2009 Grade 8 CEMC Gauss Problems/Problem 23

Problem

In the diagram, the circle is inscribed in the square. This means that the circle and the square share points $S$ , $T$ , $U$ , and $V$ , and the width of the square is exactly equal to the diameter of the circle. Rounded to the nearest tenth, what percentage of line segment $XY$ is outside the circle?

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$\text{ (A) }\ 29.3 \qquad\text{ (B) }\ 28.3 \qquad\text{ (C) }\ 33.3 \qquad\text{ (D) }\ 25.0 \qquad\text{ (E) }\ 16.7$

Solution 1

Let $s$ be the side length of the square. From the 45-45-90 triangles formed by the square, we can see that $XY = s \times \sqrt{2}$ .

The part of $XY$ that is in the circle is the diameter of the circle, which means we can subtract it from the total length of $XY$ to get the part that is outside of the circle, and then divide it by the full length of $XY$ :

$\frac{s \times \sqrt{2} - s}{s \times \sqrt{2}} \times 100\%$

$=\frac{\sqrt{2} - 1}{\sqrt{2}} \times 100\%$

Rounding this, we get $\boxed {\textbf {(A) } 29.3}\%$

~anabel.disher

Solution 2

We can set the side length of the square to a number like $1$ . This gives us $XY = \sqrt{2}$ . The part of $XY$ that is outside of the circle is then $\sqrt{2} - 1$ .