how to find values of x for which a triangle exists

Question

Given a $\triangle PQR , QR=3 , PR= x , PQ=2x$ and $\angle PQR= \theta$ calculate the values of $x$ for which the triangle exists. I don't even know where to start.

Answer 1

Hint: triangle inequality (what is the relation between the sum of lengths of any two sides and the length of the remaining side?)

Answer 2

By the triangle inequality, the sum of any possible combination of two sides must be strictly greater than or equal to the third side (for a non degenerate triangle).

So:

$2x + 3 > x \implies x > -3$ (holds trivially, ignore).

$x+3 > 2x \implies x < 3$

$x+2x > 3 \implies x > 1$

So $1 < x < 3$ is the required range.

Answer 3

The answer given by @Deepak is undoubtly the shortest possible and the one which is certainly awaited at your level.

Nevertheless, here is an answer that gives another proof for the result:

$$1<x<3$$

in the framework of a "global understanding" (see figure below).

It involves a classical result : the locus of points $P$ such that the ratio of their distances to two fixed points is constant (here $\dfrac{PQ}{PR}=2=\dfrac{2x}{x}$) is a circle (https://www.cut-the-knot.org/Curriculum/Geometry/LocusCircle.shtml)

(see commentary below)

(more generally, constant 2 can be any positive constant).

On the figure, you see a triangle $PQR$ where $RQ$ is fixed and $P$ can be any point on the circle.

Now take the extreme points :

• $P$ is in $P_1$ iff $x=1$,

• $P$ is in $P_2$ iff $x=3$,

all values of $x$ in the range $[1,3]$ being allowed, with the restriction that these extreme points give flat triangles...

Commentary: this circle has its center on line $RQ$, with diameter $P_1P_2$ where $P_1$ and $P_2$ are the two extreme points on line $RQ$ mentionned above.