Question

A square with area $3$ has a square with area $2$ inscribed in it. This creates $4$ smaller congruent right triangles. What is the ratio of the smaller leg to the larger leg in the shaded right triangle?

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$2 - \sqrt{3}$

$\sqrt{3} - \sqrt{2}$

$\sqrt{2} - 1$

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: $2ab=(a+b)^2-(a^2+b^2)=3-2=1,$ so $ab=\tfrac12.$ 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.

2023 AMC 10A Problem 11

Solution:

Problem 11 in Other Years