trigonometry

[denian]

in triangle PQR, PQ = a, QR = a + d, RP = a + 2d, and angle PQR is more than 120 degree., where a and d are positive constants. show that

2a/3 < d < a.


i tried to think, but i really dont know how to start to solve this question. hope for some clue.

thank you

[AKG]

in triangle PQR, PQ = a, QR = a + d, RP = a + 2d, and angle PQR is more than 120 degree., where a and d are positive constants. show that

2a/3 < d < a.


i tried to think, but i really dont know how to start to solve this question. hope for some clue.

thank youStart with a picture, even if it doesn't help in the end, it's nice to have something to look at, you never know how it might inspire you. Now, we want to show:
1) d < a
2) d > 2a/3

The first part is easy. Assume d \geq a. Let d = a + \epsilon, where \epsilon \geq 0, \epsilon \in \lR.

PQ + QR\ = 3a + \epsilon
PR = 3a + 2\epsilon

Therefore:

PR \geq PQ + QR. Now, it should be obvious to you (especially if you try to draw it out with a ruler) that such a triangle is impossible. Our contradiction (that our triangle forms an impossible triangle) implies that our assumption is wrong (the assumption that d \geq a), therefore, d is indeed less than a.

2) Now, we know that the \angle PQR = q satisfies:

q > 120^o

Therefore:

\cos q < -0.5

Now, set up an equation using the Cosine Law:

(a + 2d)^2 = a^2 + (a + d)^2 - 2a(a + d)\cos q
\frac{a^2 - 2ad - 3d^2}{2a(a + d)} = \cos q
\frac{a^2 - 2ad - 3d^2}{2a(a + d)} < -0.5
a^2 - 2ad - 3d^2 < -a^2 - ad
2a^2 - ad - 3d^2 < 0
(2a - 3d)(a + d) < 0

Therefore, either:

2a - 3d < 0 AND a + d > 0, OR
2a - 3d > 0 AND a + d < 0

It should be clearly impossible for "a + d < 0" to be true, therefore, we must meet the first set of conditions. Looking at "2a - 3d < 0", we get:

2a < 3d
d > 2a/3

This proves 2). Finally, we have it that 2a/3 < d < a. Q.E.D.

[denian]

Thank You!