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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)
•••  ••• •••  # Previous Puzzles of the Month + Solutions Back to Puzzle-of-the-Month page | Home  Puzzle # 125  Suggest this page to a friend Contact us Thumb up this page Share it on FaceBook Follow us on Twitter View our RSS feeds Digg this story 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:
- Achtung Minen! Italiano Français

Source of the puzzle:  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!  Previous puzzles of the month...   Back to Puzzle-of-the-Month page | Home  Information Services & Products Follow us via... Support us... • About Us • Privacy & Terms • Copyrights • Contact us • Sitemap • Press Review • Products • Features • Workshops • For Publishers • Facebook • Newsletter • RSS feeds • Twitter • Blogs • Tell a Friend • Merchandising • Link to us • Sponsorship © Archimedes' Laboratory™ | The web's best resource for puzzling and mental activitie | Introduzione | Introduction | Einführung   