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Use dodecagons to deduce an inequality about π

by RoRi

By considering dodecagons inscribed and circumscribed about a unit disk, establish the inequalities

    \[ 3 < \pi < 12(2-\sqrt{3}). \]

First, we draw some pictures of the situation for reference.

Rendered by QuickLaTeX.com

( Note: I don’t know a way to do this without using trig functions, which haven’t been introduced in the text yet. If you have an alternative approach without them, please leave a comment and let us know about it. )

Since these are dodecagons, the angle at the origin of the circle of each triangular sector is 2 \pi / 12 = \pi/6, and the angle of the right triangles formed by splitting each of these sectors in half (shown in the diagrams) is then \pi/12. Then we use the fact that

    \[ \tan \left( \frac{\pi}{12} \right) = 2 - \sqrt{3}, \]

    \[ \sin \left( \frac{\pi}{12} \right) = \frac{\sqrt{3} - 1}{2 \sqrt{2}}, \]

    \[ \cos \left( \frac{\pi}{12} \right) = \frac{\sqrt{3} + 1}{2 \sqrt{2}}. \]

Now, for the circumscribed dodecagon we have the area of the right triangle T with base 1 in the diagram on the left given by

    \[a(T) = \frac{1}{2} bh = \frac{1}{2} \cdot 1 \cdot (2 - \sqrt{3}) = 1 - \frac{\sqrt{3}}{2}. \]

Since there are 24 such triangles in the dodecahedron, we then have the area of the circumscribed dodecahedron D_c given by

    \[ a(D_c) = 24 \left(1 - \frac{\sqrt{3}}{2} \right) = 12 (2 - \sqrt{3}). \]

For the inscribed dodecagon we consider the right triangle T with hypotenuse 1 in the diagram. The length of one of the legs is then given by \sin \left( \frac{\pi}{12} \right) = \frac{\sqrt{3} - 1}{2} and the other is given by \cos \left( \frac{\pi}{12} \right). So the area of the triangle is

    \[ a(T) = \frac{1}{2} bh = \frac{1}{2} \cdot \frac{\sqrt{3}-1}{2 \sqrt{2}} \cdot \frac{\sqrt{3}+1}{2 \sqrt{2}} = \frac{2}{16} = \frac{1}{8}.\]

Since there are 24 such triangles in the inscribed dodecahedron, D_{i} we then have,

    \[ a(D_i) = 24 \cdot \frac{1}{8} = 3. \]

Since the area of the unit circle is, by definition, \pi, and it lies in between these two dodecahedrons, we have,

    \[ 3 < \pi < 12(2 - \sqrt{3}). \qquad \blacksquare \]

6 comments

  1. Anonymous says:

    No. The ABC case is similar to the 2nd (rightmost) image, above. Point B is the center of the circle. Segment BA is the solid line that’s shown and segment BC is another solid line of length 1 (not shown in the diagram), that’s a mirror image of BA but on the opposite side of the dashed line. If you draw a line from A to a point D on BC (where D falls between B and C such that BDA is a right angle), angle ABD is the same as the original angle ABC (since segment BD falls along the path of the original segment BC). The EFG case is different. EFG is similar to the 1st (leftmost) image, above. Point F is the center of the circle. Segment FE is the solid line that’s shown and segment FG is another solid line of length 1 (not shown in the diagram), that’s a mirror image of FE but on the opposite side of the dashed line. This time, though, we’re drawing a line between points E and G, such that the line is tangent to the circle (touching the end of the dashed line that’s shown at a new point H). EFH is half of the original EFG.

    • Anonymous says:

      Anyway if you do as I see (but resorting to trig) you get the same outcome for the area of the inner polygon:
      sin(15)*cos(15)*12 = 0.25*12 = 3.

      I omit dividing and multiplying by 2.

  2. rjpatrick says:

    Here’s an approach to Question 18 in Section 2.4 the uses only basic geometry:

    A dodecagon is a 12-sides polygon. Consider one of the 12 uniform triangles that make up a dodecagon inscribed in a unit circular disk. Let’s call it triangle ABC, where segments AB and BC are unit length (i.e., length of one) and angle ABC is 30 degrees (30 degrees = 360 degrees / 12). Draw a line through point A and perpendicular to segment BC. Call the intersection point on segment BC point D, where point D lies between points B and C. Consider triangle ABD. If angle ABC is 30 degrees, so is angle ABD. Angle BDA is 90 degrees, so angle BAD must be 60 degrees. For a 30-60-90 triangle, if the hypotenuse (segment AB) is length 1, the short leg (segment AD) must be length 1/2. Consider again triangle ABC, where segment AD is the height and segment BC is the base. The area of that triangle is (1/2)*1*(1/2) = 1/4. The area of 12 of those triangles (the inscribed dodecagon) is 12*(1/4) = 3. That proves pi > 3.

    Now consider the dodecagon circumscribed about the unit circular disk. Consider one of the 12 uniform triangles that make up that larger dodecagon. Let’s call it triangle EFG, where segments EF and FG are equal length (and extend past the edge of the unit circular disk) and angle EFG is 30 degrees (30 degrees = 360/12). Draw a point H that bisects segment EG. Segment FH is unit length (i.e., length of one). Consider triangle EFH. Angle EFH is 15 degrees (30 degrees / 2 = 15). Angle FHE is 90 degrees, so angle FEH is 75 degrees. For a 15-75-90 triangle, if the long leg (segment FH) is length 1, the short leg (segment EH) must be length 2-SQRT(3). Consider again triangle EFG. If segment EH is length 2-SQRT(3), segment EG must be twice that length, or 4-2*SQRT(3). If segment EG is the base and segment FH is the height, then the area of that triangle is 1/2*(4-2*SQRT(3))*1 = 2-SQRT(3). The area of 12 of those triangles is 12*(2-SQRT(3)). This proves pi < 12*(2-SQRT(3)). Combing this with the earlier proof, we get 3 < pi < 12*(2-SQRT(3)), as desired.

    • Anonymous says:

      Thanks for providing an alternative solution.
      I do not understand why: “If angle ABC is 30 degrees, so is angle ABD”.

      Wouldn’t it be 15 degrees? Since you cut the angle ABC in half with the line through point A and perpendicular to segment BC? Like you did with EFH?

      • Anonymous says:

        No. The ABC case is similar to the 2nd (rightmost) image, above. Point B is the center of the circle. Segment BA is the solid line that’s shown and segment BC is another solid line of length 1 (not shown in the diagram), that’s a mirror image of BA but on the opposite side of the dashed line. If you draw a line from A to a point D on BC (where D falls between B and C such that BDA is a right angle), angle ABD is the same as the original angle ABC (since segment BD falls along the path of the original segment BC). The EFG case is different. EFG is similar to the 1st (leftmost) image, above. Point F is the center of the circle. Segment FE is the solid line that’s shown and segment FG is another solid line of length 1 (not shown in the diagram), that’s a mirror image of FE but on the opposite side of the dashed line. This time, though, we’re drawing a line between points E and G, such that the line is tangent to the circle (touching the end of the dashed line that’s shown at a new point H). EFH is half of the original EFG.

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