In `Delta ABC, AB = 1, BC = 1, and AC = 1//sqrt2`. In `Delta MNP, MN =1, NP =1, and angle MNP = 2 angle ABC`. Find the side MP

Explanation

To solve the problem step by step, we will use the cosine rule and properties of triangles.

### Step 1: Find angle ABC in triangle ABC
We are given the sides of triangle ABC:
- \( AB = 1 \)
- \( BC = 1 \)
- \( AC = \frac{1}{\sqrt{2}} \)

Using the cosine rule:
\[ \cos(\angle ABC) = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC} \]

Substituting the values:
\[ \cos(\angle ABC) = \frac{1^2 + 1^2 - \left(\frac{1}{\sqrt{2}}\right)^2}{2 \cdot 1 \cdot 1} \]
\[ = \frac{1 + 1 - \frac{1}{2}}{2} = \frac{2 - \frac{1}{2}}{2} = \frac{\frac{4}{2} - \frac{1}{2}}{2} = \frac{\frac{3}{2}}{2} = \frac{3}{4} \]

### Step 2: Calculate \( \angle ABC \)
Let \( \theta = \angle ABC \). We have:
\[ \cos(\theta) = \frac{3}{4} \]

### Step 3: Find \( \cos(2\theta) \)
Using the double angle formula:
\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \]
Substituting \( \cos(\theta) = \frac{3}{4} \):
\[ \cos(2\theta) = 2\left(\frac{3}{4}\right)^2 - 1 = 2 \cdot \frac{9}{16} - 1 = \frac{18}{16} - 1 = \frac{18}{16} - \frac{16}{16} = \frac{2}{16} = \frac{1}{8} \]

### Step 4: Use the cosine rule in triangle MNP
In triangle MNP, we know:
- \( MN = 1 \)
- \( NP = 1 \)
- \( \angle MNP = 2\theta \)

We need to find \( MP \). Using the cosine rule:
\[ \cos(2\theta) = \frac{MN^2 + NP^2 - MP^2}{2 \cdot MN \cdot NP} \]
Substituting the known values:
\[ \frac{1}{8} = \frac{1^2 + 1^2 - MP^2}{2 \cdot 1 \cdot 1} \]
\[ \frac{1}{8} = \frac{1 + 1 - MP^2}{2} \]
\[ \frac{1}{8} = \frac{2 - MP^2}{2} \]

### Step 5: Solve for \( MP^2 \)
Cross-multiplying gives:
\[ 1 = 2 - MP^2 \]
\[ MP^2 = 2 - 1 = 1 \]
\[ MP = \sqrt{1} = 1 \]

### Final Answer
Thus, the length of side \( MP \) is \( 1 \).