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=(1-X*X-X*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 2005-10-24 13:54:11