1. **Stating the problem:** We have an equilateral triangle ABC with side length BC = 1 unit. Points I and K are midpoints of segments BC and AI respectively. Vectors are defined as $\mathbf{u_2} = \overrightarrow{IA}$ and $\mathbf{v_1} = \overrightarrow{BC}$. We need to: 1) Calculate $\frac{1}{2}(||\mathbf{u_2}||^2 + ||\mathbf{v_1}||^2)$ and deduce $||\mathbf{u_2} + \mathbf{v_1}||$. 2) Calculate $||\mathbf{u_2}|| \cdot ||\mathbf{v_1}|| \cdot \cos \theta$ where $\theta$ is the angle between $\mathbf{u_2}$ and $\mathbf{v_1}$. 3) Calculate the scalar products $\mathbf{AIS} \cdot \mathbf{BI2}$, $\mathbf{IK} \cdot \mathbf{KI}$, and $\mathbf{AI} \cdot \mathbf{AC}$. 2. **Important formulas and rules:** - For vectors $\mathbf{a}$ and $\mathbf{b}$, $||\mathbf{a} + \mathbf{b}||^2 = ||\mathbf{a}||^2 + 2\mathbf{a} \cdot \mathbf{b} + ||\mathbf{b}||^2$. - The dot product $\mathbf{a} \cdot \mathbf{b} = ||\mathbf{a}|| \, ||\mathbf{b}|| \, \cos \theta$. - Midpoint divides a segment into two equal parts. 3. **Step 1: Calculate $\frac{1}{2}(||\mathbf{u_2}||^2 + ||\mathbf{v_1}||^2)$ and deduce $||\mathbf{u_2} + \mathbf{v_1}||$** - Since ABC is equilateral with side length 1, $||\mathbf{v_1}|| = ||\overrightarrow{BC}|| = 1$. - Point I is midpoint of BC, so $I$ divides BC into two segments of length $\frac{1}{2}$. - Vector $\mathbf{u_2} = \overrightarrow{IA}$: length $||\mathbf{u_2}|| = ||\overrightarrow{IA}||$. - Coordinates (assuming B at origin (0,0), C at (1,0), A at $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$): - $I$ midpoint of BC: $I = \left(\frac{0+1}{2}, \frac{0+0}{2}\right) = (0.5, 0)$ - $A = \left(0.5, \frac{\sqrt{3}}{2}\right)$ - Then $\overrightarrow{IA} = A - I = \left(0.5 - 0.5, \frac{\sqrt{3}}{2} - 0\right) = (0, \frac{\sqrt{3}}{2})$ - So $||\mathbf{u_2}|| = \frac{\sqrt{3}}{2}$ Calculate: $$\frac{1}{2}\left(||\mathbf{u_2}||^2 + ||\mathbf{v_1}||^2\right) = \frac{1}{2}\left(\left(\frac{\sqrt{3}}{2}\right)^2 + 1^2\right) = \frac{1}{2}\left(\frac{3}{4} + 1\right) = \frac{1}{2} \times \frac{7}{4} = \frac{7}{8}$$ Next, deduce $||\mathbf{u_2} + \mathbf{v_1}||$: $$||\mathbf{u_2} + \mathbf{v_1}||^2 = ||\mathbf{u_2}||^2 + 2 \mathbf{u_2} \cdot \mathbf{v_1} + ||\mathbf{v_1}||^2$$ We need $\mathbf{u_2} \cdot \mathbf{v_1}$ to find this. 4. **Step 2: Calculate $||\mathbf{u_2}|| \cdot ||\mathbf{v_1}|| \cdot \cos \theta$** - $\mathbf{v_1} = \overrightarrow{BC} = C - B = (1,0) - (0,0) = (1,0)$ - $\mathbf{u_2} = (0, \frac{\sqrt{3}}{2})$ - Dot product: $$\mathbf{u_2} \cdot \mathbf{v_1} = 0 \times 1 + \frac{\sqrt{3}}{2} \times 0 = 0$$ - So $\cos \theta = \frac{\mathbf{u_2} \cdot \mathbf{v_1}}{||\mathbf{u_2}|| \, ||\mathbf{v_1}||} = 0$ Therefore, $$||\mathbf{u_2}|| \cdot ||\mathbf{v_1}|| \cdot \cos \theta = \frac{\sqrt{3}}{2} \times 1 \times 0 = 0$$ 5. **Step 3: Calculate scalar products $\mathbf{AIS} \cdot \mathbf{BI2}$, $\mathbf{IK} \cdot \mathbf{KI}$, and $\mathbf{AI} \cdot \mathbf{AC}$** - The problem does not define vectors $\mathbf{AIS}$ or $\mathbf{BI2}$ explicitly, so we cannot compute these without further information. - For $\mathbf{IK} \cdot \mathbf{KI}$: - $\mathbf{IK} = - \mathbf{KI}$, so $$\mathbf{IK} \cdot \mathbf{KI} = \mathbf{IK} \cdot (-\mathbf{IK}) = -||\mathbf{IK}||^2$$ - For $\mathbf{AI} \cdot \mathbf{AC}$: - $\mathbf{AI} = I - A = (0.5, 0) - \left(0.5, \frac{\sqrt{3}}{2}\right) = (0, -\frac{\sqrt{3}}{2})$ - $\mathbf{AC} = C - A = (1,0) - \left(0.5, \frac{\sqrt{3}}{2}\right) = \left(0.5, -\frac{\sqrt{3}}{2}\right)$ - Dot product: $$\mathbf{AI} \cdot \mathbf{AC} = 0 \times 0.5 + \left(-\frac{\sqrt{3}}{2}\right) \times \left(-\frac{\sqrt{3}}{2}\right) = \frac{3}{4}$$ **Final answers:** - $\frac{1}{2}(||\mathbf{u_2}||^2 + ||\mathbf{v_1}||^2) = \frac{7}{8}$ - $||\mathbf{u_2} + \mathbf{v_1}|| = \sqrt{||\mathbf{u_2}||^2 + 2 \mathbf{u_2} \cdot \mathbf{v_1} + ||\mathbf{v_1}||^2} = \sqrt{\frac{3}{4} + 0 + 1} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2}$ - $||\mathbf{u_2}|| \cdot ||\mathbf{v_1}|| \cdot \cos \theta = 0$ - $\mathbf{IK} \cdot \mathbf{KI} = -||\mathbf{IK}||^2$ - $\mathbf{AI} \cdot \mathbf{AC} = \frac{3}{4}$