A regular pentagon $ABCDE$ is given. The point $X$ is on his circumcircle, on the arc $\overarc{AE}$. Prove that $|AX|+|CX|+|EX|=|BX|+|DX|$. Remark: Here's a more general version of the problem: Prove that for any point $X$ in the plane, $|AX|+|CX|+|EX|\ge|BX|+|DX|$, with equality only on the arc $\overarc{AE}$.
Problem
Source: Gillis Olympiad 5777 (Israel National '16-'17) #5
Tags: national olympiad, geometry, pentagon, circumcircle
24.12.2016 06:37
Let the side of the pentagon be $a$. Then $AC=EC=BD=a\frac{\sqrt{5}+1}{2}.$ The solution is based on using Ptolemy's theorem to cyclic quadrilaterals. From $XABC$: $XA\cdot a+XC\cdot a=XB\cdot a\frac{\sqrt{5}+1}{2}\ (1)$ From $XEDC$: $XE\cdot a+XC\cdot a=XD\cdot a\frac{\sqrt{5}+1}{2}\ (2)$ From $XBCD$: $XB\cdot a+XD\cdot a=XC\cdot a\frac{\sqrt{5}+1}{2}\ (3)$ Adding $(1),(2)$ we get $\frac{\sqrt{5}+1}{2}(XB+XD)=XA+XE+2\cdot XC,$ which (from $(3)$) leads to $XA+XC+XE=\frac{\sqrt{5}+1}{2}(XB+XD)-XC=XB+XC.$ The proof of the general version is identical to the above solution but instead of using Ptolemy's theorem we use Ptolemy's inequality.
24.12.2016 08:24
Another generalization: Let $n$ be a positive integer. A regular polygon $A_1A_2\dots A_{2n+1}$ is given. The point $X$ is on the circumcircle, on the arc $\stackrel\frown{A_1A_{2n+1}}$. Prove that $\sum_{k=0}^n |XA_{2k+1}|=\sum_{k=1}^n |XA_{2k}|$
27.12.2016 11:39
math90 wrote: Another generalization: Let $n$ be a positive integer. A regular polygon $A_1A_2\dots A_{2n+1}$ is given. The point $X$ is on the circumcircle, on the arc $\stackrel\frown{A_1A_{2n+1}}$. Prove that $\sum_{k=0}^n |XA_{2k+1}|=\sum_{k=1}^n |XA_{2k}|$ The inequality in this case is also true.
13.05.2017 15:25
Damarliyaprak wrote: This problem can be solved easily by yahudidolleri theorem. Sincerely,damarliyaprak What is that theorem?
13.05.2017 17:52
it's a spam I guess
12.06.2018 11:54
The inequality case and the general case are unsolved here.