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.
Slope CD=(a+ac)/(1-c)
Slope AC=(2a+ac)/(2-c)
Setting the sum of these slopes to 0 and solving for c in terms of a just gives c=sqrt(2).
C=(sqrt(2), -a*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
f(a)= a(1+sqrt(2))/(1+a^2(3+2sqrt(2))
it's derivative is to messy to type out here, but setting it equal to zero and solving gives
a=sqrt(2)-1
and
f(a)=1
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 |
Long solution
