Sketch: (Might write up when I get home) Prove that X,I and D, the point of tangency of the incircle and BC are collinear, the end.

Call $(AIB) \cap AC,BC = U,V$ and $(AIC) \cap AB,BC = T,S$ and let $D$ be the foot of perp from $I$ to $BC$. Clearly $DV=DS$ and $\measuredangle XVS = \measuredangle XSV=\measuredangle BAC$ by cyclic quads so done.

My method: extend the reflections of the sides across angle bisectors to hit BC at R, S then prove that XI is perpendicular bisector of that

Let $C'$ and $A'$ be the reflections of $C$ and $A$ across $BI$ and let $B'$ and $A'_1$ be the reflections of $B$ and $A$ across $CI$. Note that $A'$ and $A'_1$ lie on $BC$ and $B'$ and $C'$ lie on $AC$ and $AB$ respectively. By the fact that $AA'_1 \perp BB'$ and $AA' \perp CC'$, we get that $\widehat{XA'A'_1}=\widehat{XA'_1A'}=\widehat{BAC}$ so $XA'=XA'_1$ and since $AB'IBA'_1$ and $AC'CA'I$ are cyclic, $I$ is the incenter of $XA'A'_1$ so $XI\perp BC$

mine hope you all enjoyed it. The shortest solution I know is: Suppose the reflection of $AB$ across $CI$ intersects $BC$ at $E$, and define $F$ similarly. Then $\angle{XEC} = \angle{BAC} = \angle{XFB}$, so $XE = XF$. But also since $E$ and $A$ are reflections across $IC$, $IE = IA$ and hence $IA = IF$. So $XI$ is the perpendicular bisector of $EF$.

What!? Let $A_B$ and $A_C$ respectively be the reflections of $A$ about $CI$ and $BI$ Notice that both of these lie on line $BC$ and $IA_B=IA=IA_C$ Obviously $\angle XA_BA_C=\angle BAC = \angle XA_CA_B$ which gives us $XA_B=XA_C$ which combined with the above fact gives us that $XI$ is the perpendicular bisector of $A_BA_C$ which implies it is perpendicular to $BC$ Sniped @above xD Nice problem!

awesomeming327. wrote: My method: extend the reflections of the sides across angle bisectors to hit BC at R, S then prove that XI is perpendicular bisector of that I did this as well

hard for a problem 1, or is it just since im bad with lengths?? Really cute though Let $B'$ and $C'$ be where the reflections hit $BC$. As $\angle XB'C'=\angle XC'B'=\angle BAC$, it suffices to show the midpoint of $B'C'$ is the intouch point. But as $CB'=CA$ and $BC'=BA$ be reflection, this is clearly true say by coordinates or anything to keep track of lengths really.

Let $BI$ meet $AC$ at $E$ and $CI$ meet $AB$ at $F$. Also, let $XF$ and $XE$ meet $BC$ at $A_1$ and $A_2$, respectively. Note that $$\measuredangle A_2A_1X = \measuredangle CA_1F = \measuredangle FAC = \measuredangle BAC,$$and similarly $\measuredangle XA_2A_1 = \measuredangle BAC$, so $\triangle XA_1A_2$ is isosceles. But from the reflections we find that $A_1I$ bisects $\angle XA_1A_2$ and $A_2I$ bisects $\angle XA_2A_1$, so $I$ is the incenter of $\triangle XA_1A_2$. Thus $XI$ bisects $\angle A_1XA_2$, which is enough to imply $\overline{XI} \perp \overline{BC}$. Edit: I think in the obtuse case $I$ is actually the $X$-Excenter oops but it should be similar

sus problem why are there config issues Let the reflection of $AB$ over $CI$ intersect $BC$ at $A_1$ and let the reflection of $AC$ over $BI$ intersect $BC$ at $A_2$. Note that $\angle XA_1A_2 = \angle A = \angle XA_2A_1$. Therefore, $XA_1A_2$ is isosceles, so it suffices to show that the foot from $I$ to $BC$, which we call $D$, is the midpoint of $A_1A_2$. By reflections, we have that $IA_1 = IA = IA_2$. Therefore, by the Pythagorean Theorem or congruent triangles or something, we are done. P.S. I initially used length chase to show the midpoint, but that runs into SO MANY config issues

Let $E,F=BI\cap AC,CI\cap AB.$ Let $B',C'$ be the reflections of $B,C$ over $CI,BI.$ Pappus on $BB'FCC'E$ gives $X$ lies on the line through $I$ and the orthocenter of $BIC$ which finishes.

Define $P$, $Y$ to be the intersection of the perpendicular line to $AC$ through $I$ with $AC$ and $AB$. Let $X_1$ be the intersection with the perpendicular line to $BC$ through $I$ with the reflection of $AB$ about $CI$. We can easily prove $X_1IF$ and $IYF$ are congruent. If $Q$ and $Z$ are defined similarly, we see $IY=IZ$, thus $IX_1=IX_2$, hence done.

khina wrote: mine hope you all enjoyed it. The shortest solution I know is: Suppose the reflection of $AB$ across $CI$ intersects $BC$ at $E$, and define $F$ similarly. Then $\angle{XEC} = \angle{BAC} = \angle{XFB}$, so $XE = XF$. But also since $E$ and $A$ are reflections across $IC$, $IE = IA$ and hence $IA = IF$. So $XI$ is the perpendicular bisector of $EF$. Here is a shorter solution: Let $DEF$ be the contact triangle. Use complex numbers wrt the incircle, so the reflection of $\overline{AB}$ is the tangent at $\tfrac{de}{f}$ and similarly the reflection of $\overline{AC}$ is the tangent at $\tfrac{df}{e}$. Thus $x=\tfrac{2}{\tfrac{f}{de}+\tfrac{e}{df}}$ and $\tfrac{x}{d}=\tfrac{ef}{e^2+f^2} \in \mathbb{R}$ so $I,D,X$ collinear; done.

Let $X$ and $Y$ be the reflections of $A$ over $CI$ and $BI$. If $T$ is the desired intersection point, then $\angle TXY = \angle TYX = \angle A$, so the foot from $T$ to $BC$ has distance $\frac{BX+BY}{2} = \frac{BC-AC+AB}{2}$ from $B$ as desired.

Very simple and beautiful too Let $E$ be the intersection of the angle bisector of $\angle ABC$ with $\overline{AC}$; let $F$ be the intersection of the angle bisector of $\angle ACB$ with $\overline{AB}$; let $L$ be the reflection of $A$ over the angle bisector of $\angle ACB$; let $K$ be the reflection of $A$ over the angle bisector of $\angle ABC$. Because of how angle bisectors work, $K$ lies on line $\overline{BC}$ such that $AB = BK$. Furthermore, line $BE$ is the perpendicular bisector of $\overline{AK}$. So, by symmetry, $\angle EKB = \angle EAB = \angle A$. Similarly, $\angle FLC = \angle A$, so $\triangle XLK$ is isosceles. Lastly, since $I$ lies on the perpendicular bisectors of $AL$ and $AK$, it also lies on the perpendicular bisector of $LK$, so it follows that $XI \perp BC$ as desired.

If $A_1C=AC$ and $A_2B=AB$ with $A_1,A_2$ on $BC$, then $XA_1=XA_2$ and $IA_1=IA_2$, so $XI\perp BC$.

Nice geo. But still it took time Let $A_B$ and $A_C$ respectively be the reflections of $A$ about $CI$ and $BI$. Let $IP \perp BC$ Observe that $AA_BA_C$ is isosceles. Now by standard length calculations we can find that $ IP$ is bisects $A_BA_C$. We are done

Let $E=\overline{BI} \cap \overline{AC}$ and $F=\overline{CI} \cap \overline{AB}$. Let $B'$ and $C'$ be the reflections of $B$ and $C$ over $\overline{CI}$ and $\overline{BI}$, respectively. Notice that $H=\overline{BB'} \cap \overline{CC'}$ is the orthocenter of $IBC$. Pappus on $BB'FCC'E$ gives $I$, $H$, and $X$ collinear, which finishes.

Let $\omega$ denote the incircle and $D$ be it's tangent at $BC$. By reflection properties, we have $\angle BFI = \angle IFX$. Thus, if we let $FX$ intersect $AC$ at $R$ then $\omega$ is the excircle of $AFR$ which implies $FX$ is tangent to $\omega$. Similarly, $EX$ is tangent to $\omega$. Thus, by Brianchon's on $BFXECD$ we have $BE$, $CF$, and $XD$ concur. But, this point of concurrence is $I$. Thus, $D, I, X$ are collinear which implies the result.

Denote the intersection of $BC$ and the two reflections as $K$ and $L$. Note \[\angle XKC = \angle XLB = \angle A, \quad d(I, XK) = d(I, AB) = d(I, AC) = d(I, XL),\] so $\triangle XKL$ is isosceles and $I$ lies on the $X$-altitude, as desired. $\blacksquare$

Let $A_1$, $A_2$ be the reflections of $A$ over $\overline{CI}$, $\overline{BI}$, respectively. Let $D=\overline{AB}\cap\overline{CI}$ and $E=\overline{AC}\cap\overline{BI}$. Claim: $I$ is the incenter of $\triangle XA_1A_2$. Proof. Notice \[\measuredangle IA_2B=\measuredangle BAI=\measuredangle IAD=\measuredangle DA_2I\]and similarly $\overline{IB}$ bisects $\angle A_2A_1X$. $\blacksquare$ Notice $\measuredangle DA_2B=\measuredangle BAD=\measuredangle EAC=\measuredangle CA_1E$ so $\triangle XA_1A_2$ is isosceles and we finish. $\square$

Let $A,B$ are reflected to $A', B'$ and $A,C$ are reflected to $A'', C'$. Denote the tangency points of the incircle and $AB, BC, CA$ are $D,E,F$ respectively. The condition implies $AC=A'C, AB=A''B, AB=B'C, BC'=BC$ where $A', A''\in BC; B'\in AC; C'\in AB$. Let $AF=AD=a, CD=CE=b, BE=BD=c, AC=a+c=A'C=A'E+EC=A'E+c, A'E=a$. Similar reason yields $A''E=a, A'E=A''E$. Now, the question is equivalent to prove $\angle{B'A'A"}=\angle{C'A"A'}$. Let $\angle{ACI}=\angle{A'CI}=\alpha, \angle{ABI}=\angle{A"BI"}=\beta$. Since $A'B'$ is the reflection of $AB$ about $CI$, $A'B'$ must intersect $AB$ on $CI$, denote the intersection as $K$. We have $\angle{KB'B}=\angle{KBB'}=\angle{KAA'}=\angle{KA'A}$. Since $AA'||BB', \angle{CA'A}=\angle{CBB'}=90-\alpha, \angle{B'BA}=90-\alpha-2\beta=\angle{B'A'A}, \angle{B'A'C}=180-2\alpha-2\beta$ By the same reason, we get $\angle{C'A''A}=90-\beta-2\alpha, \angle{C'A"A'}=180-2\alpha-2\beta=\angle{B'A'A"}$ Thus, we have $\angle{XA'A"}=\angle{XA"A'}, AE=A"E$, we know an isosceles triangle's height, angle bisector, and median to the base are the same line, we get $X,I,E$ are collinear which yields $XI\bot BC$ as desired. (I probably add too many details since I wrote basically nothing in my scratch work but i want to make sure the sol is right <<

Hmm maybe I should return to trying to bash every problem I do...

PEKKA wrote: Let $ABC$ be a triangle with incenter $I$. Suppose the reflection of $AB$ across $CI$ and the reflection of $AC$ across $BI$ intersect at a point $X$. Prove that $XI$ is perpendicular to $BC$. Let $D,E,F$ be the tengency points of the incercly.$M=EF\cap CI,N=EF\cap BI$. By Iran lemma we have that $F,M,I,D,B$ lie on the same circle the symmetry of thiw circle with respect to $CI$ is the circle $Z,M,I,E,B'$. Now we have that $\measuredangle MZE=\measuredangle EIC=90-\frac{\measuredangle C}{2}=\measuredangle DFE=\measuredangle DZE$ we get that $Z,M,D$ are collinear.SIlilary we have $D,Y,N$. $\angle ZDI=\angle MDI=\angle MBI=\angle MBN=\angle MCN=\angle ICN=\angle IDN=\angle IDY$ means that $DI$ is the angel bisector of $ZDY$ but also$I$ is the circumcenter of $ZDY$ so $DY=DZ$ witch givesw that $YZ//BC$ it is enougt to prove that $IX$ perpendicular to $YZ$. This is true because $XY,XZ$ is tengent to the incircle.

Let the reflections be $l_C$ and $l_B$ respectively. Let $l_C$ and $l_B$ intersect $BC$ at $C'$ and $B'$ respectively. Let the incirle touch $BC$ at $D$. Using angle chasing $\measuredangle(l_B,BC)=\measuredangle(l_C,BC)$. Use the fact that $AC=CC'$ and $AB=BB'$ to show that $DB'=DC'$. Now it follows that $D$, $X$, and $I$ are collinear proving the claim.

Let the reflection of $AB$ across $CI$ be $A_1B'$, and the reflection of $AC$ across $BI$ be $A_2C'$. Since we have reflected, $\angle ACI = \angle A_1CI = \angle BCI$ and similarly $\angle ABI = \angle A_2BI = \angle CBI$. So, $A_1$ and $A_2$ both lie on $BC$. Also, $\angle XA_1A_2 = \angle B'A_1C = \angle BAC$ and $\angle XA_2A_1 = \angle C'A_1B = \angle CAB$ so $\triangle XA_1A_2$ is isosceles, implying that the incentre and orthocentre are collinear. Since $A_1$ is the reflection of $A$ across $CI$, $\angle IA_1A_2 = \angle IA_1C = \angle IAC = \angle BAC/2 = \angle XA_1A_2/2$ and similarly $\angle IA_2A_1 = \angle XA_2A_1/2$, so $A_1I$ and $A_2I$ bisect $\angle XA_1A_2$ and $\angle XA_2A_1$ respectively, which means $I$ is the incentre of $\triangle XA_1A_2$. So, $XI \perp A_1A_2 \implies XI \perp BC$ and we are done.

PEKKA wrote: Let $ABC$ be a triangle with incenter $I$. Suppose the reflection of $AB$ across $CI$ and the reflection of $AC$ across $BI$ intersect at a point $X$. Prove that $XI$ is perpendicular to $BC$. Let $D$, $E$, $F$ be points of tangency of $(I)$ with $BC$, $CA$, $AB$, respectively $\hspace{0.4cm}$$E'$, $F'$ be refelections of $E$, $F$ across $BI$, $CI$ respectively We see that $E'$, $F'$ are points of tangency of $(I)$ with tangents from $X$ to $(I)$ So $IX \perp E'F'$ Next, cuz $BI$, $CI$ are perpendicular bisecter of segments $DF$, $DE$ respectively, so $DE' = EF = DF'$ Which means $\triangle DE'F'$ is an isosceles triangle at $D$, $ID \perp E'F'$ Thus,$E'F' \| BC$ and $IX \perp BC$

Very very easy problem. Took me around 5 minutes. Let $A_B$ and $A_C$ denote the reflections of $A$ across lines $BI$ and $CI$ respectively. Further let $B'$ and $C'$ be the reflections of $B$ across $CI$ and $C$ across $BI$. We can first locate all of these points. Claim : Points $A_B$ and $A_C$ lie on $\overline{BC}$ and points $B'$ and $C'$ lie on $\overline{AC}$ and $\overline{AB}$ respectively. Proof : The proof of all of these are entirely similar so we only show one of them. Let $D$ be the intersection of line $CI$ with side $AB$. Then, let $A_C'$ denote the intersection of the reflection of side $AB$ across $\overline{CI}$. Note that, $\measuredangle CDA_B' = \measuredangle ADC$ and $\measuredangle A_B'CD = \measuredangle DCA$ which since $DC$ is a common side implies that $\triangle A_B' CD \cong \triangle DCA$. Thus, $DA_B' = DA$ which implies that $A_B'$ is the reflection of $A$ across $\overline{CI}$ and $A_B'\equiv A_B$ as desired. Now, note that $IA_B = IA=IA_C$ due to reflections so $I$ lies on the perpendicular bisector of segment $A_BA_C$. Further, \[\measuredangle XA_BA_C = \measuredangle DA_BA_C = \measuredangle CAB = \measuredangle EAB = \measuredangle BA_CE = \measuredangle BA_CX\]so $\triangle XA_BA_C$ is isosceles and in particular, $X$ lies on the perpendicular bisector of segment $A_BA_C$. Thus, $\overline{XI}$ is the perpendicular bisector of segment $A_BA_C$ which implies $XI \perp BC$ as we wished to show.

Nice and easy problem. Walkthrough $1$. Let $A_{1}$ and $A_{2}$ be the reflections of $A$ across $CI$ and $BI$ respectively. It is evident that, $B-A_{1}-A_{2}-C$ because $CI$ and $BI$ are angle bisectors of $\angle{C}$ and $\angle{B}$ respectively. $2$. After some trivial angle chasing we get that the points $(A,I,A_{1},B)$ and $(A,I,A_{2},B)$ are concyclic. Also since, $A_{1}$ and $A_{2}$ are the reflections of $A$ across $CI$ and $BI$ respectively, we have $IA=IA_{1}=IA_{2}$ i.e, $I$ is the center of $(AA_{1}A_{2})$. $3$. Again angle chase to show that, $XA_{1}=XA_{2},\angle{XA_{1}A_{2}}=\angle{XA_{2}A_{1}}=180^{\circ}-\angle{A}$ and, $\angle{IA_{1}A_{2}}=\angle{IA_{2}A_{1}}=\frac{\angle{A}}{2}$. $4$. Let $IA_{1}=IA_{2}=x$,$XA_{1}=XA_{2}=y$ and replace $\angle{B}+\angle{C}$ with $\theta$. Then, $\angle{XA_{1}A_{2}}=\angle{XA_{2}A_{1}}=\theta$ and $\angle{IA_{1}A_{2}}=\angle{IA_{2}A_{1}}=90^{\circ}-\dfrac{\theta}{2}$. Now, $A_{1}A_{2}=2x\sin\left(\frac{\theta}{2}\right)=2y\cos(\theta) \implies \dfrac{x}{y}=\dfrac{\cos(\theta)}{\sin\left(\frac{\theta}{2}\right)} \implies \boxed{\dfrac{A_{1}I}{A_{1}X}=\dfrac{\sin(90^{\circ}-\theta)}{\sin\left(\frac{\theta}{2}\right)}}$. But by Sine Rule, $\dfrac{A_{1}I}{A_{1}X}=\dfrac{\sin(\angle{A_{1}XI})}{\sin(\angle{A_{1}IX})}$ and also, $\angle{A_{1}XI}+\angle{A_{1}IX}=180^{\circ}-\angle{IA_{1}X}=90^{\circ}-\dfrac{\theta}{2}=(90^{\circ}-\theta)+\left(\frac{\theta}{2}\right)$. Thus we get, $\boxed{\dfrac{\sin(\angle{A_{1}XI})}{\sin(\angle{A_{1}IX})}=\dfrac{\sin(90^{\circ}-\theta)}{\sin\left(\frac{\theta}{2}\right)}}$ with $\boxed{\angle{A_{1}XI}+\angle{A_{1}IX}=(90^{\circ}-\theta)+\left(\frac{\theta}{2}\right)}$. $5$. By "$\alpha+\gamma=\beta+\delta$ Lemma" conclude that, $\angle{A_{1}XI}=90^{\circ}-\theta$ and $\angle{A_{1}IX}=\dfrac{\theta}{2}$. We had $\angle{XA_{1}A_{2}}=\angle{XA_{2}A_{1}}=\theta$ thus, $XI \perp BC$ as desired. $\blacksquare$ ($\mathcal{QED}$)