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.
Friday, June 13, 2014
Geometry Problem 1022: Circular Sector of 90 Degrees, Squares, Metric Relations
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Let OA=R. Let S(ABC) be the area of ABC.
ReplyDeleteOD² + 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
We use complex numbers.
ReplyDeleteLet 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
Que bueno! Sois unos genios
ReplyDeleteLet AC = a, BC =b and OA = R
ReplyDeleteLet p be the height of the altitude from O to AC, q from O to BC
OD2 + OG2
= {(p+a)2 + a2/4 } + {q+b)2 +b2/4}
= {R2 + 2ap + a2} + (R2 + 2bq + b2}
= 2R2 + a2 + b2 + 2(ap + bq)
= 2R2 + (2R2 – sqrt2.ab) +2{2S(OAC) + 2(OBC)}
= 4R2 – sqrt2.ab + 4{R2/2 + S(ABC)}
= 6R2 – sqrt2.ab + 4 X (1/2)/(sqrt2).ab
= 6R2
= 3AB2
Sumith Peiris
Moratuwa
Sri Lanka