2023 AMC 10A Problem 11

Problem 11 of 25IntermediateAlgebra Geometry

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

Each corner right triangle has legs

a

and

b.

A side of the outer square gives

a+b=\sqrt3,

and a side of the inscribed square gives

a^2+b^2=2.

Subtract to find the product:

2ab=(a+b)^2-(a^2+b^2)

=3-2

=1,

ab=\tfrac12.

Then

a

and

b

are the roots of

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

namely

\tfrac{\sqrt3\pm1}{2}.

The ratio of the smaller leg to the larger is

\dfrac{\sqrt3-1}{\sqrt3+1}

=\dfrac{(\sqrt3-1)^2}{2}

=2-\sqrt3.

Thus, C is the correct answer.

Tagged: square (geometry) · Pythagorean Theorem · Vieta’s Formulas