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 Smile! "It is impossible for a man to learn what he thinks he already knows" -- Epictete Math Shortcuts The radii of circumscribed (R) and inscribed (r) circles within regular polygons of n sides, each of length x, are given by: (x/2) csc(180°/n) and (x/2) cot(180°/n)
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# Previous Puzzles of the Month + Solutions

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Puzzle # 125

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A mathematical shield

Once upon a time, Mars, the God of war, intended to test the IQ of the goddess Minerva. So, showing his shield, he told her: "Darling, on my shield, there are 3 equal circles which represent the qualities of the warrior: strength, flexibility, and decisiveness. As you can see, one of the circles has been scratched by a sword, and the resulting score is three inches long (line AB in the illustration). Can you then tell me what is the area of my shield?".
Find the very shortest way to solve this puzzle and use only basic geometry, trigonometry is not allowed!

Difficulty level: , basic geometry knowledge.
Category: Geometrical puzzle.
Keywords: inscribed circles, incircles.
Related puzzles:
- Red monad,
- Achtung Minen!

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Source of the puzzle:
© G. Sarcone.
You cannot reproduce any part of this page without prior written permission.
 The centers of the small green circles are equidistant from each other, and thus form the vertices of an equilateral triangle with midpoints A, B and C (see image below). It follows that the small triangles ABC and AC'B are also equilateral and identical to each other. The radii r of the small circles are congruent to the sides of the triangles ABC and AC'B, then r = C'B = AB. As shown in the image, the radius R of the large circle can be calculated by adding r + h + NM together. We can calculate the height h of the triangle AC'B by multiplying its side AB by √3/2: h = (AB√3)/2 = (3√3)/2 Since the heights of an equilateral triangle meet at a point (here, point M) that is two thirds of the distance from the vertex of the triangle to the base, we can also calculate MN with this simple formula: MN = h x 1/3 = (3√3)/2 x 1/3 = (3√3)/2 x 3 = √3/2 Therefore, the area of large yellow circle (which represents the shield) is: π[3 + (3√3)/2 + √3)/2]2 = π(3 + 2√3)2 ≈ 131.27 square inches The 5 Winners of the Puzzle of the Month are: John Pelot, USA - Benoît Humbert, France - Amedeo Squeglia, Italy - Paritosh Singh, India - Walter Jacobs, Belgium Congratulations!
 Math fact behind the puzzle Properties of the equilateral triangle An equilateral triangle is simply a specific case of a regular polygon, in this case with 3 sides. Equilateral triangles are triangles in which all sides are equal, and all angles are equal as well and each of them measures 60Å. With an equilateral triangle, the radius of the incircle is exactly half the radius of the circumcircle. © 2006 G. Sarcone, www.archimedes-lab.org You can re-use content from Archimedes’ Lab on the ONLY condition that you provide credit to the authors (© G. Sarcone and/or M.-J. Waeber) and a link back to our site. You CANNOT reproduce the content of this page for commercial purposes. You're encouraged to expand and/or improve this article. Send your comments, feedback or suggestions to Gianni A. Sarcone. Thanks!
 Previous puzzles of the month...
 Apr-May 2010: A chopping problem Oct-Nov 09: The Mark of Zorro July-Sept 09: radiolarian's shell May-June 09: circle vs square Jan-Feb 09: geometric mouse Sept-Oct 08: perpendicular or not... July-Aug 08: ratio of triangles May-June 08: geometry of the bees Febr-March 08: parrot sequence... Dec 07-Jan 08: probable birthdates? Oct-Nov 2007: infinite beetle path Aug-Sept 07: indecisive triangle June-July 07: Achtung Minen! April-May 07: soccer balls Febr-March 07: prof Gibbus' angle Jan 07: triangles to square Aug-Sept 2006: balance problem June-July 06: squared strip Apr-May 06: intriguing probabilities Febr-March 06: cows & chickens Dec 05-Jan 06: red monad Sept-Oct 2005: magic star Puzzle Archive
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