parmenides51 wrote: A parallelogram $ABCD$ is given. Two circles with centers at the vertices $A$ and $C$ pass through $B$. The straight line $\ell$ that passes through $D$ and crosses the circles at second time at points $X, Y$ respectively. Prove that $BX = BY$. The straight line $\ell$ that passes through $B$ and crosses the circles at second time at points $X, Y$ respectively. Prove that $DX = DY$

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Given a parallelogram $A(0,0),B(b,0),C(c,a),D(c-b,a)$. The line $BX$, $BY$ has equation $y=m(x-b)$. Point $X(\frac{b(m^{2}-1)}{m^{2}+1},-\frac{2bm}{m^{2}+1})$. Point $Y(\frac{bm^{2}+2am-b+2c)}{m^{2}+1},\frac{2m(am-b+c)}{m^{2}+1})$. $DX^{2}=DY^{2}=\frac{a^{2}(m^{2}+1)+4abm+4b^{2}m^{2}-4bcm^{2}+c^{2}(m^{2}+1)}{m^{2}+1}$.

vanstraelen wrote: parmenides51 wrote: A parallelogram $ABCD$ is given. Two circles with centers at the vertices $A$ and $C$ pass through $B$. The straight line $\ell$ that passes through $D$ and crosses the circles at second time at points $X, Y$ respectively. Prove that $BX = BY$. The straight line $\ell$ that passes through $B$ and crosses the circles at second time at points $X, Y$ respectively. Prove that $DX = DY$ you are correct, there was a typo, the two circles passed through $D$ and not $B$, in order to keep your solution, I shall keep your correction which is equivalent, thanks

Dear Mathlinkers, any synthetic proof? Sincerely Jean-Louis

here is a synthetic proof: WLOG let $AB<BC$ and let $G$ the intersection of $(A,AB)$ and $AD$. finally let $S$ the intersection of $[DC)$ and $(C,CB)$. Now let $BG$ cuts $(C,CB)$ at $T$ it's easy to prove that $T,D,C$ are collinear. $FAC=2 FBG=2BST=DCA$. $AD=CY$, $AX=CD$ so $DCY$ and $XAD$ are congruent so $DX=DY$ .

vanstraelen wrote: Given a parallelogram $A(0,0),B(b,0),C(c,a),D(c-b,a)$. The line $BX$, $BY$ has equation $y=m(x-b)$. Point $X(\frac{b(m^{2}-1)}{m^{2}+1},-\frac{2bm}{m^{2}+1})$. Point $Y(\frac{bm^{2}+2am-b+2c)}{m^{2}+1},\frac{2m(am-b+c)}{m^{2}+1})$. $DX^{2}=DY^{2}=\frac{a^{2}(m^{2}+1)+4abm+4b^{2}m^{2}-4bcm^{2}+c^{2}(m^{2}+1)}{m^{2}+1}$. I highly recommend not bashing on olympiads. Someone once said that if you bash, then you would always get a 5 if correct, 0 if incorrect (or -1).

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Highway%20to%20Geometry%204.pdf p. 18... Sincerely Jean-Louis