Let $K$ be a convex quadrangle and let $l$ be a line through the point of intersection of the diagonals of $K$. Show that the length of the segment of intersection $l\cap K$ does not exceed the length of (at least) one of the diagonals of $K$.
Problem
Source: Romania TST 2013 Day 5 Problem 2
Tags: geometry unsolved, geometry
trigocu
22.08.2017 10:36
Assume our quadrilateral is ABCD, diagonals intersect at O, l and AD intersect at L, l and BC intersect at Y. The line perpendicular to AD and passes through O intersects with AD at S and intersects with BC at T. The line perpendicular to BC and passes through O intersects with BC at M and intersects with AD at N.WLOG assume N is between A and S; T is between B and M. There are 3 possibilities about points L and Y. Case 1: If L is between S and D; Y is between B and T. DLO and BYO are obtuse angles, hence DO>LO and BO>YO, hence DB>YL Case 2: If L is between A and N; Y is between M and C. Similar to case 1, AC>YL Case 3: If L is between N and S; Y is between T and M. NST and NMT are right angles, hence NSMT is a circular, say this circle is G. Assume ray OL intersects with G at R and ray OY intersects with G at V. NT is diameter, hence NRV is obtuse angle, hence RO=<NO. Similarly, VO=<TO. WLOG, assume NO>=TO. NMT is right angle, hence OM=<OT. Due to prior inequalitis, we can say OM = NO-a, OR = NO-b, OV = NO-c where none of a,b,c is negative. Due to circle, OR*OV=ON*OM => (NO-b)(NO-c)=NO(NO-a) => NO(b+c-a)=bc => b+c-a>=0 => NO + (NO-a) >= (NO-b) + (NO-c) => NM >= RV NM =< AC due to case 1. Also RV >= LY, hence AC >= NM >= RV >= YL