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Topic: Triangle Problem (Read 456 times) |
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ThudnBlunder
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ABC is a right-angled triangle with angle ABC = 90°.
D is a point on AB such that CD bisects the angle ACB.
E is a point on BC such that AE bisects the angle CAB.
If AE = 9 cm and CD = 8[sqrt]2 cm, find the length of AC.
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| « Last Edit: Dec 19th, 2003, 11:56pm by ThudnBlunder » |
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Eigenray
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Re: Triangle Problem
« Reply #1 on: Dec 20th, 2003, 4:08am » |
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Label BE=x, EC=y, BC=a=x+y; BD=u, AD=v, AB=b=u+v; AC=c.
Okay, those labels don't make too much sense, but you're the one that had to go and make B the right angle.
By angle bisectors, x/b = y/c, and x+y=a, so x=ab/(b+c). Similarly, u=ab/(a+c).
Now x2+b2=81, so
b2(a2/(b+c)2+1)=81
b2(a2+b2+c2+2bc)=81(b+c)2
2b2c = 81(b+c) (1)
Similarly, u2+a2=128, so
2a2c = 128(a+c) (2)
Let t = angle BCA, T=tan t, S=sin t, C=cos t. Dividing (1) by (2) gives:
T2 = 81/128 (S+1)/(C+1) = 81/128 (T + (1-T)/(1+C))
Solving for C:
(81 - 128T2)/(128T2 - 81T) = C = 1/sqrt(1+T2)
Which gives the, uh, sixth degree polynomial:
214T6 - 2834T4 + 2834T3 - 2834T2 + 38 = 0.
Any rational roots of this have to be in the form T = 3m/2n, and luckily enough, T = 3/4 works, and S = 3/5, C = 4/5.
(There must be an easier way!)
Now, adding (1) and (2):
2c3 = 128a + 81b + 209c
2c2 = 128C + 81S + 209 = 128(4/5) + 81(3/5) + 209 = 360
So AC = c = sqrt(180) = 6sqrt(5).
Yay!
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ThudnBlunder
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Re: Triangle Problem
« Reply #2 on: Jan 17th, 2004, 9:19am » |
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Quote:| (There must be an easier way!) |
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This is a bit easier:
Let
angle EAB = x
and
angle BCD = y
So
sin2x = cos2y
and
y = 45 - x
AB = AC*cos2x = 9cosx
BC = AC*cos2y = 8[smiley=surd.gif]2*cosy
Dividing,
tan2x = 8[smiley=surd.gif]2*cosy/9cosx
= 8[smiley=surd.gif]2*cos(45 - x)/9cosx
= 8(cosx + sinx)/9cosx
So 9tan2x = 8(1 + tanx)
Letting t = tanx we get
(2t - 1)(2t2 + 3t + 4) = 0 => t = 1/2
and
AC = 4(1 + t)[smiley=surd.gif](1 + t2)/t
Thus AC = 6[smiley=surd.gif]5
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| « Last Edit: Jan 17th, 2004, 9:42am by ThudnBlunder » |
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