In the diagram, four circles of radius 1 with centres P, Q, R, and S are tangent to one another and to the sides of triangle ABC, as shown. What is the area of triangle ABC?
Triangle PQS is similar to triangle ABC
PRS is equilateral with a height of sqrt 3
Dropping a perpendicular to BC from P will be tangent to the circles R and S
Call the point of tangency of this segment T
Triangle QTP is right with QT = 3 TP = sqrt 3
And QP = sqrt (3^2 + (sqrt 3)^2) = sqrt 12
Using the Law of Cosines
QS^2 = QP^2 + PS^2 - 2(QP * PS) cos (QPS)
4^2 = 12 + 4 - 2(QP * PS)cos (QPS)
0 = - 2 (QP * PS) cos (QPS)
cos (QPS) = 0
So angle QPS is right
And sin (PQS) = PS/ QS = 2/4 = 1/2
So angle (PQS) = 30 ° = angle ABC
So angle (PSQ) = 60° = angle ACB
The area of PQS = PS * PQ / 2 = 2(sqrt 12) / 2 = sqrt (12)
Drop a perpendicular from Q to BC and call the intersection U
The triangle BUQ is right and we can find BU as follows
QU / sin (1/2 angle ABC) = BU /sin ( BQU)
1/ sin (15) = BQ/ sin (75)
BQ = sin 75 / sin 15 = cos 15/ sin 15 = cot 15 = 2 + sqrt 3
Similarly......drop a perpendicular from S to BC and call the intersection point V
Using a similar proceedure for finding BU, we can find VC as sqrt 3
So...... BC = BU + QR + RS + VC = 2 + sqrt 3 + 2 + 2 + sqrt 3 = 6 + 2 sqrt 3
The scale factor from ABC to PQS = (BC / QS) = [ 6+ 2 sqrt 3 ] / 4
Then the area of ABC = (scale factor)^2 (area of PQS) =
[ (6 + 2sqrt 3) / 4]^2 * sqrt (12) ≈
19.39 units^2