| Inequalities and Geometric Similarities |
Given:
AB = 4x
BC = x + 2
AC = x + 4
CD = 3x + 10
AD = 12
Find the range of values of x such that the area of triangle ABC exceeds that of triangle ACD. |
Problem Moderated by: Sasha |
| Problem Solution |
Use Heron's formula.
For any triangle XYZ, its area is equal to
√(s(s-x)(s-y)(s-z)), where s = (x + y + z)/2
For triangle ABC, s = 3x + 3
Area = √((3x+3)(2x-1)(3-x)(2x+1))
For triangle ADC, s = 2x + 13
Area = √((2x+13)(2x+1)(3-x)(x+9)}
So, we want to solve the inequality
√((3x+3)(2x-1)(3-x)(2x+1)) > √((2x+13)(2x+1)(3-x)(x+9)}
It is compulsory that each of the multiplied terms in parentheses be greater than zero. If (a+b+c)/2 - a were less than or equal to zero, we can derive a >= b + c, which violates the triangle inequality. We therefore can cancel out like terms while preserving the inequality - that is,
√((3x+3)(2x-1)) > √((2x+13)(x+9))
From triangle ABC, we are given
2x-1 > 0 -> x > 0.5 and
3-x > 0 -> x < 3
Quite conveniently, this guarantees that (2x+13)(x+9) > 1, so the special cases of √n > √m while n and m can be similtaneously between zero and one is irrelevant, and we can proceed directly to
2x2 + 31x + 117 < 6x2 + 3x - 3 ->
4x2 - 28x - 120 > 0 ->
x2 - 7x - 30 > 0
This is factorable.
(x-10)(x+3) > 0 ->
x>10 or x<-3
The intersection of this solution and the compulsory 0.5<x<3 leaves an empty set - in other words, for all allowable values of x, the area of ABC will be less than that of ACD.
Answer: Ø
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