Printable View
Hello everyone, pls help me with this problem. Thanks in advance!
Three unit circles are arranged side by side in a row. A is the point on the first circle
that is collinear with the centers of the three circles which does not touch the second circle. The lines AB and AC are tangent
to the third circle at the points B and C. Find the area of ABC.
Let O be the center of the third circle. Triangle ABO is right. From it you can find AB and angle BAO, which is half of angle BAC.
Hello, shiny718!
Quote:
Three unit circles are arranged side by side in a row.
A is the point on the first circle that is collinear with the centers of the three circles
. . and does not touch the second circle.
The lines AB and AC are tangent to the third circle at the points B and C.
Find the area of ABC.
Code:* * * * * * B * * *
* * * * o *
* * * * * *
* * * * * *
* * * *
A o * * * * o *
* * * P *
* * * * *
* * * * * *
* * * * o *
* * * * * * C * * *
. .
. . . . . . . . . . . .
Edit: too slow . . . again!
A geometric solution ( reference Soroban's solution)
Let D = intersection of BC and AO
BO =1 AO =5 AB= rad 24
Triangle ABO is a rt triangle @ B. BD is the altitude on AO
let DO = x AD = 5-x
AB^2 /BO^2 = 5-x/x
24/1 = 5-x /x whence x=1/5 and 5-x= 24/5
BD^2 = 24/5 *1/5
BD = rad 24/5
area of ABC = BD * AD = rad24 *24/25
The relationships of sides and hypot segments of a rt triangle created by BD and that of the segments to the altitude are standard theorems