2022 AMC 10A Problem 5

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}

Use the

45

–

45

–

90

triangle at a corner of the square

Write the side of the square as one hexagon side plus one corner leg

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.

Problem 5 in other years