IMO Shortlist 1969 problem 16
Dodao/la:
arhiva2. travnja 2012. ![(CZS 5)](/media/m/6/0/5/605e46b8dfbb01dda6aab6ad769599c3.png)
A convex quadrilateral
![ABCD](/media/m/9/c/e/9ce25711ba18d9663b73c3580de4bf5a.png)
with sides
![AB = a, BC = b, CD = c, DA = d](/media/m/7/7/2/772662058f70b6f31e95b5e0c595db94.png)
and angles
![\alpha = \angle DAB, \beta = \angle ABC, \gamma = \angle BCD,](/media/m/6/e/2/6e28c4b113cbf1006042c2110957d639.png)
and
![\delta = \angle CDA](/media/m/c/9/f/c9f473093ead24b7c52bf9d2c83b643d.png)
is given. Let
![s = \frac{a + b + c +d}{2}](/media/m/e/2/8/e289ec331b068a5485349cac645528bd.png)
and
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
be the area of the quadrilateral. Prove that
%V0
$(CZS 5)$ A convex quadrilateral $ABCD$ with sides $AB = a, BC = b, CD = c, DA = d$ and angles $\alpha = \angle DAB, \beta = \angle ABC, \gamma = \angle BCD,$ and $\delta = \angle CDA$ is given. Let $s = \frac{a + b + c +d}{2}$ and $P$ be the area of the quadrilateral. Prove that $P^2 = (s - a)(s - b)(s - c)(s - d) - abcd \cos^2\frac{\alpha +\gamma}{2}$
Izvor: Međunarodna matematička olimpijada, shortlist 1969