Composite Figures

The three figures below are composite figures: two-dimensional diagrams built from layered lines, circles, polygons, and other basic shapes. The mathematical challenge here is to find the total amount of colored area in the figure, given some baseline measurement. For example, assume that the large circumscribing circle in each of the examples below has area 1.

\(\qquad \qquad\)

This page is for those learning to solve and design their own intermediate and advanced composite figure challenges. If you're new to this kind of problem, check out this wiki: Basic Composite Figures.

We will discuss three high-level techniques for solving composite figures challenges:

• Iterative Inclusion and Exclusion
• Systems of Linear Equations
• Self-Similarity and Infinite Designs.

In each section, we'll demonstrate the technique by solving an interesting example and then providing a few additional examples that you can try yourself.

Lastly, once you've mastered these techniques yourself, we challenge you to design a composite figures challenge on Brilliant at the intermediate or advanced level, and to add it to this topic area so that it appears in the community posts at the bottom of the page.

Basic Geometry

To successfully solve solid composite figures problems, you frequently need to know:

• the properties of triangles, polygons, and circles
• how to find the perimeter and area of triangles; the perimeter and area of polygons; and the circumference and area of circles
• how to identify and compare congruent and similar triangles
• how to find area and perimeter relations in similar polygons

Basic Techniques

This page assumes that you're already familiar with the techniques needed to solve basic composite figure problems. Let's quickly review each technique:

Adding in Lines and Grids
This technique...

Complementary Area
This technique...

Using Symmetry
This technique...

Using Coordinate Geometry
This technique...

Existing Examples:

AMC 2008

The smallest circle has radius 2 inches, with each successive circle's radius increasing by 2 inches. Approximately what percent of the design is black?

To fully appreciate this section, you should be familiar with Solving system of linear equations using matrices.

Have you seen a composite figure problem where the solution involves some complicated series of taking certain figures here and certain figures there and adding multiples of them to each other? Let's first look at an example of a complicated problem, and how one would work it out.

In a square of side length 1, 4 quarter circles are drawn in blue. Find the area \(A\) that is common to these circles.

It is not immediately apparent how one could use a combination of shapes in order to find \(A\) by itself. As it turns out, such a combination is achieved by "4 equilateral triangles + 8 truncated sectors - 4 quarter circles - the square gives us \(A\) exactly". That is hard to visualize, and even harder to convince yourself easily, and almost impossible to come up with! How would one solve this problem then?

We are interested in finding \(A \). It is not immediately clear how we can arrive at just \(A\) by itself, so let's start off by listing several basic shapes that we can find.

1) By considering the entire square, we get that \( A + 4B + 4C = 1 \).
2) By considering 2 diagonally opposite quarter circles minus the square, we get that \( A + 2B = 2 \times \frac{\pi}{4} - 1 \).

Of course, these two equations are not useful enough to determine any of the values of \(A, B, C \), so let's try and find one more equation.

3) By considering the equilateral triangle of side length 1, along with 2 truncated sectors, we get that \( 2 \times \frac{\pi}{6} - \frac{ \sqrt{3} } { 4} = A + 2B + C \).

Now let's hope that these are sufficient:
The third minus the second equation gives \( C = 1 - \frac{\sqrt{3}}{4} - \frac{\pi}{6}\).
Twice of the second minus the first (along with the value of C) gives \( A = \frac{ \pi}{ 3} +1 - \sqrt{3}. \ _\square\)

This approach of initially listing out various observations, meant that we just needed to solve the system of equations. We know how to do this really easily, and could even simplify it by using the inverse matrix.

Let's express the equations in matrix form. We have:

\[ \begin{pmatrix}
1 & 4 & 4 \\ 1 & 2 & 0 \\ 1 & 2 & 1 \\ \end{pmatrix}
\begin{pmatrix} A \\ B \\ C \\ \end{pmatrix} = \begin{pmatrix} 1 \\ \frac{\pi}{2} - 1 \\ \frac{ \pi}{3} - \frac{ \sqrt{3} } { 4} \end{pmatrix}. \]

Hence, we obtain that

\[ \begin{pmatrix} A \\ B \\ C \\ \end{pmatrix} = \begin{pmatrix}
1 & 4 & 4 \\ 1 & 2 & 0 \\ 1 & 2 & 1 \\ \end{pmatrix}^{-1} \begin{pmatrix} 1 \\ \frac{\pi}{2} - 1 \\ \frac{ \pi}{3} - \frac{ \sqrt{3} } { 4} \end{pmatrix} = \begin{pmatrix}
-1 & -2 & 4 \\ \frac{1}{2} & \frac{3}{2} & -2 \\ 0 & -1 & 1 \\ \end{pmatrix}
\begin{pmatrix} 1 \\ \frac{\pi}{2} - 1 \\ \frac{ \pi}{3} - \frac{ \sqrt{3} } { 4} \end{pmatrix} = \begin{pmatrix} 1 + \frac{\pi}{3} - \sqrt{3} \\ -1 + \frac{ \sqrt{3}}{2} + \frac{\pi}{12} \\ 1 - \frac{ \sqrt{3} }{4} - \frac{ \pi}{6} \end{pmatrix} .\\ \]

Note: The stated combination to obtain A exactly is

\[ A = - ( A + 4B + 4C) - 2 ( A + 2B) + 4 ( A + 2B +C) . \]

Do you see how the inverse matrix tells us the exact combination? Hint: The coefficients in the first row of the inverse matrix are \( -1, -2, 4 \).

Now that you've learnt this trick, let's make quick work of these "hard" problems:

Existing Examples:

This can be done with many pictures/puzzles that have self-similarity:

For some infinite complex figure problem that can't easily be done by self-similarity, the approach requires some work to determine the area pattern. Here is an example below to start off: