Home > Shapes
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 ≥ 60^{o} ?
If so, give an example.
If not, prove it.

No Solution Yet
Submitted by K Sengupta
Rating: 4.0000 (1 votes)

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
= (a^{2} + c^{2} – b^{2} )/(2ac)
Now using (1) cos/ QPR = (a^{2}
+ c^{2} – (a^{2} + c^{2} + 2ac)/9)/(2ac)
= (8/9)(a^{2} + c^{2} )/(2ac) –
1/9
The minimum value of (a^{2} + c^{2} )/(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:
Forums (0)
Newest Problems
Random Problem
FAQ |
About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox: