# Arbelos

Arbelos
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.

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 $A_{arbelos} = A_{semicircleAC} - A_{semicircleAB} - A_{semicircleBC}$,

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

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 $(BD)^2 = (AB)(BC)$.

$\blacktriangleright$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 $(BD)^2 = (AB)(BC)$:

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.)

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.

(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

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 $R$.

 $(EM)^2 - (MF)^2$ $=$ $(EN)^2- (NF)^2$ $\left(\frac{r}{2}+R\right)^2 - \left(\frac{r}{2}-R\right)^2$ $=$ $\left(\frac{1}{2}-R\right)^2 - \left(\frac{1}{2} + R - r\right)^2$ $\left(\frac{r^2}{4} + rR + R^2\right) - \left(\frac{r^2}{4}-rR+R^2\right)$ $=$ $\left(\frac{1}{4}-R +R^2\right) - \left[\left(\frac{1}{2}+R\right)^2 - 2r\left(\frac{1}{2}+R\right) + r^2\right]$ $\frac{r^2}{4} + rR + R^2 - \frac{r^2}{4} + rR - R^2$ $=$ $\left(\frac{1}{4}-R +R^2\right) - \left(\frac{1}{4} + R + R^2 - r - 2rR + r^2\right)$ $2rR$ $=$ $\frac{1}{4} - R + R^2 - \frac{1}{4} - R - R^2 + r + 2rR - r^2$ $2rR$ $=$ $-2R + r + 2rR - r^2$ $2R$ $=$ $r - r^2 = r(1- r)$

Therefore, the diameter of the left twin circle is $r(1- r)$.

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?
Learn more in Proposition 5 of ‘‘ Book of Lemmas ’’.

### Archimedes’ Circles and 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

In addition to the Archimedes' twin circles, there is another circle inside the arbelos that is congruent to the twin circles. It is called the Bankoff Circle, the black circle in the figure above. Leon Bankoff, a dentist in Beverly Hills and also a mathematician, found the third circle and wrote it down in his article named Are The Twin Circles of Archimedes Really Twins?

The Bankoff circle and the twin circles are all identical. The diameter of the Bankoff Circle is also $r(1- r)$.

Let x be the radius of the Bankoff Circle and let A be the area of $\triangle{ABC}$.

Let $x$ be the radius of the Bankoff Circle and let $Area$ be the area of $\triangle{ABC}$.

Figure 4 - Proof of Bankoff Circle
$AB = \frac{1}{2}$ (half of the diameter of the enclosing circle)
$AC = \frac{r}{2} + R$ (the sum of the radius of circle $A$ and the radius of circle $C$)
$BC = \frac{1- r}{2} + R$ (the sum of the radius of circle $B$ and the radius of circle $C$)

According to a property found by Pappus, the altitude to the base $AB$ of $\triangle{ABC}$ is two times the radius of circle $C$.

$height = 2R$

Therefore,

The area of $\triangle{ABC}$: $Area = \frac{1}{2} \cdot base \cdot height = \frac{1}{2} \cdot AB \cdot 2R = \frac{1}{2} \cdot \frac{1}{2} \cdot 2R$
Eq. 1         $Area = \frac{R}{2}$

According to Heron's Formula, it states that the area of a triangle with sides a, b, and c is $Area = \sqrt{s(s-a)(s-b)(s-c)}$ where s is half of the perimeter of the triangle.

Let
$a = BC = \frac{1-r}{2} + R$
$b = AC = \frac{r}{2} + R$
$c = AB = \frac{1}{2}$
Then
$s = \frac{a + b + c}{2} =\frac{\frac{1- r}{2} + R + \frac{r}{2} + R+ \frac{1}{2}}{2} = \frac{1}{2} + R$
$s - a = \frac{r}{2}$
$s - b = \frac{1 - r}{2}$
$s - c = R$
$Area = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt {\left(\frac{1}{2} + R \right) \cdot (\frac{r}{2}) \cdot (\frac{1 - r}{2}) \cdot (R)}$
Eq. 2        $Area = \sqrt{\frac{R}{8} \cdot r \cdot (1 - r) \cdot (1 + 2R)}$

Because the area of $\triangle{ABC}$ is the sum of the areas of $\triangle{AOB}$, $\triangle{AOC}$, and $\triangle{BOC}$, therefore Recall that $x$ is the radius of the Bankoff Circle.

$Area = S \triangle{AOB} + S \triangle{AOC} + S \triangle{BOC}$
$Area = \frac{1}{2} \cdot x \cdot AB + \frac{1}{2} \cdot x \cdot AC + \frac{1}{2} \cdot x \cdot BC$
$Area = \frac{1}{2} \cdot x \cdot \left( AB + AC + BC \right)$
$Area = \frac{x}{2} \cdot \left( \frac{1}{2} +\frac{r}{2} + R + \frac{1- r}{2} + R\right)$
Eq. 3        $Area = \frac{x}{2} \left( 1 + 2R\right)$

We have three equations all representing the area of $\triangle ABC$, so we will combine them. Since $Area = \frac{Area^2}{Area}$, we can say that $Area= \frac{(Eq.2)^2}{Eq.3} = Eq.1.$

 $\frac{\frac{R}{8} \cdot r \cdot (1 - r) \cdot (1 + 2R)}{\frac{x}{2} \left( 1 + 2R\right)}$ $=$ $\frac{R}{2}$ $\frac {\frac{R}{4} \cdot r \cdot (1 - r) }{x}$ $=$ $\frac{R}{2}$

Solve for $x$, and then we get the answer $x = \frac{r(1- r)}{2}$.

# 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

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.”

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} = $

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} = $

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

 $y^2 + x^2 - (\lambda_1 + \lambda_2) x$ $=$ $\frac{(\lambda_1 - \lambda_2)^2 - (\lambda_1 + \lambda_2)^2}{4}$ $y^2 + \Big( x^2 - (\lambda_1 + \lambda_2) x \Big)$ $=$ $(\frac{\lambda_1 - \lambda_2}{2})^2 - (\frac{\lambda_1 + \lambda_2}{2})^2$ $y^2 +\bigg( x^2 - (\lambda_1 + \lambda_2) x +\Big ( \frac{\lambda_1 + \lambda_2}{2} \Big) ^2 \bigg)$ $=$ $(\frac{\lambda_1 - \lambda_2}{2})^2 - (\frac{\lambda_1 + \lambda_2}{2})^2 +(\frac{\lambda_1 + \lambda_2}{2})^2$ $y^2 + (x- \frac{\lambda_1 + \lambda_2}{2}) ^2$ $=$ $(\frac{\lambda_1 - \lambda_2}{2})^2$
Figure 8-Mohr's Circle

It is a circle centered at $(\frac{\lambda_1 +\lambda_2}{2}, 0)$ with radius $\frac{\lambda_1 - \lambda_2}{2}$. See the blue circle in Figure 8.

Use the same method for the other two sides and we'll get the three circles as Figure 8 shows and the shaded region is the range we want to get.

Based on the work we've done, Mohr's Circle analyzed in 3-D is indeed in the shape of an arbelos.

An interesting thing to note is that you can view the arbelos as a stretched triangle.

## 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 .

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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.

[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.

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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