2023 AMC 12A Problem 9

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?

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Solution

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,

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.