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Max angle in triangle (Posted on 2021-03-26) Difficulty: 3 of 5
D is midpoint of AC for triangle ABC. Bisectors of ∠ACB, ∠ABD are perpendicular. Find max value for ∠BAC.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Long solution | Comment 1 of 2
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

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
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