# Difference between revisions of "Circular Inversion"

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The definition of inversion tells us that <math>\overline{OP} \cdot \overline{OP'} = \overline{OQ} \cdot \overline{OQ'} = \overline{OR} \cdot\overline{OR'} = k^2</math>. From here, we obtain that <math>\frac{\overline{OP}}{\overline{OR}} = \frac{\overline{OR'}}{\overline{OP'}}</math> and <math>\frac{\overline{OQ}}{\overline{OR}} = \frac{\overline{OR'}}{\overline{OQ'}}</math> By SAS symmetry (exploiting <math>\angle ROQ</math>), the ratios tell us that: <cmath>\triangle ORP \sim \triangle OP'R', \triangle ORQ \sim \triangle OQ'R'.</cmath> | The definition of inversion tells us that <math>\overline{OP} \cdot \overline{OP'} = \overline{OQ} \cdot \overline{OQ'} = \overline{OR} \cdot\overline{OR'} = k^2</math>. From here, we obtain that <math>\frac{\overline{OP}}{\overline{OR}} = \frac{\overline{OR'}}{\overline{OP'}}</math> and <math>\frac{\overline{OQ}}{\overline{OR}} = \frac{\overline{OR'}}{\overline{OQ'}}</math> By SAS symmetry (exploiting <math>\angle ROQ</math>), the ratios tell us that: <cmath>\triangle ORP \sim \triangle OP'R', \triangle ORQ \sim \triangle OQ'R'.</cmath> | ||

− | Therefore, we have <math>\angle OPR = \angle OR'P'</math> and <math>\angle OQR = \angle OR'Q'</math>. Note that <math>\angle OPR = \angle PRQ + \angle PQR = 90^\circ + \angle PQR</math>, which must equal <math>\angle OR'P'</math>. Therefore, <math>\angle Q'R'P' = \angle OR'P' - \angle OR'Q' = 90^\circ + \angle PQR - \angle OQR</math>. But <math>\angle OQR = \angle PQR</math>. Therefore, <math>\angle Q'R'P' = 90^\circ</math>. As this holds for any <math>R</math>, all points on circle <math>C</math> will invert to a point on a circle with diameter <math> | + | Therefore, we have <math>\angle OPR = \angle OR'P'</math> and <math>\angle OQR = \angle OR'Q'</math>. Note that <math>\angle OPR = \angle PRQ + \angle PQR = 90^\circ + \angle PQR</math>, which must equal <math>\angle OR'P'</math>. Therefore, <math>\angle Q'R'P' = \angle OR'P' - \angle OR'Q' = 90^\circ + \angle PQR - \angle OQR</math>. But <math>\angle OQR = \angle PQR</math>. Therefore, <math>\angle Q'R'P' = 90^\circ</math>. As this holds for any <math>R</math>, all points on circle <math>C</math> will invert to a point on a circle with diameter <math>\overline{Q'P'}</math>. |

== 3. General Formula for the Radius of a Circle in Terms of the Radius of its Inverse Circle == | == 3. General Formula for the Radius of a Circle in Terms of the Radius of its Inverse Circle == |

## Latest revision as of 14:39, 11 May 2021

**Circular Inversion**, sometimes called **Geometric Inversion** or simply **Inversion**, is a transformation where point in the Cartesian plane is transformed based on a circle with radius and center such that , where is the transformed point on the ray extending from through .

Note that , when inverted, transforms back to . All points outside of are transformed inside , and vice versa. Points on transform to themselves, meaning . Finally, the transformation of is debated on its existence. Some call the transformation the *ideal point*, which is infinitely far away and in every direction. Others claim that this point does not have an inverse.

Geometric Inversion technically refers to many different types of inversions, however, if Geometric Inversion is used without clarification, Circular Inversion is usually assumed.

Circular Inversion can be a very useful tool in solving problems involving many tangent circles and/or lines.

# Basics of Circular Inversion

## 1. Inversion of a Circle intersecting O

The first thing that we must learn about inversion is what happens when a circle which intersects the center of the inversion, , is inverted. Let us have circle , with diameter . is chosen arbitrarily on circle . Points and represent the inversions of and , respectively. is the radius of . We seek to show that circle inverts to a line perpendicular to through .

By the definition of inversion, we have and .

We can combine the two equations to get . Rewriting this gives:

Also, since is a diameter of circle , must be right.

Now, we consider and . They share an angle - , and we know that Therefore, we have SAS similarity. Therefore, must be right. From there, it follows that all points on circle will be inverted onto the line perpendicular to at .

**Therefore, the inversion of circle becomes a line.**

Note that, if circle extends beyond , the argument still holds. All one needs to do is shuffle things around.

## 2. Inversion of a Circle not intersecting O

Now, we study the inversion of a circle not intersecting the center of inversion. Let us have circle not intersecting , the center of , the circle which we invert around. The points where intersect circle are points and , respectively. Point is arbitrary and on circle . We invert points , , and , producing , , and , respectively. We draw and . Because is a diameter, must be right. We wish to show that circle inverts to another circle.

The definition of inversion tells us that . From here, we obtain that and By SAS symmetry (exploiting ), the ratios tell us that:

Therefore, we have and . Note that , which must equal . Therefore, . But . Therefore, . As this holds for any , all points on circle will invert to a point on a circle with diameter .

## 3. General Formula for the Radius of a Circle in Terms of the Radius of its Inverse Circle

This is how circular inversion is useful in the first place - we find the radius of an inverted circle to find the radius of the original circle. Let the original circle be and the inverted circle be , with radii of and , respectively. The radius of the circle of inversion is . We draw the tangent line of circle intersecting O. We know that this is also a tangent line to circle from the result from part 2 - the tangent line, by definition, intersects circle at exactly one point, and for every intersection point, part 2 says that there will be another intersection point. Therefore, the tangent line to circle intersects circle at exactly one point, necessitating this line to be a tangent line. Call the intersections and , respectively. We have .

We have and = . We can write an equation for by dividing:

From the definition of inversion, we have . Subsituting yields:

From Power of a Point, we know that , which equals . Subsistuting gives , and solving for gives:

# Problems that use Circular Inversion

## Problem 1 (AMC12)

Source: 2017 AMC12A Problem 16

In the figure below, semicircles with centers at and and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter . The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at ?

### Solution using Circular Inversion

Let be a circle with radius of and centered at the left corner of the semi-circle (O) with radius . Extend the three semicircles to full circles. Label the resulting four circles as shown in the diagram:

has radius , has radius , and has radius . We want to find the radius of .

We now invert the four circles. inverts to a line. Given that one point is on , and all points on invert to themselves, we know that the resulting line must intersect that intersection point. also inverts to a line. has radius , and since has radius of , the resulting line must be units away from . inverts to a circle. By observing the diagram, we note that 's center must be on and be between the two inverted lines, because is tangent to and (Remeber that tangency still holds in inverted diagrams). Therefore, we must have a circle with radius that is units from .

Now, we invert . Note that is tangent to the three other original circles. So, in the inversion, must be tangent to the two lines *and* . It is then quickly seen that and have the same radius: .

Now, we can determine the radius of using the formula . , and . is just the distance from the center of the inverted circle to the center of inversion. The center of is units above the center of . Since , we use Pythagoras to learn that . We do not take the square root because our relationship formula takes .

Therefore, we have:

Here is the diagram with .