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Inscribed Circles and Hexagons | 7 Hard Geometry Problems

Updated on October 18, 2014
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TR Smith is a product designer and former teacher who uses math in her work every day.

Geometry problems involving circles inscribed in regular hexagons and regular hexagons inscribed in circles can be solved the same way as problems involving circles and equilateral triangles. The key geometric properties are that a regular hexagon can be divided into 6 equilateral triangles, and that the radius of a circle that circumscribes a regular hexagon is equal to the side length of the hexagon. Other useful math facts are the properties of 30-60-90-degree right triangles and their side length ratios. The hypotenuse is twice the length of the shorter leg, and the longer leg is sqrt(3) times the shorter leg.

One last geometric property of circles is that if a circle is tangent to a line M, and L is a line segment connecting the center of the circle to the point of tangency, then M and L are perpendicular. Knowing these mathematical and geometric properties, you can solve each of the seven challenging problems below. (See also, Inscribed Squares and Circles)


(1) Circle Area Ratio Problem with Inscribed Hexagon

Problem: A regular hexagon is inscribed in a circle. Within the hexagon is inscribed a smaller circle. What is the ratio of the area of the larger circle to the smaller circle?

Solution: The key is to notice that the radius of the larger circle is equal to the side length of hexagon, and that if you divide the hexagon into six equilateral triangles, the altitude of each triangle is the radius of the smaller circle.

If the radius of the larger circle and side length of the hexagon is R, then the radius of the smaller circle and altitude of the triangle is R*sqrt(3)/2. The areas of the larger and smaller circles are then

larger area = pi * R^2

smaller area = pi * R^2 * 3/4

By dividing the larger area by the smaller area you get a ratio of 1/[3/4] or 4:3.


(2) Seven Equal Circles Arranged in a Hexagon

Problem: Seven circles with a radius of 1 are arranged in a hexagonal formation. What is the area of the hexagon formed by the centers of the six outer circles?

Solution: The side length of the hexagon formed by the centers of the outer circles is equal to twice the radius of each circle, so the side length is 2. Knowing the side length we can calculate the area.

Since the area of a regular hexagon with a side length of X is equal to [3*sqrt(3)/4]*X^2, the area of this hexagon is

[3*sqrt(3)/4]*2^2
= [3*sqrt(3)/4]*4
= 3*sqrt(3)
≈ 5.196


(3) Overlapping Circle and Hexagon

Problem: Consider a regular hexagon with a side length of 6. Centered at one of the vertices is a circle of radius 6. Inscribed in the space where the two figures overlap is a smaller circle. This smaller circle is tangent to two adjacent sides of the hexagon and the larger circle. What is the radius of this smaller circle?

Solution: Mark the center of the smaller circle and call its radius R. In the diagram, R is the length of the two black line segments that start from the center of the smaller circle and end at points on the circumference.

With the red line segment, one of the black lines forms the altitude of a 30-60-90-degree triangle shaded in green, with the red line segment as the hypotenuse. Using properties of equilateral triangles, we can see that the red line segment is 2/sqrt(3) times the length of R.

Looking at the horizontal line, the total length of the other black line segment plus the red line segment is equal to the radius of the larger circle, 6. Thus we have the equation

R + [2/sqrt(3)]*R = 6

Solving for R gives you the solution R = 6/[1 + 2/sqrt(3)], which simplifies to 12*sqrt(3) - 18 ≈ 2.7846.


(4) Inscribed Circles in a Hexagon

Problem: A regular hexagon is circumscribed around seven equally sized circles, each with a radius of 1, as shown in the diagram above. What is the side length of the outer hexagon?

Solution: Consider just the top two circles and the top border of the regular hexagon. The red lines shown in the next diagram are radii and have a length of 1. Together with the green and purple line segments they can form the altitudes of smaller 30-60-90-degree triangles.

The hypotenuse of the triangle (green line segment) has a length of 2/sqrt(3). The shorter purple line segment is half as long, with a length of 1/sqrt(3).

The total length of the top edge of the hexagon is two purples plus two reds, so its total length is

2*[1/sqrt(3)] + 2*1
= 2/sqrt(3) + 2
≈ 3.1547


(5) Inscribed Hexagon and Circle

Problem: Consider a regular hexagon with a side length of 18. Inside this hexagon is a smaller hexagon with a side length of 12, situated so that it shares a vertex and two sides with the larger hexagon. Also inside the larger hexagon is a circle, tangent to the sides of the larger hexagon, and touching the other hexagon at a vertex. See image above. What is the radius of this circle?

Solution: Draw a line connecting opposite vertices of the larger hexagon such that the line segment goes through the center of the smaller hexagon and the circle.

The total length of this line segment is 36. The length of the red portion is 24. Call the length of the green portion R since it is the radius of the circle. Since the yellow triangle in the corner is a 30-60-90-degree triangle, the length of the black portion is [2/sqrt(3)]*R. Thus we get the equation\

36 = 24 + R + [2/sqrt(3)]*R

12 = [1 + 2/sqrt(3)]*R

12/[1 + 2/sqrt(3)] = R

R = 24*sqrt(3) - 36 ≈ 5.5692


(6) Three Hexagons in a Circle

Problem: Three hexagons with a side length of 1 are arranged as in the figure above, with a circle enclosing them so that the circle passes through the three outermost vertices of the hexagonal arrangement. What is the radius of the circle?

Solution: This problem does not require analysis of 30-60-90-degree triangles, but can be solved just by drawing some extra lines in the diagram.

If you extend six of the blue line segments to the circumference of the circle with the black line segments shown, you will form a larger regular hexagon inscribed within the circle. The length of the black line segments are all equal to 1, the side lengths of the smaller blue hexagons.

From here you can see that the side length of the larger hexagon is 1 + 1 = 2, thus the radius of the circle is also 2.


(7) Three Circles Inscribed in a Hexagon

Problem: Three equally sized circles are inscribed in a regular hexagon such that each circle is tangent to a pair of adjacent sides of the hexagon and tangent to two other circles. See figure above. What is the ratio of the perimeter of the hexagon to the combined perimeter (circumferences) of the three circles?

Solution: The ratio will be the same regardless of the measurements of the hexagon's side length or the circles' radius. Let us assume each circle has a radius of 1 and use this as a starting point to find the side length of the hexagon.

The length of each of the two parallel black line segments is twice the diameter of a circle, so its length is 4. Half its length is the diameter of one circle, so its half-length is 2.

The half-length of the upper black line segment forms one leg of a 30-60-90-degree triangle. This means the length of the green line segment is 2/sqrt(3), and the length of the blue line segment is 4/sqrt(3). This is the side length of the outer hexagon.

The total of the three circles' circumferences is 3*(2*pi*1) = 6*pi. The perimeter of the hexagon is 6*(4/sqrt(3)) = 8*sqrt(3). Therefore, the ratio of the hexagonal boundary to the circular boundary is

[8*sqrt(3)] / [6*pi]

= [4*sqrt(3)] / [3*pi]

= 12 / [3*sqrt(3)*pi]

= 4 / [sqrt(3)*pi]

≈ 4 : 5.4414

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    • snerfu profile image

      Vivian Sudhir 2 years ago from Madurai, India

      Very well done calculus-geometry. Good effort, Voted up and tweeted.

    • FitnezzJim profile image

      FitnezzJim 2 years ago from Fredericksburg, Virginia

      So, can these techniques be combined to figure out how to trisect an arbitrary line? :)

    • profile image

      Charitie 12 months ago

      Hi, I'm trying to figure out this problem: The regular hexagon, square, and circle in the image above have the same perimeter of 34.82 cm. Which shape has the largest area? Which shape has the smallest area? ( I guess you can imagine what the "image above" looks like.) Thank you.

    • calculus-geometry profile image
      Author

      TR Smith 12 months ago from Eastern Europe

      Hi Charitie,

      If several shapes all have the same perimeter, the one with the largest area is the one that is the 'most circular.' A circle is the most circular of all, so it will always have a larger area than any other shape with the same perimeter. Among the three, the square is the least circular so it will have the smallest area.

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