A certain triangle ABC has D on AB and E on AC such that AD=DB=BC=CE and AE=ED. What is the measure of angle BAC?
Scale the drawing to that AD=DB=BC=CE = 1 and AE=ED = x. Call the measure of angle BAC theta.
Side AB = 2; side BC = 1; side CA = 1 + x.
The smaller triangle, ADE, has sides 1, x, x.
In triangle ABC, by the law of cosines:
1 = 4 + (1+x)^2  4(1+x) cos theta
In triangle ADE, by the law of cosines:
x^2 = 1 + x^2  2x cos theta
From this latter equation, cos theta = 1/(2x).
Substituting this into the first equation and multiplying by 2x gives:
2x = 8x + 2x(1+x)^2  4(1+x)
Simplifying gives:
x^3 + 2x^2 + 2x  2 = 0
Not being up on the solution of cubic equations, I let Excel solve this to 0.574743073887022, and give the angle whose cosine is the reciprocal of twice this as 29.54671285 degrees.
I also wrote a UBASIC program to solve this numerically:
5 point 9
10 X=0.6
20 while X<>Px
25 Px=X
30 X=(1X*XX*X*X/2)*0.1+X*0.9
40 print X
50 wend
60 print acos(1/(2*X))
70 print acos(1/(2*X))*180/#pi
which converges to 0.5747430738870215957014669401978547746303914 for x, 0.5156874224399999344289486607112603272731115 as the radian value of theta and 29.5467128537919765555445412793451236727526806 as theta's value in degrees.
Edited on October 24, 2005, 1:57 pm

Posted by Charlie
on 20051024 13:54:11 