2023 AMC 12A Problem 9

Question

A square of area

2

is inscribed in a square of area

3,

creating four congruent triangles, as shown below. What is the ratio of the shorter leg to the longer leg in the shaded right triangle?

\dfrac{1}{5}

\dfrac{1}{4}

2-\sqrt{3}

\sqrt{3}-\sqrt{2}

\sqrt{2}-1

The outer square has side

\sqrt3

and the inner square has side

\sqrt2

If the legs are

p

and

q,

then

p+q=\sqrt3

and

p^2+q^2=2

The outer square has side

\sqrt3

and the inner square has side

\sqrt2.

Each triangle is right, with legs

p

and

q

along an outer side, so

p+q=\sqrt3,

and with hypotenuse an inner side, so

p^2+q^2=2.

Then

(p+q)^2=3

gives

2pq=1,

so

p

and

q

are the roots of

t^2-\sqrt3\,t+\tfrac12=0,

namely

\dfrac{\sqrt3\pm 1}{2}.

The ratio of shorter to longer leg is

\begin{gathered} \dfrac{\sqrt3-1}{\sqrt3+1}\\ {}=\dfrac{(\sqrt3-1)^2}{2}\\ {}=2-\sqrt3. \end{gathered}

Thus, the correct answer is C.

Problem 9 in other years