Circle Sandwich
Task
A square is inscribed in a circle which is inscribed in a square as shown below. Note that the vertices of the inner square meet the midpoints of the outer square's sides.
Consider the area of the region left by removing the interior of the small square from the interior of the big square. Is the area of the blue region more or less than \frac12 of that?
IM Commentary
The purpose of this task is to apply knowledge about triangles, circles, and squares in order to calculate and compare two different areas. This can be done in several ways: the solutions below use definition of congruence, the Pythagorean Theorem, and symmetry considerations. The symmetry argument requires auxiliary lines (the two diagonals of the inner square) while the Pythagorean Theorem can be applied to the picture as is after labeling some key points.
This task was adapted from problem #25 on the 2011 American Mathematics Competition (AMC) 8 Test. The question on the exam was different, asking which fraction, from a multiple choice list, is closest to the ratio of the shaded area to the difference of the areas of the two squares: in particular, the exam question could be answered correctly by studying the picture carefully without calculating any of the areas. For the 2011 AMC 8, which was taken by 153,485 students, the multiple choice answers for the problem had the following distribution:
Choice | Answer | Percentage of Answers |
(A)* | {\frac{1}{2}} | 31.14 |
(B) | {1} | 13.94 |
(C) | {\frac{3}{2}} | 19.95 |
(D) | {2} | 10.14 |
(E) | {\frac{5}{2}} | 10.94 |
Omit | - | 13.80 |
Of the 153,485 students: 72,648 or 47% were in 8th grade, 50,433 or 33% were in 7th grade, and the remainder were less than 7th grade.
Solutions
Solution: 1 Adding and Subtracting Areas
Let's start with the case of a circle with radius 1.
Below is a picture with the two diagonals of the smaller square drawn and some points labelled:
The diagonals divide the larger square into 4 smaller, congruent squares because they join the midpoints of the sides of the larger square and reflecting across the vertical one (or horizontal, respectively) exchanges the left side of the big square and the right side (or the top and the bottom, respectively). The sides of the small squares are radii of the circle, so the side lengths are 1 and the area of the small squares is 1^2=1.
Segment AC is a diagonal of square ABCD, and reflecting across the diagonal exchanges triangle ABC and triangle ADC and so these two triangles are congruent. That means that each of the triangles has area \frac12. So the small square (which is composed of 4 of these triangles) has area 2, and the area of the big square with the small square removed (shown in grey below) is also 2.
The area of the circle of radius r is \pi r^2, and since r=1, that is just \pi. So the blue shaded area is \pi - 2 \approx 1.14
Since all squares are similar to all squares and all circles are similar to all circles, we can make the same argument with any radius r because the areas of the figure with radius 1 will all scale by a factor of r^2. In other words, the area of the circle will be \pi r^2, the area of the big square will be 4r^2, the area of each of the 8 triangles will be \frac12 r^2, and the area of the smaller square will be 2r^2. The area of the blue region will be \pi r^2 - 2r^2=r^2(\pi-2) and if we divide the area of the blue region by the area of the grey region, we get \frac{r^2(\pi-2)}{2r^2}=\frac{\pi-2}{2} \approx \frac{1.14}{2} \gt \frac12
Solution: 2 Using the Pythagorean Theorem
Here we apply the Pythagorean Theorem to find the difference of the areas of the two squares. We first focus on finding the side length of the inner square.
We know \triangle ABC is a right triangle because we are given that the outer quadrilateral is a square. Let r denote the radius of the circle. We have |AB| = |CB| = r since the side length of the large square is twice the radius of the circle and the small square touches the large square at the midpoints of its sides. By the Pythagorean theorem |AB|^2 + |CB|^2 = |AC|^2
Instead of using the Pythagorean theorem to find |AC|, we could also find the difference of areas of the two squares directly: the inner square together with 4 triangles congruent to \triangle ABC make the outer square and the area of \triangle ABC is \frac{r^2}{2}. So the difference of areas of the squares is 2r^2.
The area of the circle of radius r is \pi r^2. So the blue shaded area is \pi r^2 - 2r^2 = (\pi -2)r^2.
Solution: 3 Using Symmetry
Below is a picture with the two diagonals of the smaller square drawn and some points labelled:
Note that \angle ADC is a right angle because the diagonals of the small square are perpendicular. So the area of \triangle ADC is \frac{1}{2} \times |DA| \times |DC|.
The area of the circle of radius r is \pi r^2. So the blue shaded area is \pi r^2 - 2r^2 = (\pi -2)r^2.
Circle Sandwich
A square is inscribed in a circle which is inscribed in a square as shown below. Note that the vertices of the inner square meet the midpoints of the outer square's sides.
Consider the area of the region left by removing the interior of the small square from the interior of the big square. Is the area of the blue region more or less than \frac12 of that?