Let $ABC$ be a triangle with $O$ as its circumcenter. A circle $\Gamma$ tangents $OB, OC$ at $B, C$, respectively. Let $D$ be a point on $\Gamma$ other than $B$ with $CB=CD$, $E$ be the second intersection of $DO$ and $\Gamma$, and $F$ be the second intersection of $EA$ and $\Gamma$. Let $X$ be a point on the line $AC$ so that $XB\perp BD$. Show that one half of $\angle ADF$ is equal to one of $\angle BDX$ and $\angle BXD$. Proposed by usjl
Problem
Source: 2024 Taiwan TST Round 2 Independent Study 1-G
Tags: Taiwan, geometry
hukilau17
27.03.2024 20:20
Complex bash with $\Gamma$ as the unit circle, so that $$|b|=|c|=1$$$$o = \frac{2bc}{b+c}$$$$d = \frac{c^2}b$$$$e = \frac{d-o}{d\overline{o}-1} = \frac{c(2b+c)}{b+2c}$$Let line $AC$ intersect $\Gamma$ again at $T$. Then $$|t|=1$$$$a = t + o - ct\overline{o} = \frac{bt-ct+2bc}{b+c}$$$$f = \frac{e-a}{e\overline{a}-1} = \frac{t(bt+2ct+c^2)}{bt+2bc+c^2}$$$$x = \frac{ct(b-d) - b(-d)(c+t)}{ct - b(-d)} = \frac{b^2t+bc^2+bct-c^2t}{b(c+t)}$$Now we find the vectors \begin{align*} a-d &= \frac{(b-c)(bt+2bc+c^2)}{b(b+c)} \\ f-d &= \frac{b^2t^2+2bct^2-2bc^3-c^4}{b(bt+2bc+c^2)} \\ b-d &= \frac{(b+c)(b-c)}b \\ x-d &= \frac{(b-c)(bt+2ct+c^2)}{b(c+t)} \\ b-x &= \frac{c(b-c)(b-t)}{b(c+t)} \end{align*}Then $$\frac{\left(\frac{a-d}{f-d}\right)}{\left(\frac{b-d}{x-d}\right)^2} = \frac{(b-c)(bt+2bc+c^2)^2(bt+2ct+c^2)^2}{(b+c)^3(c+t)^2(b^2t^2+2bct^2-2bc^3-c^4)}$$We check that this equals its conjugate and is thus real. Additionally, we have $\frac{b-x}{b-d} \in i\mathbb{R}$, so $\left(\frac{b-x}{b-d}\right)^2$ is a negative real number. Thus $$\frac{\left(\frac{a-d}{f-d}\right)}{\left(\frac{b-x}{d-x}\right)^2}$$is also real, and has opposite sign from the other ratio. So one of the ratios $$\frac{\left(\frac{a-d}{f-d}\right)}{\left(\frac{b-d}{x-d}\right)^2}, \frac{\left(\frac{a-d}{f-d}\right)}{\left(\frac{b-x}{d-x}\right)^2}$$is real and positive. Thus $\angle ADF$ equals either $2\angle BDX$ or $2\angle BXD$, depending on which one is positive. $\blacksquare$
VicKmath7
28.03.2024 15:11
If the perpendiculars at $B, C$ to $AB, AC$ meet $AC, AB$ at $V, U$, then $\Gamma$ is the circle with diameter $UV$. Since $OB, OC$ touch $\Gamma$, $(BC; ED)=-1$, so projecting through $A$ onto $\Gamma$ gives that $SUFV$ is harmonic, where $S=AD \cap \Gamma$. Since $UV$ is a diameter, this quadrilateral is a kite, i.e. $\angle ADF=\angle SDF=2\angle SVU$ or $\angle SDF=2\angle SUV$ depending on which side of $SF$ $D$ lies. Hence, it suffices to show that $\triangle SUV \sim \triangle BDX$, so we need $\frac{SU} {SV}=\frac{BD} {BX}$ since both triangles are right-angled. This is straightforward to show by length-chasing using the following three similarities: $\triangle ASU \sim \triangle ABD$, $\triangle ASV \sim \triangle ACD$ and $\triangle BXC \sim \triangle ABC$.
PROF65
28.03.2024 20:22
we will use directed angle to elude to specify the different config. the desired result then is equivalent to $2\angle BXD=2\angle BDX=ADF$ (the equality $\mod \pi$) Let $B'=AB\cap \Gamma ,C'=AC\cap \Gamma,D'=AD\cap \Gamma$ then we have $(B'C',FD')=(BC,ED )-1$ besides notice $DC'\perp DB'$ thus $DC',DB'$ are the angle bisector of $\angle ADF$ so $\angle ADF=2\angle ADB'=2\angle ADC'$. Moreover remark $ BXC\sim ABC$ then $CA.CX=CB^2=CD^2$ i.e. $CD$ is tangent to $(ADX)$ we deduce then $ \angle CXD= \angle ADC$ hence $\angle BXD=\angle BXC+\angle CXD =\angle ABC+\angle ADC=\angle B'DC +\angle ADC$ thus $2\angle BXD=2\angle B'DC +2\angle ADC= 2\angle CDC' +2\angle ADC=2 \angle ADC'=\angle ADF$ Best regards. RH HAS