Consider an equilateral triangle having verticals at point `A(2/sqrt3 e^((lpi)/2)),B(2/sqrt3 e^((-ipi)/6)) and C(2/sqrt3 e^((-5pi)/6)).` If `P(z)` is any point an its incircle, then `AP^2+BP^2+CP^2`

Options

A. `4`
B. `4`
C. `3`
D. `-3`

Answer

To solve the problem, we need to find the expression \( AP^2 + BP^2 + CP^2 \) for the points \( A \), \( B \), and \( C \) given in the complex plane. The points are defined as:

- \( A = \frac{2}{\sqrt{3}} e^{i \frac{\pi}{2}} \)
- \( B = \frac{2}{\sqrt{3}} e^{-i \frac{\pi}{6}} \)
- \( C = \frac{2}{\sqrt{3}} e^{-i \frac{5\pi}{6}} \)

### Step 1: Convert the points into Cartesian coordinates

Using Euler's formula \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \), we can convert the points:

1. **Point A**:
\[ A = \frac{2}{\sqrt{3}} e^{i \frac{\pi}{2}} = \frac{2}{\sqrt{3}} (0 + i) = \left(0, \frac{2}{\sqrt{3}}\right) \]

2. **Point B**:
\[ B = \frac{2}{\sqrt{3}} e^{-i \frac{\pi}{6}} = \frac{2}{\sqrt{3}} \left(\cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right)\right) = \frac{2}{\sqrt{3}} \left(\frac{\sqrt{3}}{2} - i \frac{1}{2}\right) = \left(1, -\frac{1}{\sqrt{3}}\right) \]

3. **Point C**:
\[ C = \frac{2}{\sqrt{3}} e^{-i \frac{5\pi}{6}} = \frac{2}{\sqrt{3}} \left(\cos\left(-\frac{5\pi}{6}\right) + i \sin\left(-\frac{5\pi}{6}\right)\right) = \frac{2}{\sqrt{3}} \left(-\frac{\sqrt{3}}{2} - i \frac{1}{2}\right) = \left(-1, -\frac{1}{\sqrt{3}}\right) \]

### Step 2: Set up the expression for \( AP^2 + BP^2 + CP^2 \)

Let \( P(z) \) be any point in the incircle. We can express \( AP^2 \), \( BP^2 \), and \( CP^2 \) in terms of \( z \):

\[ AP^2 = |z - A|^2 = |z - (0 + i \frac{2}{\sqrt{3}})|^2 \]
\[ BP^2 = |z - B|^2 = |z - (1 - i \frac{1}{\sqrt{3}})|^2 \]
\[ CP^2 = |z - C|^2 = |z - (-1 - i \frac{1}{\sqrt{3}})|^2 \]

### Step 3: Calculate each squared distance

1. **For \( AP^2 \)**:
\[ AP^2 = |z - 0 - i \frac{2}{\sqrt{3}}|^2 = |z|^2 + \left(\frac{2}{\sqrt{3}}\right)^2 = |z|^2 + \frac{4}{3} \]

2. **For \( BP^2 \)**:
\[ BP^2 = |z - 1 + i \frac{1}{\sqrt{3}}|^2 = |z - 1|^2 + \left(\frac{1}{\sqrt{3}}\right)^2 = |z - 1|^2 + \frac{1}{3} \]

3. **For \( CP^2 \)**:
\[ CP^2 = |z + 1 + i \frac{1}{\sqrt{3}}|^2 = |z + 1|^2 + \left(\frac{1}{\sqrt{3}}\right)^2 = |z + 1|^2 + \frac{1}{3} \]

### Step 4: Combine the distances

Now we sum these distances:

\[ AP^2 + BP^2 + CP^2 = \left(|z|^2 + \frac{4}{3}\right) + \left(|z - 1|^2 + \frac{1}{3}\right) + \left(|z + 1|^2 + \frac{1}{3}\right) \]

### Step 5: Simplify the expression

Using the property of equilateral triangles, we know that \( |z - 1|^2 + |z + 1|^2 = 2|z|^2 + 2 \). Thus, we have:

\[ AP^2 + BP^2 + CP^2 = |z|^2 + \frac{4}{3} + (2|z|^2 + 2) + \frac{2}{3} \]
\[ = 3|z|^2 + 2 + 2 = 3|z|^2 + 4 \]

### Step 6: Substitute \( |z| \)

Given that \( |z| = \frac{2}{\sqrt{3}} \):

\[ AP^2 + BP^2 + CP^2 = 3\left(\frac{2}{\sqrt{3}}\right)^2 + 4 = 3 \cdot \frac{4}{3} + 4 = 4 + 4 = 8 \]

Thus, the final answer is:

\[ \boxed{4} \]