Problem:
Triangle ABC has AB=9 and BC:CA=40:41. What is the largest area that this triangle can have?
Solution:
Let AB=c,AC=br, and BC=ar, with a<b. We shall show that the locus of all such points C is a circle whose center is on line AB and whose radius is (b2−a2)abc. (This circle is called a circle of Apollonius.) The radius of the circle serves as the height of the triangle of maximal area, so the desired area is
21c(b2−a2abc)
Taking a:b=40:41 and c=9, we get an answer of 820.
One way to proceed is with coordinates: Let A=(0,0),B=(c,0), and C=(x,y). Then ACBC=ba becomes
x2+y2(x−c)2+y2=ba
Squaring both sides and rearranging terms leads to
(b2−a2)x2−2b2cx+(b2−a2)y2=−b2c2
Completing the square then gives
(x−b2−a2b2c)2+y2=(b2−a2)2a2b2c2
Hence the set of all vertices C satisfying the conditions of the problem is the circle of center O=(b2−a2b2c,0) and radius b2−a2abc.
OR
Assume a<b and let K and L satisfy AK:KB=b:a=AL : LB, with K on AB and L on the extension of AB through B. Extend AC through C to P. Because AC:CB=b:a also, the Angle Bisector Theorem implies that CK bisects angle ACB and CL bisects the exterior angle BCP. It follows that angle KCL is a right angle, so C lies on the circle that has KL as a diameter. It is straightforward to calculate that KB=(a+b)ac and that BL=(b−a)ac. Therefore the radius of the circle is (b2−a2)abc, which serves as the altitude of the triangle ABC of maximal area. When c=9 and b:a=41:40, this area is 820.
where equality holds if and only if B=2π+A. Note that B>A follows from b>a. Using the trigonometric identity sin(B+A)sin(B−A)=sin2B−sin2A and then the Law of Sines again, we have
[ABC]≤21c2sin2B−sin2AsinAsinB=21c2b2−a2ab
If c=9 and b:a=41:40, we see that the maximum possible area is 820, and that this maximum is attained when B−A=2π.
.jpg)