All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
The angle bisectors (Posted on 2020-10-10) Difficulty: 3 of 5
In triangle ABC, the angle at vertex B is 120 degree. Let A1, B1, C1 be points on the sides BC, CA, AB respectively such that AA1, BB1, CC1 are bisectors of the angles of triangle ABC. Determine the angle A1B1C1.

No Solution Yet Submitted by Danish Ahmed Khan    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution computer solution Comment 1 of 1
I started with Geometer's Sketchpad, creating a 120° angle and extending legs to form a triangle and trying various combinations of the other two angles.  Regardless of the other two angles, the angle specified always came out as 90°, which is therefore the answer. It does not depend on what the other two angles are.

That got me wondering: Is that angle always 90° even if the opposite angle is not 120°? The answer was no, as other values are found from triangles without the 120° angle. Then the question is, How is the specified angle dependent on the vertex B angle?

I tried different angles at vertex B, but even keeping that angle constant, the angle A1 B1 C1 changed even for a constant angle B, as the other angles changed,

Curious, I wrote a program to do the calculations. The following table shows the size of angle A1  B1 C1 using the values of A and C.  Whichever is larger of A and C is shown at the left, and the smaller of A and C is shown at the top (B of course is 180° - A - C).

        10     20     30     40     50     60     70     80 
 10  141.70        
 20  125.37 111.25
 30  111.55  99.62  90.00
 40   99.92  90.00  82.13  75.76
 50   90.00  81.89  75.55  70.46  66.27
 60   81.36  74.88  69.90  65.94  62.71  60.00
 70   73.66  68.69  64.92  61.98  59.62  57.67  56.03
 80   66.65  63.08  60.44  58.43  56.87  55.63  54.63  53.84
 90   60.15  57.89  56.31  55.18  54.36  53.79  53.42  53.20
100   53.99  53.00  52.42  52.12  52.04  52.12  52.35
110   48.07  48.28  48.67  49.20  49.83  50.57
120   42.29  43.66  45.00  46.34  47.71
130   36.58  39.05  41.33  43.50
140   30.87  34.37  37.58
150   25.08  29.56
160   19.15

As B of course is 180 minus the sum of the two base angles, the values for a constant B are in a line running upper-right to lower-left; the values for B = 120° are the three 90.00's at 30-30, 40-20 and 50-10. As you can see, the other diagonals parallel to this do not have constant values, though some stay close together.

It was definitely good practice for learning MATLAB.

The program, a MATLAB script, assumes the base of the triangle extends from the origin to (1,0). It also uses the name Ap, for A1 , etc., the p standing for prime, which I don't know why was not used, as in A'; that would have been easier to type in this comment.

syms x y
A=[0,0]; C=[1,0];
fprintf('    ')
for a=10:10:80
    fprintf('%6d ',a)
end
fprintf('\n')
for angle1=10:10:160
    fprintf('%3d ',angle1)
    m1=tand(angle1); m1b=tand(angle1/2);
    if m1==Inf
        Eqn1=x==0; Eqn1b=y==m1b*x;
    else
        Eqn1=y==m1*x; Eqn1b=y==m1b*x;
    end
    for angle2=10:10:min([angle1,170-angle1])
        m2=tand(-angle2); m2b=tand(-angle2/2);
        Eqn2=y==m2*(x-1); Eqn2b=y==m2b*(x-1);
        Soln=solve([Eqn1 Eqn2b],x,y);
        Cp=[Soln.x Soln.y];
        Soln=solve([Eqn1b Eqn2],x,y);
        Ap=[Soln.x Soln.y];
        Soln=solve([Eqn1 Eqn2],x,y);
        B=[Soln.x Soln.y];
        ang1=angle1-180; ang2= -angle2;
        Bangle=(ang1+ang2)/2;
        mB=tand(Bangle);
        Bx=B(1); By=B(2);
        if mB==Inf || mB==-Inf
            Eqn3b=x==Bx;
        else
            Eqn3b=y==By+mB*(x-Bx);
        end
        Eqn3=y==0;
        Soln=solve([Eqn3 Eqn3b],x,y);
        Bp=[Soln.x Soln.y];
        Delta1=Bp-Cp; Delta2=Ap-Bp;
        Ang1=atand(Delta1(2)/Delta1(1)); 
        Ang2=atand(Delta2(2)/Delta2(1));
        if Delta2(1)<0 
            Ang2=Ang2+180; 
        end
        finalAngle=180+Ang1-Ang2;
        fprintf('%7.2f',finalAngle);
    end
    fprintf('\n');
end


  Posted by Charlie on 2020-10-10 15:29:57
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information