Let $A, B, C$ and $D$ be four different points lying on a common circle in this order. Assume that the line segment $AB$ is the (only) longest side of the inscribed quadrilateral $ABCD$. Prove that the inequality $AB + BD > AC + CD$ holds. (Proposed by Karl Czakler)
Problem
Source: 49th Austrian Mathematical Olympiad National Competition (Final Round, part 2, ) 31st May 2018 p2
Tags: geometric inequality, geometry, cyclic quadrilateral
WolfusA
26.05.2019 16:20
$O$-center of the circle. Take $2\alpha =\angle AOB,\ 2\gamma=\angle COD,\ 2\delta=\angle AOD,\ 2\beta=\angle BOC$ such that $\alpha+\beta+\gamma+\delta=180^\circ$. Then the inequality is $$\sin\alpha+\sin(\alpha+\delta)>\sin(\gamma+\delta)+\sin \gamma$$$$\sin\frac{\alpha-\gamma}{2}\cdot (\cos\frac{\alpha+\gamma}{2}+\cos\frac{\alpha+\gamma+2\delta}{2})>0$$$$\sin\frac{\alpha-\gamma}{2}\cdot\sin\frac{\beta}{2}\cdot\cos\frac{\delta}{2}>0$$Every number in the product is positive, so we finished. However I'm afraid that this final expression should be symmetric in $\delta$ and $\beta$. Am I right? If so, is there any mistake?
sunken rock
28.05.2019 14:36
I shall only use the triangle inequality, the angle opposite to the greatest side is the greatest one. Let $\widehat{ADB}=\alpha, \widehat{CAD}=\beta, \widehat{ABD}=\widehat{ACD}=2\gamma$; extend $(AB$ to $E|BE=BD$ and $(AC$ to $F|CF=CD$. As $\widehat{AED}=\widehat{AFD}=\gamma\implies AEFD$ cyclic and now we need to prove $AE>AF$ or $\widehat{AFE}>\widehat{AEF}\ (\ 1\ )$. But $\widehat{AFE}=\widehat{ADE}=\alpha+\gamma\ (\ 2\ )$, while $\widehat{AEF}=\widehat{DEF}+\widehat{AED}=\widehat{DAF}+\widehat{AED}=\widehat{DAC}+\widehat{AED}=\beta+\gamma\ ( \ 3\ )$. Substracting $\gamma$ from each of the relations $(2)$ and $(3)$ we need to prove $\alpha>\beta$, which is true and we are done. Best regards, sunken rock
BlazingMuddy
28.05.2019 18:13
Let $E$ be the midpoint of the arc $AD$ that does not contain $B$ and $C$. Note that this arc must be the smaller arc; otherwise $AD$ would be the longest side of $ABCD$. Let $\angle ABD = \angle ACD = 2 \alpha$. By Ptolemy's equality (using sine),
\begin{align*}
AB + BD &= BE \; \frac{\sin (2\alpha)}{\sin \alpha} \\
AC + CD &= CE \; \frac{\sin (2\alpha)}{\sin \alpha}
\end{align*}So, we are left to prove that $BE > CE$. Letting $F$ be the antipod of $E$ with respect to the common circle, the rest is just dividing cases by diagram; if $A$ and $B$ lie on different sides with respect to $EF$, obviously all possible diagram shows $BE > CE$ as $BF < CF$. Otherwise, if $A$ and $B$ lie on the same side, letting $B'$ be the mirror of $B$ with respect to $EF$, all possible diagram with $AB > CD$ shows that the order of the points in the circle are $A, B, F, B', C, D, E$, which also implies $BE > CE$. We are done.