+237 Triangle $ABC$ has altitudes $\overline{AD},$ $\overline{BE},$ and $\overline{CF}.$ If $AD = 12,$ $BE = 20,$ and $CF$ is a positive integer, then find the largest possible value of $CF.$
+130317 Area = (1/2) (AD) (BC) Area = (1/2)(BE)(AC) Area = (1/2)(CF)(AB)
2A = (12) (BC) 2A = (20)(AC) 2A = (CF)(AB)
A/6 = BC A/10 = AC 2A / CF = AB
By the Triangle Inequality :
AB + BC < AC
2A/CF + A/6 > A/10
2/CF + 1/6 > 1/10
2/CD > 1/10 -1/6
2/CD > -4/60
2/CD > -1/15
-15/2 < CD
BC + AC > AB
A/6 + A/10 > 2A / CF
CF > 2A / (A/6 + A/10)
CF > 2 / (1/6 + 1/10)
CF > 2/ [ 16/60]
CF > 120/16
CF > 15/2
AC + AB > BC
A/10 + 2A/CF > A/6
1/10 + 2/CF > 1/6
2/CF > 1/6 - 1/10
2/CF > 4/60
2/CF > 1/15
CF < 15*2
CF < 30
