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Source: ELMO Shortlist 2013: Problem G6, by Victor Wang

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#1 h Jul 23, 2013, 1:32 AM • 3 Y

Y by HamstPan38825, Adventure10, Mango247

Let $ABCDEF$ be a non-degenerate cyclic hexagon with no two opposite sides parallel, and define $X=AB\cap DE$ , $Y=BC\cap EF$ , and $Z=CD\cap FA$ . Prove that
$\frac{XY}{XZ}=\frac{BE}{AD}\frac{\sin |\angle{B}-\angle{E}|}{\sin |\angle{A}-\angle{D}|}.$ Proposed by Victor Wang

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#2 h Jul 23, 2013, 7:35 AM • 4 Y

Y by Pirkuliyev Rovsen, Adventure10, Mango247, and 1 other user

Denote $K \equiv BC \cap FA.$ Clearly $\widehat{AZD}=|\widehat{B}-\widehat{E}|$ and $\widehat{BYE}=|\widehat{A}-\widehat{D}|.$ Hence, by sine law for $\triangle ZAD$ and $\triangle BYE,$ we get:

$\frac{\sin | \widehat{B}-\widehat{E}|}{AD}=\frac{ \sin \widehat{ADC}}{AZ}=\frac{ \sin \widehat{ABK}}{AZ} \ , \ \frac{ \sin | \widehat{A}-\widehat{D}|}{BE}=\frac{ \sin \widehat{FEB}}{BY}=\frac{ \sin \widehat{BAK}}{BY}$

$\Longrightarrow \frac{ \sin | \widehat{B}-\widehat{E}|}{\sin | \widehat{A}-\widehat{D}|} \cdot \frac{BE}{AD}=\frac{ \sin \widehat{ABK}}{\sin \widehat{BAK}} \cdot \frac{BY}{AZ}=\frac{AK}{BK} \cdot \frac{BY}{AZ} \ (1).$

By Pascal theorem, $X,Y,Z$ are collinear, hence by Menelaus' theorem for $\triangle KYZ$ cut by the line $\overline{ABX},$ we get

$\frac{XY}{XZ} \cdot \frac{AZ}{AK} \cdot \frac{BK}{BY} =1 \Longrightarrow \frac{XY}{XZ}=\frac{AK}{BK} \cdot \frac{BY}{AZ} \ (2).$

Combining $(1)$ and $(2)$ gives $\frac{XY}{XZ}=\frac{ \sin | \widehat{B}-\widehat{E}|}{\sin | \widehat{A}-\widehat{D}|} \cdot \frac{BE}{AD}.$

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#3 h Jun 23, 2014, 11:45 AM • 2 Y

Y by Adventure10, Mango247

Let (*) be the statement we want to prove.
Complex bash with $(ABCDEF)$ being the unit circle. (The calculations can be done within 45 minutes with the key being to factorise expressions along the way).
Let $H$ be the point on $(ABCDEF)$ such that $CH || AF$ . Compute $h=\frac{af}{c}$ . By easy angle chasing we get $\angle DAH = |\angle B - \angle E|$ so by the Extended Sine Law, $DH=2\sin (\angle B-\angle E)$ . Now, note that if $u,v$ are points on the unit circle, $|u-v|^2=(u-v)(\frac{1}{u}-\frac{1}{v})=-\frac{(u-v)^2}{uv}$ (1).
Hence $DH^2=\frac{(d-\frac{af}{c})^2}{\frac{daf}{c}}=\frac{(cd-af)^2}{cdaf}$ . So $\sin^2 (\angle B - \angle E) = \frac{(cd-af)^2}{4cdaf}$ . Similarly, $\sin^2 (\angle A - \angle D) =\frac{(cb-ef)^2}{4cbef}$ . From (1) we can easily obtain $BE^2, AD^2$ , so putting this all together gives: $(RHS (*))^2 = \frac{(cd-af)^2(b-e)^2}{(cb-ef)^2(a-d)^2}$ . (2)

By Pascal's Theorem, $X,Y,Z$ are collinear. Now by Menelaus' Theorem on $\triangle YFZ$ and transversal $GAX$ , we get $\frac{XZ}{XY}\frac{YG}{GF}\frac{FA}{AZ}=1 \implies (LHS (*))^2=\frac{YG^2}{GF^2}\frac{FA^2}{AZ^2}$ . (3)
$ \ [ \begin{align*} YG^2 &= \left\lvert \frac{ab(e+f)-ef(a+b)}{ab-ef} - \frac{bc(e+f)-ef(b+c)}{bc-ef} \right\rvert ^2\\ &= \left\lvert \frac{ef(a-c)(b-e)(b-f)}{(ab-ef)(bc-ef)} \right\rvert ^2\\ &= -\frac{(a-c)^2(b-e)^2(b-f)^2ef}{(ab-ef)^2(bc-ef)^2}. \end{align*}$
$ \ [ \begin{align*} GF^2 &= \left\lvert \frac{ab(e+f)-ef(a+b)}{ab-ef} - f \right\rvert ^2\\ &= \left\lvert \frac{e(f-a)(f-b)}{ab-ef} \right\rvert ^2\\ &= -\frac{(f-a)^2(f-b)^2e}{f(ab-ef)^2}. \end{align*}$
Similarly, $AZ^2 = -\frac{(a-c)^2(a-d)^2f}{a(cd-af)^2}$ .
And from (1) $AF^2 = -\frac{(f-a)^2}{fa}$ .
Now substituting this all into (3), we get an orgeastic feast of cancellation and obtain
$(LHS (*))^2 = \frac{(cd-af)^2(b-e)^2}{(cb-ef)^2(a-d)^2}$ . (4)

So from (2) and (4) $(RHS (*))^2 = (LHS (*))^2$ . Since $0 \leq |\angle B - \angle E|, |\angle A - \angle D| \leq \pi$ , $RHS \geq 0$ . Clearly $LHS \geq 0$ as well, so we can square root both sides safely to get the desired result.

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