數學mc

天才A

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

[H]bunbunbunbun

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

天才A

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....

火中有火

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?)

火中有火

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

天才A

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...