For triangle PQR it is known that:
|PQ|*|PR| =8*R*r, where R and r are respectively the circumradius and the inradius of triangle PQR.
Can ∠QPR be ≥ 60^o?
If so, give an example.
If not, prove it.

Let PQ, QR and RP have lengths a, b, c respectively.

Area of triangle =     (r/2)(a + b + c)  = (1/2)PQ*PR*sin/QPR

therefore              r(a + b + c)    = 8Rr*sin/QPR

                            a + b + c       = 4(2R*sin/QPR)   = 4b

which gives                    a + c = 3b                     (1)

Using the cosine rule:     cos/QPR = (a² + c² – b²)/(2ac)

Now using (1)    cos/QPR = (a² + c² – (a² + c² + 2ac)/9)/(2ac)

                                    = (8/9)(a² + c²)/(2ac) – 1/9

The minimum value of (a² + c²)/(2ac) is 1 when a = c,

so the minimum value of  cos/QPR is 7/9, giving a maximum

value of /QPR of   cos^-1(7/9)  ~  38.94 degrees.

So /QPR < 60 degrees.

Posted by Harry on 2016-08-12 06:52:22