Let $ABC$ be a triangle with incenter $I$. The line through $I$, perpendicular to $AI$, intersects the circumcircle of $ABC$ at points $P$ and $Q$. It turns out there exists a point $T$ on the side $BC$ such that $AB + BT = AC + CT$ and $AT^2 = AB \cdot AC$. Determine all possible values of the ratio $IP/IQ$.
Problem
Source: RMM Extralist 2021 G2
Tags: geometry, RMM Shortlist, Inversion, similarities, Bashing, incenter, circumcircle
Seicchi28
19.09.2023 15:52
The only possible value is $\boxed{\{1/2, 2\}}$.
Assume WLOG $AB < AC$.
Obviously, $T$ is the touchpoint of $A$-excircle. Let $M_A$ be the touchpoint of $A$-mixtilinear incircle with $\odot(ABC)$, let $E$ be the midpoint of minor arc $\widehat{BC}$ of $\odot(ABC)$ and let $D$ be the touchpoint of the incircle of $\triangle ABC$ with $BC$. It is known that $AT$ is isogonal to $AM_A$ w.r.t. $\angle BAC$, and also $AT \times AM_A = AB \times AC$ (i.e. by considering the $\sqrt{AB\times AC}$) inversion). Therefore, $AT = AM_A$, and since $AE$ is the bisector of $\angle M_AAT$, we conclude $AE$ is the perpendicular bisector of $M_AT$. Therefore, $ED = ET = EM_A$.
By considering the pairwise radical axes of the following three circles: $\odot(BIC), \odot(IE), \odot(ABC)$, we have that the three lines $PQ, M_AE, $ and $BC$ are concurrent at a point $Y$. Let $YA$ meet $\odot(ABC)$ for the second time at $X$. Since $YX \times YA = YI^2$ and $AI \perp IY$, we have $AX \perp XI$. Therefore, $X$ is the $A$-sharkydevil point, so $BX/XC = BD/DC \implies X, D, E$ are collinear. It's also easy to see that the five points $I, X, Y, M_A, D$ are located on a circle with diameter $IY$.
To finish, let $M_AD$ cut $\odot(ABC)$ for the second time at $A'$. Since $ED = EM_A$, we have $A'X = A'D$. The angle bisectors of $\angle XA'D$ and the angle bisectors of $\angle DEM_A$ meet at the midpoint of arc $\widehat{XM_A}$ of $\odot(ABC)$. But these two lines are precisely the perpendicular bisector of $XD$ and the perpendicular bisector of $DM_A$, respectively, which meet at the center of $\odot(XIDM_AY)$, which is located on $IY$. Therefore, the center of $\odot(XIDM_AY)$ is the intersection of $IY$ and $\odot(ABC)$, namely $P$ (which comes from the assumption $AB < AC$, so $Y$ is closer to $P$ than $Q$). So, $P$ is the midpoint of $IY$. To finish off:
\[ YP \times YQ = YI^2 \implies IP \times (2IP + IQ) = 4IP^2 \implies IQ = 2IP, \]so $IP/IQ = 1/2$. For the other case ($AB > AC$), we will have $IP/IQ = 2$.
VicKmath7
21.09.2023 22:45
Amazing problem! Here is another synthetic solution.
We will show that $IQ=2IP$ if $AC>AB$. The first condition implies that $T$ is the $A$-excircle touchpoint with $BC$. If $T_a$ is the $A$-mixtilinear touchpoint, then $T$ and $T_a$ are interchanged under $\sqrt{bc}$ inversion, so $AT \cdot AT_a=AB \cdot AC$, which means that $AT=AT_a$. Hence, $T$ and $T_a$ are symmetric in $AI$.
If $M$ is the midpoint of the minor arc $BC$, then it's known that $\angle IT_aM=90^{o}$, so we know that $\angle ITM=90^{o}$. Redefine $MT \cap (ABC)=Q'$; then shooting lemma implies that $MI^2=MT \cdot MQ'$, so $\angle AIQ'=90^{o}$ and $Q=Q'$. Hence, $Q$ is the excircle sharky-devil point, so $\angle AQI_a=90^{o}$, where $I_a$ is the $A$-excenter.
Let $M_b, M_c$ be the midpoints of the minor arcs $AC, AB$ and $E, F$ are the touchpoints of the mixtilinear incircle with $AB, AC$. Then $M_c \in T_aE$ and $M_b \in T_aF$, so Pascal's theorem implies that $MT_a, BC, EF$ concur at a point $R$ (it's known that $EF$ and $PQ$ coincide).
Observe that $RITI_a$ is cyclic with diameter $RI_a$. Since $\angle ITM=\angle RTI_a=90^{o}$, we have that $\angle ITR=\angle MTI_a$, so $ITI_aR$ is harmonic. For $\sqrt{bc}$ inversion reasons, $\angle ATC=\angle ABT_a$, so $\angle ATB=\angle T_aMA=\angle RI_aT$, so $AT$ is tangent to $(ITI_aR)$ and since $A \in II_a$, $AR$ is also tangent since the quadrilateral is harmonic. Hence, $\angle ARI_a=90^{o}=\angle AQI_a$, which means that $AQI_aR$ is cyclic and since $AQMP$ is cyclic, Reim's implies that $MP \parallel RI_a$, or that $P$ is the midpoint of $RI$.
Finally, observe that $\angle IQM=\angle TIM=\angle TRI_a=\angle IRM$, so $\triangle QRM$ is isosceles and $\angle RIM=90^{o}$, so $IQ=IR=2IP$, as desired.
BlizzardWizard
07.07.2024 23:25
Complex bash with $(ABC)$ unit circle. Set $A=a^2$, etc. such that the incenter is $j=-ab-ac-bc$ and the $A$-excenter is $j_A=ab+ac-bc$. Obviously, $T$ is the extouch point on side $BC$, which can be calculated as the average of $I_A$ and its reflection over $BC$. $a^2-t=\frac{j_a+b^2+c^2-b^2c^2\overline{j_A}}2=a^2-\frac{ab+ac-bc+b^2+c^2-b^2c^2(\overline{ab+ac-bc})}2=\frac{2a^3-(b+c)a^2-(b^2+c^2)a+bc(b+c)}{2a}=\frac{(2a+b+c)(a-b)(a-c)}{2a}$ Since $AT^2=AB\cdot AC$, we have \begin{align*} (a^2-t)\overline{(a^2-t)}&=\frac{(a^2-b^2)(a^2-c^2)}{a^2bc}\\ \frac{(2a+b+c)(a-b)(a-c)}{2a}\cdot\frac{(ab+ac+2bc)(a-b)(a-c)}{2a^2b^2c^2}&=\frac{(a^2-b^2)(a^2-c^2)}{a^2bc}\\ (2a+b+c)(ab+ac+2bc)(a-b)(a-c)&=4abc(a+b)(a+c)\\ (b+c)(2a^4-(b+c)a^3-(b^2+6bc+c^2)a^2-bc(b+c)a+2b^2c^2)=0 \end{align*}Since $b+c$ cannot equal $0$ (or $B$ and $C$ would be equal). We now prove that $\frac{p-j}{q-j}+\frac{q-j}{p-j}=-\frac52$, from which it follows that the desired ratio is either $2$ or $\frac12$. We have that $p$ and $q$ are solutions to \begin{align*} \frac{z-j}{a^2-j}=-\frac{\overline z-\overline j}{\overline{a^2}-\overline j}\\ (z-j)+a^2bc(1/z-\overline j)&=0 \end{align*}By Vieta's, we have $pq=a^2bc$ and $p+q=j+a^2bc\overline j$. \begin{align*} \frac{p-j}{q-j}+\frac{q-j}{p-j}&\stackrel?=-\frac52\\ 2((p-j)^2+(q-j)^2)+5(p-j)(q-j)&\stackrel?=0\\ 2(p+q)^2-4pq-4(p+q)j+4j^2+5pq-5(p+q)j+5j^2&\stackrel?=0\\ 2(j+a^2bc\overline j)^2+a^2bc-9(j+a^2bc\overline j)j+9j^2&\stackrel?=0\\ 2(a^2+2ab+2ac+bc)^2+a^2bc-9(ab+ac+bc)(a^2+ab+ac)&\stackrel?=0 \end{align*}And the expansion of the left hand side is $2a^4-(b+c)a^3-(b^2+6bc+c^2)a^2-bc(b+c)a+2b^2c^2=0$, so we're done.
DottedCaculator
27.12.2024 19:03
By Stewart's Theorem, $(s-b)a(s-c)+bca=b^2(s-c)+c^2(s-b)$, which rearranges to $a^3-3(b-c)^2a-2(b+c)(b-c)^2=0$. By Power of a Point, $IP\cdot IQ=2Rr$, and $IP-IQ$ is twice the distance from $O$ to $AI$, which is the distance from $A$ to the midpoint of arc $BAC$, or $2R\sin\left(\frac{B-C}2\right)$. Therefore, we have \begin{align*} \frac{(IP-IQ)^2}{IP\cdot IQ}&=\frac{2R\sin^2\left(\frac{B-C}2\right)}r\\ &=\frac{R(1-\cos(B-C))}r\\ &=\frac{abc(1-\cos(B-C))}{4(s-a)(s-b)(s-c)}\\ &=\frac{abc}{4(s-a)(s-b)(s-c)}\left(1-\cos B\cos C-\sin B\sin C\right)\\ &=\frac{abc}{4(s-a)(s-b)(s-c)}\left(1-\frac{a^4-(b^2-c^2)^2}{4a^2bc}-\frac{bc}{4R^2}\right)\\ &=\frac{abc}{4(s-a)(s-b)(s-c)}\left(1-\frac{a^4-(b^2-c^2)^2}{4a^2bc}-\frac{4A^2}{a^2bc}\right)\\ &=\frac{abc}{4(s-a)(s-b)(s-c)}\left(\frac{4a^2bc-a^4+(b^2-c^2)^2-16A^2}{4a^2bc}\right)\\ &=\frac{abc}{4(s-a)(s-b)(s-c)}\left(\frac{(b-c)^2((b+c)^2-a^2)}{2a^2bc}\right)\\ &=\frac{(b-c)^2(b+c-a)(b+c+a)}{8a(s-a)(s-b)(s-c)}\\ &=\frac{(b-c)^2(a+b+c)}{a(a+b-c)(a-b+c)}\\ &=\frac{(b-c)^2(a+b+c)}{a^3-3(b-c)^2a-2(b+c)(b-c)^2+2(b-c)^2(a+b+c)}\\ &=\frac12. \end{align*} Therefore, $\frac{IP}{IQ}$ can be either $\boxed{\frac12}$ or $\boxed2$.