Dear Mathlinkers, a hint of a proof consists to consider orthologic triangles... Sincerely Jean-Louis

If the $B$-angle bisector of $\triangle ABC$ intersects $AC$ at $D$, all we have to prove is $\frac{CC_2}{DC_2}=\frac{AA_2}{DA_2}$, which is straight forward! Best regards, sunken rock

Quote: \[ \begin{gathered} \text{we know that cc}_\text{1} = \text{cc}_\text{2} = \text{bc}_\text{1} \text{ = x and aa}_\text{1} = \text{aa}_\text{2} = \text{ab}_\text{1} = y \hfill \\ \text{from general angle bisector theorm }\frac{\text{x}} {\text{x}} = \frac{{QC.\sin \text{C}_\text{2} QC}} {{QB.\sin \text{C}_\text{2} QB}} \Rightarrow \frac{{\sin \text{C}_\text{2} QB}} {{\sin \text{C}_\text{2} QC}} = \frac{{QC}} {{QB}} \hfill \\ \text{from general angle bisector theorm }\frac{\text{Y}} {\text{Y}} = \frac{{QB.\sin \text{A}_\text{2} QA}} {{QA.\sin \text{A}_\text{2} QB}} \Rightarrow \frac{{\sin \text{A}_\text{2} QB}} {{\sin \text{A}_\text{2} QA}} = \frac{{QA}} {{QB}} \hfill \\ \Rightarrow \frac{{\sin \text{C}_\text{2} QB}} {{\sin \text{C}_\text{2} QC}}\frac{{\sin \text{A}_\text{2} QA}} {{\sin \text{A}_\text{2} QB}} = \frac{{QC}} {{QA}} \to \left\{ 1 \right\} \hfill \\ \text{from general angle bisector theorm }\frac{\text{X}} {{\text{Lc}_\text{2} }} = \frac{{QC.\sin \text{C}_\text{2} QC}} {{QL.\sin \text{C}_\text{2} QB}} \hfill \\ \text{from general angle bisector theorm }\frac{{\text{La}_\text{2} }} {\text{Y}} = \frac{{QL.\sin \text{A}_\text{2} QB}} {{QA.\sin \text{A}_\text{2} QA}} \hfill \\ \Rightarrow \frac{{X.\text{La}_\text{2} }} {{Y.\text{Lc}_\text{2} }} = \frac{{QC}} {{QA}}\frac{{\sin \text{C}_\text{2} QC}} {{\sin \text{C}_\text{2} QB}}\frac{{\sin \text{A}_\text{2} QB}} {{\sin \text{A}_\text{2} QA}} = 1\text{ from}\left\{ \text{1} \right\} \hfill \\ \Rightarrow \frac{X} {Y} = \frac{{\text{Lc}_\text{2} }} {{\text{La}_\text{2} }} = \frac{{X + \text{Lc}_\text{2} }} {{Y + \text{La}_\text{2} }} = \frac{{LC}} {{LA}}\text{ but }\frac{{\text{BC}}} {{\text{BA}}} = \frac{{2X}} {{2Y}} = \frac{X} {Y} = \frac{{LC}} {{LA}} \hfill \\ \Rightarrow BQ\text{ is the bisector of }\angle CBA \hfill \\ \end{gathered} \] Image not found

[asy][asy] import graph; size(7cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -3.36, xmax = 8.38, ymin = -2.56, ymax = 5.98; draw((-2,4)--(-2,0)); draw((-2,0)--(6,0)); draw((6,0)--(-2,4)); draw((2,2)--(-2,0)); draw(circle((-0.76,2), 1.24)); draw(circle((2,0.94), 0.94)); draw((3.24,5.24)--(2,0)); draw((3.24,5.24)--(-2,2)); draw((-2,0)--(3.24,5.24),green); draw((-2,2)--(2,0),green); dot((-2,4),dotstyle); label("$A$", (-1.92,4.12), NE * labelscalefactor); dot((-2,0),dotstyle); label("$B$", (-1.92-0.8,0.12), NE * labelscalefactor); dot((6,0),dotstyle); label("$C$", (6.08,0.12), NE * labelscalefactor); dot((2,2),dotstyle); label("$M$", (2.08-0.3,2.12), NE * labelscalefactor); dot((-2,2),dotstyle); label("$A_1$", (-1.92-0.8,2.12), NE * labelscalefactor); dot((-0.21,3.11),dotstyle); label("$A_2$", (-0.14-0.3,3.22), NE * labelscalefactor); dot((2,0),dotstyle); label("$C_1$", (2.08,0.12), NE * labelscalefactor); dot((2.42,1.79),dotstyle); label("$C_2$", (2.5,1.9), NE * labelscalefactor); dot((3.24,5.24),dotstyle); label("$P$", (3.32,5.36), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Since $\triangle ABM$ and $\triangle MBC$ are isoceles, $A_1$ is the midpoint of $AB$, and $C_1$ is the midpoint of $CB$. Thus $A_1C_1$ is parallel to $AC$. Then $\angle PA_1C_1 = \angle AA_2A_1 = \angle AA_1A_2$, so $P$ is on the external angle bisector of $\triangle BA_1C_1$. Similarly, $\angle PC_1A_1 = \angle CC_2C_1 = \angle CC_1C_2$, so $P$ is on the external angle bisector of $\triangle BC_1A_1$. Thus $P$ is the $B$-excenter of $\triangle A_1BC_1$, so $P$ must be on the angle bisector of $\angle A_1BC_1$, or $\angle ABC$.

My solution:

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My solution Denote $I$ is the incenter of triangle $ABC, J$ is the intersection of $A_1A_2$ and $C_1C_2$. $M, N$ is the midpoint of $A_1A_2$ and $C_1C_2$. Because of $MA=MB=MC$ so we have $A_1, C_1$ is the midpoint of AB, BC. we obtain $A_1C_1 \| AC$ $or A_1C_1 \| A_2C_2$. That means $A_1C_1C_2A_2$ is a trapezoidal so $MN \| AC$ It's easy to show that $J, I, M, N$ are concylic so $(IN,IJ) \equiv (MN,MJ) \equiv (AC,A_1A_2) \equiv (IC,IB) (mod \pi)$ So $B, I, J$ are collinear. So we done!

Outline of my proof : It's enough to proof $\angle {A_1TC_1=45^0}$ Using cosine rule we'll get $A_1C_1$ in a closed from in terms of $a,b$ Now using sine rule in $A_1BT$ we'll get $A_1T$ in terms of $a,b$ ,similarly we'll also get $C_1T$ as we've all three sides of $A_1TC_1$ in terms of $a,b$, now using cosine rule we can prove $\angle {A_1TC_1}=45^0$

An easy corollary of the proof: Let $D$=angle bisector of$ B\cap AC,A_1C_1 \cap A_2C_2 =X,BA_2 \cap A_1D=Y,BC_2 \cap C_1D=Z$ Then $X,Y,Z$ are collinear.

also you can use of menellause theorem for triangles:$CLB$,and line $QC_{2}C_{1}$,$ALB$,and line $QA_{1}A_{2}$.

use ceva theorem in sin mode in triangle BA1C1 and point P.we know A1 is midpoint of AB and C1 is midpoint of CB.

Here is yet another proof. (EDIT: It seems that this is the same as sunken_rock's sketch, so... to qualify sunken_rock's sketch:) Define $X=A_1A_2\cap C_1C_2$. Let $D$ be the foot of the angle bisector from $B$ to $AC$, and let $D'=A_1C_1\cap BM$. Considering the homothety centered at $X$ sending $\triangle XA_2C_2$ to $\triangle XA_1C_1$, it suffices to prove that $D$ is sent to $D'$, or in other words, $\tfrac{DC_2}{DA_2}=\tfrac{D_1C_1}{D_1A_1}$. Since $\triangle MBC$ and $\triangle MAC$ are isosceles, $C_1$ and $A_1$ are midpoints of their respective sides, so by a combination of equal tangents and isosceles triangles we know that $\tfrac{D_1C_1}{D_1A_1}=\tfrac{BC_1}{BA_1}=\tfrac{CC_2}{AA_2}$. However, we already know that $\tfrac{D_1C_1}{D_1A_1}=\tfrac{DC}{DA}$, so we must also have $\tfrac{D_1C_1}{D_1A_1}=\tfrac{DC-CC_2}{DA-AA_2}=\tfrac{DC_2}{DA_2}$ as desired. $\blacksquare$

let $ A1A2 $ and $ C1C2 $ intersect at point $ D $, $ BD $ intersect $ AC $ at point $ F $. $ EF $ parallels to $ BC $ and intersect with $ DC1 $ at $ E $, $ FG $ parallels to $ AB $ and intersect $ A1D $ at $ G $. suppose $ BC $ = a, $ AB $=c, $ AC $=b. $ EF $=x, $ FG $=y. so $ CC1 $= $CC2 $ = a/2, $ FG $= $ FA2 $ = b/2. $ EF $/ $ BC1 $= $ FD $ / $ DB $ = $ FG $/ $ BA1 $, so x + y= b - 1/2* (a+c) and y/(c/2)=x/(a/2). solve the equation, we get: x=a*(2b-a-c)/(2(a+c)) and y=c(2b-a-c)/(2(a+c)),. $ CF $ = a/2 + x, $ AF $ = c/2 + y, so $ CF $/ $ AF $ = a/c. $BD $ bisect $\angle CBA $.

Easily falls to cartesian coordinates. Just set the origin at B, with A and C on the axes. Note that A1 and C1 are just the midpoints of AB and BC respectively. The coordinates of A2 and C2 can be computed using the fact that AA1 = AA2, CC1 = CC2. Find the coordinates of the intersection K of A1A2 and C1C2, and you're done.

We use barycentrics wrt $ABC.$ We have $A_1=(1:1:0)$ and $C_1=(0:1:1).$ Note that $AA_2=\frac{c+\frac{b}{2}-\frac b2}{2}=\frac c2.$ Similarly, $CC_2=\frac a2,$ so $A_2=(2b-c:0:c)$ and $C_2=(a:0:2b-a).$ Now let $P=(x:y:z).$ Then $P\in A_1A_2$ gives $$0=\begin{vmatrix} x & y & z\\ 1 & 1 & 0\\ 2b-c & 0 & c\\ \end{vmatrix}=cx-cy-z(2b-c)\implies y=\left(1-\frac{2b}c\right)z+x,$$and $P\in C_1C_2$ gives $$0=\begin{vmatrix} x & y & z\\ 0 & 1 & 1\\ a & 0 & 2b-a\\ \end{vmatrix}=(2b-a)x+ay-az\implies y=\left(1-\frac{2b}{a}\right)x+z,$$therefore, $$\left(1-\frac{2b}c\right)z+x=\left(1-\frac{2b}{a}\right)x+z\implies\frac xz=\frac{\left(1-\frac{2b}c\right)-1}{\left(1-\frac{2b}{a}\right)-1}=\frac{a}{c},$$so $BP$ bisects $\angle ABC,$ as desired.

Sharygin Geometry Olympiad 2012 - Problem 8 wrote: Let $BM$ be the median of right-angled triangle $ABC (\angle B = 90^{\circ})$. The incircle of triangle $ABM$ touches sides $AB, AM$ in points $A_{1},A_{2}$; points $C_{1}, C_{2}$ are defined similarly. Prove that lines $A_{1}A_{2}$ and $C_{1}C_{2}$ meet on the bisector of angle $ABC$. Might be similar to the proofs above, but posting my solution anyways because I enjoyed the problem : $\angle BAC \stackrel{\text{say}} = \theta \Longrightarrow \angle ACB = 90 - \theta$ $A_1A_2 \cap C_1C_2 = T$ Since $M$ is the ,midpoint of the hypotenuse $AC$ of $\triangle ABC$, $MA = MB = MC$ or $M$ is the circumcentre of $\triangle ABC$ Thus, $A_2 , C_2$ are midpoints of $AB, BC$ $\therefore$ By $\text{Midpoint Theorem}$ : $A_2C_2 \parallel BC$ $$CC_1 = CC_2 \Longrightarrow \angle CC_1C_2 = \angle CC_2C_1 = \angle C_1C_2A_2 = 45 + \frac {\theta}{2} ----- (i)$$ $$AA_2 = AA_1 \Longrightarrow \angle AA_2A_1 = \angle AA_1A_2 =\angle A_1A_2C_2 = 90 - \frac{\theta}{2} -----(ii)$$ From (i) and (ii), $C_2C_1$ and $A_2A_1$ are the exterior angle bisectors of $\angle BA_2C_2$ and $\angle BC_2A_2$. Thus, $T$ is the $B - \text{excenter}$ of $\triangle BA_2C_2$. The incenter of $\triangle A_2BC_2$ lies on $BT$ and so, $BP$ bisects $\angle ABC$

Incredible problem. Let $A_1A_2$ and $B_1B_2$ concur at $T$. I claim that $T$ is the $B$-excenter of $\triangle A_1BC_1$, or equivalently, that $A_1A_2$ bisects $\angle AA_1C_1$, and similarly for $C$. Note that since triangles $\triangle MAB$ and $\triangle MBC$ are isosceles, we have that $A_1$ and $C_1$ are the midpoints of $AB$ and $BC$, respectively, so $A_1C_1 \parallel AC$. Now $$\angle AA_1A_2 =\angle AA_2A_1 = \angle A_2A_1C_1,$$as desired.

Let $A_1A_2$ and $C_1C_2$ concur at $P$. The key claim is that $P$ is the $B$ excenter of $\triangle A_1BC_1$. I will show this by proving that $A_1A_2$ bisects $\angle AA_1C_1$ and $C_1C_2$ bisects $\angle CC_1A_1$. First, note that since $\triangle MAB$ and $\triangle MCB$ are isosceles, it is clear that $A_1$ and $C_1$ are the midpoints of $AB$ and $CB$ respectively. Hence, $A_1C_1 \parallel AC$. Let $E$ denote the incenter of $\triangle AMB$, $X$ denote the intersection of $A_1C_1$ and $AE$, and $Y$ denote the intersection of $AE$ and $A_1A_2$. Then, since $AE$ is an angle bisector and $\angle AA_1A_2=\angle AA_2A_1$, we have that $AE \perp A_1A_2$. Since $A_1C_1 \parallel AC$, \[\angle A_1AE=\angle A_2AE=\angle A_1XE.\]This means that $\triangle AA_1Y \cong \triangle XA_1Y$, or $\angle AA_1Y=\angle XA_1Y$. Hence, $A_1A_2$ is the angle bisector of $\angle AA_1C_1$. We can similarly show that $C_1C_2$ bisects $\angle CC_1A_1$, which means that $P$ is the $B$ excenter of $\triangle A_1BC_1$. This implies the fact that $PB$ bisects $\angle ABC$ and we are done.

Since $MA = MB = MC$, we know $A_1$ is the midpoint of $AB$ and $C_1$ is the midpoint of $CB$. By equal tangents, $AA_1 = AA_2$ and $CC_1 = CC_2$. Let $M_b$ be the midpoint of arc $AC$ not containing $B$. Now, we finish with Trig Ceva's on $BA_1C_1$. Indeed, notice \[ \frac{\sin A_1BM_b}{\sin M_bBC_1} \cdot \frac{\sin BC_1C_2}{\sin C_2EA_1} \cdot \frac{\sin C_1A_1A_2}{\sin A_2A_1B} \]\[ = \frac{\sin 45^{\circ}}{\sin 45^{\circ}} \cdot \frac{\sin (180^{\circ} - CEC_2)}{\sin (C_2EM + MEA_1)} \cdot \frac{\sin (C_1A_1M + MA_1A_2)}{\sin (180^{\circ} - ADA_2)} \]\[ = \frac{\sin (90 + \frac{C}{2})}{\sin (\frac{C}{2} + (90^{\circ} - C))} \cdot \frac{\sin (C + \frac{A}{2})}{\sin (90^{\circ} + \frac{A}{2})} \]\[ = \frac{\sin (90^{\circ} + \frac{C}{2})}{\sin (90^{\circ} - \frac{C}{2})} \cdot \frac{\sin (A + C - \frac{A}{2})}{\sin (90^{\circ} - \frac{A}{2})} \]\[ = \frac{\sin(90^{\circ} - \frac{A}{2})}{\sin (90^{\circ} - \frac{A}{2})} = 1 \]by parallel lines and properties of isosceles triangles. $\blacksquare$ Remarks: The only part of my solution that requires creativity is realizing Trig Ceva's works. (Overall, I feel like the set-up of this problem prompts contestants to try bash methods.)

Notice that $\triangle ABM$ is isosceles with $BM=AM$ so $A_1$ is the midpoint of $\overline{AB}.$ Similarly, $C_1$ is the midpoint of $\overline{AC.}$ Hence, $\overline{A_1C_1}\parallel\overline{AC}$ and $\angle AA_1A_2=\angle AA_2A_1=\angle A_2A_1C_1.$ Similarly, $\overline{C_1C_2}$ bisects $\angle A_CC_1C$ and $\overline{A_1A_2},\overline{C_1C_2},$ and the angle bisector of $\angle ABC$ concur at the $B$-excenter of $\triangle A_1BC_1.$ $\square$

First, it is easy to see that $C_1$ is the midpoint of $BC$ and $A_1$ is the midpoint of $BA$. Notice that $\overline{C_1C_2}$ bisects $\angle CC_1A_1$ using basic angle chasing. Similarly, $\overline{A_1A_2}$ bisects $\angle AA_1C_1$. Let $X$ be the intersection point of lines $C_1C_2$ and $A_1A_2$. Then, a homothety at $B$ with scale factor $2$ takes $X$ to the $B$-excenter, so the result is obvious.

Subjective Rating (MOHs)

$ $ I forgot $P$ was the excenter whoops. Harmonic not necessary

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