Triangle ABC has altitudes AD, BE, and CF. If AD = 12 and BE = 15 is a positive integer, then find the largest possible value of CF?
I have looked at similar problems to this but they are not correct. This needs to be solved relatively quickly if possible.
The area of triangle ABC is given by
[ABC] = 1/2 * AD * BC = 1/2 BE * AC = 1/2 * CF * AB
Since AD = 12 and BE = 15, we have
[ABC] = 1/2 * 12 * BC = 1/2 * 15 * AC = 1/2 * CF * AB
Dividing these equations, we get
BC/AC = CF/AB
Since BC and AC are positive integers, CF must also be a positive integer. The largest possible value of CF is thus AB, which is equal to
AB = 15 * BC / AC = 15 * 12 / 15 = 12
Therefore, the largest possible value of CF is 12.
Obtuse scalene triangle.
Sides: a = 362.892 b = 290.314 c = 73.809
Area: T = 2177.353
Perimeter: p = 727.014
Semi-perimeter: s = 363.507
Angle ∠ A = 168.274° = 168°16'27″ = 2.937 rad
Angle ∠ B =9.357° = 9°21'25″ = 0.163 rad
Angle ∠ C =2.369° = 2°22'8″ = 0.041 rad
Height: ha = 12
Height: hb = 15
Height: hc = 59 = altitude CF, which is the maximum positive integer possible.
Suppose that the area of the triangle is A, then (half base times height)
A = AB*CF/2, so AB = 2*A/CF,
A = BC*AD/2, so BC = 2*A/AD = 2*A/12,
A = CA*/BE/2, so CA = 2*A/BE = 2*A/15.
For the triangle to exist,
AB + BC > CA, so 1/CF + 1/12 > 1/15,
BC + CA > AB, so 1/12 + 1/15 > 1/CF,
CA + AB > BC, so 1/15 + 1/CF > 1/12.
The third of these conditions leads to CF < 60 meaning that the largest integer value for CF is 59.
One of the other two conditions gives a minimum for CF.