Question

Square $ABCD$ has side length $1.$ Points $P,$ $Q,$ $R,$ and $S$ each lie on a side of $ABCD$ such that $APQCRS$ is an equilateral convex hexagon with side length $s.$ What is $s?$

$\dfrac{\sqrt{2}}{3}$

$\dfrac{1}{2}$

$2 - \sqrt{2}$

$1 - \dfrac{\sqrt{2}}{4}$

$\dfrac{2}{3}$

Answer: C

Concepts:square (geometry)special right triangle rationalizing denominator

Difficulty rating: 1540

Solution:

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.

2022 AMC 10A Problem 5

Answer: C

Difficulty rating: 1540

Solution:

Problem 5 in Other Years