2022 AMC 10A Problem 5

Consider the diagram: Since $AP = QC = s,$ we know that $PB = BQ.$ This shows that $\triangle PBQ$ is an isosceles right triangle with hypotenuse $PQ=s.$ Using the Pythagorean theorem, we get that $PB = \dfrac{s}{\sqrt{2}}.$ We also know that $$1 = AB = AP + PB = s + \dfrac{s}{\sqrt{2}}.$$ This equation simplifies to $$1 = (1 + \dfrac{1}{\sqrt{2}})s$$ Which implies that $$s = \dfrac{1}{1 + \dfrac{1}{\sqrt{2}}} = \dfrac{\sqrt{2}}{\sqrt{2} + 1}.$$ We can rationalize this fraction to get $$\dfrac{\sqrt{2}}{\sqrt{2} + 1} \cdot \dfrac{\sqrt{2} - 1}{\sqrt{2} - 1} = 2 - \sqrt{2}.$$ Thus, C is the correct answer.