correctly Ugggh..... I wasted so much of my time for this typoed question

Plzz post all the china tst problems

Seriously though a wrong problem in a TST? [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9.894692020606424cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.354731655688832, xmax = 7.539960364917593, ymin = -0.7763733830788486, ymax = 6.811266093862853; /* image dimensions */ /* draw figures */ draw(circle((0.88,1.74), 2.720588171701112), linewidth(2.)); draw((-0.8955722309517441,-0.32129649800145077)--(2.92,3.54), linewidth(2.)); draw((1.992302314281582,-0.7428176658071035)--(-1.8074932661395424,2.1630602137457644), linewidth(2.)); draw((-0.5440607395423072,-0.19206567194751634)--(0.88,1.74), linewidth(2.)); draw((-0.5282840890116909,6.575108705606808)--(0.11019085908220025,0.6965192481830965), linewidth(2.)); draw((0.11920917712785317,4.352048491861037)--(0.11019085908220025,0.6965192481830965), linewidth(2.)); draw((-0.8955722309517441,-0.32129649800145077)--(1.992302314281582,-0.7428176658071035), linewidth(2.)); draw((xmin, 405.3449019177147*xmin-43.96878371872008)--(xmax, 405.3449019177147*xmax-43.96878371872008), linewidth(2.)); /* line */ /* dots and labels */ dot((0.88,1.74),dotstyle); label("$O$", (0.9434990178466431,1.906643278812699), NE * labelscalefactor); dot((2.92,3.54),dotstyle); label("$E$", (2.994212389993052,3.7181067575420244), NE * labelscalefactor); dot((0.11920917712785317,4.352048491861037),dotstyle); label("$A$", (0.19157078139295974,4.521302828299366), NE * labelscalefactor); dot((-1.8074932661395424,2.1630602137457644),linewidth(4.pt) + dotstyle); label("$B$", (-1.739517644044909,2.29969667514076), NE * labelscalefactor); dot((-0.8955722309517441,-0.32129649800145077),dotstyle); label("$C$", (-0.8337859046802448,-0.14407009333370682), NE * labelscalefactor); dot((1.992302314281582,-0.7428176658071035),dotstyle); label("$D$", (2.0543020944259482,-0.5713020458642081), NE * labelscalefactor); dot((-1.3515327485456432,0.9208818578721568),linewidth(4.pt) + dotstyle); label("$I$", (-1.2781071353119668,1.0521793737516967), NE * labelscalefactor); dot((2.4561511571407912,1.3985911670964484),linewidth(4.pt) + dotstyle); label("$J$", (2.5328018812601103,1.530679160585858), NE * labelscalefactor); dot((-0.5282840890116909,6.575108705606808),linewidth(4.pt) + dotstyle); label("$F$", (-0.4578217864534032,6.5720162004457725), NE * labelscalefactor); dot((-0.5440607395423072,-0.19206567194751634),linewidth(4.pt) + dotstyle); label("$G$", (-0.4749110645546233,-0.05862370282760659), NE * labelscalefactor); dot((0.11019085908220025,0.6965192481830965),linewidth(4.pt) + dotstyle); label("$H$", (0.17448150329173967,0.830018758435836), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] @above below yeah it will be very easy to complex bash

When the typing error will be sorted out, that would still not be a very elegant question.

there is some typo in the statement we define $I,J$ why?

Sorry about the typo, it is now corrected.

I draw diagram and this is not correct... Are there any more typos? Or could anyone provide a correct diagram?

Had some difficulty translating. It should be correct now

Any solution Mathlinkers

Let $X$ be the orthocenter of $\triangle ACD.$ Observe that it's trivial that $G$ is the midpoint of the line connecting the centroids $G_1, G_2$ of $\triangle AEB, ACD$. By the Euler line property, it's hence clear that $OG$ bisects $FX.$ Therefore, we can replace $OG$ with $OM,$ where $M$ is the midpoint of $FX.$ We will make use of the following lemma: $\mathbf{Lemma.}$ $H \in OX,$ and furthermore, $\frac{OH}{HX} = \frac{1}{2\cos{2 \angle DAC}}.$ $\mathbf{Proof.}$ Easy by Law of Sines bashing in $\triangle OCX, \triangle ODX.$ $\blacksquare$ Let $P_1 = FH \cap AX,$ $P_2 = OM \cap AX.$ We wish to show that $P_1 = P_2.$ By Menelaus in $\triangle OAX$ with transversal $FP_1H,$ we know that: $$\frac{AP_1}{P_1X} * \frac{XH}{HO} * \frac{OF}{FA} = -1,$$which implies that: $\frac{AP_1}{P_1X} = \frac{1}{2 \cos{2 \angle DAC} }* \frac{FA}{FO}.$ By Menelaus in $\triangle AFX$ with transveral $OP_2M,$ we get that: $$\frac{AO}{OF} * 1 * \frac{XP_2}{P_2A} = -1,$$which implies that: $\frac{AP_2}{P_2X} = \frac{AO}{FO}.$ Therefore, it only suffices to prove that: $$\frac{AO}{FA} = \frac{1}{-2 \cos{\angle BAF}},$$ but this is well-known ($AH = 2R \cos{\angle A}$ in general). $\square$ (P.S: Is there a synthetic solution for this problem? If not, then where's all the synthetic geo on 2019 China TST )

Of course, there is one. But this is way too complicated for P1. Since I prefer a triangle geo problem and the labelling in the problem is messy, I will restate the problem as follows. Quote: Let $ABC$ be a triangle with circumcenter $O$ and circumcircle $\Omega$ such that $BC = \min\{BC, CA, AB\}$. Points $B_1, C_1$ are placed on $\Omega$ such that $AB_1 = AC_1 = BC$. Let $M_B, M_C$ be midpoints of $BB_1, CC_1$ and let $G$ be the centroid of $\triangle AM_BM_C$. Let $H_1$ be the orthocenter of $\triangle AB_1C_1$ and $K$ be the intersection of lines $BC_1$ and $CB_1$. Finally, let $X$ be the intersection of lines $KH_1$ and $OG$. Prove that $AX\perp BC$. Now we begin our solution. Let $H$ be the orthocenter of $\triangle ABC$. Since $\angle ABK = \angle ACK = \angle A$, a well known lemma$\color{blue} ^{[1]}$ gives $O, K, H$ are colinear. Let $L$ be the midpoint of $HH_1$. Then using vector with $O=0$, we easily get $L = \tfrac{2A+B+C+B_1+C_1}{2}$ and $G=\tfrac{2A+B+C+B_1+C_1}{6}$ thus $O, G, L$ are colinear . Moreover, it's evident that $A, H_1, O$ are colinear which means it suffices to show that $AK\parallel HH_1$. Let $M, M_1$ be midpoints of $BC, B_1C_1$ and $A'$ be the antipode of $A$ w.r.t. $\Omega$. Notice that $MM_1$ is $A'$-midline of $\triangle A'HH_1$, we get $HH_1\parallel MM_1$. Moreover, if $A_1 = BB_1\cap CC_1$ then $MM_1$ is Newton's line of complete quadrilateral formed by $\{BB_1, CC_1, BC_1, CB_1\}$ thus it bisects $KA_1$. Now notice that $MM_1$ is $A_1$-midline of $\triangle A_1AK$ thus $AK\parallel MM_1\parallel HH_1$ and we are done. [1] Here is how to prove the lemma. Let $BK, CK$ intersect $AC, AB$ at $E, F$ respectively. Let $M_b, M_c$ be midpoints of $AB, AC$. Let $G'$ be the centroid of $\triangle ABC$. Then notice that $EA=EB$ $\implies M_c, O, E$ are colinear. Similarly $M_b, O, F$ are colinear. Hence we are done by Pappus theorem on $\overline{BM_cF}$ and $\overline{CM_bE}$.

Solved with nukelauncher. We use complex numbers with unit circle $(ABCDE)$. We're going to calculate $M$ as the point on $OG$ with $AM \perp CD$, and show $F$, $M$, $H$ are collinear (this works because both definitions give a unique $M$). First, $AB = AE = CD$ means $be = a^2$ and $bc = ad$, so $e = \frac{a^2}{b}$ and $d = \frac{bc}{a}$. Then $g = \frac{2a + b + c + d + e}{6}$, and $$h = \frac{ce(b + d) - bd(c + e)}{ce - bd} = \frac{a^3bc+a^2bc^2-b^3c^2-a^2b^2c}{a^3c-b^3c} = \frac{b(a^2 + ac + bc)}{a^2 + ab + b^2}$$after cancelling out $c(a - b)$. Also $f = a + b + e = \frac{a^2 + ab + b^2}{b}$. This means \begin{align*} f-h &= \frac{(a^2+ab+b^2)^2-b^2(a^2+ac+bc)}{b(a^2+ab+b^2)}\\ &=\frac{a^4+b^4+2a^2b^2+2a^3b+2ab^3-ab^2c-b^3c}{b(a^2+ab+b^2)}\\ &=\frac{(a + b)(a^3 + a^2b + ab^2 + b^3 - b^2c)}{b(a^2 + ab + b^2)}. \end{align*} Now we get $m$ and calculate $m - f$. From $AM \perp CD$, we get $m - cd\overline{m} = a - \frac{cd}{a}$ by setting the feet from $M$ and $A$ to $CD$ equal. Also, since $\frac{m}{g} = \frac{\overline{m}}{\overline{g}}$, then $$\overline{m} = m\cdot \frac{\overline{g}}{g} = m \cdot \frac{2/a + 1/b + 1/c + 1/d + 1/e}{2a + b + c + d + e} = m \cdot \frac{2abc + a^2c + a^2b + a^3 + b^2c}{ac(2a^2b + ab^2 + abc + b^2c + a^3)}.$$Plugging this in and using $d = \frac{bc}{a}$ gets $$a - \frac{bc^2}{a^2} = m\left(1 - \frac{bc^2}{a}\cdot \frac{2abc + a^2c + a^2b + a^3 + b^2c}{ac(2a^2b + ab^2 + abc + b^2c + a^3)}\right),$$which means $$m = \frac{(a^3 - bc^2)(2a^2b + ab^2 + abc + b^2c + a^3)}{a^2(2a^2b + ab^2 + abc + b^2c + a^3) - bc(2abc + a^2c + a^2b + a^3 + b^2c)}.$$But the denominator factors as \begin{align*}a^2(a(a + b)^2 + bc(a + b)) - bc(c(a + b)^2 + a^2(a + b)) &= (a + b)(a^2(a^2 + ab + bc) - bc(a^2 + ac + bc)) \\ &= (a + b)(a^3(a + b) - bc^2(a + b)) \\ &= (a + b)^2(a^3 - bc^2),\end{align*}while the numerator also factors as $$(a^3 - bc^2)(a(a + b)^2 + bc(a + b)) = (a^3 - bc^2)(a + b)(a^2 + ab + bc).$$So $m = \frac{a^2 + ab + bc}{a + b}$, and then $$m-f=\frac{a^2+ab+bc}{a+b}-\frac{a^2+ab+b^2}b=\frac{bc^2-a^3-a^2b-ab^2-b^3}{b(a+b)}.$$So then \[\frac{m-f}{f-h}=-\frac{a^2+ab+b^2}{(a+b)^2}.\]This is its own conjugate, so it's real and we're done.