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#1 h Apr 4, 2025, 5:35 AM • 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 $(I)$ .

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#2 h Apr 5, 2025, 12:45 PM • 1 Y

Y by PikaPika999

bump0ppppppppp

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#3 h Apr 6, 2025, 11:26 AM • 2 Y

Y by aidenkim119, PikaPika999

BUMPPPPPP
Why it didn’t proposed for imo p3/6

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#4 h Apr 6, 2025, 11:29 AM • 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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#5 h Apr 6, 2025, 11:55 AM

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\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

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#9 h Apr 6, 2025, 12:11 PM

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You have to write $ on the end of the alphabet

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#19 h Apr 6, 2025, 12:22 PM • 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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#26 h Apr 6, 2025, 7:53 PM • 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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#28 h Apr 8, 2025, 2:16 AM

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Any synthetic proof?

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#29 h Apr 8, 2025, 6:53 AM • 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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#30 h Apr 8, 2025, 11:44 AM

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That is very interestuung!!

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#31 h Apr 9, 2025, 3:33 PM

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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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#33 h Apr 9, 2025, 11:23 PM

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Thank you everyone for your contribution

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#34 h Apr 10, 2025, 6:23 AM

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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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