Let $r_A,r_B,r_C$ rays from point $P$. Define circles $w_A,w_B,w_C$ with centers $X,Y,Z$ such that $w_a$ is tangent to $r_B,r_C , w_B$ is tangent to $r_A, r_C$ and $w_C$ is tangent to $r_A,r_B$. Suppose $P$ lies inside triangle $XYZ$, and let $s_A,s_B,s_C$ be the internal tangents to circles $w_B$ and $w_C$; $w_A$ and $w_C$; $w_A$ and $w_B$ that do not contain rays $r_A,r_B,r_C$ respectively. Prove that $s_A, s_B, s_C$ concur at a point $Q$, and also that $P$ and $Q$ are isotomic conjugates. PS: The rays can be lines and the problem is still true.
Problem
Source: Brazilian National Olympiad, P3 2020
Tags: ray, circles, tangency, Hard Geometry
myh2910
16.03.2021 23:02
Are they isotomic conjugates wrt $\triangle XYZ$? It seems they're in fact isogonal conjugates wrt $\triangle XYZ$ by Delta 7.5.
myh2910
16.03.2021 23:15
[asy][asy] size(12cm); pointpen=black; pathpen=deepblue; real a=110, b=210, c=330; pair P=origin, A=2*dir(a), B=2*dir(b), C=2*dir(c), X=1.5*dir((b+c)/2), Y=-1.8*dir((c+a)/2), Z=1.7*dir((a+b)/2), P1=2*foot(P, Y, Z)-P, A1=2*foot(A, Y, Z)-A, P2=2*foot(P, Z, X)-P, B1=2*foot(B, Z, X)-B, P3=2*foot(P, X, Y)-P, C1=2*foot(C, X, Y)-C, Q=extension(P1, A1, P2, B1); D(circumcircle(foot(X, P, B), foot(X, P, C), 2*X-foot(X, P, C)), dotted); D(circumcircle(foot(Y, P, C), foot(Y, P, A), 2*Y-foot(Y, P, A)), dotted); D(circumcircle(foot(Z, P, A), foot(Z, P, B), 2*Z-foot(Z, P, B)), dotted); D(P--A, EndArrow); D(P--B, EndArrow); D(P--C, EndArrow); D(X--Y--Z--cycle, red); DPA(Q--P1^^Q--P2^^Q--P3, orange); D("P", P, NE); D("X", X, X); D("Y", Y, Y); D("Z", Z, Z); D("Q", Q, NW); MP("r_A", A, A); MP("r_B", B, B); MP("r_C", C, C); MP("s_A", P1, P1); MP("s_B", P2, P2); MP("s_C", P3, P3); [/asy][/asy] It's easy to notice that $PX$ is angle bisector of lines $r_B$ and $r_C$, etc. So, $PX$ and $r_A$ are isogonal conjugates wrt $\angle YPZ$. We can also notice that $s_A$ is the reflection of line $r_A$ wrt $YZ$, so by Delta 7.5, $s_A, s_B, s_C$ are concurrent on the isogonal conjugate of $P$ wrt $\triangle XYZ$.
hydo2332
17.03.2021 21:18
Deleted post.
myh2910
18.03.2021 00:50
Delta 7.5 wrote: Let $P$ and $Q$ be two Isogonal Conjugates WRT $\Delta ABC$. Then, the reflection of the line $AP$ WRT internal angle bisector of $\angle BPC$ and the reflection of the line $AQ$ WRT internal angle bisector of $\angle BQC$ are symmetric to each other WRT $\overline{BC}$
myh2910 wrote:
Denote by $P'$ and $Q'$ the reflections of points $P$ and $Q$ wrt line $BC$. Since $P$ and $Q$ are isogonal conjugates wrt $\triangle ABC$, \begin{align*}&\angle Q'BC=\angle CBQ=\angle PBA\quad\text{and}\quad\angle Q'CB=\angle BCQ=\angle PCA\\\implies&A\text{ and }Q'\text{ are isogonal conjugates wrt }\triangle BPC\\\implies&PQ'\text{ is the reflection of line }AP\text{ wrt the internal angle bisector of }\angle BPC\text{ and }\\&QP'\text{ is the reflection of line }AQ\text{ wrt the internal angle bisector of }\angle BQC\end{align*}Hence, the result follows. $\blacksquare$
bruckner
18.03.2021 02:28
I'm giving here only the main steps of my solution:
1) Define $Q_A$ as the intersection of $s_B$ and $s_C$. By simple angle chasing, one can prove that lines $XP$, $XQ_A$ are isogonal in $XYZ$. Therefore, our strategy will be to prove that the same happens for vertex $Y$ (ie, $YP$, $YQ_A$ are isogonal too).
2) This is equivalent to proving $\angle XPY + \angle XQ_AY= 180^{\circ}+ Z$, which is simply $\angle XPY + \angle XQ_AY + X +Y=360^{\circ}$
3) It is then pretty natural to consider the reflection $T$ of $Q_A$ over $XY$, since we only need to prove $\angle PYT=Y$. But, since we already know that $\angle TXP = X$, this is the same as proving that $T$ and $Z$ are isogonal conjugates in $PXY$.
4) But it is easy by angle chasing to get that $PZ$ and $r_C$ are symmetric by the bissector of $\angle XPY$, and therefore it must be true that $Z$ and $T$ are isogonal, as wanted.