Let's assume ABCD is concave. Then we can assume D is in the interior of triangle ABC. Then one of the angles ADB, BDC, CDA is >= 120 deg. We again assume BDC >= 120. Because DB and CD >=5 we have from cosint thm CD>= (5^2+5^2+5^2)^1/2 = 5*3^1/2 = 75^1/2 > 72^1/2 = 6*2^1/2 =the length of the diagonal, but we know that the longest segment that can be inside a square is the diagonal, so we got a contadiction, so ABCD is convex. For the second part consider the triangles ABC and BCD. Each of them has an angle >= 60 and they have all their angles <=120 (we proved this above), so each one of them has an angle with its sine>= (3^1/2/)2 and all their sides are >=5, but if you compute 2*(the area of a triangle with angle 60 and sides 5) you get a little over 21. The sum of the areas of our 2 triangles is, however >= what we calculated > 21, so it's finished. I would have enjoyed participating in that selection test (the problems weren't too hard), but unfortunately this year I completely messed up at the final round so..

cool stuff,grobber!! even i think this year's selection tests were easier ,4th one in particular. do u have solutions to all of them.?in case it is so,i'll ask for some of the solutions thru private messages,u wudnt mind,wud u? best regards

Sure, I don't mind, but I don't actually have that many solutions. I solved less than half of all the problems (in all the tests), but I could think about them..

Hey I just want to tell my opinion about the 4th selection test.It wasn't that easy , first of all because the first problem could easily trick you into taking M=A , which costed you 4 points out of 7 (It said M included or equal to A , M<>A). The second problem (this one) was probably the hardest geometry problem out of all the selection tests (except perhaps for the one in the first test , in which almost no-one dared to use computations). And the combinatorics problem was clearly the most difficult out of all the tests , since no student at all managed to solve it or do any real progress with it (the highest mark was 2 , obtained by 2-3 students). I think this was probably the most difficult selection test except the 2nd , which was really challenging. So , all in all , I don't think it was that easy. (P.S. I took part in the actual selection test)

oh i am soory abt this miscommunication. i meant the 5th selection test which was "easier " as compared to the other tests.4th one was difficult,no doubt cheers

First of all we observe that no angle formed with 3 of the 4 points can be greater of equal with $120^\circ$, because otherwise if we suppose that WLOG $\angle ABC\geq 120^\circ$, then from $AB\geq 5$ and $BC\geq 5$ we deduce $AC\geq 5\sqrt 3 >6\sqrt 2$ contradiction. Therefore if the quadrilateral $ABCD$ is not convex, then one of the 4 points lies inside the triangle formed by the other 3. Suppose WLOG that $D\in$int$[ABC]$. But then one of the angles $\angle ADB$, $\angle BDC$ and $\angle CDA$ would be, by the Pigeonhole Principle, greater or equal than $120^\circ$, contradiction. Thus $ABCD$ is a convex quadrilateral. Now because each angle of the triangle $ABC$ is smaller than $120^\circ$ and there is at least one angle, say $\angle ABC$, which is greater than $60^\circ$ it follows that \[ \sin \angle ABC \geq \frac {\sqrt 3}2 \Rightarrow \] \[ \sigma[ABC]=\frac 12 AB\cdot BC\cdot \sin \angle ABC \geq \frac {\sqrt3}4 \cdot 25 > \frac {21}2 \Leftrightarrow 625 > 12 \cdot 49 = 588 \] Analogously one can prove that $\displaystyle \sigma[ACD]\geq \frac {21}2$, and thus $\sigma[ABCD]>21$.

Prove convexity as done above. Now, knowing that our quadrilateral is convex, we have that all sides and diagonals have length at least $5$ and at most $6 \sqrt 2.$ Let our quadrilateral be called $ABCD,$ and $x = |AC|$ with $5 \le x \le 6 \sqrt2.$ We claim that if we let $B', D'$ be so that $AB' = B'C = CD' = D'A = 5$, then $[ABCD] \ge [AB'CD'].$ Indeed, it suffices to show that the distance from $B$ to $AC$ is at least the distance from $B'$ to $AC.$ Let $\omega_1, \omega_2$ be the circles of radius $5$ centered at $A, C$ respectively. We know that $B$ is not in the interior of either $\omega_1$ or $\omega_2.$ Therefore, it's clear from drawing a diagram that the foot from $B$ to $AC$ must be not be on segment $AC$, and that it must lie at least $\frac{x}{2}$ away from both vertices. This means that $|AB^2 - AC^2| \ge |(\frac{3x}{2})^2 - (\frac{x}{2})^2| = |2x^2| \ge 50.$ However, we have that $|AB^2 - AC^2| \le |(6 \sqrt2)^2 - 5^2| = 47$, which is a contradiction. $\square$