A triangle has side lengths $7,$ $8$, and $9.$ There are exactly two lines that simultaneously bisect the perimeter and area of the triangle. Let $\theta$ be the acute angle between these two lines. Find $\tan \theta.$
+128474 
Let the triangle be ABC
We can find the area of ABC thusly
Semi-perimeter = (9 + 8 + 7) / 2 = 12
Area = sqrt [ 12 (12 -9)(12 -8) (12-7) ] = sqrt [ 12 * 3 * 4 * 5 ] = 12sqrt (5)
The height can be figured as
12sqrt (5) = (1/2) (8) (h)
(3)sqrt (5) = h
Line CD divides the perimeter in half
Line CE divides the area in half
CF = the height
Angle EDC is the acute angle (theta) between these two segments
angle EDC = angle DCF - angle ECF
tan DCF = DF / CF = 3 / ( 3sqrt (5)) = 1/sqrt (5)
tan ECF = Ef/ CF = 2 / (3sqrt (5) )
tan (EDC) = tan [ DCF - ECF] = [ tan DCF - tan ECF ] / [ 1 + tan(DCF)* tan(ECF] ] =
tan (EDC) = [ 1/sqrt (5) - 2/ (3sqrt 5 ) ] / [ 1 + (1/sqrt 5)(2/(3sqrt 5)) ] =
[ 1 / (3sqrt 5 ] / [ 1 + 2 / 15] =
15 / [ 17 * 3 sqrt 5 ] =
15 / (51*sqrt 5 ) =
(5 / (17*sqrt 5) =
sqrt (5) / 17
