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# Spiral Similarity - Complex Number 1 # What is spiral similarity?

Spiral similarity, for our purpose, will be a beautiful motion in the plane. It is a combination of rotation and dilation (expansion or contraction by a constant factor.

Suppose AB is a (finite) segment in the plane. CD be another segment. Assume AB and CD to be disjoint.

It is possible to send AB onto CD using a combination of rotation and dilation. To accomplish this, we need to find:

• Center of rotation
• Angle of rotation
• Ratio of dilation

# How to find center of rotation (and dilation)

Join AC and BD. Suppose they intersect at X. Draw the circles CXD and AXB. Let them intersect at O.

O could be same as X or could be a different point.

Claim: O is the center of rotation and dilation, that sends AB to CD. Join OA, OX, OB, OC, OD.

Pause and think: Can you find some cyclic quadrilaterals in this picture. Clearly ABXO is cyclic. This implies $\angle OBX = \angle OAX$. This is because they are subtended on the circumference by the same segment OX. (Angles subtended by a chord at the circumference are equal, by property of cyclic quadrilaterals).

Similarly CDXO is cyclic implying $\angle ODX = \angle OCX$

Pause and think: Can you find a pair of similar triangles?

Hint: Equiangular triangles are similar.

$\Delta OBD \sim \Delta OAC$

This is because we have shown pair of corresponding angles to be equal. Hence the two triangles are equiangular.

Pause and imagine:

This implies $\angle AOC = \angle BOD$. We may imagine this as following:

• OA is turning into OC;
• OB is turning into OD

Since the angles (of rotation) $\angle AOC, \angle BOD$ are equal, hence this works out nicely.

Moreover, OA grows into OC and OB grows into OD by same constant factor. This is true because $\frac {OA}{OC} = \frac{OB}{OD}$. This follows from the similarity of triangles.

Hence A goes to C and B goes to D, via rotation and dilation, centered at O.

# Rotation and dilation is known spiral similarity.

Complex numbers are algebraic tools to encode this motion in the plane.

(This is a supplemental note to live lecture session of Complex Number module of Cheenta I.S.I., C.M.I. Entrance Program and Math Olympiad Program)

# What is spiral similarity?

Spiral similarity, for our purpose, will be a beautiful motion in the plane. It is a combination of rotation and dilation (expansion or contraction by a constant factor.

Suppose AB is a (finite) segment in the plane. CD be another segment. Assume AB and CD to be disjoint.

It is possible to send AB onto CD using a combination of rotation and dilation. To accomplish this, we need to find:

• Center of rotation
• Angle of rotation
• Ratio of dilation

# How to find center of rotation (and dilation)

Join AC and BD. Suppose they intersect at X. Draw the circles CXD and AXB. Let them intersect at O.

O could be same as X or could be a different point.

Claim: O is the center of rotation and dilation, that sends AB to CD. Join OA, OX, OB, OC, OD.

Pause and think: Can you find some cyclic quadrilaterals in this picture. Clearly ABXO is cyclic. This implies $\angle OBX = \angle OAX$. This is because they are subtended on the circumference by the same segment OX. (Angles subtended by a chord at the circumference are equal, by property of cyclic quadrilaterals).

Similarly CDXO is cyclic implying $\angle ODX = \angle OCX$

Pause and think: Can you find a pair of similar triangles?

Hint: Equiangular triangles are similar.

$\Delta OBD \sim \Delta OAC$

This is because we have shown pair of corresponding angles to be equal. Hence the two triangles are equiangular.

Pause and imagine:

This implies $\angle AOC = \angle BOD$. We may imagine this as following:

• OA is turning into OC;
• OB is turning into OD

Since the angles (of rotation) $\angle AOC, \angle BOD$ are equal, hence this works out nicely.

Moreover, OA grows into OC and OB grows into OD by same constant factor. This is true because $\frac {OA}{OC} = \frac{OB}{OD}$. This follows from the similarity of triangles.

Hence A goes to C and B goes to D, via rotation and dilation, centered at O.

# Rotation and dilation is known spiral similarity.

Complex numbers are algebraic tools to encode this motion in the plane.

(This is a supplemental note to live lecture session of Complex Number module of Cheenta I.S.I., C.M.I. Entrance Program and Math Olympiad Program)

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