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Source: LMAO 2024 P3

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177 posts

#1 h Jun 1, 2024, 6:54 PM • 9 Y

Y by AlexCenteno2007, Rijul saini, GeoKing, kiyoras_2001, Tastymooncake2, anantmudgal09, khina, Supercali, AlephG_64

Let $\Delta_0$ be an equilateral triangle with incircle $\omega$ . A point on $\omega$ is reflected in the sides of $\Delta_0$ to obtain a new triangle $\Delta_1$ . The same point is then reflected over the sides of $\Delta_1$ to obtain another triangle $\Delta_2$ . Prove that the circumcircle of $\Delta_2$ is tangent to $\omega$ .

Proposed by Siddharth Choppara

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1260 posts

#2 h Jun 2, 2024, 4:20 AM • 7 Y

Y by AlexCenteno2007, Rijul saini, GeoKing, Siddharth03, Rg230403, ZVFrozel, Tastymooncake2

My favourite from LMAO this year.

Suppose $\Delta_0=ABC$ , $\Delta_1=P_AP_BP_C$ , and $\Delta_2=Q_AQ_BQ_C$ . Invert at $P$ , and we'll denote the inverted image of any point $X$ by $X'$ . Since $P$ is the Fermat point of $P_AP_BP_C$ , it remains the Fermat point of $P_A'P_B'P_C'$ (angles at $P$ are preserved). Further $P_A'$ becomes center of circle $(B'C'P)$ and so on. Also being a reflection originally, $Q_A'$ becomes center of circle $(P_C'P_B'P)$ and so on, which are just centers of the three equilateral triangles constructed externally. Finally, since the sides $BC,CA,AB$ become circles passing through $P$ with centers $P_A,P_B,P_C$ , the incircle becomes a common tangent to them, say $\ell$ .

We have to prove that $(Q_A'Q_B'Q_C')$ is also tangent to $\ell$ .

Let the radii of the above three circle be $a,b,c$ respectively. Let $G$ be centroid of $P_A'P_B'P_C'$ , and suppose point $D$ is outside $P_A'P_B'P_C'$ such that $DP_B'P_C'$ is equilateral. By properties of the Fermat point, $P_A'D=a+b+c$ . Then by basic proportionality from midpoint of $P_B'P_C'$ , $GQ_A'= \frac{a+b+c}{3}$ . Since this is symmetric in $a,b,c$ , $G$ is the center of $(Q_A'Q_B'Q_C')$ and has radius $\frac{a+b+c}{3}$ . But, since $P_A,P_B,P_C$ are a distance of $a,b,c$ respectively away from $\ell$ (and on the same side of $\ell$ ), and $G$ is their centroid, $G$ is a distance of $\frac{a+b+c}{3}$ away from $\ell$ , so $(Q_A'Q_B'Q_C')$ is tangent to $\ell$ as well, as required.

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#3 h Jun 2, 2024, 6:35 AM • 4 Y

Y by mathscrazy, Siddharth03, GeoKing, ZVFrozel

Let $\Delta_0=\triangle ABC$ , $\Delta_1=\triangle A_1B_1C_1$ and $\Delta_2=\triangle A_2B_2C_2$ where symmetry is preserved in naming. Also, $\triangle A_0B_0C_0$ is the intouch triangle of $\Delta_0$ . Let $P_A$ be the reflection of $P$ in $B_0C_0$ and define $P_B$ and $P_C$ analogously. Finally, let $M_A$ , $M_B$ and $M_C$ be the small arc midpoints of $\triangle A_0B_0C_0$ .

Let $Q$ be the Anti-Steiner point of $P$ , which is known to be on $P_A M_A$ and $\omega$ . We claim $Q$ is the point of tangency. Firstly, let's show:

Lemma wrote:

Let $\angle BAC=60^\circ$ and $\gamma=(ABC)$ . If $X$ and $Y$ are the reflections of $A$ in the tangents at $B$ and $C$ to $\gamma$ , then $XY$ , $BC$ and the $A-$ angle bisector concur.

Proof.

Back to the problem. We get $PM_C$ , $A_1B_1$ and $A_0B_0$ concur at, say, $K$ .

Claim: $PQP_CC_2$ is cyclic.
Proof. $\angle PC_2P_C=\angle PKA_0=\angle M_CKA_0=\angle PA_0M_C=\angle PQM_C=\angle PQP_C.$ Similarly, we get $PQP_BB_2$ and $PQP_AA_2$ are cyclic.

Claim: $QA_2B_2C_2$ is cyclic.
Proof. $\angle A_2B_2C_2=\angle A_2B_2P+\angle PB_2C_2=\angle B_1C_1P+\angle PA_1B_1$ and
$\angle C_2QP=\angle C_2P_CP=\angle A_1KP=\angle B_1A_1P+\angle A_1PM_C$ hence
$\angle C_2QA_2=\angle C_2QP+\angle PQA_2=\angle B_1A_1P+\angle PC_1B_1+\angle A_1PM_C+\angle M_APC_1=\angle B_1A_1P+\angle PC_1B_1.$ To see the two circles are tangent, it suffices to show $\angle QB_2C_2+\angle C_2QP=\angle QM_AP.$ To this end.

Remark:

This post has been edited 1 time. Last edited by aleksijt, Jun 2, 2024, 6:38 AM

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#4 h Jun 3, 2024, 2:38 PM • 1 Y

Y by GeoKing

a fun property

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177 posts

#5 h Jun 3, 2024, 3:03 PM • 1 Y

Y by GeoKing

ZVFrozel wrote:

a fun property

And even if $P$ doesn't lie on the incircle of $\Delta_0$ , we still have that $(\Delta_0)$ is tangent to $(\Delta_2)$ .
The original problem can also be generalized by taking $P$ on any circle $\omega$ concentric with $(\Delta_0)$ intersecting the sides of $\Delta_0$ . Then, we can prove that $\omega$ makes the same angle with both the sides of $\Delta_0$ and the circle $(\Delta_2)$ .

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This post has been edited 2 times. Last edited by Siddharth03, Jun 3, 2024, 3:06 PM
Reason: typo

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184 posts

#6 h Feb 18, 2025, 6:35 PM • 1 Y

Y by trigadd123

Here's a complex numbers proof for the general case when $P$ is any point in the plane of $\Delta_0$ . We will show that the circumcircles of $\Delta_0$ and $\Delta_2$ are tangent in the Steiner point $Q$ of line $OP$ (note that this is the problem posed by @above).

Suppose that the vertices $A,B,C$ of $\Delta_0$ have affixes $1,\varepsilon,$ and $\varepsilon^2$ . Let $D,E,F$ be the vertices of $\Delta_1$ (in the obvious order). In this case, $D$ is the symmetric of $P$ w.r.t. $BC$ , so $d=\frac{\overline{p}(\varepsilon - \varepsilon^2)+\overline{\varepsilon}\varepsilon^2 - \varepsilon\overline{\varepsilon^2}}{\overline{\varepsilon}-\overline{\varepsilon^2}}=-\overline{p}-1.$ Notice that, by rotational symmetry, if a point $x$ is obtained from $p$ through a process involving side $BC$ (i.e. $x=f(p)$ ), the point obtained from $p$ by the same process but involving side $CA$ will be $y=\varepsilon f\left(\frac{p}{\varepsilon}\right)$ . Similarly, we get $\begin{cases} d=-\overline{p}-1\\ e=-\varepsilon^2\overline{p} - \varepsilon\\ f=-\varepsilon\overline{p} - \varepsilon^2\end{cases}$ Let $X,Y,Z$ be the vertices of $\Delta_2$ (in the obvious order). $X$ is the reflection of $P$ across $EF$ , so $x=\frac{\overline{p}(e-f)+\overline{e}f-e\overline{f}}{\overline{e}-\overline{f}}=\frac{p\overline{p}-1-\overline{p}(\overline{p}+1)}{p-1}.$ If $O_A$ is the reflection of $O$ across $BC$ , one can find the $Q$ as the intersection of $DO_A$ with $(ABC)$ . Quick computations give $q=\frac{\overline{p}}{p}$ .

Let $X'$ be the second intersection of $QX$ with $(ABC)$ . Via the intersection formula we have $x'=\frac{q-x}{q\overline{x}-1}=\frac{\frac{\overline{p}}{p}-\frac{p\overline{p}-1-\overline{p}(\overline{p}-1)}{p-1}}{\frac{\overline{p}}{p}\cdot\frac{p\overline{p}-1-p(p-1)}{\overline{p}-1}-1}=\frac{\frac{p\overline{p}-\overline{p}-p^2\overline{p}+p+p\overline{p}^2-p\overline{p}}{p(p-1)}}{\frac{p\overline{p}^2-\overline{p}-p^2\overline{p}+p\overline{p}-p\overline{p}+p}{p(\overline{p}-1)}}=\frac{\overline{p}-1}{p-1}\cdot\frac{(p\overline{p}-1)(\overline{p}-p)}{(p\overline{p}-1)(\overline{p}-p)}=\frac{\overline{p}-1}{p-1}.$ Similerly (using the rotation symmetry argument) we get $\begin{cases} x'=\frac{\overline{p}-1}{p-1}\\ y'=\frac{\overline{p}-\varepsilon^2}{p-\varepsilon}\\ z'=\frac{\overline{p}-\varepsilon}{p-\varepsilon^2}\end{cases}$
Our aim is now to prove that $\triangle X'Y'Z'$ and $\triangle XYZ$ are homothetic, with homothety center $Q$ . But $\left|\frac{x'-q}{x-q}\right|=\left|\frac{\frac{(p\overline{p}-1)(\overline{p}-p)}{p(p-1)}}{\frac{\overline{p}-1}{p-1}-\frac{\overline{p}}{p}}\right|=|p\overline{p}-1|.$ By applying the symmetry trick we see that the expression $\frac{QX}{QX'}$ is constant, finishing the proof.

This post has been edited 2 times. Last edited by AndreiVila, Feb 18, 2025, 6:37 PM

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#7 h Mar 3, 2025, 5:09 PM • 1 Y

Y by Funcshun840

Let $ABC$ be $\Delta_0$ with contact triangle $DEF$ , incenter $I$ and $P$ be a point on $\omega$ . Let $A_1B_1C_1$ be $\Delta_1$ where $A_1,B_1,C_1$ are reflections of $P$ over $BC,AC,BA$ respectively. Let $A_2,B_2,C_2$ be $\Delta_2$ where $A_2,B_2,C_2$ are reflections of $P$ over $B_1C_1,A_1C_1,B_1A_1$ respectively. Let $D',E',F'$ be the reflections of $D,E,F$ over $I$ . Finally, let $X$ be the anti-Steiner of $P$ wrt $DEF$ .

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair A = (-4.79,-8.01); pair B = (-0.72,-0.97); pair C = (-8.85,-0.97); pair D = (-4.79,-0.97); pair E = (-6.82,-4.49); pair F = (-2.76,-4.49); pair I = (-4.79,-3.32); pair P = (-6.99,-2.51); pair Dp = (-4.79,-5.66); pair Ep = (-2.76,-2.15); pair Fp = (-6.82,-2.14); pair X = (-3.27,-5.11); pair B_1 = (-8.45,-3.35); pair A_1 = (-6.99,0.56); pair C_1 = (1.07,-7.17); pair A_2 = (-7.97,-4.97); pair C_2 = (-9,-1.76); pair B_2 = (-3.92,0.69); pair L = (-5.61,-4.49); pair M = (-4.05,-2.26); pair N = (-5.54,1.09); pair O = (-8.59,-4.18); import graph; size(15.11cm); pen xfqqff = rgb(0.5,0,1); pen ubqqys = rgb(0.29,0,0.51); draw(A--B); draw(B--C); draw(C--A); draw(circle(I, 2.35), red); draw(A--D); draw(B--E); draw(F--C); draw(C_1--B_1, green); draw(C_1--A_1, blue); draw(A_1--B_1); draw(circle((-5.55,-2.42), 3.52), red); draw(X--P, xfqqff); draw(F--D, blue); draw(D--E, ubqqys); draw(Dp--P, green); draw(F--E, green); draw(E--Dp); draw(P--Ep, blue); draw(I--X); draw(circle((-4.36,-4.49), 1.25)); draw(M--L, xfqqff); draw(C_2--P); draw(N--A_2); draw(O--B_2); draw(O--X, linetype("4 4")); draw(N--X, linetype("4 4")); draw(N--O, ubqqys); draw(A_1--P); draw(P--B_1); dot("$A$", A, dir((-16.000, -47.000))); dot("$B$", B, dir(0)); dot("$C$", C, dir((-17.000, 13.000))); dot("$D$", D, dir((5.000, 8.000))); dot("$E$", E, dir((-8.000, -33.000))); dot("$F$", F, dir((2.000, -21.000))); dot("$I$", I, dir((4.000, -23.000))); dot("$P$", P, dir((3.000, -29.000))); dot("$D'$", Dp, dir((-15.000, -18.000))); dot("$E'$", Ep, dir((-5.000, -21.000))); dot("$F'$", Fp, dir((-8.000, 12.000))); dot("$X$", X, dir((13.000, -16.000))); dot("$B_1$", B_1, dir((-30.000, -2.000))); dot("$A_1$", A_1, dir((6.000, 8.000))); dot("$C_1$", C_1, dir((5.000, 7.000))); dot("$A_2$", A_2, dir((-18.000, -21.000))); dot("$C_2$", C_2, dir((-29.000, -1.000))); dot("$B_2$", B_2, dir((6.000, 7.000))); dot("$L$", L, dir((-13.000, -21.000))); dot("$M$", M, dir((5.000, 7.000))); dot("$N$", N, dir((4.000, 8.000))); dot("$O$", O, dir((-12.000, -21.000))); [/asy]$

Let $L = EF \cap B_1C_1$ .
Claim 1: $L \in PD'$ .
Proof: Lets first prove that $B_1E \parallel C_1F$ .

$\angle B_1EF = \angle B_1EP + \angle PEF = 2\angle PFE + \angle PEF = \angle PFE + 120^\circ = \angle PFA + 60^\circ = \angle AFC_1 + 60^\circ = \angle EFC_1$ and we get $B_1E \parallel C_1F$ .

With this, we can ratio chase,
$\frac{B_1E}{EL} = \frac{F_C1}{FL} \implies \frac{PE}{EL} = \frac{PF}{FL}$ so by angle bisector theorem, we must have $P-L-D'$ , as desired. $\square$

Claim 2: $LD'XI$ is cyclic.
Proof: By angle chase,
$\angle ILD' = 180^\circ - 2\angle LD'I = 180^\circ - \angle PD'D = 180^\circ - \angle PID/2 = 180^\circ - \angle ID'X = 180^\circ-\angle IXD'$ as desired. $\square$

Claim 3: $L$ lies on the perpendicular bisector of segment $PX$ .
Proof: By angle chase,
$\angle LPI = \angle LD'I = \angle LXI$ Since $\angle IPX = \angle IXP$ , the result easily follows. $\square$ .

Claim 4: $X \in \Delta_2$ .
Proof: Let $M = DF \cap PE'$ , the proof of the above claims for $M$ is analogous.
By angle chase,

$\angle A_2XB_2 = \angle A_2XP + \angle PXB_2= \angle PLB_1 + \angle PMA_1 = 180^\circ - \angle B_1PL - \angle LB_1P + 180^\circ - \angle A_1PM - \angle MBA_1P = 360^\circ - \angle B_1PL - \angle A_2C_2P - \angle A_1PM - \angle B_2C_2P$
But $\angle B_1PL + \angle A_2C_2P = 180^\circ$ because $\angle D'FE' = 60^\circ$ and $\angle A_1FB_1 = 120^\circ$ by Fermat point.
Therefore,

$360^\circ - \angle B_1PL - \angle A_2C_2P - \angle A_1PM - \angle B_2C_2P = 180^\circ-\angle A_2C_2B_2$
which gives us the desire result. $\square$

Let $PA_2 \cap (A_2B_2C_2) = N \neq A_2$ and $PB_2 \cap (A_2B_2C_2) = O \neq B_2$
Claim 5: $ON \parallel ED$ .
Proof: Notice we only need to show $PC_1 \perp ON$
If we take the following homothety:
$\mathbb{H}(P, \frac{1}{2}): A_1 \rightarrow A_1', B_1 \rightarrow B_1', C_1 \rightarrow C_1', N \rightarrow N', O \rightarrow O'$
We see $PC_1$ passes thorough the center of $\triangle PB'_2C'_2$ , which immediately implies, $PC_1 \cap O'N'$ and the result follows. $\square$ .
Claim 6: $O-E-X$
Proof: By angle-chase,
$\angle NOX = \angle NA_2X = \angle PA_2X = \angle ILX = \angle ID'X = \angle DD'X = \angle DEX$ Combining this with claim 5 and using phantom points, we get the desire result. $\square$

By claim 5 and 6, we know $(A_2B_2C_2)$ and $\omega$ have equilateral triangles which are homothetic with center $X$ and since we proved on claim 4 that $X$ lies on $(A_2B_2C_2)$ and $\omega$ , by the homothety tangency trick, we can conclude $(A_2B_2C_2)$ and $\omega$ are tangent.

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