GeometryDifficulty 65-10037 tagged problems

Spiral Similarity

Spiral Similarity is the habit of spotting the rotation-and-scaling that maps one segment onto another. 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

$$z \mapsto O + k\,(z - O), \quad k = \rho\,(\cos\theta + i\sin\theta): \ \text{rotation by } \theta \text{ and scaling by } \rho \text{ about center } O$$

A spiral similarity is a single rotation-plus-scaling about a fixed center that carries one segment onto another, tying together similar triangles hidden in a configuration.

Why It Works

1
Identify the plane with the complex numbers. Multiplying by $k = \rho(\cos\theta + i\sin\theta)$ rotates a vector by $\theta$ and scales it by $\rho = |k|$; adding a fixed center $O$ makes it a spiral similarity about $O$.
2
Given two segments $AB$ and $CD$ that are not parallel and not equal in length, there is a UNIQUE spiral similarity sending $A\mapsto C$ and $B\mapsto D$; its ratio is $\dfrac{CD}{AB}$ and its angle is the angle between the segments.
3
Key fact: the spiral similarity taking $A\mapsto C$, $B\mapsto D$ shares its center with the one taking $A\mapsto B$, $C\mapsto D$ — so one center governs both pairings, which is why spiral similarities link two similar triangles in a figure.
4
The center is the second intersection of circles $(A, C, X)$ and $(B, D, X)$ where the segment lines meet at $X$, and triangles $\triangle OAB \sim \triangle OCD$ for the center $O$.

Worked Examples

Example 1

Consider the spiral similarity centered at the origin $O = (0,0)$ that rotates by $90^\circ$ and scales by a factor of $2$. Where does it send $A = (1,0)$ and $B = (2,1)$, and what is the length $A'B'$ of the image segment?

1
In complex form the map is $z \mapsto k z$ with $k = 2(\cos 90^\circ + i\sin 90^\circ) = 2i$.
2
Image of $A = 1$: $A' = 2i\cdot 1 = 2i = (0, 2)$. Image of $B = 2 + i$: $B' = 2i(2 + i) = 4i + 2i^2 = -2 + 4i = (-2, 4)$.
3
Original length $AB = |B - A| = |(2+i) - 1| = |1 + i| = \sqrt{2}$.
4
Image length $A'B' = |B' - A'| = |(-2 + 4i) - 2i| = |-2 + 2i| = 2\sqrt{2}$, which is exactly $2\cdot AB$ — confirming the scale factor $2$ (and the segment has turned $90^\circ$).

Answer:

$A' = (0,2)$, $B' = (-2,4)$, and $A'B' = 2\sqrt{2} = 2\cdot AB$.

Warm-up

A spiral similarity centered at the origin sends the point $(3,0)$ to $(0,6)$. Find its rotation angle and scale factor, and the image of $(1,1)$.

1
In complex form the map is $z \mapsto kz$. From $3 \mapsto 6i$ we get $k = \dfrac{6i}{3} = 2i$.
2
So $k = 2i = 2(\cos 90^\circ + i\sin 90^\circ)$: the rotation angle is $90^\circ$ and the scale factor is $|k| = 2$.
3
Image of $(1,1) = 1 + i$: $k(1 + i) = 2i(1 + i) = 2i + 2i^2 = -2 + 2i = (-2, 2)$.

Answer:

Angle $90^\circ$, scale factor $2$; $(1,1) \mapsto (-2, 2)$.

Contest level

A spiral similarity sends $A = (0,0)$ to $B = (-1,-1)$ and $C = (1,0)$ to $D = (0,0)$. Find its center, rotation angle, and scale factor.

1
Write the map as $f(z) = kz + t$ (a spiral similarity is an affine map with complex multiplier $k$). From $f(A) = B$ and $f(C) = D$: $k = \dfrac{B - D}{A - C} = \dfrac{(-1 - i) - 0}{0 - 1} = 1 + i$.
2
So $|k| = \sqrt2$ (scale factor) and $\arg k = 45^\circ$ (rotation angle), since $1 + i = \sqrt2(\cos 45^\circ + i\sin 45^\circ)$.
3
The center is the fixed point $O = \dfrac{t}{1 - k}$ where $t = B - kA = -(1 + i)$. Thus $O = \dfrac{-(1 + i)}{1 - (1 + i)} = \dfrac{-(1 + i)}{-i} = \dfrac{1 + i}{i} = (1 + i)(-i) = 1 - i = (1, -1)$.
4
Check: $f(1 - i) = (1 + i)(1 - i) - (1 + i) = (1 - i^2) - (1 + i) = 2 - 1 - i = 1 - i$ — the center is indeed fixed.

Answer:

Center $(1, -1)$, rotation $45^\circ$, scale factor $\sqrt2$.

Olympiad / Challenge

Spiral-similarity lemma: lines $AB$ and $CD$ meet at $P$, and the circles $(PAC)$ and $(PBD)$ meet again at $Q$. Prove $Q$ is the center of the spiral similarity sending $A \mapsto B$ and $C \mapsto D$ (equivalently $A \mapsto C$, $B \mapsto D$).

1
Let $Q$ be the second intersection of circles $(PAC)$ and $(PBD)$. We show $\triangle QAC \sim \triangle QBD$ (correspondence $A \leftrightarrow B$, $C \leftrightarrow D$); the direct similarity fixing $Q$ and carrying $\triangle QAC$ to $\triangle QBD$ is then exactly the spiral similarity at $Q$ sending $A \mapsto B$, $C \mapsto D$. We use directed angles $\angle(\ell, m)$ between lines, taken mod $180^\circ$. Note $A, B, P$ lie on line $AB$ and $C, D, P$ lie on line $CD$, so $\angle(PA, PC) = \angle(PB, PD) = \angle(AB, CD)$.
2
Angle equality 1 (the apex angles): since $P, A, Q, C$ are concyclic on $(PAC)$, the chord $AC$ gives $\angle(QA, QC) = \angle(PA, PC)$; since $P, B, Q, D$ are concyclic on $(PBD)$, the chord $BD$ gives $\angle(QB, QD) = \angle(PB, PD)$. As $\angle(PA, PC) = \angle(PB, PD)$ (same pair of lines), $\angle(QA, QC) = \angle(QB, QD)$, i.e. $\angle AQC = \angle BQD$.
3
Angle equality 2 (the base angles at $A$ and $B$): on $(PAC)$ the chord $QC$ is seen from $A$ and $P$, so $\angle(AQ, AC) = \angle(PQ, PC)$; on $(PBD)$ the chord $QD$ is seen from $B$ and $P$, so $\angle(BQ, BD) = \angle(PQ, PD)$. Because $C, D, P$ are collinear, $\angle(PQ, PC) = \angle(PQ, PD)$, hence $\angle(AQ, AC) = \angle(BQ, BD)$, i.e. $\angle QAC = \angle QBD$.
4
Two equal angle pairs ($\angle AQC = \angle BQD$ and $\angle QAC = \angle QBD$) give $\triangle QAC \sim \triangle QBD$ by AA. So the spiral similarity centered at $Q$ with ratio $\frac{QB}{QA} = \frac{QD}{QC}$ and rotation angle $\angle(QA, QB)$ sends $A \mapsto B$ and $C \mapsto D$. By the pairing-swap fact (the center taking $A \mapsto B$, $C \mapsto D$ also takes $A \mapsto C$, $B \mapsto D$), the same point $Q$ centers the spiral similarity $A \mapsto C$, $B \mapsto D$.

Answer:

$Q$ (the second intersection of $(PAC)$ and $(PBD)$) is the spiral-similarity center; it is the same center for both pairings.

Going Deeper

Generalization: spiral similarity is the most general direct (orientation-preserving) similarity of the plane with a fixed point — it is multiplication by a complex number $k$ about a center, so a homothety ($k$ real) and a pure rotation ($|k| = 1$) are its two degenerate ends. The center taking $A \mapsto B$, $C \mapsto D$ is identical to the one taking $A \mapsto C$, $B \mapsto D$, which is what makes it pair up two similar triangles in a single figure.
Where it appears: the spiral-similarity lemma is a staple of hard olympiad geometry — it explains the Miquel point of a complete quadrilateral, links the two circles in a 'two circles meeting at $P$' configuration, and turns 'these two segments are related by a rotation' into concyclicity. The complex-number computation (find $k$ and the fixed point) is the fast route on coordinates.
Pitfall: a UNIQUE spiral similarity sending $A \mapsto C$, $B \mapsto D$ exists only when $AB$ and $CD$ are NOT parallel and NOT equal-and-parallel (otherwise the construction degenerates to a translation/homothety with no finite center). Also the center maps $A \to C$ and $B \to D$ — pairing it as $A \to D$, $B \to C$ instead gives a DIFFERENT spiral similarity; keep the correspondence straight.

Spot the Signal

Use it when two similar triangles in a figure share a center and you need the linking map.
You can describe the hard part as spotting the rotation-and-scaling that maps one segment onto another, 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 fixed center and the rotation angle and ratio that pair the segments.
Name the relation that makes Spiral Similarity 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

Assuming a plain rotation or homothety when both a rotation and a scaling are needed.
Applying Spiral Similarity because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

When two similar triangles share a vertex, look for the spiral that links them.

Try it on:

Use a spiral similarity to relate lengths or angles in an olympiad configuration.

Back to techniques

GeometryDifficulty 65-10037 tagged problems

Spiral Similarity

Practice this technique Match my level

The Key Result

$$z \mapsto O + k\,(z - O), \quad k = \rho\,(\cos\theta + i\sin\theta): \ \text{rotation by } \theta \text{ and scaling by } \rho \text{ about center } O$$

A spiral similarity is a single rotation-plus-scaling about a fixed center that carries one segment onto another, tying together similar triangles hidden in a configuration.

Why It Works

1
Identify the plane with the complex numbers. Multiplying by $k = \rho(\cos\theta + i\sin\theta)$ rotates a vector by $\theta$ and scales it by $\rho = |k|$; adding a fixed center $O$ makes it a spiral similarity about $O$.
2
Given two segments $AB$ and $CD$ that are not parallel and not equal in length, there is a UNIQUE spiral similarity sending $A\mapsto C$ and $B\mapsto D$; its ratio is $\dfrac{CD}{AB}$ and its angle is the angle between the segments.
3
Key fact: the spiral similarity taking $A\mapsto C$, $B\mapsto D$ shares its center with the one taking $A\mapsto B$, $C\mapsto D$ — so one center governs both pairings, which is why spiral similarities link two similar triangles in a figure.
4
The center is the second intersection of circles $(A, C, X)$ and $(B, D, X)$ where the segment lines meet at $X$, and triangles $\triangle OAB \sim \triangle OCD$ for the center $O$.

Worked Examples

Example 1

1
In complex form the map is $z \mapsto k z$ with $k = 2(\cos 90^\circ + i\sin 90^\circ) = 2i$.
2
Image of $A = 1$: $A' = 2i\cdot 1 = 2i = (0, 2)$. Image of $B = 2 + i$: $B' = 2i(2 + i) = 4i + 2i^2 = -2 + 4i = (-2, 4)$.
3
Original length $AB = |B - A| = |(2+i) - 1| = |1 + i| = \sqrt{2}$.
4
Image length $A'B' = |B' - A'| = |(-2 + 4i) - 2i| = |-2 + 2i| = 2\sqrt{2}$, which is exactly $2\cdot AB$ — confirming the scale factor $2$ (and the segment has turned $90^\circ$).

Answer:

$A' = (0,2)$, $B' = (-2,4)$, and $A'B' = 2\sqrt{2} = 2\cdot AB$.

Warm-up

A spiral similarity centered at the origin sends the point $(3,0)$ to $(0,6)$. Find its rotation angle and scale factor, and the image of $(1,1)$.

1
In complex form the map is $z \mapsto kz$. From $3 \mapsto 6i$ we get $k = \dfrac{6i}{3} = 2i$.
2
So $k = 2i = 2(\cos 90^\circ + i\sin 90^\circ)$: the rotation angle is $90^\circ$ and the scale factor is $|k| = 2$.
3
Image of $(1,1) = 1 + i$: $k(1 + i) = 2i(1 + i) = 2i + 2i^2 = -2 + 2i = (-2, 2)$.

Answer:

Angle $90^\circ$, scale factor $2$; $(1,1) \mapsto (-2, 2)$.

Contest level

A spiral similarity sends $A = (0,0)$ to $B = (-1,-1)$ and $C = (1,0)$ to $D = (0,0)$. Find its center, rotation angle, and scale factor.

1
Write the map as $f(z) = kz + t$ (a spiral similarity is an affine map with complex multiplier $k$). From $f(A) = B$ and $f(C) = D$: $k = \dfrac{B - D}{A - C} = \dfrac{(-1 - i) - 0}{0 - 1} = 1 + i$.
2
So $|k| = \sqrt2$ (scale factor) and $\arg k = 45^\circ$ (rotation angle), since $1 + i = \sqrt2(\cos 45^\circ + i\sin 45^\circ)$.
3
The center is the fixed point $O = \dfrac{t}{1 - k}$ where $t = B - kA = -(1 + i)$. Thus $O = \dfrac{-(1 + i)}{1 - (1 + i)} = \dfrac{-(1 + i)}{-i} = \dfrac{1 + i}{i} = (1 + i)(-i) = 1 - i = (1, -1)$.
4
Check: $f(1 - i) = (1 + i)(1 - i) - (1 + i) = (1 - i^2) - (1 + i) = 2 - 1 - i = 1 - i$ — the center is indeed fixed.

Answer:

Center $(1, -1)$, rotation $45^\circ$, scale factor $\sqrt2$.

Olympiad / Challenge

1
Let $Q$ be the second intersection of circles $(PAC)$ and $(PBD)$. We show $\triangle QAC \sim \triangle QBD$ (correspondence $A \leftrightarrow B$, $C \leftrightarrow D$); the direct similarity fixing $Q$ and carrying $\triangle QAC$ to $\triangle QBD$ is then exactly the spiral similarity at $Q$ sending $A \mapsto B$, $C \mapsto D$. We use directed angles $\angle(\ell, m)$ between lines, taken mod $180^\circ$. Note $A, B, P$ lie on line $AB$ and $C, D, P$ lie on line $CD$, so $\angle(PA, PC) = \angle(PB, PD) = \angle(AB, CD)$.
2
Angle equality 1 (the apex angles): since $P, A, Q, C$ are concyclic on $(PAC)$, the chord $AC$ gives $\angle(QA, QC) = \angle(PA, PC)$; since $P, B, Q, D$ are concyclic on $(PBD)$, the chord $BD$ gives $\angle(QB, QD) = \angle(PB, PD)$. As $\angle(PA, PC) = \angle(PB, PD)$ (same pair of lines), $\angle(QA, QC) = \angle(QB, QD)$, i.e. $\angle AQC = \angle BQD$.
3
Angle equality 2 (the base angles at $A$ and $B$): on $(PAC)$ the chord $QC$ is seen from $A$ and $P$, so $\angle(AQ, AC) = \angle(PQ, PC)$; on $(PBD)$ the chord $QD$ is seen from $B$ and $P$, so $\angle(BQ, BD) = \angle(PQ, PD)$. Because $C, D, P$ are collinear, $\angle(PQ, PC) = \angle(PQ, PD)$, hence $\angle(AQ, AC) = \angle(BQ, BD)$, i.e. $\angle QAC = \angle QBD$.
4
Two equal angle pairs ($\angle AQC = \angle BQD$ and $\angle QAC = \angle QBD$) give $\triangle QAC \sim \triangle QBD$ by AA. So the spiral similarity centered at $Q$ with ratio $\frac{QB}{QA} = \frac{QD}{QC}$ and rotation angle $\angle(QA, QB)$ sends $A \mapsto B$ and $C \mapsto D$. By the pairing-swap fact (the center taking $A \mapsto B$, $C \mapsto D$ also takes $A \mapsto C$, $B \mapsto D$), the same point $Q$ centers the spiral similarity $A \mapsto C$, $B \mapsto D$.

Answer:

$Q$ (the second intersection of $(PAC)$ and $(PBD)$) is the spiral-similarity center; it is the same center for both pairings.

Going Deeper

Generalization: spiral similarity is the most general direct (orientation-preserving) similarity of the plane with a fixed point — it is multiplication by a complex number $k$ about a center, so a homothety ($k$ real) and a pure rotation ($|k| = 1$) are its two degenerate ends. The center taking $A \mapsto B$, $C \mapsto D$ is identical to the one taking $A \mapsto C$, $B \mapsto D$, which is what makes it pair up two similar triangles in a single figure.
Where it appears: the spiral-similarity lemma is a staple of hard olympiad geometry — it explains the Miquel point of a complete quadrilateral, links the two circles in a 'two circles meeting at $P$' configuration, and turns 'these two segments are related by a rotation' into concyclicity. The complex-number computation (find $k$ and the fixed point) is the fast route on coordinates.
Pitfall: a UNIQUE spiral similarity sending $A \mapsto C$, $B \mapsto D$ exists only when $AB$ and $CD$ are NOT parallel and NOT equal-and-parallel (otherwise the construction degenerates to a translation/homothety with no finite center). Also the center maps $A \to C$ and $B \to D$ — pairing it as $A \to D$, $B \to C$ instead gives a DIFFERENT spiral similarity; keep the correspondence straight.

Spot the Signal

Use it when two similar triangles in a figure share a center and you need the linking map.
You can describe the hard part as spotting the rotation-and-scaling that maps one segment onto another, 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 fixed center and the rotation angle and ratio that pair the segments.
Name the relation that makes Spiral Similarity 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

Assuming a plain rotation or homothety when both a rotation and a scaling are needed.
Applying Spiral Similarity because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

When two similar triangles share a vertex, look for the spiral that links them.

Try it on:

Use a spiral similarity to relate lengths or angles in an olympiad configuration.