Problem B. 4915. (December 2017)

B. 4915. Given any points \(\displaystyle A_1\), \(\displaystyle A_2\), \(\displaystyle A_3\), \(\displaystyle A_4\), \(\displaystyle A_5\) and \(\displaystyle P\) in the plane, let \(\displaystyle k_i\) denote the number of ways it is possible to select \(\displaystyle i\) points out of \(\displaystyle A_1\), \(\displaystyle A_2\), \(\displaystyle A_3\), \(\displaystyle A_4\), \(\displaystyle A_5\) such that the convex hull of the selected points should contain \(\displaystyle P\). Show that \(\displaystyle k_3=k_4\).

(5 pont)

Deadline expired on January 10, 2018.

Sorry, the solution is available only in Hungarian. Google translation

Megoldásvázlat: Legyen \(\displaystyle h^{A_j}\) azon \(\displaystyle P\)-t tartalmazó háromszögek száma, amiknek nem csúcsa \(\displaystyle A_j\). Világos, hogy \(\displaystyle k_3=(h^{A_1}+\ldots+h^{A_5})/2.\) Vegyük észre, hogy \(\displaystyle h^{A_j}\) értéke \(\displaystyle 2,\) ha \(\displaystyle P\) az \(\displaystyle A_j\) elhagyása után megmaradt négy pont konvex burkába esik, egyébként \(\displaystyle 0\). Így \(\displaystyle h^{A_1}+\ldots+h^{A_5}=2k_4.\) Az állítás az eddigiekből következik.

Statistics:

91 students sent a solution.
5 points:	69 students.
4 points:	13 students.
3 points:	3 students.
2 points:	5 students.
1 point:	1 student.

Problems in Mathematics of KöMaL, December 2017