 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Sixty Degree Decision (Posted on 2016-08-11) 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 ≥ 60o?
If so, give an example.
If not, prove it.

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 1 of 1
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 Please log in:
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