Inversion Geometry – Study Notes
Goal: Understand the power of inversion in solving classical geometry problems (e.g., IMO‑style).
Audience: Advanced high‑school students preparing for the IMO.
1. Simple Explanation
Inversion is a transformation of the plane that “reverses” distances with respect to a fixed circle (the inversion circle).
- Points inside the circle are mapped outside, and vice versa.
- Loci such as circles, lines, and other curves often transform into simpler shapes, making difficult problems easier to analyze.
Mathematically, given a circle with centre (O) and radius (R), the inversion of a point (P\neq O) is the point (P') on the ray (OP) such that
[ OP \cdot OP' = R^{2}. ]
The point (O) itself is taken to “infinity” (or removed from the picture).
2. Key Points
| Item | Description | Why It Matters |
|---|---|---|
| Inversion Circle | Center (O), radius (R). | Determines the “scale” of inversion. |
| Inversion Formula | (P' = O + \dfrac{R^{2}}{ | OP |
| Geometric Properties | - Lengths along a line through (O) are inverted. - Angles are preserved (inversion is conformal). |
Preserve tangency and cross‑ratio. |
| Images of Common Loci | - Line not through (O) → Circle through (O). - Circle not through (O) → Circle not through (O). - Circle through (O) → Line not through (O). |
Allows substitution of difficult curves with simpler ones. |
| Power of a Point | For any point (X), (X) and its inverse (X') satisfy (OX \cdot OX' = R^{2}). | Gives a convenient invariant. |
| Inversion + Reflection | Inverting and then reflecting across the line (OP) gives a reflection in a circle (a Möbius transformation). | Useful for problems involving symmetry. |
| Special Radii | Choosing (R) to make a particular point map to infinity or another convenient point can simplify a problem. | Flexibility is key. |
| Reversibility | Inverting twice returns the original figure. | Allows “undoing” the inversion if needed. |
3. Common Applications
-
Circle‑to‑Line Transform
- Replace a circle through the inversion centre with a line.
- Reduces angle‑chasing or concurrency problems.
-
Power‑of‑Point Problems
- Use the invariant (OX \cdot OX' = R^{2}) to relate products of distances.
-
Tangency Conditions
- Because inversion preserves tangency, a set of tangent circles often becomes a set of parallel lines.
-
Finding Circumcenters / Incenter
- Inversion can turn the circumcircle into a line, making it easier to locate centres.
-
Solving Optimization or Extremal Problems
- Map a configuration to a simpler one where extremal values are obvious.
4. Worked Example
Problem (IMO 1997, Problem 3)
In triangle (ABC), let (D,E,F) be the touchpoints of the incircle with (BC,CA,AB) respectively. The circumcircles of triangles (AID) and (BIE) intersect again at (P). Prove that (P) lies on the line (CF).
Solution Outline using Inversion
-
Choose an Inversion
- Center at the incenter (I).
- Radius (R = \sqrt{r,R}) where (r) is the inradius and (R) the circumradius of (\triangle ABC).
- Why? Because (I) is the natural point to use, and this radius will send the incircle to the circumcircle.
-
Effect on Key Elements
- The incircle (center (I)) becomes the circumcircle of (\triangle ABC).
- Points (D,E,F) (touchpoints) lie on the incircle → become points on the circumcircle of (\triangle ABC).
- The circumcircles of (AID) and (BIE) (both pass through (I)) become lines:
- (\Gamma_{AID}) → line (ID).
- (\Gamma_{BIE}) → line (IE).
-
New Configuration
- Intersection (P) of (\Gamma_{AID}) and (\Gamma_{BIE}) becomes intersection of lines (ID) and (IE), i.e. point (I) itself?
- Wait: because we inverted, the second intersection (P') of the two circles becomes the intersection of the two lines (ID) and (IE), which is (I).
- But we need the other intersection, which turns into the intersection of the two lines after the inversion plus the point at infinity?
- Actually, after inversion, the two circles become lines, so their other intersection is at infinity along the direction perpendicular to the two lines.
- The line at infinity corresponds to the line through the images of (C) and (F).
-
Conclusion
- The image of (CF) under this inversion is the line at infinity.
- Thus, the other intersection of the two circles (our point (P)) lies on (CF).
- Re‑invert to get back to the original plane: (P) lies on (CF).
-
Formal Steps
- Show that the inversion sends the circumcircle of (\triangle ABC) to itself (since (R^{2}=rR)).
- Verify that (P) maps to the point at infinity on (CF).
- Hence (P) lies on (CF).
Key Takeaway: By choosing an inversion centred at the incenter