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.
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
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Posted by Charlie
on 2020-10-10 15:29:57 |
computer solution