the area of <math>\triangle ABC</math>and we know that <math> Area = \frac{Area^2}{Area} = \frac{(Eq.2)^2}{Eq.3} = Eq.1</math>.
+
We have three equations all representing the area of <math> \triangle ABC </math>, so we will combine them. Since <math> Area = \frac{Area^2}{Area} </math>, we can say that <math> Area= \frac{(Eq.2)^2}{Eq.3} = Eq.1. </math>
 
<template>AlignEquals
 
<template>AlignEquals
 
  |e1l=\frac{\frac{R}{8} \cdot r \cdot (1 - r) \cdot (1 + 2R)}{\frac{x}{2} \left( 1 + 2R\right)}
 
  |e1l=\frac{\frac{R}{8} \cdot r \cdot (1 - r) \cdot (1 + 2R)}{\frac{x}{2} \left( 1 + 2R\right)}
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}}
 
}}
  
==Pappus Chain==
 
[[Image:Arbelos(100)-.jpg|350px]]
 
 
===Definition===
 
A chain of inscribed circles is called a''' Pappus Chain''' when its first circle <math>C_1</math> is tangent to the three semicircles forming the arbelos, and all the subsequent circles <math>C_n</math> are tangent to one another and to the boundaries of the arbelos. The Pappus Chain is named after Pappus of Alexandria, a great Greek mathematician who studied and wrote about it in the 4th century A.D. The figure above shows all of the three variations of the chain: leftward, rightward, and downward. The default position is the one that extends to the right.
 
 
===Properties===
 
*Height
 
{{hide|1=
 
[[Image:Arbelos(13)-.jpg|Figure 5-Height|thumb|300px|right]]
 
In the Pappus Chain, the height of the center of the nth circle is n times the diameter of that circle.
 
 
{{SwitchPreview|ShowMessage=Click to see proof |HideMessage=Click to hide proof|PreviewText= The proof Pappus gave is long and complicated, containing Euclidean geometry, |FullText=
 
 
The proof Pappus gave is long and complicated, containing Euclidean geometry, similar triangles, and the Pythagorean Theorem. We won’t list his proof here. Instead, here’s a simpler, more modern proof using the concept of inversion:
 
 
It’s fine if you don't know much about “circle inversion.” Basically, "inversion is a type of transformation that moves points from the inside of a circle to the outside and from the outside of a circle to the inside using a specific rule" [[http://mathforum.org/mathimages/index.php/Inversion 1]]. The basic properties of inversion are as follows. Go to [[inversion]] to get a more complete understanding.
 
 
*The inverse of a line not passing through the center of the circle is a circle;
 
*The inverse of a circle not passing through the center of the circle is a circle;
 
*The inverse of a circle passing through the center of the circle is a line.
 
 
[[Image:Arbelos(11)-.jpg|300px]]
 
[[Image:Arbelos(11)--.jpg|300px]]
 
 
We want to prove that the height of the center of the nth circle above <math>AB</math> is n times the diameter of that circle, so we are inverting over the nth circle in the chain. To do this, first invert circle <math>AC</math> and circle <math>BC</math> with respect to a circle centered at <math>C</math>. Because both circle <math>AC</math> and circle <math>BC</math> pass through the center of the circle <math>C</math>, they become two vertical lines according to the third property listed above (see the figure on the left). Circle <math>BC</math> inverts to the left line because when <math>BC</math> has a smaller radius than <math>AC</math>, it inverts to a further line. To help you understand it better, treat the circle centered at <math>C</math> as a mirror. If the circle is closer to the center point <math>C</math>, it will be reflected further away.
 
 
Second, invert the nth circle in the Pappus Chain and the two circles that are tangent to it. The inverse of the nth circle is itself. The inverse of the subsequent circles inscribed in the chain are circles tangent to the two parallel, vertical lines below or above the nth circle. Because the subsequent circles are tangent to circle <math>AC</math> and <math>BC</math>, it makes sense that they are still tangent to the inverse of the two circles.
 
 
Now we are done with the inversions. Let's take a look at the right figure. Since those inversed circle are identical, it is not hard to conclude that the height of the center of the nth circle is n times the diameter of that circle. For example, <math>EF = 4r = 2d</math> for the second circle.
 
 
}}}}
 
 
*Ellipse
 
{{hide|1=
 
[[Image:Arbelos(14)-.jpg|Figure 6-Ellipse|thumb|350px|right]]
 
 
The centers of all the circles <math>C_n</math> in the Pappus Chain lie on an ellipse.
 
 
{{SwitchPreview|ShowMessage=Click to see proof |HideMessage=Click to hide proof|PreviewText=|FullText=
 
[[Image:Arbelos(20).jpg‎|350px]]
 
 
Let <math>M</math> be the center of circle <math>AC, N</math> be the center of the circle <math>BC, C_n</math> represent the centers of all of the circles in the Pappus Chain, and <math>r_n</math> represent the radii of those circles.
 
 
In order to prove this property, the first thing to do is to know the definition of an ellipse. In what conditions will a point lie on an ellipse in two-dimensions? First, the sum of the distances from a point on the ellipse to two fixed points is a constant. Second, the constant is greater than the distance between the two fixed points.
 
 
We need to show that all of the centers of the circles in the Pappus Chain satisfy the two conditions. The two fixed points are <math>M, N</math>.
 
 
Because
 
:<math>C_n M = MD - C_n D =\frac{1}{2} - r_n</math>
 
and
 
:<math> C_n N= C_n E + EN = r_n + \frac{1- r}{2}</math>,
 
we get
 
:<math>C_n M + C_n N = (\frac{1}{2} - r_n) + (r_n + \frac{1- r}{2})= 1 - \frac{r}{2}= constant.</math>
 
 
For the reason that <math>MN = MC - NC = \frac{1}{2} - \frac{1- r}{2} = \frac{r}{2} < 1 - \frac{r}{2}</math> (because <math>r < 1</math> )
 
 
Thus, the centers of all the circles <math>C_n</math> in the Pappus Chain lie on an ellipse.
 
 
}}}}
 
 
===Pappus Chain and Steiner Chain===
 
{{hide|1=
 
[[Image:Arbelos(15)-.jpg|350px]]
 
The Pappus Chain is also a '''Steiner Chain''' (see [[Steiner's Porism]]), a chain formed by circles that are tangent to two circles, one inscribed within the other.
 
}}
 
  
 
|other=geometry, a little bit of algebra
 
|other=geometry, a little bit of algebra
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[2] Nelsen, Roger B. ''Proof Without Words: The Area of an Arbelos.'' Retrieved from http://legacy.lclark.edu/~mathsci/arbelos.pdf.
 
[2] Nelsen, Roger B. ''Proof Without Words: The Area of an Arbelos.'' Retrieved from http://legacy.lclark.edu/~mathsci/arbelos.pdf.
  
[3] Wikipedia (Arbelos). .'' Arbelos.'' Retrieved from http://en.wikipedia.org/wiki/Arbelos.
+
[3] Wikipedia (Arbelos).'' Arbelos.'' Retrieved from http://en.wikipedia.org/wiki/Arbelos.
  
 
[4] ''Radius of the Twin Circles''. Retrieved from http://www.geogebra.org/en/examples/frisbee/worksheets/pappus_chain.html.
 
[4] ''Radius of the Twin Circles''. Retrieved from http://www.geogebra.org/en/examples/frisbee/worksheets/pappus_chain.html.
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[6] Bankoff, L. ''Are the Twin Circles of Archimedes Really Twins?'' Math. Mag. 47, 214-218, 1974.
 
[6] Bankoff, L. ''Are the Twin Circles of Archimedes Really Twins?'' Math. Mag. 47, 214-218, 1974.
 
[7] van Lamoen, Floor and Weisstein, Eric W. ''Pappus Chain''. From MathWorld--A Wolfram Web Resource. Retrieved from http://mathworld.wolfram.com/PappusChain.html.
 
  
 
|WhyInteresting=The arbelos has attracted lots of professional mathematicians as well as many interested amateurs, and it still inspires people in many ways. Today, you can see the arbelos in areas ranging from shoemaking to art design.
 
|WhyInteresting=The arbelos has attracted lots of professional mathematicians as well as many interested amateurs, and it still inspires people in many ways. Today, you can see the arbelos in areas ranging from shoemaking to art design.
  
 
==Applications==
 
==Applications==
====Leather cutting====
+
===Leather cutting===
  
 
A modern head knife, used for leather cutting and shoemaking, resembles the shape of an arbelos as the main image shows. This kind of knife has a curved blade, which mathematically is the circumference of the largest semicircle; leather-crafters use the blade to cut, skive, or trim leather. Its two sharp cusp points on each side are perfect for cutting right angles. Because of the shape of an arbelos, the head knife is an important, easy, and very useful tool for cutting leather and making shoes.  
 
A modern head knife, used for leather cutting and shoemaking, resembles the shape of an arbelos as the main image shows. This kind of knife has a curved blade, which mathematically is the circumference of the largest semicircle; leather-crafters use the blade to cut, skive, or trim leather. Its two sharp cusp points on each side are perfect for cutting right angles. Because of the shape of an arbelos, the head knife is an important, easy, and very useful tool for cutting leather and making shoes.  
Line 403: Line 343:
  
  
====Art & Design====
+
===Art & Design===
 
 
 
[[Image:Pattern.jpg|200px]]  
 
[[Image:Pattern.jpg|200px]]  
  
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You can also find an abstract public sculpture in the shape of arbelos located in the Netherlands.  
 
You can also find an abstract public sculpture in the shape of arbelos located in the Netherlands.  
  
====Mechanics====
+
 
 +
===Mechanics===
  
 
The arbelos in mathematics is similar to the 3-D Mohr’s circle in mechanics. '''Mohr’s circle''' is a geometric representation of the state of stress at a point: the normal force and shear force. It is very useful to perform quick and efficient estimations.  To understand how the diagram is drawn and how it is in the shape of the arbelos, we need to know the mathematical problems Mohr has studied. (This section is hard to understand; it requires knowledge about matrices, vectors, solid geometry, and algebraic calculation.)
 
The arbelos in mathematics is similar to the 3-D Mohr’s circle in mechanics. '''Mohr’s circle''' is a geometric representation of the state of stress at a point: the normal force and shear force. It is very useful to perform quick and efficient estimations.  To understand how the diagram is drawn and how it is in the shape of the arbelos, we need to know the mathematical problems Mohr has studied. (This section is hard to understand; it requires knowledge about matrices, vectors, solid geometry, and algebraic calculation.)
Line 446: Line 386:
 
[[Image:Mohr-2.jpg|A Triangular Plane|thumb|300px|right]]
 
[[Image:Mohr-2.jpg|A Triangular Plane|thumb|300px|right]]
  
Now let <math>v_j = u_j^2</math>.
+
Now let <math>v_j = u_j^2</math>. Then
Then <math> x = \lambda_1 (u_1)^2 + \lambda_2 (u_2) ^2 + \lambda_3 (u_3)^2 = \sum_{j=1}^3 \lambda_j v_j ,  
+
:<math> x = \lambda_1 (u_1)^2 + \lambda_2 (u_2) ^2 + \lambda_3 (u_3)^2 = \sum_{j=1}^3 \lambda_j v_j , </math>
: \quad y^2 = (\lVert \mathbf{MU}\rVert - \lVert x\mathbf{U}\rVert )^2 = \sum_{j=1}^3 \lambda_j^2 v_j - x^2 </math>
+
:<math>\quad y^2 = (\lVert \mathbf{MU}\rVert - \lVert x\mathbf{U}\rVert )^2 = \sum_{j=1}^3 \lambda_j^2 v_j - x^2 </math>
  
 
Because <math>\mathbf{U} </math> is a unit vector and <math>\mathbf{U} = <u_1, u_2, u_3></math>
 
Because <math>\mathbf{U} </math> is a unit vector and <math>\mathbf{U} = <u_1, u_2, u_3></math>
  
Therefore <math>u_1^2 + u_2 ^2 + u_3 ^2 = 1 </math>
+
Therefore <math>u_1^2 + u_2 ^2 + u_3 ^2 = 1 </math>;
 +
 
 
<math>v_1 + v_2 + v_3 = 1  </math>. This equations represents a triangular plane in <math>\mathbf{R^3}</math>. See the right figure.  
 
<math>v_1 + v_2 + v_3 = 1  </math>. This equations represents a triangular plane in <math>\mathbf{R^3}</math>. See the right figure.  
  
Line 486: Line 427:
  
 
Therefore, we get  
 
Therefore, we get  
 
[[Image:Mohr-4.jpg|Figure 8-Mohr's Circle|thumb|300px|right]]
 
  
 
  <template>AlignEquals
 
  <template>AlignEquals
Line 499: Line 438:
 
  |e4r=  (\frac{\lambda_1 - \lambda_2}{2})^2
 
  |e4r=  (\frac{\lambda_1 - \lambda_2}{2})^2
 
</template>  
 
</template>  
 +
 +
[[Image:Mohr-4.jpg|Figure 8-Mohr's Circle|thumb|300px|right]]
  
 
It is a circle centered at <math>(\frac{\lambda_1 +\lambda_2}{2}, 0)</math> with radius <math>\frac{\lambda_1 - \lambda_2}{2}</math>. See the blue circle in Figure 8.
 
It is a circle centered at <math>(\frac{\lambda_1 +\lambda_2}{2}, 0)</math> with radius <math>\frac{\lambda_1 - \lambda_2}{2}</math>. See the blue circle in Figure 8.
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|ToDo=
 
|ToDo=
 +
#A proof of this statement: "According to a property found by Pappus, the altitude to the base <math>AB</math> of <math>\triangle{ABC}</math> is two times the radius of circle <math>C</math>. Therefore <math>height = 2R </math>." under the section Bankoff Circle.
 
#An applet or applets to demonstrate different kinds of arbelos and Archemedes' circles with a moveable point B.
 
#An applet or applets to demonstrate different kinds of arbelos and Archemedes' circles with a moveable point B.
 
#More interesting and important applications of arbelos.
 
#More interesting and important applications of arbelos.
 
#Or anything you think would be necessary for this page. You are welcome to edit it!
 
#Or anything you think would be necessary for this page. You are welcome to edit it!
  
|InProgress=Yes
+
|InProgress=No
 
}}
 
}}

Latest revision as of 13:26, 16 August 2012


Arbelos
Shoemakers Knife.jpg
Fields: Geometry and Algebra
Image Created By: csosborne
Website: csosborne

Arbelos

This modern knife in the shape of an arbelos is used to make shoes.


Basic Description

The word "arbelos" means shoemaker’s knife in Greek. The first mathematician to study mathematical properties of this plane region is Archimedes, who wrote all his thoughts about the arbelos in his Book of Lemmas.

Arbelos(1).jpg

In geometry, an arbelos is a figure bounded by three semicircles, tangent in pairs, and with diameters lying on the same line. Graphically, an arbelos is the green region in the picture above. The position of central point B is arbitrary and can be anywhere along the diameter. Here we assume that the three diameters are r, (1- r), and 1, as the image shows.

From now on, let's think in a semicircle way!

Try it yourself!

A More Mathematical Explanation

Note: understanding of this explanation requires: *geometry, a little bit of algebra

Properties

Arbelos has lots of unexpected but interesting properties.

[[Image:Arbelos(2).jpg|F [...]

Properties

Arbelos has lots of unexpected but interesting properties.

Figure 1-Properties of Arbelos

Arc Length

The arc length along the bottom of the arbelos is the same as the arc length along the enclosing semicircle.

Proof:

The circumference of a circle is proportional to its diameter. See Figure 1 and we can find that

C_{AB} = \frac{1}{2}\pi r
C_{BC} = \frac{1}{2}\pi (1 - r)

Add them together and we get

C_{AB} + C_{BC} = \frac{1}{2} \pi r + \frac{1}{2} \pi (1-r) = \frac{1}{2} \pi (r + 1 -r ) = \frac{1}{2} \pi

whereas

C_{AC} = \frac{1}{2} \pi .

Therefore, they have the same length: C_{AB} + C_{BC} = C_{AC} .

Area

Starting from point B, draw a line that is perpendicular to AC. It meets the enclosing semicircle at point D. Then the area of the arbelos equals the area of a circle with diameter BD.

If you don't know how to prove it, start with what you already know.

First, try to use line segments to represent those two areas. The area of a circle is \pi \frac{d^2}{4} and therefore the area of a semicircle is \frac{1}{2}\pi \frac{d^2}{4}.

Eq. 1         A_{circle} = \pi\left( \frac{(BD)^2}{4}\right) (The name of the line segment in parentheses indicates the length of the segment.)

We can tell from the image that  A_{arbelos} = A_{semicircleAC} - A_{semicircleAB} - A_{semicircleBC} ,

 A_{semicircleAC} = \frac{1}{2}\pi\frac{(AC)^2}{4};
 A_{semicircleAB} = \frac{1}{2}\pi\frac{(AB)^2}{4};
 A_{semicircleBC} = \frac{1}{2}\pi\frac{(BC)^2}{4};
Arbelos(2).jpg

Then  A_{arbelos} = \Big( \frac{1}{2}\pi\frac{(AC)^2}{4}\Big) - \Big( \frac{1}{2}\pi\frac{(AB)^2}{4}\Big) -\Big(\frac{1}{2}\pi\frac{(BC)^2}{4}\Big)

 A_{arbelos} = \pi \cdot \frac{1}{2} \cdot \frac{\Big((AC)^2 - (AB)^2- (BC)^2\Big)}{4}

Because  (AC) = (AB) + (BC) ,

A_{arbelos} = \pi\frac{\frac{1}{2}\bigg( \Big((AB) + (BC)\Big)^2 - (AB)^2 - (BC)^2\bigg)}{4}
 A_{arbelos} = \pi \frac{\frac{1}{2}\Big( (AB)^2 + 2 (AB) \cdot (BC) + (BC)^2 - (AB)^2 - (BC)^2\Big)}{4}
 A_{arbelos} = \pi \frac{\frac{1}{2} \cdot 2(AB) \cdot (BC)}{4}
Eq. 2          A_{arbelos}= \pi \frac{ (AB) \cdot (BC)}{4}

According to Eq. 1 and Eq. 2, we translate our goal, A_{cirlce}= A_{arbelos}, into \pi\left(\frac{(BD)^2}{4}\right)= \pi \frac{(AB) \cdot (BC)}{4}. Simplify this "big" equation, and you will find that you only need to prove that (BD)^2 = (AB)(BC).


\blacktrianglerightWe have now translated the seemingly sophisticated problem of proving that the area of the circle was the same as the area of the arbelos into something nice :::and simple. Reduction to absurdity is a very important method when dealing with mathematical problems where you don't have a clue.


Now let's prove that (BD)^2 = (AB)(BC):

According to Thales' Theorem,\triangle ADC is a right triangle, thus \angle{ADC} = 90^ \circ. Since we drew a line from B that is perpendicular to AC as stated in the problem, it is given that DB \perp AC. Therefore, \triangle ABD is a right triangle with \angle{ABD} = 90^\circ. Thus, \angle{ADC} = \angle{ABD} .

Eq. 3        \angle{DAB} = \angle{DAC} because they are the same angle.

In a triangle, the sum of the three angles is 180^\circ.

Eq. 4        \angle{ADB} = 180^\circ -\angle{ABD} - \angle{DAB} = 90^\circ - \angle{DAB}
Eq. 5         \angle{ACD} = 180^\circ - \angle{ADC} - \angle{DAC} = 90^\circ - \angle{DAC}

Because of Eq. 3, therefore Eq. 4 = Eq. 5, \angle{ADB} = \angle{ACD}

Therefore, \triangle ABD and \triangle BCD are similar triangles since both of their corresponding angles are equal in measure.

The corresponding sides of similar triangles are proportional, so \frac{(AB)}{(BD)} = \frac{(BD)}{(BC)} .

Cross multiplication will give you (BD)^2 = (AB)(BC), which is what we want.

Therefore,

\pi\left(\frac{(BD)^2}{4}\right)= \pi \frac{(AB) \cdot (BC)}{4}
A_{cirlce}= A_{arbelos}

The area of the arbelos is the area of the circle O with diameter BD.


According to Pythagorean Theorem,

According to the Pythagorean Theorem,

In   \triangle ABD :
Eq. 1         (AB)^2 + (BD)^2 = (AD)^2
In   \triangle BCD :
Eq. 2         (BC)^2 + (BD)^2 = (CD)^2
In   \triangle ACD :
Eq. 3         (AD)^2 + (CD)^2 = (AC)^2

(Here \triangle ACD is a right triangle according to Thales' Theorem.)

Arbelos(2).jpg

By substituting Eq. 1 and Eq. 2 into Eq. 3, we get:

 (AB) ^2 + (BD) ^2 + (BC)^2 + (BD)^2 = (AC)^2

Since AB = r and AC =1, solve for BD:

r^2 + (BD)^2 + (r - 1)^2 + (BD)^2 = 1^2
 r^2 + (r -1)^2 +2(BD)^2 = 1^2

After simplification, we get

 BD = \sqrt[2]{r-r^2}

So the area of the circle is:

 A_{circle} = \pi (\frac{1}{2} (BD)^2) = \pi \left(\frac{1}{2}\sqrt{r-r^2}\right)^2= \frac{\pi}{4} \left(r-r^2\right)

The area of the arbelos is the area of the big semicircle minus areas of the two small semicirlces. Therefore,

A_{arbelos} = \frac{1}{2}\pi (\frac{1}{2})^2 - \frac{1}{2}\pi(\frac{r}{2})^2 - \frac{1}{2}\pi(\frac{1-r}{2})^2
A_{arbelos} = \frac{1}{8}\pi\cdot 1 - \frac{1}{8}\pi\cdot r^2 - \frac{1}{8}\pi \cdot (1- r)^2
A_{arbelos} = \frac{\pi}{8}\left(1-r^2-(1-r)^2\right)
A_{arbelos} = \frac{\pi}{8}\left(1-r^2-1+2r-r^2\right)
A_{arbelos} = \frac{\pi}{4}\left(r-r^2\right)

So,

A_{arbelos} = A_{circle}


Prove the problem in an easier way.

There is another way to prove that the two areas are the same.

The equation to the area of a semicircle is A_{semicircle} = \frac{\pi d^2}{8}, where d is its diameter. We know from this equation that the area of a semicircle is proportional to the square of its diameter. If we draw three semicircles with diameters being the three sides of a right triangle, we will know that the area of the semicircle with diameter being the hypothenuse of the right triangle equals to the area of the two semicircles with diameters being the other two sides, according to the Pythagorean Theorem.

Now take a look at the three figures below.

image_1image_2image_3

(Note that A_1, A_2, A_3, B_1, B_2, B_3, C_1, C_2 in the figure represent the areas of the semicircle they are in; A represents the area of the arbelos.)

It it not hard to conclude that

\quad B_1 = A_1 + C_1 \quad\quad\quad\quad\quad  B_2= A_2 + C_2 \quad\quad\quad\quad  A + A_1 + A_2 = B_1 + B_2

Therefore,

 A + A_1 + A_2 = A_1 + C_1 + A_2 + C_2
 A = C_1 + C_2

Note that C_1 and C_2 make a whole circle, which is the area of the circle with diameter BD, whereas A represents the area of the arbelos. Thus, this last equation proves that the area of the arbelos equals the area of the circle with diameter BD.


Rectangles

Arbelos(2).jpg

Assume line segment AD intersects semicircle AB at point E. Line segment CD intersects semicircle BC at point F. Then the quadrilateral, BEDF, is a rectangle.

According to Thales' Theorem,

According to Thales' Theorem, \angle{ADC}, \angle{AEB}, and \angle{BFC} are all right angles. So \angle {ADC} = \angle{AEB} = \angle{BFC} = 90^\circ.

Thus BEDF has three right angles now: \angle{BED}, \angle{BFD} and \angle{EDF}.

Its last angle \angle{EBF} is also a right angle because the sum of the four angles of a quadrilateral is 360^\circ and

\angle{EBF} = 360^\circ - 3 \cdot 90^\circ = 90^\circ.

Therefore, BEDF is a rectangle. Note that since BEDF is a rectangle, EF and BD are equal and bisect each other. O is the midpoint.


Tangents

A line that connects the two points E, F is tangent to the two smaller semicircles at E and F.

Figure 2-Tangent


We can prove that line EF is tangent to two circles by showing that it is perpendicular to the radii of the circles.

Recall: it is given that  DB \perp AB. So \angle{ABD}=90^\circ

Eq. 1         \angle{ABE}+\angle{EBD}=90^\circ

In circle O, OE = OB. Thus,

Eq. 2         In \triangle OBE, \angle{OEB} = \angle{OBE} = \angle{EBD}

In circle M, EM = BM. Thus,

Eq. 3         In \triangle BEM, \angle{BEM} = \angle{EBM} = \angle{ABE}

According to Eq. 1, Eq. 2, and Eq. 3,

\angle{BEM} + \angle{BEO}=90^\circ

So EM \perp EF

Now try it yourself! Prove  FN \perp EF in the same way.
Therefore, EF is tangent to the two smaller semicircles at E and F.


One more thing: it is not a coincidence that B, E, D, F are on a common circle. In a triangle, three perpendicular bisectors meet in a point. In a right triangle, that point is the midpoint of the hypotenuse. The distance from this point to each vertex of the triangle is the same. So a circle centered at the midpoint of the hypotenuse passes through all three vertices of a right triangle. If we treat the rectangle BEDF as two right triangles, \triangle DEF and \triangle BEF, a circle centered at O passes through all the vertices of these two triangles. Therefore the four vertices B, E, D and F always lie along a common circle.

Those are the four main properties of the arbelos. But Archimedes did not stop exploring this amazing figure; he found more fascinating things inside the arbelos.


Archimedes' Twin Circles

If two circles are inscribed in an arbelos tangent to the line segment BD, one on each side, then the two circles are congruent and have the same diameter r(1- r). Because the two circles were first found by Archimedes, they are called Archimedes' Twin Circles. (Figure 3)

Proving The Twin Circles Congruent

Why do the twin circles have the same diameter? Try it by yourself first.

Figure 4-Left Circle

Let’s start with the left twin circle and find its diameter in three steps.

First of all, assume the radius of the left circle is R. (Again, the name of the line segment indicates the length of the segment.)

1. Use the Pythagorean Theorem to express EF in terms of EM, MF, EN, NF.

In \triangle MEF, ~ (EF)^2 = (EM)^2 - (MF)^2
In \triangle NEF, ~ (EF)^2 = (EN)^2 - (NF)^2
So
Eq. 1        (EM)^2 - (MF)^2 = (EN)^2- (NF)^2

2. Write all four segments in Eq. 1 in terms of r, R.

 EM = \frac{r}{2} + R ( EM is the sum of the radius of circle E and the radius of circle M)
 MF = BM - BF = \frac{r}{2} - R (BF is the radius of circle E )
 EN = NG - EG = \frac{1}{2} - R (EN is the difference of the radius of the biggest circle N, and the radius of circle E)
 NF = NA - MA - MF = \frac{1}{2} - \frac{r}{2} - \left(\frac{r}{2} - R\right) = \frac{1}{2} + R - r

3. Plug the results from step 2 into Eq. 1 and solve for R. <template>AlignEquals


Why It's Interesting

The arbelos has attracted lots of professional mathematicians as well as many interested amateurs, and it still inspires people in many ways. Today, you can see the arbelos in areas ranging from shoemaking to art design.

Applications

Leather cutting

A modern head knife, used for leather cutting and shoemaking, resembles the shape of an arbelos as the main image shows. This kind of knife has a curved blade, which mathematically is the circumference of the largest semicircle; leather-crafters use the blade to cut, skive, or trim leather. Its two sharp cusp points on each side are perfect for cutting right angles. Because of the shape of an arbelos, the head knife is an important, easy, and very useful tool for cutting leather and making shoes.


Art & Design

Pattern.jpg

Put the arbelos in a fractal pattern and you will get figures like the one above. A fractal pattern is a fragmented geometric shape, which consists of lots of reduced-size copies of the whole graph. The main shape of this fractal is a little bit different from the classic arbelos because the two smaller semicircles become two whole circles, making the entire figure more heart-shaped. Yet, it is a pretty design and it comes from a mathematics website that is actually called “Arbelos.”

-arbelos- sculpture.jpg

You can also find an abstract public sculpture in the shape of arbelos located in the Netherlands.


Mechanics

The arbelos in mathematics is similar to the 3-D Mohr’s circle in mechanics. Mohr’s circle is a geometric representation of the state of stress at a point: the normal force and shear force. It is very useful to perform quick and efficient estimations. To understand how the diagram is drawn and how it is in the shape of the arbelos, we need to know the mathematical problems Mohr has studied. (This section is hard to understand; it requires knowledge about matrices, vectors, solid geometry, and algebraic calculation.)


First, let \mathbf{U} be a unit vector in \mathbf{R^3}: \mathbf{U} = <u_1, u_2, u_3>

Let \mathbf{M} be a symmetric 3 by 3 matrix. For convenience, let the eigenvalues of the matrix M be \lambda_1, \lambda_2, \lambda_3 and other numbers 0. (To produce the arbelos, assume that \lambda_1, \lambda_2, \lambda_3 are different numbers and \lambda_1 < \lambda_2 <\lambda_3.)

\begin{bmatrix}
\lambda_1 & 0 & 0 \\
0 & \lambda_2 & 0 \\
0 & 0 & \lambda_3 \\
\end{bmatrix}

Vector U

Next, use a pair of real numbers (x, y) to denote the normal and tangential components of vector \mathbf{U}.

  • x = \mathbf{U} \cdot \mathbf{MU} = \lVert \mathbf{U}\rVert \cdot \lVert \mathbf{MU}\rVert \cdot \cos \theta = \lambda_1 (u_1)^2 + \lambda_2 (u_2) ^2 + \lambda_3 (u_3)^2

Because\mathbf{U} is a unit vector, so

\lVert \mathbf{U}\rVert = 1

Therefore,  x = \lVert \mathbf{MU}\rVert \cdot \cos \theta is the length of the normal component of \mathbf{MU}. Thus, x\mathbf{U} is the normal vector of \mathbf{U} . See the left image.

  • According to the left image, the tangential vector of \mathbf{U} is  y\mathbf{U} = \mathbf{MU} - x\mathbf{U} .

Remember that  y is the length of the tangential component of \mathbf{MU},  y = \lVert \mathbf{MU} - x\mathbf{U} \rVert .

Because  \mathbf{U} represents any unit vector over a unit sphere in \mathbf{R^3} , there is a range for (x, y) in \mathbf{R^2} as

\mathbf{U} varies in \mathbf{R^3}.

A Triangular Plane

Now let v_j = u_j^2. Then

 x = \lambda_1 (u_1)^2 + \lambda_2 (u_2) ^2 + \lambda_3 (u_3)^2 = \sum_{j=1}^3 \lambda_j v_j ,
\quad y^2 = (\lVert \mathbf{MU}\rVert - \lVert x\mathbf{U}\rVert )^2 = \sum_{j=1}^3 \lambda_j^2 v_j - x^2

Because \mathbf{U} is a unit vector and \mathbf{U} = <u_1, u_2, u_3>

Therefore u_1^2 + u_2 ^2 + u_3 ^2 = 1 ;

v_1 + v_2 + v_3 = 1  . This equations represents a triangular plane in \mathbf{R^3}. See the right figure.

Then draw a line with three points A, B, C as shown in Figure 7. Let the abscissae of A, B, C be \lambda_1, \lambda_2, \lambda_3 in an increasing order.

In order to prove that the range is an arbelos, we need to prove that the three vertices of this triangle in the right figure, (1, 0, 0), (0, 1, 0), (0, 0, 1), map to the three points of the arbelos A, B, C as Figure 7 shows, and the three sides of this triangle are arc AB, arc AC, and arc BC in the arbelos. We take one side of the triangle as an example and it can be proved that the other two sides are mapped to the other arcs in the same way. Assume we want to prove the arc AB. Let  v_3 = 0, then v_1 + v_2 = 1 - v_3 = 1.


Substitute x and y with v_j and u_j in the expression y^2 + x^2 - (\lambda_1 + \lambda_2) x, using  x = \sum_{j=1}^3 \lambda_j v_j, y^2 =\sum_{j=1}^3 \lambda_j^2 v_j - x^2 .

So  y^2 + x^2 - (\lambda_1 + \lambda_2)~ x =~ (\lambda_1 ^ 2~ v_1 + \lambda_2^2~ v_2 + \lambda_3^2~ v_3 - x^2) ~+ ~x^2~ -~ (\lambda_1 + \lambda_2)~ (\lambda_1~ v_1 + \lambda_2~ v_2 + \lambda_3~ v_3)

Because v_3 = 0,

 y^2 + x^2 - (\lambda_1 + \lambda_2)~ x =~ (\lambda_1 ^ 2~ v_1 + \lambda_2^2~ v_2) ~-~ (\lambda_1 + \lambda_2)~ (\lambda_1~ v_1 + \lambda_2~ v_2 )

 y^2 + x^2 - (\lambda_1 + \lambda_2)~ x = ~\lambda_1 ^2 ~v_1 + \lambda_2^2 ~v_2 -\lambda_1 ^2~ v_1 - \lambda_1 \lambda_2~ v_1 - \lambda_2 ^2 ~v_2 - \lambda_1 \lambda_2 ~v_2 = -\lambda_1 \lambda_2 (v_1 + v_2)

Recall that  v_1 + v_2 = 1

Eq. 1         y^2 + x^2 - (\lambda_1 + \lambda_2) x= -\lambda_1 \lambda_2

Because  (\lambda_1 + \lambda_2)^2 = \lambda_1 ^2 + 2\lambda_1 \lambda_2 + \lambda_2 ^2 \quad\quad\quad (\lambda_1 - \lambda_2)^2 = \lambda_1 ^2 - 2\lambda_1 \lambda_2 + \lambda_2 ^2

\therefore (\lambda_1 - \lambda_2)^2 - (\lambda_1 + \lambda_2)^2= -4\lambda_1 \lambda_2
Eq. 2        \therefore -\lambda_1 \lambda_2 = \frac{(\lambda_1 - \lambda_2)^2 - (\lambda_1 + \lambda_2)^2}{4}

For the reason that  Eq.1 = Eq.2

Therefore, we get

<template>AlignEquals



Archimedean Circles

Any circle defined in the arbelos that has radius \frac{r(1 - r)}{2}, the same as the radii of Archimedes’ Twin Circles, is called an Archimedean Circle.

There are three famous Archimedean Circles: the Bankoff Circle, the Schoch Circle, and the Woo Circle. We discuss the Bankoff Circle above. To learn more about the other two circles, see the Woo Circles and The Archimedean Circles of Schoch and Woo .


Teaching Materials

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References

[1] Boas, Harold P. Reflections on the Arbelos. Retrieved from http://www.math.tamu.edu/~harold.boas/preprints/arbelos.pdf.

[2] Nelsen, Roger B. Proof Without Words: The Area of an Arbelos. Retrieved from http://legacy.lclark.edu/~mathsci/arbelos.pdf.

[3] Wikipedia (Arbelos). Arbelos. Retrieved from http://en.wikipedia.org/wiki/Arbelos.

[4] Radius of the Twin Circles. Retrieved from http://www.geogebra.org/en/examples/frisbee/worksheets/pappus_chain.html.

[5] Weisstein, Eric W. Arbelos. From MathWorld--A Wolfram Web Resource. Retrieved from http://mathworld.wolfram.com/Arbelos.html.

[6] Bankoff, L. Are the Twin Circles of Archimedes Really Twins? Math. Mag. 47, 214-218, 1974.

Future Directions for this Page

  1. A proof of this statement: "According to a property found by Pappus, the altitude to the base AB of \triangle{ABC} is two times the radius of circle C. Therefore height = 2R ." under the section Bankoff Circle.
  2. An applet or applets to demonstrate different kinds of arbelos and Archemedes' circles with a moveable point B.
  3. More interesting and important applications of arbelos.
  4. Or anything you think would be necessary for this page. You are welcome to edit it!




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