Construct a convex polygon such that each of its sides has the same length as one of its diagonals and each diagonal has the same length as one of its sides, or prove that such a polygon does not exist.
Problem
Source: Canada Repêchage 2018/4
Tags: geometry
ythomashu
09.04.2018 16:10
It doesn't exist.
GCampos
09.04.2018 16:11
Prove that it doesn't.
ythomashu
09.04.2018 16:30
A convex quadrilateral will always have a diagonal longer than all of its sides. Adding sides equal to the length of the diagonal will create longer diagonals
wu2481632
09.04.2018 16:43
ythomashu wrote: A convex quadrilateral will always have a diagonal longer than all of its sides. Adding sides equal to the length of the diagonal will create longer diagonals Prove this.
Math-Ninja
09.04.2018 16:45
wu2481632 wrote: ythomashu wrote: A convex quadrilateral will always have a diagonal longer than all of its sides. Adding sides equal to the length of the diagonal will create longer diagonals Prove this. Prove that you have to prove this.
GCampos
09.04.2018 16:46
Math-Ninja wrote: wu2481632 wrote: ythomashu wrote: A convex quadrilateral will always have a diagonal longer than all of its sides. Adding sides equal to the length of the diagonal will create longer diagonals Prove this. Prove that you have to prove this. Prove that you have to prove that you have prove this.
Math-Ninja
09.04.2018 16:50
@above, prove that you have to prove that you have to prove that you have prove this.
ythomashu
09.04.2018 16:59
Math-Ninja wrote: wu2481632 wrote: ythomashu wrote: A convex quadrilateral will always have a diagonal longer than all of its sides. Adding sides equal to the length of the diagonal will create longer diagonals Prove this. Prove that you have to prove this. he can't because I was actually wrong. take a equilateral triangle stuck to an obtuse triangle with an angle of 120 or more degrees. anyway, we will move to this existing thread that is non-spammy
InCtrl
10.04.2018 03:40
@above that isn't convex... anyway this is the first time i see HSO being spammy cross posting my solution from canada forum InCtrl wrote:
Consider the diagonal with the shortest length and the side with the greatest length. Let them be $\overline{AB}, \overline{CD}$ respectively. Let $AD \cap BC=X$. By the convexity condition and the fact that $AB$ is a diagonal, there is a vertex $P$ inside $\triangle ABX$ such that $AP, PB$ are edges/diagonals.
Now consider $\triangle PAB$. Since $AB$ is the shortest length among all edges/diagonals, $\angle APB\le 60^\circ$. It is easy to see that, therefore, $\angle DPC<60^\circ$. Therefore, in $\triangle PCD$, side $CD$ is not the longest side, contradicting maximality.
The case where $A=D$ has to be considered, but is very similar and left as an exercise.
Diagram is attached. https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvMy81L2U1OWQ5OGRhN2Y0YmQ0NTU1MDgwNDRlZTkzNTc5NzcyM2I1MzMzLnBuZw==&rn=U2NyZWVuIFNob3QgMjAxOC0wMi0wMiBhdCAxMC40MC4yOCBQTS5wbmc=
Smita
10.04.2018 03:54
Let the polygon have n sides then the no. of diagonal are n(n-3)/2 equating it with n give n=0 or n=5