Friday, June 13, 2014

Geometry Problem 1022: Circular Sector of 90 Degrees, Squares, Metric Relations

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1022.

Online Math: Geometry Problem 1022: Circular Sector of 90 Degrees, Squares, Metric Relations

2 comments:

  1. Let OA=R. Let S(ABC) be the area of ABC.

    OD² + OG²
    = (AC² + R² − 2×AC×R cos∠OCD) + (BC² + R² − 2×BC×R cos∠OCG)
    = (AC² + R² + 2×AC×R sin∠OCA) + (BC² + R² + 2×BC×R sin∠OCB)
    = AC² + BC² + 2R² + 2R (AC sin∠OCA + BC sin∠OCB)
    = AC² + BC² + 2R² + 4 [S(OAC) + S(OBC)]
    = AC² + BC² + 2R² + 4 [S(OAB) + S(ABC)]
    = AC² + BC² + 2R² + 2R² + 4×S(ABC)
    = AC² + BC² + 4R² + 2×AC×BC sin∠ACB
    = AC² + BC² + 4R² + 2×AC×BC sin135°
    = AC² + BC² + 4R² − 2×AC×BC cos135°
    = AC² + BC² + 4R² − 2×AC×BC cos∠ACB
    = AB² + 4R²
    = 3 AB²

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  2. We use complex numbers.
    Let C = cos a + isin a be a point on the unit circle. Then A = 0 + 1i and B = 1 + 0i.
    Then AB² = 2, so we have to prove that OD² + OG² = 6.

    D = C -i(A - C) = cos a - sin a + 1 + i(coa s + sin a).
    G = C + i(B - C) = cos a - sin a + i(sin a - cos a + 1).

    OD² + OG² = (cos a - sin a + 1)² + (cos a + sin a)² + (cos a - sin a)² + (sin a - cos a + 1)²
    = 4(cos²a + sin²a) + 2
    = 6

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