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Iranian geometry configuration
Assassino9931   0
33 minutes ago
Source: Al-Khwarizmi Junior International Olympiad 2025 P7
Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$, such that $CD$ is not a diameter of its circumcircle. The lines $AD$ and $BC$ intersect at point $P$, so that $A$ lies between $D$ and $P$, and $B$ lies between $C$ and $P$. Suppose triangle $PCD$ is acute and let $H$ be its orthocenter. The points $E$ and $F$ on the lines $BC$ and $AD$, respectively, are such that $BD \parallel HE$ and $AC\parallel HF$. The line through $E$, perpendicular to $BC$, intersects $AD$ at $L$, and the line through $F$, perpendicular to $AD$, intersects $BC$ at $K$. Prove that the points $K$, $L$, $O$ are collinear.

Amir Parsa Hosseini Nayeri, Iran
0 replies
+1 w
Assassino9931
33 minutes ago
0 replies
Inequality, inequality, inequality...
Assassino9931   0
34 minutes ago
Source: Al-Khwarizmi Junior International Olympiad 2025 P6
Let $a,b,c$ be real numbers such that \[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]Find the smallest possible value of $a^2 + b^2 + c^2$.

Binh Luan and Nhan Xet, Vietnam
0 replies
Assassino9931
34 minutes ago
0 replies
Prime sums of pairs
Assassino9931   0
36 minutes ago
Source: Al-Khwarizmi Junior International Olympiad 2025 P5
Sevara writes in red $8$ distinct positive integers and then writes in blue the $28$ sums of each two red numbers. At most how many of the blue numbers can be prime?

Marin Hristov, Bulgaria
0 replies
Assassino9931
36 minutes ago
0 replies
help me solve this problem. Thanks
tnhan.129   0
38 minutes ago
Find f:R+ -> R such that:
(x+1/x).f(y) = f(xy) + f(y/x)
0 replies
tnhan.129
38 minutes ago
0 replies
Line passing through orthocenter
pokmui9909   4
N 39 minutes ago by Tkn
Source: KMO 2024 P7
In an acute triangle $ABC$, let a line $\ell$ pass through the orthocenter and not through point $A$. The line $\ell$ intersects line $BC$ at $P(\neq B, C)$. A line passing through $A$ and perpendicular to $\ell$ meets the circumcircle of triangle $ABC$ at $R(\neq A)$. Let the feet of the perpendiculars from $A, B$ to $\ell$ be $A', B'$, respectively. Define line $\ell_1$ as the line passing through $A'$ and perpendicular to $BC$, and line $\ell_2$ as the line passing through $B'$ and perpendicular to $CA$. Prove that if $Q$ is the reflection of the intersection of $\ell_1$ and $\ell_2$ across $\ell$, then $\angle PQR = 90^{\circ}$.
4 replies
pokmui9909
Nov 9, 2024
Tkn
39 minutes ago
Inequality with number of divisors
MathLuis   12
N an hour ago by Grasshopper-
Source: Iberoamerican MO 2024 Day 1 P1
For each positive integer $n$, let $d(n)$ be the number of positive integer divisors of $n$.
Prove that for all pairs of positive integers $(a,b)$ we have that:
\[ d(a)+d(b) \le d(\gcd(a,b))+d(\text{lcm}(a,b)) \]and determine all pairs of positive integers $(a,b)$ where we have equality case.
12 replies
MathLuis
Sep 21, 2024
Grasshopper-
an hour ago
No three collinear
USJL   1
N an hour ago by Photaesthesia
Source: 2025 Taiwan TST Round 3 Mock P6
Given a positive integer $n\geq 3$. A convex polygon is said to be $n$-good if it contains $n$ lattice points where any three of them are not collinear.

(a) Show that there exists an $n$-good convex polygon with area at most $4n^2$.
(b) Show that there exists a constant $c>0$ so that any $n$-good convex polygon has area at least $cn^2$.

Proposed by usjl
1 reply
USJL
Apr 26, 2025
Photaesthesia
an hour ago
Concentric Circles
MithsApprentice   61
N an hour ago by endless_abyss
Source: USAMO 1998
Let ${\cal C}_1$ and ${\cal C}_2$ be concentric circles, with ${\cal C}_2$ in the interior of ${\cal C}_1$. From a point $A$ on ${\cal C}_1$ one draws the tangent $AB$ to ${\cal C}_2$ ($B\in {\cal C}_2$). Let $C$ be the second point of intersection of $AB$ and ${\cal C}_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects ${\cal C}_2$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$.
61 replies
MithsApprentice
Oct 9, 2005
endless_abyss
an hour ago
My FE handout
FEcreater   77
N 2 hours ago by NicoN9
After graduated from IMO, I planned to LaTeX a complete FE handout.

However, because of my laziness, I have only finished a part of this handout. :oops:

Hope this note will help some of AoPS users ~ :yup:

Anyway, Happy Lunar New Year
77 replies
FEcreater
Feb 15, 2018
NicoN9
2 hours ago
2014 JBMO Shortlist G2
parmenides51   6
N 2 hours ago by tilya_TASh
Source: 2014 JBMO Shortlist G2
Acute-angled triangle ${ABC}$ with ${AB<AC<BC}$ and let be ${c(O,R)}$ it’s circumcircle. Diameters ${BD}$ and ${CE}$ are drawn. Circle ${c_1(A,AE)}$ interescts ${AC}$ at ${K}$. Circle ${{c}_{2}(A,AD)}$ intersects ${BA}$ at ${L}$ .(${A}$ lies between ${B}$ and ${L}$). Prove that lines ${EK}$ and ${DL}$ intersect at circle $c$ .

by Evangelos Psychas (Greece)
6 replies
parmenides51
Oct 8, 2017
tilya_TASh
2 hours ago
Disjoint Pairs
MithsApprentice   42
N 2 hours ago by endless_abyss
Source: USAMO 1998
Suppose that the set $\{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1\leq i\leq 999$) so that for all $i$, $|a_i-b_i|$ equals $1$ or $6$. Prove that the sum \[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|  \] ends in the digit $9$.
42 replies
MithsApprentice
Oct 9, 2005
endless_abyss
2 hours ago
Beautiful problem
luutrongphuc   13
N Apr 10, 2025 by ItzsleepyXD
(Phan Quang Tri) Let triangle $ABC$ be circumscribed about circle $(I)$, and let $H$ be the orthocenter of $\triangle ABC$. The circle $(I)$ touches line $BC$ at $D$. The tangent to the circle $(BHC)$ at $H$ meets $BC$ at $S$. Let $J$ be the midpoint of $HI$, and let the line $DJ$ meet $(I)$ again at $X$. The tangent to $(I)$ parallel to $BC$ meets the line $AX$ at $T$. Prove that $ST$ is tangent to $(I)$.
13 replies
luutrongphuc
Apr 4, 2025
ItzsleepyXD
Apr 10, 2025
Beautiful problem
G H J
G H BBookmark kLocked kLocked NReply
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luutrongphuc
50 posts
#1 • 1 Y
Y by PikaPika999
(Phan Quang Tri) Let triangle $ABC$ be circumscribed about circle $(I)$, and let $H$ be the orthocenter of $\triangle ABC$. The circle $(I)$ touches line $BC$ at $D$. The tangent to the circle $(BHC)$ at $H$ meets $BC$ at $S$. Let $J$ be the midpoint of $HI$, and let the line $DJ$ meet $(I)$ again at $X$. The tangent to $(I)$ parallel to $BC$ meets the line $AX$ at $T$. Prove that $ST$ is tangent to $(I)$.
This post has been edited 1 time. Last edited by luutrongphuc, Apr 7, 2025, 1:49 AM
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aidenkim119
33 posts
#2 • 1 Y
Y by PikaPika999
bump0ppppppppp
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whwlqkd
99 posts
#3 • 2 Y
Y by aidenkim119, PikaPika999
BUMPPPPPP
Why it didn’t proposed for imo p3/6
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aidenkim119
33 posts
#4 • 1 Y
Y by PikaPika999
whwlqkd wrote:
BUMPPPPPP
Why it didn’t proposed for imo p3/6

Solved but i dont know how to type this in latex sorry
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whwlqkd
99 posts
#5
Y by
\angle:
$\angle$
\triangle:
$\triangle$
\perp:
$\perp$
\times:
$\times$
\cap:
$\cap$
etc
(You can search the latex code more)
If you want to write $\LaTeX$, you have to write dollar sign before and after the code.
Some example of latex:
$1+1=2$
$2\times 5=10$
$3-(1-2)=4$
$\frac{3}{67}$
etc
Click the text, then you can see the latex code
This post has been edited 3 times. Last edited by whwlqkd, Apr 6, 2025, 12:01 PM
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whwlqkd
99 posts
#9
Y by
You have to write $ on the end of the alphabet
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aidenkim119
33 posts
#19 • 1 Y
Y by whwlqkd
Use six point line and polepolar to erase useless points

Then use pascal to change the question

Then easy calaulation finishes it
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hukilau17
288 posts
#26 • 1 Y
Y by PikaPika999
No one's actually going to post a solution? All right, here goes.

Complex bash with the incircle of $\triangle ABC$ as the unit circle, and let it touch $AC,AB$ at $E,F$ respectively, and let $\triangle ABC$ have circumcenter $O$, so
$$|d|=|e|=|f|=1$$$$a = \frac{2ef}{e+f}$$$$b = \frac{2df}{d+f}$$$$c = \frac{2de}{d+e}$$$$o = \frac{2def(d+e+f)}{(d+e)(d+f)(e+f)}$$$$h = a+b+c-2o = \frac{2(d^2e^2+d^2ef+d^2f^2+de^2f+def^2+e^2f^2)}{(d+e)(d+f)(e+f)}$$$$j = \frac{h}2 = \frac{d^2e^2+d^2ef+d^2f^2+de^2f+def^2+e^2f^2}{(d+e)(d+f)(e+f)}$$Now we find the coordinate of $S$. Since $S$ lies on line $BC$ we have
$$\overline{s} = \frac{2d-s}{d^2}$$Since line $SH$ is tangent to the circumcircle of $\triangle BHC$, we have
$$\frac{(b-c)(h-s)}{(b-h)(c-h)} \in \mathbb{R} \implies \frac{d(h-s)}{ef} \in i\mathbb{R}$$$$\frac{d\left[2(d^2e^2+d^2ef+d^2f^2+de^2f+def^2+e^2f^2) - s(d+e)(d+f)(e+f)\right]}{ef(d+e)(d+f)(e+f)} = -\frac{ef\left[2d^2(d^2+de+df+e^2+ef+f^2) - (2d-s)(d+e)(d+f)(e+f)\right]}{d^3(d+e)(d+f)(e+f)}$$$$2d^4(d^2e^2+d^2ef+d^2f^2+de^2f+def^2+e^2f^2) - d^4s(d+e)(d+f)(e+f) = -2d^2e^2f^2(d^2+de+df+e^2+ef+f^2) + 2de^2f^2(d+e)(d+f)(e+f) - e^2f^2s(d+e)(d+f)(e+f)$$$$s = \frac{2d^4(d^2e^2+d^2ef+d^2f^2+de^2f+def^2+e^2f^2) + 2d^2e^2f^2(d^2+de+df+e^2+ef+f^2) - 2de^2f^2(d+e)(d+f)(e+f)}{d^4(d+e)(d+f)(e+f) - e^2f^2(d+e)(d+f)(e+f)}$$We simplify this to get
$$s = \frac{2d(d^5e^2+d^5ef+d^5f^2+d^4e^2f+d^4ef^2+2d^3e^2f^2-de^3f^3-e^4f^3-e^3f^4)}{(d+e)(d+f)(e+f)(d^2+ef)(d^2-ef)}$$Next we find the coordinate of $X$. We have
$$d - j = \frac{d^3e+d^3f+d^2ef-e^2f^2}{(d+e)(d+f)(e+f)}$$and so
$$x = \frac{d-j}{d\overline{\jmath}-1} = -\frac{d-j}{d(\overline{d}-\overline{\jmath})} = -\frac{d^3e+d^3f+d^2ef-e^2f^2}{e^2f+ef^2+def-d^3}$$Now we solve the rest of this problem in reverse. We know $T$ doesn't lie on line $BC$, so if the line $ST$ is tangent to the unit circle, it must be the other tangent to the unit circle passing through $S$ (besides line $BC$). So letting the other tangent through $S$ touch the unit circle at $U$, we have
$$s = \frac{2du}{d+u}$$and so
$$u = \frac{ds}{2d-s}$$Now
$$2d - s = \frac{2d\left[(d^4-e^2f^2)(d+e)(d+f)(e+f) - (d^5e^2+d^5ef+d^5f^2+d^4e^2f+d^4ef^2+2d^3e^2f^2-de^3f^3-e^4f^3-e^3f^4)\right]}{(d+e)(d+f)(e+f)(d^2+ef)(d^2-ef)}$$which we simplify to
$$2d - s = -\frac{2d(-d^6e-d^6f-d^5ef+2d^3e^2f^2+d^2e^3f^2+d^2e^2f^3+de^4f^2+de^3f^3+de^2f^4)}{(d+e)(d+f)(e+f)(d^2+ef)(d^2-ef)}$$So
$$u = -\frac{d^5e^2+d^5ef+d^5f^2+d^4e^2f+d^4ef^2+2d^3e^2f^2-de^3f^3-e^4f^3-e^3f^4}{-d^5e-d^5f-d^4ef+2d^2e^2f^2+de^3f^2+de^2f^3+e^4f^2+e^3f^3+e^2f^4}$$Then let the tangent to the unit circle at $U$ meet the tangent line to the unit circle parallel to $BC$ at $V$. We want to show that $V$ lies on line $AX$ -- then it will follow that $V=T$ and that $ST$ is tangent to the unit circle at $U$. Now
$$v = \frac{2(-d)u}{-d+u} = -\frac{2d(d^5e^2+d^5ef+d^5f^2+d^4e^2f+d^4ef^2+2d^3e^2f^2-de^3f^3-e^4f^3-e^3f^4)}{-d^6e-d^6f+d^5e^2+d^5f^2+d^4e^2f+d^4ef^2+4d^3e^2f^2+d^2e^3f^2+d^2e^2f^3+de^4f^2+de^2f^4-e^4f^3-e^3f^4}$$Then we find the vectors
$$a-x = \frac{2ef(e^2f+ef^2+def-d^3) + (e+f)(d^3e+d^3f+d^2ef-e^2f^2)}{(e+f)(e^2f+ef^2+def-d^3)} = \frac{d^3e^2+d^3f^2+d^2e^2f+d^2ef^2+2de^2f^2+e^3f^2+e^2f^3}{(e+f)(e^2f+ef^2+def-d^3)}$$and
\begin{align*}
a-v &= \frac{2ef(-d^6e-d^6f+d^5e^2+d^5f^2+d^4e^2f+d^4ef^2+4d^3e^2f^2+d^2e^3f^2+d^2e^2f^3+de^4f^2+de^2f^4-e^4f^3-e^3f^4) + 2d(e+f)(d^5e^2+d^5ef+d^5f^2+d^4e^2f+d^4ef^2+2d^3e^2f^2-de^3f^3-e^4f^3-e^3f^4)}{(e+f)(-d^6e-d^6f+d^5e^2+d^5f^2+d^4e^2f+d^4ef^2+4d^3e^2f^2+d^2e^3f^2+d^2e^2f^3+de^4f^2+de^2f^4-e^4f^3-e^3f^4)} \\
&= \frac{2(d^6e^3+d^6e^2f+d^6ef^2+d^6f^3+2d^5e^3f+2d^5e^2f^2+2d^5ef^3+3d^4e^3f^2+3d^4e^2f^3+4d^3e^3f^3-2de^4f^4-e^5f^4-e^4f^5)}{(e+f)(-d^6e-d^6f+d^5e^2+d^5f^2+d^4e^2f+d^4ef^2+4d^3e^2f^2+d^2e^3f^2+d^2e^2f^3+de^4f^2+de^2f^4-e^4f^3-e^3f^4)}
\end{align*}Now there's only one way that the numerator of $a-v$ could conceivably factor so that $\frac{a-x}{a-v}$ is real, and so we conveniently discover the factorization
$$a-v = \frac{2(d^3e+d^3f+d^2ef-e^2f^2)(d^3e^2+d^3f^2+d^2e^2f+d^2ef^2+2de^2f^2+e^3f^2+e^2f^3)}{(e+f)(-d^6e-d^6f+d^5e^2+d^5f^2+d^4e^2f+d^4ef^2+4d^3e^2f^2+d^2e^3f^2+d^2e^2f^3+de^4f^2+de^2f^4-e^4f^3-e^3f^4)}$$Then
$$\frac{a-x}{a-v} = \frac{-d^6e-d^6f+d^5e^2+d^5f^2+d^4e^2f+d^4ef^2+4d^3e^2f^2+d^2e^3f^2+d^2e^2f^3+de^4f^2+de^2f^4-e^4f^3-e^3f^4}{2(d^3e+d^3f+d^2ef-e^2f^2)(e^2f+ef^2+def-d^3)}$$This is equal to its conjugate and thus real. $\blacksquare$
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aidenkim119
33 posts
#28
Y by
Any synthetic proof?
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WLOGQED1729
47 posts
#29 • 1 Y
Y by Bluecloud123
Fantastic Problem! Here’s my synthetic proof.
First WLOG, we can assume that $AB<AC$
Part 1 Simplify the problem
Let $(I)$ tangent to $AB,AC$ at $F,E$, respectively.
Let $L \neq D$ be a point on $(I)$ s.t. $SL$ is tangent to $(I)$ and define $D’$ as the antipode of $D$ wrt. $(I)$
Let $T’$ be the intersection between $SL$ and the tangent line of $(I)$ at $D’$
If we can prove that $A,T’,X$ are collinear, we can conclude that $T’=T$ and we’re done.
Next, by pole-polar duality we know that poles are collinear if and only if its polars are concurrent.
Thus, we can just prove that $D’L$, $EF$ and the tangent of $(I)$ at $X$ are concurrent.
This is equivalent to show that there exists an involution on $(I)$ which swaps $(D’,L),(E,F)$ and $(X,X)$.

Part 2 Breakdown the problems into different parts
Since $D$ lies on $(I)$, an involution swapping $(D', L), (E, F), (X, X)$ on $(I)$
is equivalent to an involution on the pencil from $D$ swapping $(DD', DL), (DE, DF), (DX, DX)$.
Let $DL, DX, DD'$ intersect $EF$ at $L', X', K$, respectively.
Projecting this pencil onto line $EF$, we seek an involution on $EF$ that swaps $(K, L'), (E, F), (X', X')$.
Let $AK, AX, AL'$ intersect $BC$ at $M, Y, T$, respectively.
Projecting through $A$ onto line $BC$, this reduces to showing that there exists an involution on $BC$ that swaps $(M, T), (C, B), (Y, Y)$.
We claim that $HY$ bisects $\angle BHC$ and $TH,MH$ are isogonal conjugate wrt. $\angle BHC$ and will prove in the next section. If this is true, we get the desired involution.

Part 3 Solving sub problem 1
We’re going to prove that $TH,MH$ are isogonal conjugate wrt. $\angle BHC$
Recall the well known lemma which is used in 2005 G6, $AK$ bisects $BC$. We deduce that $M$ is the midpoint of $BC$.
Thus, our goal is to show that $HT$ is H-symmedian of $\triangle BHC$ which is equivalent to showing that $(S,T;B,C)=-1$.
Let $SL$ intersects $AB,AC$ at $P,Q$. Consider tangential quadrilateral $PQCB$, it is well known that $PC,QB,LD,EF$ are concurrent. So, $P,L’,C$ are collinear and $Q,L’,B$ are collinear.
By well known harmonic configuration, we conclude that $(S,T;B,C)=-1$, as desired.

Part 4 Solving sub problem 2
We’re going to prove that $HY$ bisects $\angle BHC$
Let the line through $H$ parallel to $EF$ intersects $BC$ at $R$.
First, we’ll show that our goal is equivalent to showing that $DJ \perp RI$
Suppose we’ve already shown that $DJ \perp RI$, we conclude that polar of point $R$ wrt. $(I)$ is $DJ$
We then apply the same trick as Part 3. Let $RX$ intersects $AB,AC$ at $R_1,R_2$, respectively.
Consider the tangential quadrilateral $R_1R_2CB$ and recall the well known harmonic configuration, we can conclude that $(R,Y;B,C)=-1$.
By trivial angle chasing, we know that $HR$ externally bisects $\angle BHC$. Thus, $HY$ internally bisects $\angle BHC$, we’re done.

Now, we focus on our goal proving that $DJ \perp RI$.
This is equivalent to $\angle IDJ =\angle IRD$. Let $H’,I’$ be the reflections of $H,I$ wrt. $BC$, respectively.
Observe that $\angle IDJ = \angle II’H = \angle HH’I = \angle AH’I$. So, our new goal is to show that $\angle IRD =\angle AH’I$
It is well known that $H’$ lies on $(ABC)$ and $A’=AI \cap (ABC)$ is the circumcenter of $\triangle BIC$.
Note that $A’$ is midpoint of arc $BC$ not containing $A$ and $H’$ lies on $(ABC)$, we can easily show that $A’H$ externally bisects $\angle BH’C$.
Since we already have that $RH$ externally bisects $\angle BHC$, we deduce that $RH’$ externally bisects $\angle BH’C$. Thus, $R,H’,A’$ are collinear.
Finally, consider an inversion $\phi$ wrt. $(BIC)$ centered at $A’$.
Let $AI$ intersects $BC$ at $Z$. We know that $\phi$ swaps $H’\leftrightarrow R$ and $Z \leftrightarrow A$.
Note that $$\angle IRD = \angle IRA’ - \angle H’RZ =\angle H’IA’ - \angle H’AZ = \angle H’IA’ - \angle H’AI = \angle AH’I $$Thus, we’re done. $\blacksquare$
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This post has been edited 1 time. Last edited by WLOGQED1729, Apr 8, 2025, 6:55 AM
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aidenkim119
33 posts
#30
Y by
That is very interestuung!!
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pingupignu
49 posts
#31
Y by
Nice problem! Here's another solution using DDIT and trigonometry. Firstly we can delete $S$ and $T$ as follows:
Let $Z = BC \cap AX$ and $S' \in BC$ such that $S'T$ is the other tangent from $T$ to $(I)$. From Dual of Desargues Involution theorem we have the following reciprocal pairs on the pencil through $T$:
$$(T \infty_{BC}, TS'), (TB, TC), (TA, TD)$$Projecting it to $BC$ gives
$$(S', \infty_{BC}), (B, C), (Z, D)$$are reciprocal pairs of some involution on $BC$, so that $S'B \cdot S'C = S'Z \cdot S' D$. We need to show $S'H$ is tangent to $(BHC)$ $\iff$ $(BHC), (ZHD)$ are tangent $\iff$ $\frac{BZ}{ZC} \cdot \frac{BD}{BC} = (\frac{BH}{CH})^2$ $\iff$ $\frac{c}{b} \cdot \frac{\sin \angle BAX}{\sin \angle XAC} \cdot \frac{s-b}{s-c} = (\frac{\cos B}{\cos C})^2$.

From the solution from #29, if we let $Q = DX \cap EF$, then $AQ$ passes through the foot of internal angle bisector of $\angle BHC$ onto $BC$. Hence we deduce (letting $Y$ be said foot)
$$\frac{BY}{YC} = \frac{BH}{HC} \implies \frac{c}{b} \cdot \frac{\sin \angle BAQ}{\sin \angle QAC} = \frac{\cos B}{\cos C}$$
We can see that $$\frac{\sin \angle FAX}{\sin \angle EAX} = (\frac{FX}{EX})^2 = (\frac{FQ}{QE})^2 \cdot (\frac{ED}{DF})^2 = (\frac{\sin \angle BAQ}{\sin \angle QAC})^2 \cdot (\frac{\cos \frac{C}{2}}{\cos \frac{B}{2}})^2 = (\frac{b \cos B \cos \frac{C}{2}}{c \cos C \cos \frac{B}{2}})^2$$
And
$$\frac{c}{b} \cdot \frac{\sin \angle BAX}{\sin \angle XAC} \cdot \frac{s-b}{s-c} = \frac{c}{b} \cdot (\frac{b \cos B \cos \frac{C}{2}}{c \cos C \cos \frac{B}{2}})^2 \cdot \frac{BI}{CI} \cdot \frac{\sin \angle BID}{\sin \angle DIC}$$$$= (\frac{\cos B}{\cos C})^2 \cdot \frac{b}{c} \cdot \frac{\sin \frac{C}{2}}{\sin \frac{B}{2}} \cdot \frac{\cos \frac{B}{2}}{\cos \frac{C}{2}} \cdot (\frac{\cos \frac{C}{2}}{\cos \frac{B}{2}})^2 = \frac{b}{c} \cdot \frac{\sin C}{\sin B} \cdot (\frac{\cos B}{\cos C})^2 = (\frac{\cos B}{\cos C})^2,$$as desired. $\blacksquare$
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luutrongphuc
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Thank you everyone for your contribution
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ItzsleepyXD
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Maybe non DDIT solution but a lot of projective spam.
Define point - Redefine point

Lemma

Claim 1

Claim 2

Claim 3

Claim 4

Finished
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