Let $ABC$ be an acute triangle with circumcenter $O$. Let $A'$ be the center of the circle passing through $C$ and tangent to $AB$ at $A$, let $B'$ be the center of the circle passing through $A$ and tangent to $BC$ at $B$, let $C'$ be the center of the circle passing through $B$ and tangent to $CA$ at $C$. a) Prove that the area of triangle $A'B'C'$ is not less than the area of triangle $ABC$. b) Let $X, Y, Z$ be the projections of $O$ onto lines $A'B', B'C', C'A'$. Given that the circumcircle of triangle $XYZ$ intersects lines $A'B', B'C', C'A'$ again at $X', Y', Z'$ ($X' \neq X, Y' \neq Y, Z' \neq Z$), prove that lines $AX', BY', CZ'$ are concurrent.
Problem
Source: 2024 Vietnam National Olympiad - Problem 3
Tags: geometry
buratinogigle
05.01.2024 08:51
a) Denote $[X]$ the area of polygon $X$. Since $AA'\parallel OB'$ we get $$[A'OB']=[AOB']=\frac{1}{2}[AOBB']$$Similarly, $$[B'OC']=\frac{1}{2}[OBC'C],\, [C'OA']=\frac{1}{2}[OCA'A].$$From these we get $$2[A'B'C']=[AB'BC'CA']=S+[A'AC]+[B'BA]+[C'CB]$$where $[ABC]=S$. We have $$[A'AC]=\frac{1}{2}AC\cdot d(A',AC)=\frac{b}{2}\cdot\frac{b\cot A}{2}=\frac{b^2(b^2+c^2-a^2)}{16S}.$$Similarly, $$[B'BA]=\frac{c^2(c^2+a^2-b^2)}{16S},\, [C'CB]=\frac{a^2(a^2+b^2-c^2)}{16S}.$$Thus, $$[A'AC]+[B'BA]+[C'CB] =\frac{(b^2(b^2+c^2-a^2)+c^2(c^2+a^2-b^2)+a^2(a^2+b^2-c^2)}{16S}\ge S$$because $$16S^2=2(b^2c^2+c^2a^2+a^2b^2)-a^4-b^4-c^2\le b^2(b^2+c^2-a^2)+c^2(c^2+a^2-b^2)+a^2(a^2+b^2-c^2).$$(because it is equivalent to $ a^+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2$). We are done with part a) b) It is easily seen that $(A'), (B'), (C')$ have a common point $W$ (the second Brocard point). Angle chasing $$\angle OC'Z=\angle WCB=\frac{1}{2}\angle WC'B=\angle WC'B'.$$Similarly at the vertices $B'$ and $A'$ we get that $O$ and $W$ are isogonal conjugates in the triangle $A'B'C'$ so $X',Y',Z'$ are orthogonal projections of $W$ on $A'B', B'C', C'A' $. Combining with $AW \perp A'B'$ (because $A$ and $W$ are reflection in $A'B'$) so $AX'$ goes through $W$. Similarly, we get $AX', BY', CZ'$ concur at $W$.
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wassupevery1
05.01.2024 13:25
Let $S_{XYZ}$ denote the area of triangle $XYZ$, $R=OA=OB=OC, a=BC, b=CA, c=AB$.
Since $OA' \parallel C'C$, it follows that $S_{OA'C'} = S_{OA'C}$.
Now, triangles $OA'C$ and $BAC$ are similar, so $S_{OA'C} = \frac{OC^2}{BC^2} S_{ABC} = \frac{R^2}{a^2} S_{ABC}.$
Proving similarly, we get that $S_{OA'B'} = \frac{R^2}{b^2}S_{ABC}, S_{OA'B'} = \frac{R^2}{c^2}S_{ABC}$.
This means $$\frac{S_{A'B'C'}}{S_{ABC}} = \frac{S_{OA'B'}+S_{O'B'C}+S_{OC'A'}}{S_{ABC}}=R^2 \left( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right).$$We want to prove that this ratio is at least 1, so what we need is equivalent to $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \geq \frac{1}{R^2} = \frac{16S^2}{a^2b^2c^2}$.
Note that $16S^2=2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)$, so we now need to prove $$a^2b^2+b^2c^2+c^2a^2 \geq 2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4),$$which is equivalent to $a^4+b^4+c^4 \geq a^2b^2+b^2c^2+c^2a^2$. But this is obvious by AM-GM, so we are done.
Call the circles with centers $A', B', C'$ $\omega_a, \omega_b, \omega_c$ respectively.
Let $D$ be the intersection of $\omega_a$ and $\omega_b$. It follows that $\angle BAD = \angle ACD = \angle CBD$, implying $(BDC)$ is tangent to $AC$, or $D \in \omega_c$.
Let $M, N, P$ be the midpoints of $AD, BD, CD$ respectively, and $O'$ be the circumcenter of triangle $MNP$.
We prove the following key claim:
Claim. $D, O$ are isogonal conjugates wrt triangle $A'B'C'$.
A homothety at $D$, factor $2$ maps $\triangle MNP$ to $\triangle ABC$, hence it maps $O'$ to $O$, implying $O'$ is the midpoint of $MO$. But note that $MNP$ is the pedal triangle of $D$ wrt triangle $A'B'C'$, hence $D, O$ are isogonal conjugates wrt triangle $A'B'C'$.
This claim already implies $X' \equiv M, Y' \equiv N, Z' \equiv P$, and hence $AX', BY', CZ'$ concur at $D$.
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buratinogigle
06.01.2024 06:09
buratinogigle wrote: a) Denote $[X]$ the area of polygon $X$. Since $AA'\parallel OB'$ we get $$[A'OB']=[AOB']=\frac{1}{2}[AOBB']$$Similarly, $$[B'OC']=\frac{1}{2}[OBC'C],\, [C'OA']=\frac{1}{2}[OCA'A].$$From these we get $$2[A'B'C']=[AB'BC'CA']=S+[A'AC]+[B'BA]+[C'CB]$$where $[ABC]=S$. We have $$[A'AC]=\frac{1}{2}AC\cdot d(A',AC)=\frac{b}{2}\cdot\frac{b\cot A}{2}=\frac{b^2(b^2+c^2-a^2)}{16S}.$$Similarly, $$[B'BA]=\frac{c^2(c^2+a^2-b^2)}{16S},\, [C'CB]=\frac{a^2(a^2+b^2-c^2)}{16S}.$$Thus, $$[A'AC]+[B'BA]+[C'CB] =\frac{(b^2(b^2+c^2-a^2)+c^2(c^2+a^2-b^2)+a^2(a^2+b^2-c^2)}{16S}\ge S$$because $$16S^2=2(b^2c^2+c^2a^2+a^2b^2)-a^4-b^4-c^2\le b^2(b^2+c^2-a^2)+c^2(c^2+a^2-b^2)+a^2(a^2+b^2-c^2).$$(because it is equivalent to $ a^+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2$). We are done with part a) b) It is easily seen that $(A'), (B'), (C')$ have a common point $W$ (the second Brocard point). Angle chasing $$\angle OC'Z=\angle WCB=\frac{1}{2}\angle WC'B=\angle WC'B'.$$Similarly at the vertices $B'$ and $A'$ we get that $O$ and $W$ are isogonal conjugates in the triangle $A'B'C'$ so $X',Y',Z'$ are orthogonal projections of $W$ on $A'B', B'C', C'A' $. Combining with $AW \perp A'B'$ (because $A$ and $W$ are reflection in $A'B'$) so $AX'$ goes through $W$. Similarly, we get $AX', BY', CZ'$ concur at $W$. An easy extension from my proof: $$[A'B'C']-[ABC]=\frac{a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2}{16S}\ge 0.$$ Proof. Since $2[A'B'C']=[AB'BC'CA']=S+[A'AC]+[B'BA]+[C'CB]$, $$[A'B'C']-[ABC]=\frac{1}{2}([A'AC]+[B'BA]+[C'CB]-S)=\frac{1}{2}\left(\frac{b^2(b^2+c^2-a^2)+c^2(c^2+a^2-b^2)+a^2(a^2+b^2-c^2)}{16S}-S\right)$$$$=\frac{b^2(b^2+c^2-a^2)+c^2(c^2+a^2-b^2)+a^2(a^2+b^2-c^2)-(2(b^2c^2+c^2a^2+a^2b^2)-a^4-b^4-c^2)}{32S}$$$$=\frac{a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2}{16S}.$$Done! Consequence. If we construct similarly, let $A''$ be the center of the circle passing through $B$ and tangent to $AC$ at $A$, let $B''$ be the center of the circle passing through $C$ and tangent to $BA$ at $B$, let $C''$ be the center of the circle passing through $A$ and tangent to $CB$ at $C$. Then $$[A'B'C']=[A''B''C'']=\frac{a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2}{16S}+S=\frac{a^2b^2+b^2c^2+c^2a^2}{16S}\ge S.$$Apply the fundamental inequality $xa^2+yb^2+zc^2\ge 4\sqrt{xy+yz+zx}S$, we get $$[A'B'C']=[A''B''C'']\ge\frac{\sqrt{a^2b^2+b^2c^2+c^2a^2}}{4}\ge S.$$So we get some stronger results as above.
LLL2019
03.04.2024 05:40
Alternative solution for (1): It is $[A'B'C']=\frac{R^2}{2}\sin A\sin B\sin C \sum_{cyc} (\cot A+\cot B)(\cot A + \cot C)\geq 2R^2\sin A\sin B\sin C \sum_{cyc} \cot A\cot B = [ABC]$
sami1618
16.12.2024 21:05
Nice Problem ! Lemma: The reflections of $A$, $B$, and $C$ about the lines $A'B'$, $B'C'$, and $A'C'$, respectively, coincide at some point $P$. Proof: First, observe that \[ \angle AA'C + \angle BB'A + \angle CC'B = 2\angle A + 2\angle B + 2\angle C = 360^\circ. \]Thus, we must also have \[ \angle A'AB' + \angle B'BC' + \angle C'CA' = 360^\circ. \]Notice that $A'A = A'C$, $B'B = B'A$, and $C'C = C'B$. Thus, triangles $A'AB'$, $B'BC'$, and $C'CA'$ can be "rearranged" into triangle $A'B'C'$, proving the lemma. Part A: $[A'B'C'] \geq [ABC]$ Proof: A consequence of the lemma is that \[ [A'AB'BC'C] = 2[A'B'C']. \]The segments from $O$ to the vertices of the hexagon divide it into three pairs of congruent triangles, each similar to $ABC$. Thus, \[ [A'B'C'] = \frac{1}{2}[A'AB'BC'C] = [A'OA] + [B'OB] + [C'OC] = [ABC] \left( \frac{R^2}{a^2} + \frac{R^2}{b^2} + \frac{R^2}{c^2} \right)\geq \]\[ [ABC] \left( \frac{R^2}{ab} + \frac{R^2}{bc} + \frac{R^2}{ca} \right)= [ABC] \left( \frac{R^2 (a+b+c)}{abc} \right) \]Substituting $R = \frac{abc}{4[ABC]}$ and $a+b+c = \frac{2[ABC]}{r}$ gives \[ [ABC] \left( \frac{R^2 (a+b+c)}{abc} \right) = [ABC] \left( \frac{R}{2r} \right) \geq [ABC]. \]Thus, $[A'B'C'] \geq [ABC]$, as desired. See here for almost the same inequality. Part B: $AX'$, $BY'$, and $CZ'$ are concurrent. Proof: Notice that $A$, $B$, and $C$ are the reflections of $P$ about the respective sides of $A'B'C'$. As $O$ is the circumcenter of $ABC$, we must have that $O$ is the isogonal conjugate of $P$ with respect to $A'B'C'$. Thus, $X'Y'Z'$ is the pedal triangle of $P$, and hence $AX'$, $BY'$, and $CZ'$ all concur at $P$, finishing the problem.
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