Equilateral triangles $ ABQ$, $ BCR$, $ CDS$, $ DAP$ are erected outside of the convex quadrilateral $ ABCD$. Let $ X$, $ Y$, $ Z$, $ W$ be the midpoints of the segments $ PQ$, $ QR$, $ RS$, $ SP$, respectively. Determine the maximum value of \[ \frac {XZ+YW}{AC + BD}. \]
Problem
Source: China Girls Math Olympiad 2008, Problem 4
Tags: inequalities, geometry unsolved, geometry
shy10
31.08.2008 04:32
Complex Numbers
CCMath1
29.08.2009 16:22
This is Problem four in CGMO 2008 proposed by XiongBing lemma:if $ {\triangle QBA }$and $ {\triangle PAD}$are Equilateral triangles denote $ {E,H,A,X}$ are the midpoints of segment $ {BA,AD,EH,QP}$ Then we have :$ {XT=\frac{\sqrt{3}}{4}DB}$;proof: denote the midpoint of $ {QA,PA}$ are $ {A',A''}$ We can easily get :$ {\triangle A'XE\cong \triangle A''HX \cong \triangle AHE}$ $ { \Longrightarrow}$$ {\triangle HXE}$ is an equilateral triangle ,the result easily follows. Then we begin to solve this problem: denote $ {E,H,F,G,T,U}$ are the midpoints of segments :$ {AB,AD,BC,DC,EH,FG}$; Easy to get :$ {XZ\le XT+TU+ZU}$; with these midpoints we can get :$ {TU=EF=\frac{1}{2}AC}$;and by our lemma :$ {TX=ZU=\frac{\sqrt{3}}{4}BD}$;So we get :$ {XZ\le\frac{\sqrt{3}+1}{2}AC}$; with the same way ,we similarly get that :$ {ZW\le\frac{\sqrt{3}+1}{2}BD}$ FINALLY:we get the inequality:$ {\frac {XZ+WY}{AC+BD}\le\frac{\sqrt{3}+1}{2}}$; So the maximum value of $ {\frac {XZ+WY}{AC+BD}}$ is $ {\frac{\sqrt{3}+1}{2}}$; the equality reach when $ {ABCD}$ is a square (just like the image shows) .
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