數學知識交流 － Geometric problem (1)

Let me trytry...

2012-05-29 21:06:53 補充：


圖片參考：http://imgcld.yimg.com/8/n/HA00016323/o/701205290049813873409750.jpg

http://i1099.photobucket.com/albums/g395/jasoncube/f77d25ca.png ∵ OA₁ = OA₂ = OB₁ = OB₂ = OC₁ = OC₂∴ We can easily prove that AB₂ = AC₁, BC₂ = BA₁, CA₂ = CB₁Let A₁A₂ = B₁B₂ = C₁C₂ = x, BC₂ = BA₁ = y. AB₂ + B₂B₁ + B₁C = 5 ⇒ AC₁ + x + CA₂ = 5⇒ (3 - x - y) + x + (4 - x - y) = 5 ⇒ y = 1 - x/2 --- (1) ∆AB₂C₁ + ∆A₁BC₂ + ∆A₂B₁C + 4 = ∆ABC⇒ ½(3 - x- y)²sin∠CAB + ½y² + ½(4 - x - y)²sin∠ACB + 4 = ½(3×4)⇒ (3 - x - y)²(4/5) + y² + (4 - x - y)²(3/5) + 8 = 12Sub. (1)⇒ ⅘(3 - x - 1 + x/2)² + (1 - x/2)² + ⅗(4 - x- 1 + x/2)² + 8 = 12⇒ ⅘(2 - x/2)² + (1 - x/2)² + ⅗(3 -x/2)² = 4⇒ ⅘(4 - 2x + x²/4) + (1 - x + x²/4) + ⅗(9 -3x + x²/4) =4⇒ (⅕ + ¼ + 3/20)x² - (8/5 + 1 + 9/5)x + (16/5+ 1 + 27/5) = 4⇒ ⅗x² - (22/5)x + 28/5 = 0⇒ x = {22/5 ±√[(22/5)² - 4(⅗)(28/5)]}/[2(⅗)]⇒ x = [22/5 ±√(148/25)]/(6/5)⇒ x = (22/5 ± ⅕√37)/(6/5)⇒ x = (22 ± √37)/6∵ (22 + √37)/6 > 3 (rej.)∴ x = (22 - √37)/6
FINISH!

2012-05-29 21:26:47 補充：
CORRECTION:

Last 4 lines:

⇒ x = (22/5 ± ⅖√37)/(6/5)

⇒ x = (11 ± √37)/3

∵ (11 + √37)3 > 3 (rej.)

∴ x = (11 - √37)/3