Let $A_0A_1A_2A_3A_4A_5$ be a convex hexagon inscribed in a circle. Define the points $A_0'$, $A_2'$, $A_4'$ on the circle, such that \[ A_0A_0' \parallel A_2A_4, \quad A_2A_2' \parallel A_4A_0, \quad A_4A_4' \parallel A_2A_0 . \] Let the lines $A_0'A_3$ and $A_2A_4$ intersect in $A_3'$, the lines $A_2'A_5$ and $A_0A_4$ intersect in $A_5'$ and the lines $A_4'A_1$ and $A_0A_2$ intersect in $A_1'$. Prove that if the lines $A_0A_3$, $A_1A_4$ and $A_2A_5$ are concurrent then the lines $A_0A_3'$, $A_4A_1'$ and $A_2A_5'$ are also concurrent.
Problem
Source: Romanian IMO TST 2005 - day 3, problem 1
Tags: trigonometry, geometry, trapezoid, geometry proposed
19.04.2005 21:44
Just sine law+(trigo form of) Ceva. I believe everyone solved this problem.
19.04.2005 22:28
mecrazywong wrote: Just sine law+(trigo form of) Ceva. I believe everyone solved this problem. Quite the opposite!
19.04.2005 23:23
Valentin Vornicu wrote: mecrazywong wrote: Just sine law+(trigo form of) Ceva. I believe everyone solved this problem. Quite the opposite! Valentin, you mean Ceva doesn't work or someone failed to solve it?
19.04.2005 23:37
I mean many students failed to solve it.
20.04.2005 12:08
I don't see any reason unless they haven't heard about the trigo form of ceva
01.05.2005 01:36
can anyone post more details?
01.05.2005 13:40
perfect_radio wrote: can anyone post more details? Just angle chasing+trigo form of Ceva. No other techniques are needed.
01.05.2005 22:52
i understood that from your previous posts. to you the solution may seem obvious, but to me it's quite the opposite
02.05.2005 17:18
Ok, here are some more datails: $\frac{A_2A_3'}{A_3'{A_4}}=\frac{\sin{A_2A_3A_3'}}{\sin{A_3A_0A_4}}\frac{\sin{A_3A_4A_2}}{\sin{A_3'A_3A_4}}$. But $\angle{A_2A_3A_0'}=\angle{A_2A_5A_4'}$ from isosceles trapeziums. Use the other similar relations and Ceva trigonometric form to complete the solution.
02.05.2005 21:08
Thanks Sailor for explaining this in details
02.05.2005 23:15
Sailor wrote: Ok, here are some more datails: $\frac{A_2A_3'}{A_3'{A_4}}=\frac{\sin{A_2A_3A_3'}}{\sin{A_3A_0A_4}}\frac{\sin{A_3A_4A_2}}{\sin{A_3'A_3A_4}}$. But $\angle{A_2A_3A_0'}=\angle{A_2A_5A_4'}$ from isosceles trapeziums. Use the other similar relations and Ceva trigonometric form to complete the solution. From my calculations i get: $\frac{A_2A_3'}{A_3'{A_4}}=\frac{\sin{A_2A_3A_3'}}{\sin{A_3A_2A_4}}\frac{\sin{A_3A_4A_2}}{\sin{A_3'A_3A_4}}$. is $A_3A_2A_4=A_3A_0A_4$ ??? i can't get $\angle{A_2A_3A_0'}=\angle{A_2A_5A_4'}$ from isosceles trapeziums
03.05.2005 11:24
Of course, $\angle{A_3A_2A_4}=\angle{A_3A_0A_4}$ because the equal to half the $arc{A_3A_4}$, but $arc{A_0'A_2}=arc{A_0A4}=arc{A_2A_4'}$ and hence $\angle{A_2A_3A_0'}=\angle{A_2A_5A_4'}$.
23.12.2010 21:50
Sailor wrote: Ok, here are some more datails: $\frac{A_2A_3'}{A_3'{A_4}}=\frac{\sin{A_2A_3A_3'}}{\sin{A_3A_0A_4}}\frac{\sin{A_3A_4A_2}}{\sin{A_3'A_3A_4}}$. But $\angle{A_2A_3A_0'}=\angle{A_2A_5A_4'}$ from isosceles trapeziums. Use the other similar relations and Ceva trigonometric form to complete the solution. Can write more detail?
11.10.2024 09:06
A complete solution?