Question

If+x²+x+1,2x+1,x²-1 are the sides of a triangle, then its largest angle is


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Answer 1

Answer:

For the third of these side lengths to be possible (i.e. positive), we see that:

x2−1>0⟹x>1 or x<−1 .

For the second of these side lengths to be positive, we see that:

2x−1>0⟹x>12 .

Together, we see that x>1 is required. Note that when x>1 , x2+x+1>12+1+1>0 so the first side length will also be positive. Next, note that when x>1 , we can multiply both sides by x to see that x2>x . This implies that x2+x+1>x+x+1=2x+1 , and we see that the first length is longer than the second. Furthermore, when x>1 we also see that x2+x+1−(x2−1)=x+2>0 . So the first length is greater than the third. Finally, x2−1+2x+1−(x2+x+1)=x−1>0 , so the triangle inequality is satisfied and the three sides do form a triangle.

Now that we know which side is the longest, we are left to find the angle opposite this side which will be the largest angle.

We expect this angle to be written as a function of x .

(x2+x+1)2=(2x+1)2+(x2−1)2−2(2x+1)(x2−1)cosϕ

Some algebra shows that all the x terms cancel out giving:

cosϕ=−12

Since the angles of a triangle must be between zero and one hundred eighty degrees, the only valid solution to this equation is:

120 degrees = 23π radians

Here's a fast way to do the algebra:

First, move a term from the Right Hand Side (RHS) to the Left Hand Side (LHS):

(x2+x+1)2−(x2−1)2=(2x+1)2−2(2x+1)(x2−1)cosϕ

Factor the LHS using the difference of squares:

(2+x)(2x2+x)=(2x+1)2−2(2x+1)(x2−1)cosϕ

Factor out the x on the LHS:

x(2+x)(2x+1)=(2x+1)2−2(2x+1)(x2−1)cosϕ

Divide by the common factor of 2x+1 (which we can do since 2x+1>0 is required to be a triangle).

x(2+x)=2x+1−2(x2−1)cosϕ

Combine the terms that don't have a ϕ :

x2−1=−2(x2−1)cosϕ

Divide by the common factor of x2−1 which is possible since x2−1>0 is required for this to be a triangle:

1=−2cosϕ⟹cosϕ=−12