
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Suppose $\angle A=75^{\circ}$ and $\angle C=45^{\circ}$. Then $\angle B=60^{\circ}$. Points $M$ and $N$ lie on the circle with diameter $AC$. Triangle $BMN$ is similar to triangle $BCA$:

$$MNA = \angle MCA \quad (\text{they subtend the same arc}), \quad \angle BNM = 90^{\circ} - \angle MNA = 90^{\circ} - \angle MCA = \angle A = 75^{\circ}$$

$$NMC = \angle NAC (\text{they subtend the same arc}), \quad \angle BMN = 90^{\circ} - \angle NMC = 90^{\circ} - \angle NAC = \angle C = 45^{\circ}$$

The similarity coefficient $k_{1}$ of triangles $BMN$ and $BCA$ is the ratio of corresponding sides:

$$
k_{1}=\frac{BN}{BA}=\cos \angle B = \frac{1}{2}.
$$

Similarly, triangle $MAP$ is similar to triangle $ABC$ with similarity coefficient $k_{2}=\cos \angle A = \cos 75^{\circ}$, and triangle $CNP$ is similar to triangle $ABC$ with similarity coefficient:

$$
k_{3}=\cos \angle C = \frac{\sqrt{2}}{2}.
$$

The areas of similar triangles are proportional to the squares of their similarity coefficients:

$$
S_{BMN} = k_{1}^{2} \cdot S_{ABC} = \frac{1}{4} S_{ABC}, \quad S_{AMP} = k_{2}^{2} \cdot S_{ABC}, \quad S_{CNP} = k_{3}^{2} \cdot S_{ABC} = \frac{1}{2} S_{ABC}.
$$

Then:

$$
S_{MNP} = \left(1 - \frac{1}{4} - \frac{1}{2} - \cos^{2} 75^{\circ}\right) \cdot S_{ABC} \rightarrow S_{MNP} : S_{ABC} = \left(\frac{1}{4} - \frac{1 + \cos 150^{\circ}}{2}\right) = \frac{\sqrt{3} - 1}{4}.
$$

Present the answer in LaTex format: \boxed{Your answer}
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