1. **Problem statement:** Given points $A=1$, $A'=-1$, $B=i$, $B'=-i$ in the complex plane, and a point $M$ with complex number $z$ different from $O=0$, $A$, $A'$, $B$, $B'$, define points $M_1(z_1)$ and $M_2(z_2)$ such that triangles $BMM_1$ and $AMM_2$ are right-angled isosceles with angles $(\overrightarrow{M_1B}, \overrightarrow{M_1M}) = (\overrightarrow{M_2M}, \overrightarrow{M_2A}) = \frac{\pi}{2}$. Find all points $M$ such that triangle $OM_1M_2$ is equilateral. 2. **Step 1: Express $M_1$ and $M_2$ in terms of $z$.** - Since $B = i$, $M = z$, and $BMM_1$ is right isosceles with angle $\pi/2$ at $M_1$, $M_1$ is obtained by rotating vector $\overrightarrow{BM}$ by $\pi/2$ around $B$. - Vector $\overrightarrow{BM} = z - i$. - Rotation by $\pi/2$ corresponds to multiplication by $e^{i\pi/2} = i$. - So, $\overrightarrow{M_1B} = i (z - i)$, thus $z_1 = M_1 = B + \overrightarrow{M_1B} = i + i(z - i) = i + i z - i^2 = i + i z + 1 = 1 + i + i z$. Similarly for $M_2$: - $A = 1$, $M = z$, triangle $AMM_2$ is right isosceles with angle $\pi/2$ at $M_2$. - Vector $\overrightarrow{M_2A} = A - M_2$ and $\overrightarrow{M_2M} = M - M_2$ satisfy $(\overrightarrow{M_2M}, \overrightarrow{M_2A}) = \pi/2$. - This means $\overrightarrow{M_2M} = i \overrightarrow{M_2A}$. - So, $z - z_2 = i (1 - z_2)$. - Rearranged: $z - z_2 = i - i z_2$. - Group $z_2$ terms: $z_2 - i z_2 = z - i$. - $z_2 (1 - i) = z - i$. - Therefore, $z_2 = \frac{z - i}{1 - i}$. 3. **Step 2: Condition for triangle $OM_1M_2$ to be equilateral.** - Points are $O=0$, $M_1 = 1 + i + i z$, $M_2 = \frac{z - i}{1 - i}$. - Triangle $OM_1M_2$ equilateral means $|M_1| = |M_2| = |M_1 - M_2|$ and angles between vectors are $\pi/3$. 4. **Step 3: Use complex rotation property for equilateral triangle.** - Vector $\overrightarrow{OM_2} = z_2$, $\overrightarrow{OM_1} = z_1$. - For equilateral triangle, $z_1 = z_2 e^{i \pi/3}$ or $z_2 = z_1 e^{i \pi/3}$. - Check $z_1 = z_2 e^{i \pi/3}$: $$1 + i + i z = \frac{z - i}{1 - i} e^{i \pi/3}$$ 5. **Step 4: Solve for $z$.** - Multiply both sides by $1 - i$: $$(1 - i)(1 + i + i z) = (z - i) e^{i \pi/3}$$ - Compute left side: $(1 - i)(1 + i) + (1 - i) i z = (z - i) e^{i \pi/3}$ - Note $(1 - i)(1 + i) = 1 - i^2 = 1 + 1 = 2$. - Also, $(1 - i) i = i - i^2 = i + 1$. - So left side: $2 + (i + 1) z = (z - i) e^{i \pi/3}$. 6. **Step 5: Rearrange and isolate $z$.** - $2 + (i + 1) z = z e^{i \pi/3} - i e^{i \pi/3}$ - Bring $z$ terms to one side: $(i + 1) z - z e^{i \pi/3} = - i e^{i \pi/3} - 2$ - Factor $z$: $$z \left( i + 1 - e^{i \pi/3} \right) = - i e^{i \pi/3} - 2$$ - Therefore, $$z = \frac{- i e^{i \pi/3} - 2}{i + 1 - e^{i \pi/3}}$$ 7. **Step 6: Simplify and find numeric value (optional).** - $e^{i \pi/3} = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$. - Substitute and simplify to get explicit $z$. **Final answer:** $$z = \frac{- i \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) - 2}{i + 1 - \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right)}$$ This $z$ gives the point $M$ such that triangle $OM_1M_2$ is equilateral. --- "slug": "equilateral triangle", "subject": "complex geometry", "desmos": {"latex": "z_1=1+i+i z, z_2=\frac{z - i}{1 - i}", "features": {"intercepts": true, "extrema": true}}, "q_count": 1