## BdMO National 2013: Secondary 10, Higher Secondary 9

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

### BdMO National 2013: Secondary 10, Higher Secondary 9

Six points $A$, $B$, $C$, $D$, $E$, $F$ are chosen on a circle anticlockwise. None of $AB$, $CD$, $EF$ is a diameter. Extended $AB$ and $DC$ meet at $Z$, $CD$ and $FE$ at $X$, $EF$ and $BA$ at $Y$. $AC$ and $BF$ meets at $P$, $CE$ and $BD$ at $Q$ and $AE$ and $DF$ at $R$. If $O$ is the point of intersection of $YQ$ and $ZR$. Find the angle $XOP$.

sowmitra
Posts: 155
Joined: Tue Mar 20, 2012 12:55 am
Location: Mirpur, Dhaka, Bangladesh

### Re: BdMO National 2013: Secondary 10, Higher Secondary 9

In the follwing solution,
$[V_1V_2V_3V_4V_5V_6]$ means that, the Hexagon $V_1V_2V_3V_4V_5V_6$ is cyclic, and, the vertices are connected in this order.
$\mathcal P(\ell_1, \ell_2, \ell_3)$ means that, the lines $\ell_1, \ell_2, \ell_3$ are concurrent at a point.
$\mathcal L(P_1, P_2, P_3)$ means that, the points $P_1, P_2, P_3$ are co-linear.

Now, for the proof:

From Pascal's Theorem , we get,
$[ACEFBD] \rightarrow \mathcal L(AC \cap FB, CE\cap BD, DA\cap EF)$
$\rightarrow \mathcal L(P, Q, DA\cap EF)$
$\rightarrow \mathcal P(PQ, DA, EF)$
$\rightarrow PQ\cap EF=AD\cap EF$ $(1)$
Similarly,
$[CEABDF] \rightarrow \mathcal L(CE\cap BD, EA\cap DF, AB\cap FC)$
$\rightarrow \mathcal L(Q, R, AB\cap CF)$
$\rightarrow QR\cap AB=AB\cap CF$ $(2)$
$[CAEBFD] \rightarrow \mathcal L(CA\cap BF, AE\cap FD, EB\cap DC)$
$\rightarrow \mathcal L(R, P, BE\cap DC)$
$\rightarrow RP\cap DC=BE\cap DC$ $(3)$

But,
$[DABEFC]\rightarrow \mathcal L(AD\cap EF, AB\cap CF, CD\cap BE).$

$\therefore \mathcal L(AD\cap EF, AB\cap CF, CD\cap BE) \rightarrow \mathcal L(RP\cap EF, QR\cap CF, RP\cap BE)$ [$(1),(2)$ & $(3)$] $\rightarrow \mathcal L(RP\cap XY, QR\cap YZ, RP\cap ZX)$
$\therefore \triangle PQR$ and $\triangle XYZ$ are line-persective.
So, from, Desargues's Theorem, they are point-perspective.

$\therefore PX, QY,$ and, $RZ$ are concurrent, i.e,
$\mathcal P(PX, QY, RZ)$
$\rightarrow \mathcal L(P, X, QY \cap RZ)$
$\rightarrow \mathcal L(P, X, O)$
i.e, $P, X$ and $O$ are co-linear.

$\therefore \angle XOP=\boxed{180^\circ}$, and, we're done. $$\square$$

BTW, finally, my $100^{th}$ Post....
"Rhythm is mathematics of the sub-conscious."
Some-Angle Related Problems;

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### Re: BdMO National 2013: Secondary 10, Higher Secondary 9

To understand this solution you need to know about