AcademiaAgriculture & FarmingHumanitiesSocial SciencesSTEM

Inscribed Squares and Triangles: Challenging Geometry Problems With Solutions

Updated on November 29, 2016
calculus-geometry profile image

TR Smith is a teacher and creator who uses mathematics in her line of work every day.

Abstract geometry problems involving squares inscribed in triangles and triangles inscribed in squares frequently appear on standardized tests such as the GMAT, GRE, SAT, ACT, IIT-JEE, and AIEEE. These questions are meant to test students' ability to apply basic math knowledge to real-world situations, recognize important patterns, and come up with a logical sequence of steps to solve a math or engineering problem in real life. Here are some examples of classic and hard geometry problems with squares and triangles nested inside one another or adjacent to each other.

(1) Equilateral Triangle Inside a Square

The diagram above shows an equilateral triangle inscribed within a square such that one of the triangle's vertices coincides with a corner of the square, while the other two triangular vertices lie on the square's sides. The triangle is symmetrically angled in such a way that it creates the triangles colored in yellow and red. The two yellow triangles are identical. What is the ratio of the area of the red triangle (A) to the area of one of the yellow triangles (B)?

Solution: It doesn't matter how big the surrounding square is, so suppose its side length is 1. The red triangle (A) is an isosceles right triangle. Let's say its legs have length X. In that case, the yellow triangle is a right triangle with leg lengths 1 and 1 - X. Both triangles have the same hypotenuse since the blue triangle is equilateral.

The area of the red triangle is (1/2)X^2 and the area of the yellow triangle is (1/2)(1 - X). The ratio of the red to yellow is (X^2)/(1 - X).

The fact that their hypotenuses are equal implies that X^2+ X^2 = 1^2 + (1-X)^2, an equation that simplifies to X^2 = 2(1 - X). Plugging this into the ratio expression above gives us (X^2)/(1 - X) = 2(1 - X)/(1- X) = 2. Therefore, the red triangle is twice as big as a yellow triangle.

(2) Triangle-Square House Inscribed in a Circle

The triangle and square in the diagram above have side lengths equal to 1. What is the radius of the circle that circumscribes the house shape?

Solution: There are two ways to solve this problem: with an elegant pictorial solution and with algebra. First we work out the boring algebraic solution.

The center of the circle is inside the square somewhere along the house's line of symmetry. Any line segment from the circumference to the center is a radius with length R. Suppose that the center of the circle is X units below the triangle. The height of the triangle is sqrt(3)/2, therefore the distance from the top of the house to the center of the circle is

R = sqrt(3)/2 + X

The distance from the bottom corner of the house can be found as the hypotenuse of a right triangle with side lengths 1/2 and 1 - X. This gives

R^2 = 5/4 - 2X + X^2

Combining the two equations for R gives us

[sqrt(3)/2 + X]^2 = 5/4 - 2X + X^2
3/4 + sqrt(3)*X = 5/4 - 2X
1/2 = (2 + sqrt(3))X
X = 1/(4 + 2*sqrt(3))
X = 1 - sqrt(3)/2

Using this value of X we get R = 1.

Now for the pictorial solution. Since the height of the square is 1 and the height of the triangle is sqrt(3)/2, the total height of the house is 1 + sqrt(3)/2. Draw a triangle with a side length of 1 at the base of the house as in the diagram below. The distance between the top of this new triangle and the top of the house is 1.

The line segment descending from the top of the house lies along a diameter and has a length of 1. It intersects a line segment of the same length that also has one end on the circumference of the circle. Therefore, this other line segment lies on a diameter and the point where they meet is the center of the circle. In other words, these line segments are radii, so the radius of the circle is 1. (Yet another pictorial proof uses the fact that the top vertex of the house and the bottom two vertices are among the vertices of a regular dodecagon inscribed in the circle.)

(3) Square Inside Equilateral Triangle

In the diagram above, the large outer equilateral triangle has a side length of 1. What is the side length of the square inscribed within it?

Solution: Suppose the side length of the square is T. The top left corner of the square lies along the left side of the surrounding triangle and partitions it into two segments of unequal length. Using properties of 30-60-90 triangles, the length of the lower segment is T*2/sqrt(3). The length of the upper segment is T. Therefore, the total side length of the surrounding triangle is T*2/sqrt(3) + T.

From the statement of the problem we know the side length of the surrounding triangle is also 1. This gives us the equation 1 = T*2/sqrt(3) + T. Finally, when we solve this for T we get T = 2*sqrt(3) - 3 ≈ 0.4641.

(4) Fitting Two Equilateral Triangles in a Square

The squares shown in the diagram above each have a side length of 1. Inscribed in each square is a pair of equal-sized equilateral triangles, but one square has a bigger pair than the other. Which square has the larger triangles?

Solution: The triangles with the longer side length (and therefore larger area) are in the first square. To see this, first notice that the altitudes of the triangles in the first square lie along the square's diagonal. If the square has a side length of 1, then the altitude of each triangle is sqrt(2)/2 = 1/sqrt(2). The side length of each triangle is [2/sqrt(3)]*[1/sqrt(2)] = sqrt(2/3), or approximately 0.8165. See solution diagram below.

Now for the second square, look at the red and pink line segments and their lengths in the solution diagram above. The side length of the green triangle (and equivalently, the blue triangle) is equal to the sum of 1/2 - 1/(2*sqrt(3)) and 1/sqrt(3), which equals 1/2 + 1/(2*sqrt(3). This is approximately 0.7887.

(5) Tilted Square in an Equlilateral Triangle

A square is inscribed in an equilateral triangle so that the diagonal of the square lies along the altitude of the triangle. If the triangle has a side length of 1, what is the side length of the square? How does this answer compare to the solution of Problem 3?

Solution: Call the side length of the square W. In the diagram below, you can see that the left corner of the square divides the left side of the triangle into two pieces with lengths sqrt(2)*W and sqrt(2/3)*W.

This gives us the equation 1 = sqrt(2)*W + sqrt(2/3)*W. Solving for W gives us W = (3/4)sqrt(2) - (1/4)sqrt(6), or W ≈ 0.4482. The tilted square inscribed in the triangle has a slightly smaller side length than the rectified square.


    0 of 8192 characters used
    Post Comment

    No comments yet.

    Click to Rate This Article