D is midpoint of AC for triangle ABC. Bisectors of ∠ACB, ∠ABD are perpendicular. Find max value for ∠BAC.
I assume there's an easier way, but here's my approach:
Using coordinates, B=(0,0), D=(1,a), A=(2,2a) which are on a line with slope a, so C is on a line with slope -a. C=(c,-ac) for some c.
The bisector of B is just the x-axis so the perpendicular is a vertical line through C. This means the two sides of ∠ACD have opposite slopes.
Setting the sum of these slopes to 0 and solving for c in terms of a just gives c=sqrt(2).
Now to find ∠BAC = arctan(slope CA) - arctan(slope AB)
Slope CA simplifies to a(3+2sqrt(2))
Slope AB = a
using the arctan difference formula gives
∠BAC = arctan [2a(1+sqrt(2))/(1+a^2(3+2sqrt(2))]
Since arctangent is an increasing function, we just need to maximize the argument.
Using calculus to maximize
it's derivative is to messy to type out here, but setting it equal to zero and solving gives
Finally arctan(1) = 45 degrees.
Remarks: At this maximum, the arctangents of CA ad AB are 67.5 and 22.5 respectively and also ACB is an isosceles right triangle.
Posted by Jer
on 2021-03-26 11:09:27