Maths Olympiad Prep

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Geometry Difficulty 8.0 National olympiad, round 2 Find the answer

We define a binary operation \star in the plane as follows: Given two points AA and BB in the plane, C=ABC = A \star B is the third vertex of the equilateral triangle ABC oriented positively. What is the relative position of three points I,M,OI, M, O in the plane if I(MO)=(OI)MI \star (M \star O) = (O \star I)\star M holds?

A number or a short expression. Spacing and $ signs are ignored.

Solution

Given the binary operation \star defined in the plane as follows: for any two points AA and BB, C=ABC = A \star B is the third vertex of the equilateral triangle ABCABC oriented positively.

We aim to determine the relative position of three points II, MM, and OO such that:
I(MO)=(OI)M. I \star (M \star O) = (O \star I) \star M.

To solve this, consider the properties of the operation \star:

1. Equilateral Triangles: The operation \star produces the third vertex of an equilateral triangle oriented positively. This implies that if P=XYP = X \star Y, the triangle XYPXYP is equilateral with a counterclockwise orientation.

2. Orientation and Triangle Properties:
- For the operation I(MO)I \star (M \star O), let N=MON = M \star O, meaning INI \star N finds the third point of the equilateral triangle completing vertex II with base NN.
- Similarly, (OI)M(O \star I) \star M results in a point where the triangles are also equilateral.

3. Properties of Rotations:
- Each \star operation corresponds geometrically to a rotation of the plane by 6060^\circ counterclockwise about the point II, followed by translating the point MM.

For these expressions to be equal, IMO\triangle IMO must satisfy specific geometric properties:

- Isosceles Triangle: For both paths of \star operations, the configurations lead to a requirement: each point must subtend the same base with an equal angle, which implies the isosceles nature with OI=OMOI = OM.

- **Angle IOM=2π3 \angle IOM = \frac{2\pi}{3} **: The operations must satisfy rotation symmetry to maintain equality, implying a rotation by 120120^\circ to cycle through each vertex.

Thus, these geometric constraints firmly conclude:

I(MO)=(OI)M if and only if IMO is positively oriented, is isosceles with OI=OM, and IOM=2π3. I \star (M \star O) = (O \star I) \star M \text{ if and only if } \triangle IMO \text{ is positively oriented, is isosceles with } OI = OM \text{, and } \angle IOM = \frac{2\pi}{3}.

The configuration ensuring the operation equivalence is:
The triangle IMO is positively oriented, isosceles with OI=OM, and IOM=2π3. \boxed{\text{The triangle } \triangle IMO \text{ is positively oriented, isosceles with } OI = OM, \text{ and } \angle IOM = \frac{2\pi}{3}.}

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