求解幾題數學問題

16.
解說圖:
http://imgur.com/RCL8tTU
Sol:
設圓O半徑為R, 故 OA = OC = R
設 AC = X
因為O,A,B,C皆在圓O'上,由托勒密定理得
18X + 6R = 37R
X = 31R / 18
X^2 = 961*R^2/324 .....(1式)

因為 ∠ABC , ∠AOC 對應相同的AC弧
故 圓周角∠ABC = 圓周角∠AOC
推得 cos∠ABC = cos∠AOC
由餘弦定理得
(6^2+37^2-X^2)/(2*6*37) = (R^2+R^2-X^2)/(2*R*R)
(1405-X^2)R^2 = 222(2R^2-X^2)
1405*R^2 - X^2*R^2 = 444*R^2 - 222*X^2
(1式)代入上式得
1405*R^2 - 961*R^4/324 = 444*R^2 - 222*961*R^2/324
等式左右同乘 324/R^2 , 等式恆成立
455220 - 961*R^2 = 143856 - 213342
R^2 = 524706/961 = 546
圓O面積 = π*R^2 = 546π
Ans: 546π

2.
解說圖:
http://imgur.com/WooYpYQ
Sol:
自BC做延長線,交DG於P
自AB做延長線,交DG於Q
因為DG平行於AC,故內錯角相等:
∠BAC = ∠BQP , ∠BCA = ∠BPQ
令∠BAC = ∠BQP = α , ∠BCA = ∠BPQ = β
設正方形ABDE邊長為a, 正方形BCFG邊長為b
因為 ΔABC 相似於 ΔQBP , 故得
α對應邊 / β對應邊 = BC / AB = BP / BQ
b / a = b*cotβ / a*cotα
cotα = cotβ
故得證 α = β
( 因為α,β皆在0~π的區域,且cot函數在此區域為1對1函數 )


2014-03-26 18:04:56 補充：
因為α = β , 所以 AB = BC

參考下面的網址看看

http://phi008780520.pixnet.net/blog

(1) r=O'C=?; R=OC=?; Area(OC)=? x=OAB, 2x=OO'B => sin(x)=9/r, cos(x)=√(r^2-81)/ry=BAC, 2y=BO'C => sin(y)=3/r, cos(y)=√(r^2-9)/rΔABO': AO'B=4x+2y => 37/2r=sin(2x+y)=sin(2x)*cos(y)+sin(y)*cos(2x)=2sin(x)*cos(x)*cos(y)+sin(x)*[1-2*sin^2(x)]={18√[(r^2-81)(r^2-9)]+3(r^2-2*81)}/r^3=> 0=r^2(335r^2-176904)r=22.979809=>22.98x=asin(9/r)=23.057(deg)y=asin(3/r)=7.50134(deg)2(x+y)=61.1169(deg)ΔACO': AO'C=4x+4yAC=2*r*sin(2x+2y)=40.24托勒密定理: R=OA=OCOB*AC+BC*OA=AB*OC => R(AB-BC)=OB*ACR=18*40.24/(37-6)=23.37Area=pi*R^2=1715.8(長度^2)......ans


2014-03-26 15:41:58 補充：
(2) 座標化:

A=(0,0), B=(c, 0), C=(c*cosA,c*sinA)

D=C+[c*cos(90+A),c*sin(90+A)]

=(c*cosA-c*sinA,c*sinA+c*cosA)

G=C+[a*cos(90-C),a*sin(90-C)]

=(c*cosA+a*sinC,c*sinA+a*cosC)

Slope(DG)={c*sinA+c*cosA-c*sinA-a*cosC}/分母

=(c*cosA-a*cosC)/分母

=0(水平)

=> c*cosA=a*cosC

So c=a, A=C.........ans

2014-03-27 09:53:52 補充：
圖形在

https://www.facebook.com/