Given an acute scalene triangle $ABC$ inscribed in circle $(O)$. Let $H$ be its orthocenter and $M$ be the midpoint of $BC$. Let $D$ lie on the opposite rays of $HA$ so that $BC=2DM$. Let $D'$ be the reflection of $D$ through line $BC$ and $X$ be the intersection of $AO$ and $MD$. a) Show that $AM$ bisects $D'X$. b) Similarly, we define the points $E,F$ like $D$ and $Y,Z$ like $X$. Let $S$ be the intersection of tangent lines from $B,C$ with respect to $(O)$. Let $G$ be the projection of the midpoint of $AS$ to the line $AO$. Show that there exists a point with the same power to all the circles $(BEY),(CFZ),(SGO)$ and $(O)$.
Problem
Source: Vietnam TST 2019 Day 1 P3
Tags: geometry, geometric transformation, reflection
Pathological
30.03.2019 23:39
(a) Let $A'$ be the point diametrically opposite to $A$ on $(O),$ so that $M$ is the midpoint of $BC$ and of $HA'.$ Let $D_1$ be the foot from $A$ to $BC.$ Observe that $\angle BDC = \angle BD'C = 90.$
By Menelaus' Theorem in $\triangle AHA'$ with transversal $DMX,$ we get that $\frac{AX}{A'X} = \frac{AD}{DH}.$ Since $D_1D^2 = D_1D'^2 = D_1B * D_1C = D_1H * D_1A,$ it's well-known by a property of harmonic ranges that $(A, H; D, D') = -1.$ Therefore, we know that $\frac{AD}{DH} = \frac{AD'}{HD'}.$ Therefore, as $\frac{AX}{A'X} = \frac{AD'}{HD'},$ we know that $HA' || D'X.$ Thus, as $AM$ is an $A-$median in $\triangle AHA',$ it must also be an $A-$median in $\triangle AD'X$ by homothety about $A,$ i.e., $AM$ bisects $D'X.$
(b) We claim that the desired point is just the symmedian point $K.$ Let's first show that $K$ is on the radical axis of $(O), (ADX),$ from which it follows that $K$ has equal powers w.r.t. $(O), (ADX), (BEY), (CFZ)$ by symmetry.
Let $T = AK \cap (O).$ It suffices to show that $ADTX$ is cyclic, since from then it would follow that $AT$ is the radical axis of $(O), (ADX),$ and so we'd be done as $K \in AT.$ Let $H_a$ be the reflection of $H$ over $BC$ (well-known that this is on $(O)$). It suffices to show that $T$ is the center of the spiral similarity sending $H_aD$ to $A'X.$ Since we already know $\angle TH_aD = \angle TA'X,$ we just need that $\frac{d_1}{d_2} = \frac{H_aD}{A'X},$ where $d_1, d_2$ denote the distances from $T$ to $AH, AO$ respectively. However, as $AM$ is the $A-$median of $\triangle AHA',$ and $AT, AM$ are isogonal conjugates w.r.t $\angle HAA',$ we know that $AT$ is the $A-$symmedian of $\triangle AHA',$ and so hence $\frac{d_1}{d_2} = \frac{AH}{AA'} = \frac{HD'}{A'X} = \frac{H_aD}{A'X},$ where we used that $HA' || D'X$ and $HD' = H_aD.$ Therefore, we've shown that $ADTX,$ and so we've obtained that $K$ has equal powers w.r.t $(O), (ADX), (BEY), (CFZ),$ with the second two following by symmetry.
It now suffices only to show that $K$ has equal power w.r.t $(O)$ and $(SGO).$
Let $Y$ be the foot of the altitude from $O$ to $AS,$ and let $Z$ be the midpoint of $AY.$ It's then easily seen that $ZOGS$ is cyclic. Therefore, since $A, Z, K, T, S$ are all on $AK,$ it suffices by Power of the Point to show that $ZK * KS = AK * KT.$ We can do this by some length computations. All computations from here on are directed.
Set $W = AK \cap BC.$ Notice that it's well-known that $Y$ is the midpoint of $AT,$ and so we have $AZ = \frac14 AT.$ Thus, we need to show that:
$$(AK - \frac14 AT) *(KS) = AK * KT \Leftrightarrow 1 - \frac{AT}{4 AK} = 1 - \frac{ST}{SK},$$
so that we want to show that $\frac{ST}{SK} = \frac{AT}{4AK}.$
Now, let's WLOG assume that $AK = k$ and $KW = 1.$ Since $(A, W; K, S) = -1,$ we know now that $WS = \frac{k+1}{k-1}.$ Now, since $(A, T; W, S) = -1,$ we have that $WT = \frac{k+1}{2k-1}$ and $TS = \frac{k(k+1)}{(k-1)(2k-1)}.$
Therefore, $\frac{AT}{4AK} = \frac{k + 1 + \frac{k+1}{2k-1}}{4k} = \frac{2k(k+1)}{4k(2k-1)} = \frac{k+1}{4k-2}.$
Also, $\frac{ST}{SK} = \frac{\frac{k(k+1)}{(k-1)(2k-1)}}{1 + \frac{k+1}{k-1}} = \frac{k+1}{4k-2},$ which matches what we got for $\frac{AT}{4AK}.$
Therefore, we've shown that $K$ has equal power to $(O), (SGO),$ and so hence it has equal powers w.r.t $(O), (ADX), (BEY), (CFZ), (SGO),$ as desired.
$\square$
khanhnx
01.04.2019 16:06
Here is my solution for this problem Solution a) Let $K$ $\equiv$ $AO$ $\cap$ ($O$) ($K$ $\ne$ $A$); $T$ $\equiv$ $AH$ $\cap$ $BC$, $N$ $\equiv$ $AM$ $\cap$ $D'X$ Since: $\overline{AD'}$ . $\overline{AD}$ = $\overline{AH}$ . $\overline{AT}$, we have: $\dfrac{\overline{AD'}}{\overline{AT}}$ = $\dfrac{\overline{AH}}{\overline{AD}}$ = $\dfrac{\overline{AD'} - \overline{AH}}{\overline{AT} - \overline{AD}}$ = $\dfrac{\overline{HD'}}{\overline{DT}}$ So: $\dfrac{\overline{HD'}}{\overline{DD'}}$ = $\dfrac{\overline{HD'}}{2 \overline{DT}}$ = $\dfrac{\overline{AH}}{2 \overline{AD}}$ = $\dfrac{\overline{OM}}{\overline{AD}}$ = $\dfrac{\overline{MX}}{\overline{DX}}$ or $D'X$ $\parallel$ $HK$ But: $M$ is midpoint of $HK$ then $N$ is midpoint of $D'X$ Hence: $AM$ passes through midpoint of $D'X$
Attachments:
khanhnx
01.04.2019 18:09
b) Let $L$ be symmedian point of $\triangle$ $ABC$; $U$ is reflection of $C$ through $O$; $Q$ be midpoint of $AB$; $W$ $\equiv$ $CL$ $\cap$ ($O$) ($W$ $\ne$ $C$); $I$ be intersection of tangents at $A$, $B$ of ($O$); $J$ $\equiv$ $BU$ $\cap$ $CA$; $V$ $\equiv$ $AU$ $\cap$ $AB$; $R$ be a point on the opposite ray of $UC$ which satisfies $R$ $\in$ ($I$; $IB$) We have: ($JB$; $JA$) $\equiv$ $\dfrac{\pi}{2}$ $-$ ($CA$; $CB$) $\equiv$ $\dfrac{\pi - (CA; CB)}{2}$ $\equiv$ $\dfrac{(IB; IA)}{2}$ (mod $\pi$) So: $J$ $\in$ ($I$; $IB$) Similarly: $V$ $\in$ ($I$; $IB$) But: $AJ$ $\perp$ $AU$, $BV$ $\perp$ $BU$ then: $I$, $J$, $V$ are collinear Since: $JV$ is anti - parallel line of $\triangle$ $ABC$, we have: $\triangle$ $ABC$ $\stackrel{-}{\sim}$ $\triangle$ $VJC$ Hence: $\triangle$ $BQF$ $\stackrel{-}{\sim}$ $\triangle$ $JIR$ So: ($QO$; $QZ$) $\equiv$ ($QI$; $QF$) $\equiv$ $\dfrac{\pi}{2}$ $-$ ($QF$; $QB$) $\equiv$ $\dfrac{\pi}{2}$ $-$ ($IJ$; $IR$) $\equiv$ ($RI$; $RU$) (mod $\pi$) or $I$, $Q$, $Z$, $R$ lie on a circle But: $U$ is orthocenter of $\triangle$ $VJC$ then: $\overline{CZ}$ . $\overline{CR}$ = $\overline{CW}$ . $\overline{CI}$ or $I$, $W$, $Z$, $R$ lie on a circle Hence: $I$, $W$, $Q$, $Z$, $R$ lie on a circle So: ($FC$; $FZ$) $\equiv$ ($QO$; $QZ$) $\equiv$ ($RI$; $RZ$) $\equiv$ ($WC$; $WZ$) (mod $\pi$) or $W$ $\in$ ($CFZ$) Then: $C$ - symmedian of $\triangle$ $ABC$ is radical axis of ($O$) and ($CFZ$) Similarly: $B$ - symmedian of $\triangle$ $ABC$ is radical axis of ($O$) and ($BEY$) Hence: $L$ is radical center of ($O$), ($BEY$) and ($CFZ$) or $P_{L / (O)}$ = $P_{L / (BEY)}$ = $P_{L / (CFZ)}$ Let $S'$ , $T'$ be midpoint of $AS$, $AT$; $G'$ $\equiv$ $AO$ $\cap$ ($BOCS$) ($G'$ $\ne$ $O$); $P$ $\equiv$ $GS'$ $\cap$ ($SGO$) ($P$ $\ne$ $G$) Since: ($AT$; $AM$) $\equiv$ ($AS$; $AG'$) $\equiv$ $-$ ($AG'$; $AS$) (mod $\pi$), we have: $\triangle ATM$ $\stackrel{-}{\sim}$ $\triangle$ $AG'S$ But: $T'$, $G$ are midpoint of $AT$, $AG'$, respectively so: $\triangle TMT'$ $\stackrel{-}{\sim}$ $\triangle$ $G'SG$ Then: ($T'T$; $T'M$) $\equiv$ $-$ ($GG'$; $GS$) $\equiv$ $-$ ($PS$; $PO$) (mod $\pi$) or $\triangle TMT'$ $\stackrel{-}{\sim}$ $\triangle$ $SOP$ Hence: $\dfrac{T'T}{TM}$ = $\dfrac{PS}{SO}$ or $\overline{T'T}$ . $\overline{OS}$ = $\overline{TM}$ . $\overline{PS}$ Since: $BC$, $OS$ are radical axises of (($O$); ($BOCS$)), (($BOCS$); ($SGO$)), we have: $M$ is radical center of ($O$), ($BOCS$) and ($SGO$) So: $M$ lies on radical axis of ($SGO$) and ($O$) It's well - known that: $L$, $T'$, $M$ are collinear Then: $\overrightarrow{T'M}$ . $\overrightarrow{OP}$ = ($\overrightarrow{T'T}$ + $\overrightarrow{TM}$) ($\overrightarrow{OS}$ + $\overrightarrow{SP}$) = $\overrightarrow{T'T}$ . $\overrightarrow{OS}$ + $\overrightarrow{T'T}$ . $\overrightarrow{SP}$ + $\overrightarrow{TM}$ . $\overrightarrow{OS}$ + $\overrightarrow{TM}$ . $\overrightarrow{SP}$ = $\overline{T'T}$ . $\overline{OS}$ $-$ $\overline{TM}$ . $\overline{PS}$ = 0 or $T'M$ $\perp$ $OP$ Hence: $T'M$ is radical axis of ($SGO$) and ($O$) or $P_{L / (SGO)}$ = $P_{L / (O)}$ Therefore: $P_{L / (O)}$ = $P_{L / (BEY)}$ = $P_{L / (CFZ)}$ = $P_{L / (SGO)}$ or symmedian point $L$ of $\triangle$ $ABC$ has equal power to the circles ($O$), ($BEY$), ($CFZ$), ($SGO$)
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G81928128
10.09.2019 18:43
Part a)Let $A'$ be the point diametrically opposite $A$ on $(ABC)$. It's well known that $H,M,A'$ collinear, and that $AM$ is the median of $\triangle AHA'$. It suffices to show $D'X || HM \Leftrightarrow X(H,A';M,D') = -1 \Leftrightarrow (H,A;D,D') = -1 \Leftrightarrow WH \cdot WA = (WD')^2$, where $W$ is the midpoint of $DD'$, i.e. the foot of perpendicular from $A$ onto $BC$.
But $WH \cdot WA = - WB \cdot WC = -WD \cdot WD'$ (recall that the reflection of $H$ across $BC$ lies on $(ABC)$), so we have $D'X || HM \Rightarrow \triangle AD'X \sim \triangle AHA'$, so $AM$ is also the median of $\triangle AD'X$.
Part b)We claim that the point in question is the symmedian point $K$ of $\triangle ABC$
First halfWe claim that $K$ is on the radical axis of $(ABC)$ and $(ADX)$ (and so $(BEY),(CFZ)$ by symmetry).
Let $(AXD),(ABC)$ meet again at $U$. It suffices to show that $K\in AU \Leftrightarrow AU$ is the symmedian of $\triangle ABC$.
Invert about $A$ with radius $\sqrt{AB\cdot AC}$, then reflect across the angle bisector of $\angle BAC$. This sends $B\leftrightarrow C,(ABC)\leftrightarrow BC$. We want to show $D\leftrightarrow X \Leftrightarrow AD \cdot AX = AB \cdot AC$ (since $AH$ and $AO$ are isogonal). Let the foot of perpendicular from $B$ to $AC$ be $E$; note $E\in (BDCD')$.
Now, we have $AD \cdot AX = AD \cdot \frac{AD'}{AH} \cdot AA' = AE \cdot AC \cdot \frac{AA'}{AH} = \frac{AB}{AA'} \cdot AC \cdot AA' = AB \cdot AC$ (as $\triangle AEH \sim \triangle ABA'$). Thus, $D\leftrightarrow X$, so $(ADX) \leftrightarrow DX$.
But since $M=DX\cap BC$, it maps to the second intersection of $(ADX)$ and $(ABC)$, i.e. $U$, so $AU,AM$ are isogonal as desired.
Second halfNow, we will show that $K$ is on the radical axis of $(ABC)$ and $(SGO)$.
Invert about $(ABC)$. Then $S\leftrightarrow M$; let $N$ be the point such that $G\leftrightarrow N$, so $(A,A';G,N)=-1$. Then $(SGO)\leftrightarrow MN$, which is the radical axis of the two circles, so it suffices to show that $K,M,N$ collinear.
Now let $P=BC\cap AS$; $C(A,B;K,S)=-1 \Rightarrow (A,P;K,S) = -1$. Let $Q = BC\cap AO$. It suffices to show $(A,Q;O,N)=-1$ as that would imply $PQ,OS,NK$ concurrent, and $PQ,OS$ intersect at $M$.
Now, invert about perform the inverflection as in the previous part. Again, $B\leftrightarrow C$; $O$ gets sent to the reflection of $A$ across $BC$ (call this $R$), so $A'\leftrightarrow W$ (as defined in part a)). $Q'$ gets sent to $AW \cap (ABC)$, i.e. the reflection of $H$ across $BC$, $H'$. Also, $P$ gets sent to $AM\cap (ABC)$; call this $M'$. Suppose $S$ gets sent to $S'$. Then since $(A,U;P,S)=-1$, then $(\infty,M;M',S')=-1$, i.e. $S'$ is the reflection of $M'$ across $M$ (recall that inversion preserves cross-ratios). So the image of the midpoint of $AS$ would be the reflection of $A$ cross $S'$; call this $T$. $G$ gets sent to a point $G'$ on $AW$ such that $\angle ATG' = 90$. Let $N'$ be the image of $N$. Then $(A,A';G,N) = (\infty,W;G',N')=-1$, so $N'$ is the reflection of $G'$ across $W$, i.e. across $BC$. Now the condition we want to prove, $(A,Q;O,N) = -1 \Leftrightarrow (\infty,H';R,N') = -1 \Leftrightarrow H'$ is the midpoint of $RN'$.
At this point, I caved and bashed with coordinates, but I later found a synthetic finish. Let $N''$ be the reflection of $R$ across $H'$, and $G''$ be the reflection of $N''$ across $W$, so we want to show $G'=G''$, i.e. $\angle ATG'' = 90$. Note that $A,H$ are reflections of $R,H'$ across $BC$, so the image of reflecting $N''$ across $W$ (which is $G''$) is just the reflection of $A$ over $H$. In short, $H$ is the midpoint of $AG''$. Since $S'$ is the midpoint of $AT'$, it suffices to show $\angle AS'H = 90$. Moreover, $\angle BS'C = \angle CM'B = \angle CH'B = \angle BHC$, so $S\in (HBC)$, which is just the reflection of $(ABC)$ over $BC$ (and so also contains $R$). So if $AM$ intersects $(HBC)$ again at $V$, $V$ is just the reflection of $A$ across $M$, so $RV || BC \perp AR \Rightarrow \angle AS'H = \angle VRH = 90$, as desired.
alexiaslexia
16.05.2021 09:31
Weirdly bashable TST $\#3$. In its core, this Solution tries to $\textit{demystify}$ the absurd $X, G$ and $S$ the problem gave us as $\textsf{floating points}$, before determining their uses to establish $K$ as the radical center of all the circles. According to vanilla conventions in the above posts, let $K$ be the symmedian point of $\triangle ABC$.
$\color{red} \spadesuit$ $\color{red} \boxed{\textbf{Proof.}}$ $\color{red} \spadesuit$ We strengthen the Claim: let $X' = D'M \cap AO$. Then, we will prove that $I_A$ is the Miquel Point of the complete quad $DD'X'X$. Of course, this directly implies that $(ADI_AX)$ is a cyclic quad. To that end, we prove that $(DD'MI_A)$ and $(XX'MI_A)$ are cyclic, an arguably easier way to prove that $I_A$ also passes through $(ADX)$ and $(AD'X')$. $\color{red} \rule{25cm}{0.2pt}$ First, to proving $(DD'MI_A)$ concylic. Now, we go further by defining $H_A$ to be the reflection of $H$ to $BC$, and $Q_A$ to be the harmonic conjugate of $P_A$ with respect to $BC$. We Claim that $(Q_AD'MI_AD)$ are all cyclic. Projecting $(P_A,Q_A;B,C) = -1$ from $H_A$ to $(ABC)$ yields \[ (A; \overline{H_AQ_A} \cap (ABC);B,C) = -1\]which implies $Q_A,H_A$ and $I_A$ collinear. After that, noting that $Q_AD$ is a tangent of the circle $(BP_CD'P_BCD)$, we know that \[ Q_AH_A \cdot Q_AI_A = Q_AB \cdot Q_AC = Q_AD'^2 \]implying that $(Q_AD'I_AD)$ concyclic by inversion centered in $Q_A$ with radius $Q_AD'$. Finally, by standard projective techniques, since $M$ is the midpoint of $BC$ (also the center of the circle $(BD'C)$), we know that \[ \angle Q_AD'M = \angle Q_ADM = 90^{\circ} \]implying $(Q_AD'MD)$ cyclic. $\color{red} \rule{25cm}{0.2pt}$ Second, to proving $(XX'MI_A)$ concyclic. Here we cheese the fact by computing length ratios: we can reduce the needed \[ \angle DXI_A = \angle D'X'I_A \ \text{or} \ \triangle DXI_A \sim \triangle D'X'I_A \]into proving \[ \dfrac{DI_A}{D'I_A} = \dfrac{DX}{D'X'} \]If this holds true, then the two triangles are indeed similar by $SAS$ similarity. However, we know that \[ \dfrac{DX}{D'X'} = \dfrac{MX}{MD'} = \dfrac{MX}{MD} \]as $MD' = MD$. Also, since $Q_AI_A$ is an angle bisector of $\angle DI_AD'$, \[ \dfrac{DI_A}{D'I_A} = \dfrac{DH_A}{H_AD} = \dfrac{D'H}{HD} \]since $D,H_A$ are reflections of $D',H$ towards $BC$. We are done by observing that $HM \parallel D'X$ from $\color{green} \diamondsuit$ $ \boxed{\textbf{Short Preamble}}$ $\color{green} \diamondsuit$. $\blacksquare$ $\blacksquare$ $\color{red} \rule{4.6cm}{2pt}$ $\color{red} \clubsuit$ $\color{red} \boxed{\textbf{Linear Power Bash.}}$ $\color{red} \clubsuit$ $\color{red} \rule{4.6cm}{2pt}$ We know that $(ADX), (BEY), (CFZ)$ and $(O)$ have radical center $K$. Here, we prove that $K$ has the same power towards $(O)$ and $(S_AG_AO)$; with $S_A, G_A$ replacing $S,G$'s definitions. $\color{red} \rule{25cm}{0.2pt}$ $\color{red} \spadesuit$ $\color{red} \boxed{\textbf{Proof.}}$ $\color{red} \spadesuit$ First, note that $\text{Pow}_A(O) = 0$ and $\text{Pow}_{S_A}(S_AG_AO) = 0$. As we'd like to prove $\text{Pow}_{K}(O) - \text{Pow}_{K}(S_AG_AO) = 0$, it remains to prove \[ \dfrac{\text{Pow}_A(S_AG_AO)}{\text{Pow}_{S_A}(O)} = \dfrac{AK}{KS_A} \]by power linearity. Moreover, aside from $I_A$, we also draw $I_B$ for later usage. We compute the LHS first: \begin{align*} \text{LHS} &= \dfrac{R \cdot \frac{AS_A \cos{(\angle I_AAO)}}{2}}{R^2 \tan^2{A}} \\ &= \dfrac{R \cdot R \tan{A} \frac{\sin{C}}{2 \sin{x}} \cos{(I_AAO=90-C-x)}}{R^2\tan^2{A}} \\ &= \dfrac{\sin{C}}{\tan{A}} \cdot \dfrac{\sin{(C+x)}}{2\sin{x}} = \dfrac{\sin{C}}{\tan{A}} \cdot \dfrac{\sin{\angle(ACI_A)}}{2\sin{\angle(BCI_A)}} \end{align*}(letting $x = \angle BAI_A$) and the RHS next: \begin{align*} \text{RHS} &= \dfrac{AB \cdot \sin{(\angle ABK)}}{BS_A \cdot \sin{(\angle S_ABK)}} \\ &= \dfrac{2R\sin{C}}{R\tan{A}} \cdot \dfrac{\sin{(\angle ACI_B)}}{\sin{(\angle BAI_B)}} \end{align*}This might seem like a dead end, but we will prove that both of those equations are equal to $\dfrac{\sin{C}}{\tan{A}} \cdot \dfrac{b}{a}$. Redraw the midpoint $M_A,M_B$ of $BC$ and $AB$. The conclusion follows from the final observation that $\triangle AI_BB \sim \triangle M_BCB$ and $\triangle AI_AB \sim \triangle ACM_A$ (so the sine ratios are easily represented by side ratios.) We are done. $\blacksquare$ $\blacksquare$ $\blacksquare$
Most of the meat of this Motivation is to motivate the existence of $I_A$ and why it almost magically appears on the radical center of $(ADX)$ and $(O)$.
$\color{green} \rule{25cm}{2pt}$
$\color{red} \text{Sec 1: Circular Argument Multiplication.}$ On the first time attempting this problem, I was extremely sleepy and my brain power wasn't ready to be strained --- but after seeing the relatively simple statement on part (a), I was down for one more (sub)problem.
Basically, my first diagram was akin to Fig 1 (without the addition of $A'$.) While I had already conjured the obvious
\[ (H,A;D,D') = -1 \]in mind, I had not gotten any ideas about the beauty/characterization of $X$. Sure, it is in the intersection of $\overline{DM} \cap \overline{AO}$, but what does $M$ and $O$ have to do with the earlier four points? And then, came the idea to explicitly draw $\overline{OM}$, because it's easily proven to be parallel with $\overline{AD'HD}$.
Since we are asked to prove $D'L = LX$ where $L = \overline{AM} \cap \overline{DX}$, by Menelaus on $\triangle AMD$ with transversal $\overline{D'LX}$, that statement is equivalent to
\[ \dfrac{DA}{AD'} = \dfrac{MD}{XM} \]By the harmonic earlier, it's equivalent to
\[ \dfrac{DH}{HD'} = \dfrac{MD}{MX} \Rightarrow HM \parallel D'X \]Here I finally got the insight to draw $A' = \overline{HM} \cap (ABC)$, and we can just simply show $D'X \parallel A'H$. Since $HM = MA'$ seems a good candidate to project $(H,A;D,D')$ into, I did exactly that and part (a) is done.
$\color{green} \rule{25cm}{2pt}$
$\color{red} \text{Sec 2: The big absence of} \ D'.$ Next, as the easier circle to analyze between $(BEY)$ and $(SGO)$ is the former one, I tried to find the (second) intersection of $(ADX)$ and $(O)$. If this is doable, then the cevians will intersect at one point, and the final aim would be to consider the power of this point relative to $(SGO)$.
However, the absence of $D'$ in the second statement bothered me: why bother having $(ADX) \cap (O)$ when $(ADX)$ can be replaced with its counterpart? If this is so (and the difference between $D$ and $D'$ is only whether it lies inside or outside $\triangle ABC)$ then $(ADX), (AD'X')$ and $(O)$ is coaxial!
So, defining $X'$ seemed like a perfect way to neutralize the dominance of $(ADX)$, and it quickly shifted into finding the Miquel Point of the aforementioned complete quad. Since $(X'XM)$ is very dependent on the definitions of $D$ and $D'$, it seemed better to analyze the circle $(D'MD)$ first.
Since $S$ is already defined for us first, the first good candidate of the second intersection, is, of course, $I_A$. While this point is hardly related with any of the other points, the existence of $D,D'$ may warrant a new helping point. The fact that gave away $Q_A$'s construction was that if that is constructed as the conjugate of $P_A$, it seemed like a quick transform away from $(A,I_A;B,C)$ --- and this is exactly the case.
$\color{cyan} \rule{25cm}{2pt}$
$\color{blue} \text{Sec 2.5: Second Pair - Finding the Right Ratio.}$ Next is proving that $I_A$ takes $DD'$ to $XX'$. I originally planned to include $H_A$ to $A'$, and compare the ratio of $DH_A/H_AD'$ with $XA'/A'X'$, but $\textbf{all}$ the lengths
\[ |I_AX|, |I_AX'|, |XA'|, |A'X'| \]is too obscure to be (comfortably) computable. So, this brought the idea to individually consider the more defined $|DX|$ and $|D'X'|$ --- and this way of computing was more promising, since we know that $D'M = MD$. After simplifying the expression, we are quite done by citing part (a).
\[\begin{tabular}{p{12cm}}
$\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\
\end{tabular} \\ \]Notes.Notes. The most difficult part about the trig bash was the randomness of the angles being stuck with the (sym)medians; particularly $\angle I_AAO$. Since we wish to compare $\cos{(\angle I_AAO)}$, I thought tying it with $\angle BAO$ was the way to go, since any semblance of $\cos$ is usually better shifted into $\sin$ for more sides/angle sines ratio (by sine Lemma.)
The quantity
\[ \dfrac{\sin{(C+x)}}{\sin{x}} \]is also intimidating at first sight. To simplify this, it's usually a good idea to find (or artificially construct) some chord of $(ABC)$ relating to these strange angles, and turn it into
\[ \dfrac{2R \cdot \sin{(C+x)}}{2R \cdot \sin{x}} \]for further examination.