Take a projective transformation mapping $\omega$ to a circle and $AC\cap BD$ to its center. $ABCD$ is mapped to a rectangle and $B$-symmedian is mapped to $B$-symmedian again. Now we are left to prove $GI$,$HJ$ and $B$-symmedian are parallel,and $B$-symmedian is in fact perpendicular line from $B$ to $AC$. Now we can do coordinate bash.

Sorry if this sounds silly, but what does $AA \cap CD$ mean (more specifically, what is line $AA$?)?

AnonymousBunny wrote: Sorry if this sounds silly, but what does $AA \cap CD$ mean (more specifically, what is line $AA$?)? It's the line tangent to the circle at $A$. This is just an overall badly phrased problem imo...$B$-symmedian of what triangle? Anyway we can harmonic bash this problem, just note(prove) that $D(E,H;G,F)=B(F,A;I,E)=-1$.

So which triangle is the $ B $-symmedian taken with respect to?

Wolstenholme wrote: So which triangle is the $ B $-symmedian taken with respect to? triangle ABC

Lemma 1: Let $ K $ = $ AA \cap CC $ with respect to $ \omega $. Then $ BK $ is the $ B $-symmedian of triangle $ ABC $. Proof: Let the reflection of $ BK $ about the bisector of $ \angle{ABC} $ meet $ AC $ at $ M $. Then: \[ \frac{AM}{CM} = \frac{BM * \frac{\sin{ABM}}{\sin{BAM}}}{BM * \frac{\sin{CBM}}{\sin{BCM}}} = \frac{\sin{ABM} * \sin{BAK}}{\sin{CBM} * \sin{BCK}} = \frac{\sin{CBK} * \sin{BAK}}{\sin{BCK} * \sin{ABK}} = \frac{CK * BK}{BK * AK} = 1 \] which implies the desired result. Lemma 2: $ AICG $ is a harmonic quadrilateral Proof: Taking perspective at $ B $ and projecting the quadrilateral onto line $ EF $ it suffices to show that $ (E, X; A, F) = -1 $ where $ X = BI \cap EF $. Now let $ Y = AI \cap BF $. It is well-known that $ (E, X; A, F) = -1 $ if and only if $ Y, J, E $ are collinear (this can be proven by subsequent applications of Ceva and Menelaus). But by Pascal's Theorem on $ AABCDI $ we have that $ Y, J, E $ are collinear as desired. Now, by a well-known result, this implies that $ K \in IG $. And now by Pascal's theorem on $ AABGID $ we have that $ K \in HJ $ as well so the three lines in the problem all concur at $ K $ as desired.

Projective transformation and done

Apply Pascal for $\left( \begin{array}{l} A\ D\ G \\ I\ B\ A \\ \end{array} \right)$ $\Rightarrow EF, IG, HJ$ concur at $S$ Apply Pascal for $\left( \begin{array}{l} C\ D\ G \\ I\ B\ C \\ \end{array} \right)$ $\Rightarrow SC$ is tangent to $\omega$, hence $SB$ is the symmedian of $\triangle{ABC}$ Q.E.D

infiniteturtle wrote: Projective transformation and done Can you please write a complete solution using projective transformation ?

XmL wrote: AnonymousBunny wrote: Sorry if this sounds silly, but what does $AA \cap CD$ mean (more specifically, what is line $AA$?)? It's the line tangent to the circle at $A$. This is just an overall badly phrased problem imo...$B$-symmedian of what triangle? Anyway we can harmonic bash this problem, just note(prove) that $D(E,H;G,F)=B(F,A;I,E)=-1$. Why is this enough please to prove parallelism ? I've proven it but can't see the link between them, thank you .

By Pascal's theorem on $GBAADI$, $H,J$ and $AA \cap IG$ are collinear. By Pascal's theorem on $GBCCDI$, $E,F$ and $CC \cap IG$ are collinear. Thus, $AA,CC, HJ, IG$ are collinear, but as line joining $B$ and $AA \cap CC$ is $B-$symmedian of $\triangle ABC$, we are done. $\square$

Darn, this is only a G4? Here is a less ingenious solution using the Generalized Pingpong Lemma: Let $\mathcal C$ be a conic and let $\ell$ be a line meeting $\mathcal C$ at $A,B$. Also, let $P_1,P_2,P_3,P_4$, and let $f_i(P) = P_iP\cap \mathcal C$ for $P\in \mathcal C$. Then, $f_4\circ f_3 \circ f_2\circ f_1 = \mathrm{id} \iff f_4\circ f_3 \circ f_2\circ f_1$ has a fixed point $P\not \in \{A,B\}\iff$ there is an involution swapping $(P_1,P_3)$, $(P_2,P_4)$, and $(A,B)$. The proof is a straight application of Desargue's Involution Theorem Clearly it suffices to show that $GI$, $HJ$, $AA$, and $CC$ concur. Let $BI$ meet $AA = EF$ at $X$. Since $FB\cap \omega = C$, $EC\cap \omega = D$, $FD\cap \omega = I$, $XI\cap \omega = B$, we see that by Generalized Pingpong the involution with fixed points $A, F$ swaps $X,E$. Hence $(F,A;X,E) = -1$, and so \[ -1 = (F,A;X,E) \stackrel{B}{=} (C,A;I,G), \]so $AICG$ is harmonic, or $IG$, $CC$, $AA$ concur, say at $P$. Then, by Pascal on $AADIGB$, $AA\cap IG = P$, $AD\cap BG = H$, and $AB\cap ID = J$ are collinear, so $HJ$ passes through $P$ as well, or that $GI$, $HJ$ meet at $P$ on the $B$-symmedian. $\blacksquare$ Edit: 850th post! Nothing like using nuclear weapons on a bird to celebrate

When taking a projective transformation, why can we only make ABCD a rectangle and not a square?

anser wrote: When taking a projective transformation, why can we only make ABCD a rectangle and not a square? It is well known that there is a projective transformation fixing a circle and mapping a point inside the circle to another point inside the circle ! Apply this for the intersection of $AC$ and $BD$ so $ABCD$ is mapped to a rectangle ( but may not to a square always !) Now one may apply this for this problem since tangency and intersection are preserved !

bcp123 wrote: Take a projective transformation mapping $\omega$ to a circle and $AC\cap BD$ to its center. $ABCD$ is mapped to a rectangle and $B$-symmedian is mapped to $B$-symmedian again. Now we are left to prove $GI$,$HJ$ and $B$-symmedian are parallel,and $B$-symmedian is in fact perpendicular line from $B$ to $AC$. Now we can do coordinate bash. The use of coordinate bash can be avoided by converse of Reim's theorem and angle chasing.

By Pascal's theorem on $(IDAABG)$, we have that $JH$ and $IG$ meet on $AA$. Now, we use complex bashing to show that this intersection is $AA \cap CC$ by showing that $I$, $G$, and $AA \cap CC$ are collinear. This clearly proves the statement. Let $ABCD$ be the unit circle. Then, $F = \frac{a^2(b+c) - bc(2a)}{a^2 - bc}$. Then, $\frac{g-d}{\frac{1}{g} - \frac{1}{d}} = - gd = \frac{d-f}{\bar{d}-\bar{f}} = \frac{d-\frac{a^2(b+c)-2abc}{a^2-bc}}{\frac{1}{d}-\frac{2a-b-c}{a^2-bc}} $. Therefore, $g = \frac{da^2 - ba^2-ca^2 + 2abc-bcd}{2ad-bd-cd-a^2+bc}$. Then, $\frac{g- \frac{a+c}{2ac}}{g-\frac{2}{a+c}} = \frac{(da^2-ba^2-ca^2+2abc-bdc)(ba^2-da^2-ca^2+2acd-bdc)(c-a)}{(2ad-bd-cd-a^2+bc)(2ab-bd-cb-a^2+dc)(a-c)} = \frac{-(da^2-ba^2-ca^2+2abc-bdc)(ba^2-da^2-ca^2+2acd-bdc)}{(2ad-bd-cd-a^2+bc)(2ab-bd-cb-a^2+dc)}$. Notice that this is reflexive for $b$ and $d$, and so, $g, i, \frac{2ac}{a+c}$ are collinear.

Easily solved by Pascal theorem

The important thing to prove is $(A, C; G, I) = -1$. The key is to focus on triangle $CEF$. Invert about $C$ with radius $\sqrt{CE \cdot CF}$ and reflect about the bisector of $\angle ECF.$ Then the problem becomes Inverted Problem wrote: Let cyclic quadrilateral $ABCD$ have circumcircle $\omega$. The tangent to $\omega$ at $A$ meets $CD$ and $CB$ at $E$ and $F$ respectively. The circumcircles of $\triangle CDF$ and $\triangle CBE$ meet $EF$ again at points $G$ and $I$ respectively. Prove that $A$ is the midpoint of $\overline{GI}$ Using directed lengths, this is a simple exercise with Power of a Point. We have \begin{align*} EA^2 &= ED \cdot EC = EG \cdot EF \iff \\ EA^2 &= (EA + AG)(EA + AF) \\ &= EA^2 + EA \cdot AF + AG \cdot EA + AG \cdot AF \\ &\iff EA \cdot AF + AG \cdot EA + AG \cdot AF = 0 \end{align*}and similarly, $$FA \cdot AE + AI \cdot AE + AI \cdot FA = 0.$$Subtracting the two gives $$AG \cdot EA - AI \cdot AE + AG \cdot AF - AI \cdot FA = (EA + AF)(AG + AI) = EF \cdot (AG + AI) = 0$$so $AG + AI = 0$ and thus $A$ is the midpoint of $\overline{GI}$. $\blacksquare$ Finishing up the problem is easy; simply apply Pascal on $JDAABH$ to get that $JH$ and $GI$ meet on $EF$, which implies the conclusion.

This problem dies to two applications of Pascal's Theorem. Let $K = IG \cap HJ$. By Pascal on $AADIGB$, the points $AA \cap IG, H, J$ are collinear, so $IG \cap HJ$ lies on $AA$. By Pascal on $CCDIGB$, the points $CC \cap IG, E, F$ are collinear, so $AA,CC,$ and $IG$ are concurrent. Now $K$ lies on both $AA$ and $CC$, so $K$ is on the $B-$symmedian, as desired. Remarks: The solution is inspired by the fact that why would you label a point $AA \cap CD$ in a problem if it doesn't involve Pascal

Oops why am I so bad at recognizing applications of Pascal's . Solved with MathJams. Let $P = AA \cap CC$, which obviously lies on the $B$-Symmedian of $ABC$. We claim $P$ is the desired concurrency point. Claim: $P$ lies on $GI$. Proof. By Pascal's on $CCBIGD$, we know $CC \cap GI$, $CB \cap ID = F$, and $BG \cap DC = E$ are collinear, so $CC, GI, EF$ are concurrent. This implies $CC \cap EF = CC \cap AA = P$ lies on $GI$, as desired. $\square$ To finish, we note Pascal's on $AADIGB$ implies $AA \cap GI = P$, $AD \cap GB = H$, and $DI \cap AB = J$ are collinear, which obviously suffices. $\blacksquare$ Remark: I actually first spotted the second application of Pascal's in my solution, which further motivated my conjecture that $(G, I; A, C) = -1$.

Let $K=\overline{AA}\cap\overline{GI}.$ Pascal on $AADIGB$ and $CCBGID$ implies $K$ lies on $\overline{HJ}$ and $\overline{CC},$ respectively. $\square$

... Employ a projective transformation which maps $\overline{AC} \cap \overline{BD}$ to the center of $\omega$ and fixing $\omega$, so $ABCD$ becomes a rectangle (using the tangent definition of the $B$-symmedian). This sends the $B$-symmedian to the line through $B$ parallel to $\overline{EF}$, so we want to prove that the other two lines are parallel as well. We could do this by coordbash, but we will not. Since $$\angle BDC=\angle BAC=90^\circ-\angle BAF=\angle EAB,$$$BDEF$ is cyclic, so by Reim's $\overline{GI} \parallel \overline{EF}$. Now, we are done by Reim's again if we can prove that $BDHJ$ is cyclic, or $\angle BJD=\angle BHD$. By a grade-school fact about angles formed by intersecting chords, this is equivalent to $\widehat{AI}=\widehat{AG}$, which follows by symmetry since $AGCI$ is a kite. $\blacksquare$

Let $K = AA \cap CC$ denote the intersection of the tangents at $A$ and $C$. Observe then that $BK$ is the $B$-symmedian of $\triangle ABC$. Then we make the following two claims: Claim 1: Let $K' = GI \cap AA$. Then $J, H, K'$ are collinear. Proof: By Pascal's theorem on $ABGIDA$, we have that \begin{align*} J &= AB \cap ID \\ H &=BG \cap DA \\ K' &= GI \cap AA, \end{align*}are collinear. $\square$ Claim 2: Let $K'' = IG \cap CC$. Then $E, F, K''$ are collinear. Proof: By Pascal's theorem on $CDIGBC$, we have that \begin{align*} E &= CD \cap GB \\ F &= DI \cap BC \\ K'' &= IG \cap CC, \end{align*}are collinear. $\square$ Now from the above two claims, we find that $K' = K'' = AA \cap CC = K$, as desired. $\blacksquare$

Then $AA \cap CC = T$ and the $B$-symmedian can be written as $BT$, however $T$ lies at the point at infinity along $BD$ which is parallel to $GI$ so $GI$, $HJ$ and $BT$ concur at the point at infinity.