Similarity Simplifies

When you see that two figures are similar to each other, everything becomes simpler. Suddenly, you realize that if you understand one figure, you know everything about the other, except for its absolute size.

I was observing some students solving this problem from the 1966 AHSME:

Circle I is circumscribed about a given square and circle II is inscribed in the given square. If r is the ratio of the area of circle I to that of circle II, what is r?

All of the students solved the problem using the same basic strategy: Compute the radii of the two circles in terms of the length of the side of the square. Use this information to compute the areas of each circle in terms of the side length of the square. Take the ratio.

Here’s their strategy carried out. Let be the side length of the square. The diameter of Circle I is the diagonal of the square. By the Pythagorean theorem, the diagonal has length . Therefore the radius of Circle I is and its area is . The diameter of Circle II is the side length of the square, so its radius is . Therefore, its area is . The ratio of the area of Circle I to that of Circle II is which simplifies to .

That’s certainly a valid solution and draws on the Pythagorean theorem and the formula for the area of circle in terms of its radius, both beautiful mathematical facts. But one notion the solution ignores is similarity.

Since all circles are similar to each other, the ratio of the area of Circle I to that of Circle II is equal to the square of the similarity factor, which you can find by taking the ratio of any two corresponding lengths. For example, we could take the square of the ratio of their diameters. The diameters of Circle I and Circle II are and , respectively. Their ratio is . Therefore the ratio of their areas is . Done.

Similarity relieves you of having to apply a circle area formula and thereby simplifies the computation. The simplification is even more dramatic if one imagines what may have happened had the original problem specified an unwieldy side length for the square. Suppose the original question were this instead:

Circle I is circumscribed about a given square and circle II is inscribed in the given square. If the square has side length 987, what is the ratio of the area of circle I to that of circle II?

Would you proceed to solve this using the same strategy that the students followed, only using the specific number 987 throughout the computation? I hope not (unless you really want to give yourself a computational workout).

If you’re savvy, you’d replace 987 with a variable (as urged in Avoid Computations – Use Variables!). But even if you don’t do that, if you exploit the similarity, you would still spare yourself a great deal of computation. Try it and see!

By the way, there’s another level of similarity in this problem: All figures representing this problem will be similar to each other. That’s why the original problem had a specific answer despite not providing information that would fix the absolute sizes of the figures involved. This observation might also suggest to you to use a variable for the side length of the square, or, at the very least, make you realize that you can replace 987 with any number you wish and you won’t affect the answer.

To further test your understanding of similarity, consider the same problem only this time, instead of using circles, circumscribe and inscribe a square tilted by 45 degrees relative to the given square. Quick, what is the ratio of the area of the larger to the smaller square?

Problem 3 from the 1984 AIME has similarity written all over it and if you exploit similarity, you can crack this problem without ever invoking any kind of area formula at all. For your convenience, here’s the problem:

A point  is chosen in the interior of such that when lines are drawn through  parallel to the sides of , the resulting smaller triangles , , and  in the figure, have areas 4, 9, and 49, respectively. Find the area of .

All those parallel lines produce four mutually similar triangles. Also, by parallel transport, we can see that the sides of the bigger triangle are equal to the sums of the corresponding sides of the three smaller triangles. By similarity, corresponding lengths of the smaller triangles are in the proportion 2 : 3 : 7, and hence corresponding lengths of all four triangles are in the proportion 2 : 3 : 7 : 2 + 3 + 7. Therefore, the area of the larger triangle, by similarity, must be . Notice how similarity enables us to completely avoid invoking any triangle area formula?

If you’d like more practice using similarity, try problem 16 on the 2006 AMC 12A, problem 8 on the 2000 AIME I, and problem 8 on the 2011 AIME I.

As one last illustration of similarity, here is a beautiful proof of the Pythagorean theorem which I learned from John Conway. Let be a right triangle. Suppose we erect similar figures on the sides of in such a way that the 3 sides of the triangle are corresponding parts. Then the proportions of the areas of the 3 similar figures would be in proportion to the squares of the lengths of the sides of . Therefore, the Pythagorean theorem is equivalent to showing that the sum of the areas of the two smaller figures is equal to that of the third. A typical illustration of the Pythagorean theorem erects 3 squares on the sides.

But, instead of squares, draw in the altitude to the hypotenuse as above. Do you see the 3 similar right triangles whose hypotenuses are the 3 sides of the original triangle? And since the bigger one is made up exactly of the smaller two, its area most certainly is the sum of the areas of the smaller two, therefore, the Pythagorean theorem is true!