In $\triangle {ABC}$, tangents of the circumcircle $\odot {O}$ at $B, C$ and at $A, B$ intersects at $X, Y$ respectively. $AX$ cuts $BC$ at ${D}$ and $CY$ cuts $AB$ at ${F}$. Ray $DF$ cuts arc $AB$ of the circumcircle at ${P}$. $Q, R$ are on segments $AB, AC$ such that $P, Q, R$ are collinear and $QR \parallel BO$. If $PQ^2=PR \cdot QR$, find $\angle ACB$.
Problem
Source: 2024 CTST P8
Tags: geometry, ratio
LoloChen
11.03.2024 07:33
$\angle ACB= \frac{\pi}{4}$
Let tangents of $\odot {O}$ at $A, C$ intersects at ${Z}$ and $BZ$ cuts $AC$ at ${E}$. Let rays $ED, FE$ cut $\odot {O}$ at $S, T$ respectively. Construct $A'$ on $\odot {O}$ such that $A'A \parallel OB$.
Claim: $PSX, STZ, TPY$ are collinear respectively and $(B, C; R, A)=\frac{\sqrt5 -1}{2}$.
Sejfried Theorum PropertySince the conclusion is totally projective, there exists a projective transformation that sends $\triangle {ABC}$ into an equilateral triangle and fixes the circumcircle. Then ${D}$ becomes the midpoint of $BC$ and $\angle PDB=\angle SDB=\frac{\pi}{3}$, so $PSX$ are collinear. Furthermore, $\frac{BP}{PC}=\frac{BS}{SC}=\frac{AP}{PB}$, combining Ptolemy Theorum, $\frac{AP}{PB}=\frac{\sqrt5 -1}{2}$. So $(B, C; R, A)=\frac{\sqrt5 -1}{2}$. Before the projective transformation, the collinear and crossratio property still holds.
Back to the problem:
The condition imply $\frac{PQ}{PR}=\frac{\sqrt5 -1}{2}$, that is $(Q, R; P, \infty _{OB})=\frac{\sqrt5 -1}{2}$. $$(B, C; R, A')=(Q, R; P, \infty _{OB})=\frac{\sqrt5 -1}{2}=
(B, C; R, A)$$So $A=A'$. Simple angle chase gives $\angle ACB= \frac{\pi}{4}$.
PROF65
11.03.2024 20:25
LoloChen wrote: In $\triangle {ABC}$, tangents of the circumcircle $\odot {O}$ at $B, C$ and at $A, B$ intersects at $X, Y$ respectively. $AX$ cuts $BC$ at ${D}$ and $CY$ cuts $AB$ at ${F}$. Ray $DF$ cuts arc $AB$ of the circumcircle at ${P}$. $Q, R$ are on segments $AB, AC$ such that $P, Q, R$ are collinear and $QR \parallel BC$. If $PQ^2=PR \cdot QR$, find $\angle ACB$. $ \angle ACB= \frac{\pi}{3}$ is a solution not $\frac{\pi}{4}$
GeoKing
12.03.2024 11:41
@above its typo actually $QR \parallel BO$
LLL2019
12.03.2024 11:48
GeoKing wrote: @above its typo actually $QR \parallel BO$ Ravi how are you doing these days
internationalnick123456
12.03.2024 12:04
$BX$ intersects $AC$ at $I$. $AD$ cuts $(O)$ again at $Z$, $L$ is the Lemoine point of $\Delta ABC$. We have $(AD,LX)= (CF,LY) = -1$, so $AC,DF,XY$ are concurrent $\Rightarrow I\in DF\Rightarrow I, D, P$ are collinear. Therefore, $A(BI,DP)=B(AI,DP)\Rightarrow (BC,ZP)= (AB,CP)$ $\Rightarrow \frac{ZB}{ZC}:\frac{PB}{PC}=\frac{CA}{CB}:\frac{PA}{PB}$ Nevertheless, easy to see that $\frac{ZB}{ZC}=\frac{AB}{AC}$, hence, $\frac{AC.PB}{AB.PC} =\frac{BC.PA}{CA.PB}$. By applying Ptoleme's theorem for convex cyclic quadrilateral $APBC$ we have $\frac{BC.PA}{CA.PB}=\frac{PC.AB}{CA.PB}-1$ $\Rightarrow \frac{AC.PB}{AB.PC}=\frac{PC.AB}{CA.PB}-1\Rightarrow (BC,PA) = \frac{1}{(BC,PA)}-1$. Otherwise, from $PQ^2 = PR.QR$, we get $\frac{PQ}{PR}=\frac{PR}{PQ}-1$. And by taking note that $(BC,PA)>0$, we obtain $$(BC,PA)=\frac{PQ}{PR}$$$\Rightarrow A(BC,PA) = A(QR,P\infty)\Rightarrow A(QR,PY)=A(QR,P\infty)$. This leads to $AY\parallel QR$, or $AY\parallel OB$ $\Leftrightarrow \widehat{AOB}=90^\circ\Leftrightarrow \widehat{ACB}=45^\circ$.
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zhihanpeng
12.03.2024 13:13
Disclaimer: I didn't know nothing about trilinear coordinates before I went onto 数之谜 and got EMOTIONAL DAMAGE when I saw many people crushing this problem easily T_T Now I know I'm very bad at geometry So I learned a bit trilinear coordinates and since no one is posting about them here, imma do it for you- p.s. I'm a beginner at this so if there are any technical mistakes please tell me... All the formulas can be found here: https://en.wikipedia.org/wiki/Trilinear_coordinates
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