touch points adn equality of BM, M is midpoint

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touch points adn equality of BM, M is midpoint

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Source: Azerbaijan NMO 2023. Senior P5

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#1 h Aug 24, 2023, 7:53 PM

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The incircle of the acute-angled triangle $ABC$ is tangent to the sides $AB, BC, CA$ at points $C_1, A_1, B_1,$ respectively, and $I$ is the incenter. Let the midpoint of side $BC$ be $M.$ Let $J$ be the foot of the altitude drawn from $M$ to $C_1B_1.$ The tangent drawn from $B$ to the circumcircle of $\triangle BIC$ intersects $IJ$ at $X.$ If the circumcircle of $\triangle AXI$ intersects $AB$ at $Y,$ prove that $BY = BM.$

This post has been edited 1 time. Last edited by falantrng, Aug 24, 2023, 7:53 PM

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#2 h Aug 24, 2023, 9:08 PM • 1 Y

Y by OronSH

Let $N, M_A$ be the midpoints of $\widehat{BAC}, \widehat{BC}$ respectively; then $NB$ is tangent to $(BIC)$ , and $I, J, N$ are collinear because the ratio of the distances from $J, N$ to $AI$ is the ratio of the distances from $M, N$ to $AI$ which is $MM_A : MN$ which is the ratio of (the distance from $A$ to $EF$ ) : $AI$ . Thus, $N = X$ , so $Y = (ANI) \cap AB \implies \angle{AYI} = 180^{\circ} - \angle{ANI}$ , and thus it suffices to show that $\angle{ANI} = \angle{BMI}$ .

If $M_B, M_C$ are the midpoints of $\widehat{AC}, \widehat{AB}$ then $\triangle{IBC} \sim \triangle{IM_CM_B} \implies \angle{BMI}$ is equal to the angle between $M_BM_C$ and the $I$ -median in $\triangle{IM_BM_C}$ , which is indeed equal to $\angle{ANI}$ since $AN \parallel M_BM_C$ and $I$ is the orthocenter of $\triangle{M_AM_BM_C} \implies IN$ passes through the midpoint of $M_BM_C$ . $\square$

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#3 h Oct 22, 2023, 8:29 PM • 3 Y

Y by Mirhabib, ismayilzadei1387, Ismayil_Orucov

$X'$ is the intersection of tangent drawn from $B$ to the circumcircle of $\triangle BIC$ and $(ABC) ;$ $M'$ is the feet of altitude drawn from $M$ to $AI ;$ $L$ is the center of $(BIC).$
By angle chasing we get $X'B=X'C \implies X',M,L$ are collinear.
$\triangle LM'M \sim LAX' \implies \frac{MM'}{AX'}=\frac{LM}{LX'} \implies \frac{JJ'}{AX'}=\frac{LM}{LX'} \ \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \ \triangle AC_1I \sim \triangle X'CL \implies \frac{IJ'}{AI}=\frac{LM}{LX'}$ So $\frac{IJ'}{JJ'}=\frac{AI}{AX'} \implies \triangle IJ'J \sim \triangle IAX' \implies I,J,X' \text{ are collinear} \implies X' \equiv X$ Since $\angle YAI=\angle YXI= \angle BXL$ we obtain $\angle YXB=\angle IXL.$ Also we have $\angle YBX=\angle ILX.$ Therefore
$\triangle XYB \sim XIL \implies BY=\frac{IL \times BX}{LX}=\frac{BL\times BX}{LX} = BM$ so we are done

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#4 h Oct 23, 2023, 2:27 AM

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The main claim is that $X$ is the midpoint of arc $BAC$ .

We do this via phantom points. Let $X'$ be the midpoint of arc $BAC$ . Then I claim $N = IX' \cap B_1C_1$ is point $J$ . Notice that $IX'$ meets $(ABC)$ at the $T_A$ point, and it is well known that $AT_A$ is isogonal with $AZ$ , where $Z$ is the $A$ extouch point. We are also given that $\angle IMB = \angle AZB$ by homothety. Now quick angle chasing gives $\angle IMB = x, \angle T_AAB = \angle CAZ = x - \angle C, \angle AX'T_A = \angle ACT_A = \angle ACB + \angle BCT_A = \angle C + \angle BAT_A = x$ . Now let $IX'$ meet $BC$ at $K$ . We have $\angle IKC + \angle ACK = \angle KX'A + \angle CAX'$ , giving $\angle IKC = 90 + \angle \frac A2 - x - \angle C$ , and finally we have $\angle IKC + \angle INM + \angle NMK = 180$ , giving $\angle INM = 90 - x$ . Now since $\angle AX'I = \angle B_1NI$ by homothety, we have $\angle BNM = 90$ , as desired, so $N = J$ . Now $IJX'$ are collinear and $BX'$ is a tangent to $(BIC)$ , so $X' = X$ .

Now we know that $\angle IXA = \angle IYB$ , as well as angle $\angle IXA = \angle IMB$ from the proof of the previous claim, so we are done by symmetry.

This post has been edited 1 time. Last edited by ezpotd, Oct 23, 2023, 2:29 AM

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