GeometryDifficulty 70-10021 tagged problems

Inversion

Inversion is the habit of transforming circles and lines through reciprocal distances to simplify difficult configurations. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$\text{Inversion of radius } r \text{ about } O: \quad P \mapsto P' \text{ with } OP \cdot OP' = r^2 \text{ on ray } OP$$

Inversion sends each point to a new point on the same ray whose distance is the reciprocal (scaled by $r^2$), swapping lines through the center with circles and simplifying tangency problems.

Why It Works

1
Fix a center $O$ and radius $r$. Each point $P \neq O$ maps to $P'$ on ray $OP$ with $OP\cdot OP' = r^2$, so points near $O$ go far out and vice versa.
2
A line NOT through $O$ inverts to a circle through $O$, and a circle through $O$ inverts to a line not through $O$; a line through $O$ maps to itself.
3
A circle not through $O$ inverts to another circle not through $O$.
4
Tangency is preserved, and the key length identity: if $A,B$ map to $A',B'$, then $A'B' = \dfrac{r^2 \cdot AB}{OA\cdot OB}$ — used to translate distances after inverting.

Worked Examples

Example 1

Under inversion centered at $O$ with radius $r = 6$, a point $A$ lies at distance $OA = 4$ from $O$. Find the distance $OA'$ of its image.

1
Inversion preserves the ray and satisfies $OA \cdot OA' = r^2$.
2
Substitute: $4 \cdot OA' = 6^2 = 36$.
3
Solve: $OA' = \dfrac{36}{4} = 9$.

Answer:

$OA' = 9$

Warm-up

Under inversion centered at $O$ with radius $r = 10$, two points $A$ and $B$ satisfy $OA = 5$, $OB = 4$, and $AB = 3$. Find the distance $A'B'$ between their images.

1
Inversion distorts distances by the rule $A'B' = \dfrac{r^2\cdot AB}{OA\cdot OB}$.
2
Substitute $r = 10$, $AB = 3$, $OA = 5$, $OB = 4$: $A'B' = \dfrac{100\cdot 3}{5\cdot 4} = \dfrac{300}{20}$.
3
Compute: $A'B' = 15$.

Answer:

$A'B' = 15$

Contest level

A circle $\omega$ does NOT pass through the inversion center $O$. Its image under inversion of radius $r$ is another circle. If $O$ is at distance $d$ from the center of $\omega$ and $\omega$ has radius $\rho$ (with $d > \rho$, so $O$ is outside $\omega$), express the ratio of the image circle's radius to $\rho$ in terms of the power $d^2 - \rho^2$.

1
A circle not through $O$ inverts to a circle. To get its radius, invert the two points where the line through $O$ and the center of $\omega$ meets $\omega$: these lie at distances $d - \rho$ and $d + \rho$ from $O$, and they are diametrically opposite on $\omega$, so their images are diametrically opposite on $\omega'$. The image diameter is therefore $r^2\bigl|\tfrac{1}{d-\rho} - \tfrac{1}{d+\rho}\bigr| = r^2\cdot\dfrac{2\rho}{|d^2-\rho^2|}$, giving $\rho' = \dfrac{r^2\,\rho}{|d^2 - \rho^2|}$, where $d^2 - \rho^2 = \text{pow}_O(\omega)$ is the power of $O$ with respect to $\omega$.
2
Therefore $\dfrac{\rho'}{\rho} = \dfrac{r^2}{|d^2 - \rho^2|}$.
3
The image radius scales by $r^2$ divided by the power of the center — large power (far-away circle) gives a small image, matching the 'points near $O$ go far, far points come close' intuition.

Answer:

$\dfrac{\rho'}{\rho} = \dfrac{r^2}{|d^2 - \rho^2|}$ (the power of $O$ controls the scale).

Olympiad / Challenge

Ptolemy's INEQUALITY: for any four points $A, B, C, D$ in the plane, $AC\cdot BD \le AB\cdot CD + AD\cdot BC$, with equality iff $ABCD$ is a cyclic quadrilateral (in order). Prove it by inverting at $A$.

1
Invert with center $A$ and any radius $r$; let $B', C', D'$ be the images of $B, C, D$. The distance distortion gives $B'C' = \dfrac{r^2\,BC}{AB\cdot AC}$, $C'D' = \dfrac{r^2\,CD}{AC\cdot AD}$, and $B'D' = \dfrac{r^2\,BD}{AB\cdot AD}$.
2
The ordinary triangle inequality on $B', C', D'$ reads $B'D' \le B'C' + C'D'$ (with equality iff $C'$ lies on segment $B'D'$, i.e. $B', C', D'$ are collinear in that order).
3
Substitute the three distorted distances and multiply through by $\dfrac{AB\cdot AC\cdot AD}{r^2}$ to clear denominators: $BD\cdot AC \le BC\cdot AD + CD\cdot AB$.
4
Equality holds exactly when $B', C', D'$ are collinear; since a line not through $A$ is the inverse image of a circle through $A$, this is precisely the condition that $A, B, C, D$ are concyclic — recovering Ptolemy's equality as the equality case.

Answer:

$AC\cdot BD \le AB\cdot CD + AD\cdot BC$, equality iff $ABCD$ is cyclic (inversion turns it into the triangle inequality).

Going Deeper

Generalization: inversion is the key non-affine transformation in the plane — it is conformal (preserves angles) and maps the family of 'lines and circles' to itself (a 'generalized circle'). Treating lines as circles of infinite radius unifies the four cases (line/circle through or not through $O$) into one statement: generalized circles map to generalized circles.
Where it appears: inversion linearizes tangency. Configurations with many mutually tangent circles (Steiner chains, the Arbelos, Apollonius' problem, Descartes circle theorem) collapse to parallel lines or simple circles after inverting at a well-chosen tangency point. It also gives slick proofs of Ptolemy and the existence of the nine-point circle.
Pitfall: the center $O$ has NO image (it goes to infinity) — never invert at a point that is one of the points you care about unless you intend it to vanish. Also, the center of $\omega$ does NOT map to the center of the image circle $\omega'$; computing $\omega'$ means inverting the points of $\omega$, not its center. Assuming 'center maps to center' is the single most common inversion blunder.

Spot the Signal

Look for problems where the key step is transforming circles and lines through reciprocal distances to simplify difficult configurations.
You can describe the hard part as transforming circles and lines through reciprocal distances to simplify difficult configurations, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the diagram relation that calls for inversion, then rewrite the givens around it.
Name the relation that makes Inversion legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Inversion just because the surface looks familiar; verify the required condition first.
Applying Inversion because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let inversion reveal the geometric configuration; then compute only what remains.

Try it on:

Practice a contest problem where the key step is transforming circles and lines through reciprocal distances to simplify difficult configurations.

Back to techniques

GeometryDifficulty 70-10021 tagged problems

Inversion

Practice this technique Match my level

The Key Result

$$\text{Inversion of radius } r \text{ about } O: \quad P \mapsto P' \text{ with } OP \cdot OP' = r^2 \text{ on ray } OP$$

Inversion sends each point to a new point on the same ray whose distance is the reciprocal (scaled by $r^2$), swapping lines through the center with circles and simplifying tangency problems.

Why It Works

1
Fix a center $O$ and radius $r$. Each point $P \neq O$ maps to $P'$ on ray $OP$ with $OP\cdot OP' = r^2$, so points near $O$ go far out and vice versa.
2
A line NOT through $O$ inverts to a circle through $O$, and a circle through $O$ inverts to a line not through $O$; a line through $O$ maps to itself.
3
A circle not through $O$ inverts to another circle not through $O$.
4
Tangency is preserved, and the key length identity: if $A,B$ map to $A',B'$, then $A'B' = \dfrac{r^2 \cdot AB}{OA\cdot OB}$ — used to translate distances after inverting.

Worked Examples

Example 1

Under inversion centered at $O$ with radius $r = 6$, a point $A$ lies at distance $OA = 4$ from $O$. Find the distance $OA'$ of its image.

1
Inversion preserves the ray and satisfies $OA \cdot OA' = r^2$.
2
Substitute: $4 \cdot OA' = 6^2 = 36$.
3
Solve: $OA' = \dfrac{36}{4} = 9$.

Answer:

$OA' = 9$

Warm-up

Under inversion centered at $O$ with radius $r = 10$, two points $A$ and $B$ satisfy $OA = 5$, $OB = 4$, and $AB = 3$. Find the distance $A'B'$ between their images.

1
Inversion distorts distances by the rule $A'B' = \dfrac{r^2\cdot AB}{OA\cdot OB}$.
2
Substitute $r = 10$, $AB = 3$, $OA = 5$, $OB = 4$: $A'B' = \dfrac{100\cdot 3}{5\cdot 4} = \dfrac{300}{20}$.
3
Compute: $A'B' = 15$.

Answer:

$A'B' = 15$

Contest level

1
A circle not through $O$ inverts to a circle. To get its radius, invert the two points where the line through $O$ and the center of $\omega$ meets $\omega$: these lie at distances $d - \rho$ and $d + \rho$ from $O$, and they are diametrically opposite on $\omega$, so their images are diametrically opposite on $\omega'$. The image diameter is therefore $r^2\bigl|\tfrac{1}{d-\rho} - \tfrac{1}{d+\rho}\bigr| = r^2\cdot\dfrac{2\rho}{|d^2-\rho^2|}$, giving $\rho' = \dfrac{r^2\,\rho}{|d^2 - \rho^2|}$, where $d^2 - \rho^2 = \text{pow}_O(\omega)$ is the power of $O$ with respect to $\omega$.
2
Therefore $\dfrac{\rho'}{\rho} = \dfrac{r^2}{|d^2 - \rho^2|}$.
3
The image radius scales by $r^2$ divided by the power of the center — large power (far-away circle) gives a small image, matching the 'points near $O$ go far, far points come close' intuition.

Answer:

$\dfrac{\rho'}{\rho} = \dfrac{r^2}{|d^2 - \rho^2|}$ (the power of $O$ controls the scale).

Olympiad / Challenge

1
Invert with center $A$ and any radius $r$; let $B', C', D'$ be the images of $B, C, D$. The distance distortion gives $B'C' = \dfrac{r^2\,BC}{AB\cdot AC}$, $C'D' = \dfrac{r^2\,CD}{AC\cdot AD}$, and $B'D' = \dfrac{r^2\,BD}{AB\cdot AD}$.
2
The ordinary triangle inequality on $B', C', D'$ reads $B'D' \le B'C' + C'D'$ (with equality iff $C'$ lies on segment $B'D'$, i.e. $B', C', D'$ are collinear in that order).
3
Substitute the three distorted distances and multiply through by $\dfrac{AB\cdot AC\cdot AD}{r^2}$ to clear denominators: $BD\cdot AC \le BC\cdot AD + CD\cdot AB$.
4
Equality holds exactly when $B', C', D'$ are collinear; since a line not through $A$ is the inverse image of a circle through $A$, this is precisely the condition that $A, B, C, D$ are concyclic — recovering Ptolemy's equality as the equality case.

Answer:

$AC\cdot BD \le AB\cdot CD + AD\cdot BC$, equality iff $ABCD$ is cyclic (inversion turns it into the triangle inequality).

Going Deeper

Generalization: inversion is the key non-affine transformation in the plane — it is conformal (preserves angles) and maps the family of 'lines and circles' to itself (a 'generalized circle'). Treating lines as circles of infinite radius unifies the four cases (line/circle through or not through $O$) into one statement: generalized circles map to generalized circles.
Where it appears: inversion linearizes tangency. Configurations with many mutually tangent circles (Steiner chains, the Arbelos, Apollonius' problem, Descartes circle theorem) collapse to parallel lines or simple circles after inverting at a well-chosen tangency point. It also gives slick proofs of Ptolemy and the existence of the nine-point circle.
Pitfall: the center $O$ has NO image (it goes to infinity) — never invert at a point that is one of the points you care about unless you intend it to vanish. Also, the center of $\omega$ does NOT map to the center of the image circle $\omega'$; computing $\omega'$ means inverting the points of $\omega$, not its center. Assuming 'center maps to center' is the single most common inversion blunder.

Spot the Signal

Look for problems where the key step is transforming circles and lines through reciprocal distances to simplify difficult configurations.
You can describe the hard part as transforming circles and lines through reciprocal distances to simplify difficult configurations, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the diagram relation that calls for inversion, then rewrite the givens around it.
Name the relation that makes Inversion legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Inversion just because the surface looks familiar; verify the required condition first.
Applying Inversion because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let inversion reveal the geometric configuration; then compute only what remains.

Try it on:

Practice a contest problem where the key step is transforming circles and lines through reciprocal distances to simplify difficult configurations.