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標題: 數學mc [打印本頁]

作者: 天才A 時間: 3-1-2012 19:23
標題: 數學mc
2001 年CE MC 第54題
the coordinates of Q are (0,4)
the coordinates of P are (3,0)

The inscribed circle of triangle OPQ touches PQ at R.
Find the coordinates of R.

答案係 (9/5 , 8/5)

我一開始唔果時唔明..
結果去維基搵多左少少應該唔洗識既野
跟住我先計到...
但係我諗應該冇可能咁樣
應該有個簡單既方法...

我既方法:
搵左個circle 既 radius 先
by 維基
r= (2* area of triangle OPQ)/(perimeter)
即係 r= (2*3*4/2)/(3+4+5)= 1
之後考慮直角triangle IRP (I係個內接圓中心)
角RPI = 1/2角QPO (角平分線)
之後用Sin law 搵RP
5-RP=QR
搵QR同RP既比
最後用比例搵返粒點既座標

作者: [H]bunbunbunbun 時間: 6-1-2012 04:17
本帖最後由 [H]bunbunbunbun 於 5-1-2012 20:18 編輯

well u know radius is 1 la
since tangents from a point are of equal length
R dissect QR 3:2
then coordinates of R = (2xQ+3xR)/5

作者: 天才A 時間: 6-1-2012 20:48

well u know radius is 1 la
since tangents from a point are of equal length
R dissect QR 3:2
then coo ...
[H]bunbunbunbun 發表於 6-1-2012 04:17

但係個1 理論上唔應該咁搵...

應該用tangent property去搵....
應該話
成題都係用tangent property去搵....

作者: 火中有火 時間: 1-2-2012 17:11
先係radius...唔使樓主1樓個方法咁辛苦...

再來...
(x-1)^2+(y-1)^2=1
and
x/3+y/4=1

~~(咁睇係咪快+爽先?~~)

作者: 火中有火 時間: 1-2-2012 17:13

well u know radius is 1 la
since tangents from a point are of equal length
R dissect QR 3:2
then coo ...
[H]bunbunbunbun 發表於 6-1-2012 04:17

好想知你點睇個3:2的比例出來

作者: 天才A 時間: 1-2-2012 17:26

好想知你點睇個3:2的比例出來
火中有火發表於 1-2-2012 17:13

都係用 tangent property 搵返出來...

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