practice problems pending

1)


P(x) = x^4- 4x^3 + 6x^2 - 4x - 2005.

Find the integer part of the product of all non-real roots of (P).

S1)
P(x) = 0
=> (x-1)^4 = 2006
=> (x-1) = K.(2006)^(1/4) where K = -+1, -+i
So non real roots are
x = 1 +-i.(2006)^(1/4)
Multiply:
1 + 2006^(1/2)
45^2 = 2025
45^2 - 44^2 = 89
So 2006^(1/2) = 44.something
Answer: 45


2)

First we will prove this:
In the incircle trisects the median then the sides are in the ratio 5:10:13 where 10 maps to the base where median is landing.
Once we do that, since BC = 20, other sides will become 10 and 26.

Then apply Heron's formula.
s = 28
s - a = 28 - 20 = 8
s - b = 28 - 10 = 18
s - c = 28 - 26 = 2
[ABC] = sqrt(28.18.8.2) = sqrt(64.7.2.9) = 24.sqrt(14)
m + n =38

Ok coming to the proof of the first part.

Using power of the point theorem:
If DF is the tangent from D, then DF^2 = DY.DX = d/3 * 2d/3 = 2d^2/9
Similarly Tangent from A = AE then AE^2 = AX.AY = DF^2
=> AE^2 = DF^2
But AE = s - a from incircle properties.
And DF = (s-c) - a/2 = a+b-c/2 - a/2 = (b-c)/2
s-a = |b-c|/2
WLOG, b > c
=> (b+c-a)/2 = (b-c)/2
=> c = a/2
=> a = 2c

Applying Stewart's theorem:
m.a.n + d.a.d = b.m.b + c.n.c
a.a^2/4 + a.d^2 = b^2.a/2 + c^2.a/2 where d is the median length.
=> a^2 + 4.d^2 = 2b^2 + 2c^2
=> d^2 = [2b^2 + 2c^2 -a^2]/4 = (2b^2 - 2c^2)/4 = (b^2 - c^2)/2
AE^2 = 2d^2/9
=> (b-c)^2/4 = (b^2 - c^2)/9
=> (b^2 + c^2 - 2bc)/4 = (b^2 - c^2)/9
Divide all by c^2 and b/c = k
=>
k^2 + 1 - 2k = 4/9(k^2 - 1)
9k^2 - 18k + 9 = 4k^2 - 4
5k^2 - 18k + 13 = 0
5k^2 - 5k - 13k + 13 = 0
5k(k-1) -13(k-1) = 0
=> k = 1,13/5
k = 1 => b = c but a = 2c => triangle inequality won't hold.
k = 13/5 = b/c => 5b = 13c
a = 2c
b = 13c/5
So side ratio is:
2:13/5:1 = 10:13:5






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