In trapezoid $ABCD$ with $AB || CD$, $\omega_1$ and $\omega_2$ are two circles with diameters $AD$ and $BC$, respectively. Let $X$ and $Y$ be two arbitrary points on $\omega_1$ and $\omega_2$, respectively. Show that the length of segment $XY$ is not more than half the perimeter of $ABCD$. Proposed by Mahdi Etesami Fard
Problem
Source: Iranian Geometry Olympiad 2016 Medium 1
Tags: geometry, trapezoid, perimeter
dovakin123
26.05.2017 14:54
Let $C_1,C_2$ be centers of the circles with the same index. The shortest distance between two pints is a straight line so $XY\le XC_1+C_1C_2+C_2Y=AD/2+BC/2+\frac{AB+CD}{2}$, and we are done. The fact that $C_1C_2= \frac{AB+CD}{2}$ can be shown easily by joining one of the diagonals and applying mid-point theorem.
Mahdi_Mashayekhi
27.12.2021 09:41
Let E and F be midpoints of AD and BC. We know XY ≤ XE + EF + FY. XE = AD/2 and EF = AB+CD/2 and FY = BC/2. we're Done.
Chevrolet23
04.05.2022 20:06
we shoulld determine centere of circles O1 and O2 . we should look quadriarterial O1O2XY we know from quadriarterial inequality in this quadriarterial O1X+O1O2+O2Y>XY .
lian_the_noob12
31.08.2023 18:18
$$XY\le XO_1+O_1O_2+O_2Y=\frac{AD}{2}+\frac{AB+CD}{2}+\frac{BC}{2}=\frac{AB+BC+CD+AD}{2} \blacksquare$$