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 (b2βa2)abcβ. (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
21βc(b2βa2abcβ)
Taking a:b=40:41 and c=9, we get an answer of 820β.
One way to proceed is with coordinates: Let A=(0,0),B=(c,0), and C=(x,y). Then ACBCβ=baβ becomes
x2+y2β(xβc)2+y2ββ=baβ
Squaring both sides and rearranging terms leads to
(b2βa2)x2β2b2cx+(b2βa2)y2=βb2c2
Completing the square then gives
(xβb2βa2b2cβ)2+y2=(b2βa2)2a2b2c2β
Hence the set of all vertices C satisfying the conditions of the problem is the circle of center O=(b2βa2b2cβ,0) and radius b2βa2abcβ.
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)acβ and that BL=(bβa)acβ. Therefore the radius of the circle is (b2βa2)abcβ, which serves as the altitude of the triangle ABC of maximal area. When c=9 and b:a=41:40, this area is 820β.
where equality holds if and only if B=2Οβ+A. Note that B>A follows from b>a. Using the trigonometric identity sin(B+A)sin(BβA)=sin2Bβsin2A and then the Law of Sines again, we have
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