pohoatza wrote: Let $ABC$ be a triangle such that $b>c$ and denote by $D,E,F$ the intersections of the its incircle with the sides $BC$, $AC$, $AB$ respectively. Consider the circle $w=C(M)$ with the diameter $[BC]$. Denote the intersections $\{\begin{array}{c}X\in w\cap (NF\\\ Y\in w\cap (NE\\\ N \in AM \cap EF\end{array}$. Prove that $\frac{NX}{NY}=\frac{AB}{AC}$. Proof I. The well-known properties $\|\boxed{\begin{array}{c}IN\perp BC\\\\ X\in CI\ ,\ Y\in BI\end{array}}\|$ $\implies$ $\{\begin{array}{c}m(\widehat{YXI})=\frac{B}{2}\ ,\ m(\widehat{XYI})=\frac{C}{2}\\\\ m(\widehat{NIX})=90^{\circ}-\frac{C}{2}\ ,\ m(\widehat{NIY})=90^{\circ}-\frac{B}{2}\end{array}\|$ $\implies$ $\frac{NX}{NY}=\frac{IX}{IY}\cdot\frac{\sin \widehat{NIX}}{\sin \widehat{NIY}}=$ $\frac{\sin \widehat{XYI}}{\sin\widehat{YXI}}\cdot \frac{\sin \widehat{NIX}}{\sin \widehat{NIY}}=$ $\frac{\sin\frac{C}{2}}{\sin\frac{B}{2}}\cdot\frac{\cos\frac{C}{2}}{\cos\frac{B}{2}}$ $\implies$ $\frac{NX}{NY}=\frac{\sin C}{\sin B}$ $\implies$ $\boxed{\ \frac{NX}{NY}=\frac{AB}{AC}\ }$. Remark. $\frac{NF}{NE}=\frac{AC}{AB}$. Proof II. Denote the intersection $Z\in BX\cap CY$. Observe that $Z\in ID$ and the point $I$ is the orthocenter of the triangle $BZC$, i.e. the triangle $XDY$ is the orthic triangle of $\triangle BZC$. From the well-known property "the ray $[DI$ is the bisector of the angle $\widehat{XDY}$" obtain $\frac{NX}{NY}=\frac{DX}{DY}$. Prove easily that $\{\begin{array}{c}m(\widehat{DXY})=B\\\ m(\widehat{DYX})=C\end{array}\|$ $\implies$ $\triangle XDY\sim\triangle BAC$ $\implies$ $\frac{DX}{DY}=\frac{AB}{AC}$. In conclusion, $\frac{NX}{NY}=\frac{AB}{AC}$. A general remark. If you know the above mentioned remarkable properties, then this very nice Pohoatza's problem is easily ! Virgil Nicula wrote: A similar problem. Let $C(I)$ be the incircle of $\triangle ABC$. Denote the second intersections $M$, $N$, $P$ of the circumcircle $C(O)$ of $\triangle ABC$ with the lines $AI$, $BI$, $CI$ respectively. Prove that $AM\perp NP$ and $R\in OM\cap NP\Longrightarrow$ $\frac{RN}{RP}=\frac{AB}{AC}$.

Virgil Nicula wrote: $X\in CI\ ,\ Y\in BI$ Please, Virgil Nicula, I have no idea how this comes!

cancer wrote: Virgil Nicula wrote: $X\in CI\ ,\ Y\in BI$ Please, Virgil Nicula, I have no idea how this comes! See http://www.mathlinks.ro/Forum/viewtopic.php?t=155800

What? I can't see how it's related... are there any tangent circles?

See the definitions of the points $X$, $Y$ from the above enunciation !!.

Virgil Nicula wrote: See the definitions of the points $X$, $Y$ from the above enunciation !!. I already see! Unless the incircle and $w$ are tangent??

NO ! 1. $N\in AM\cap EF$. 2. $(NE$ and $(NF$ are rays (semiline) ! 3. The circle $w$ has the center in the middle $M$ of the side $[BC]$ and the length of the its radius is equal to $MB$ ! Sorry, this is the all ! Now please, ask you Pohoatza to explain away ...

pohoatza wrote: Let $ABC$ be a triangle, let $E, F$ be the tangency points of the incircle $\Gamma(I)$ to the sides $AC$, respectively $AB$, and let $M$ be the midpoint of the side $BC$. Let $N = AM \cap EF$, let $\gamma(M)$ be the circle of diameter $BC$, and let $X, Y$ be the other (than $B, C$) intersection points of $BI$, respectively $CI$, with $\gamma$. Prove that $$\frac {NX} {NY} = \frac {AC} {AB}.$$ Author: Cosmin Pohoata We have some well-known results in this configuration: 1. $X$, $Y$ lie on line $EF$. 2. $N$ lies on $ID$, where $D$ is the tangency point of $\Gamma(I)$ to the side $BC$. Now, denote by $P$ the intersection of $AI$ with $BC$. It's easily to see that $\angle NIX=\angle PIC$, therefore triangles $NIX$ and $PIC$ are similar. It implies $\frac{NX}{NI}=\frac{PC}{PI}$. Similarly, we conclude that $\frac{NI}{NY}=\frac{PI}{PB}$. Hence, we have $\frac{NX}{NY}=\frac{NX}{NI}\cdot\frac{NI}{NY}=\frac{PC}{PI}\cdot\frac{PI}{PB}=\frac{PC}{PB}=\frac{AC}{AB}$. This completes our solutions.

Dear April, Your proof is Cosmin's. It was introduced in an aritcile of his.

Let $D$ be the tangency point of $BC$ and $(I)$ and $\mathcal {K}\equiv EF\cap BC$. Also, let $X',Y'$ respectively be the intersections of $BI,CI$ with $EF$. It is well known that $(\mathcal {K}DBC) = - 1$ $\Longrightarrow$ $(X'\mathcal {K},X'D,X'B,X'C) = - 1$. In the other hand, since $X'$ lies on the perpendicular bisector of $DF$, therefore, $X'B$ is the bisector of $\angle \mathcal {K}X'D$, thus, $BX'\perp CX'$, which implies $X'\equiv X$. With the same argument, we can also conclude that $Y'\equiv Y$. As the consequence, $E,F,X,Y$ are collinear. Let $A'$ be the intersection of $ID$ and the line $\triangle$ passes through $A$ parallel to $BC$. Let $N'\equiv ID\cap EF$ and $\mathcal {K'}$ be the intersection of $EF$ with $\triangle$. Since $\angle AA'I = 90^{\circ}$ $\Longrightarrow$ $A'\in (AEF)$. Therefore, $A'N'$ is the bisector of $\angle EA'F$, but $A'N'\perp A'\mathcal {K'}$ $\Longrightarrow$ $(A'E,A'F,A'N',A'\mathcal {K'}) = - 1$, which implies $(EFN'\mathcal {K'}) = - 1$ $\Longrightarrow$ $A(CBN'\mathcal {K'}) = - 1$. But $A\mathcal {K'}\|BC$, therefore, $AN'$ passes through $M$, which implies $N'\equiv N$ $\Longrightarrow$ $I,D,N$ are collinear. Now, it is easy to prove that $\triangle YDX$ $\cong$ $\triangle BAC$ (a.a) and $DI$ is the bisector of $\angle YDX$, then $\frac {NX}{NY} = \frac {DX}{DY} = \frac {AC}{AB}$. $\square$

Well, the same trick ( $MR$, $BI$, $FE$ and $\gamma$ intersect in $X$) appears 7 years later at USAJMO. (http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3478583&sid=5997de18ba862fa0672276d599ecf7cb#p3478583). I first saw this trick in one of Cosmin's book. ($R$ is midpoint of $AC$)

USAJMO #6 is my problem too, so no copyright issues I hope . By the way, the problem didn't mean to have a part (a) initially, but we figured the problem might be too difficult without isolating the known lemma (which unfortunately turned out not to be so well known, as there weren't more than 30 students who solved the first part).

Dear mathlinkers, any ideas for the similar problem of Virgil Nicula? Sincerely Jean-Louis

Dear Virgil and Mathlinkers, are you sure of your result (similar problem )? Toying with the figure in question, I have a doubt? Sincerely Jean-Louis

Lemma: $NI \perp BC$ (This is lemma 7 here) Proof: Let the line passing through $A$ parallel to $BC$ be $l$. Let $AN \cap \Gamma(I) \equiv P, Q$ such that $AP < AQ$. Let $FP \cap QE \equiv S$ and $FQ \cap PE \equiv T$. We have that $PEQF$ is a harmonic quadrilateral. Applying Brokard to $PEQF$ we have $NI \perp TS$, so it is enough to show that $T, A, S, Z$ are collinear. By Brokard, $ST$ is the polar of $N$, and by tangency $N$ is on the polar of $A$, thus La Hire yields $A$ is on the polar of $N$. It follows that $T, A, S$ collinear. Also, $PEQF$ is harmonic so $(Z, N; E, F)= -1$. Hence, $N$ is on the polar of $Z$, and La Hire again yields the desired. Also, by the Right Angles on Intouch Chord Lemma $X, Y \in \gamma(M)$. Now we length chase a little. By the Ratio Lemma, $\frac{NX}{NY} = \frac{XI}{YI} \frac{\sin \angle XIN}{\sin \angle YIN}$. By PoP or just similar triangles, this is $\frac{CI}{BI} \frac{\sin \angle BID}{\sin \angle CID}$. It's easy to see with lengths around the incircle and LoS that $\frac{CI}{BI} = \frac{\sin \frac{B}{2}}{\sin \frac{C}{2}}$ and $\frac{\sin \angle BID}{\sin \angle CID} = \frac{\cos \frac{B}{2}}{\cos \frac{C}{2}}$. Then by the double angle formula for sine we have $\frac{NX}{NY} = \frac{\sin B}{\sin C} = \frac{AC}{AB}$ as requested.

It is well-known that $X, Y$ lie on $FE$. It is well-known that $NI \perp BC$. We have $\angle XIN = 90^{\circ} - \frac{\angle B}{2}$ and $\angle NIY = 90^{\circ} - \frac{\angle C}{2}$. From Ratio Lemma, we have $$\frac{NX}{NF} = \frac{XI}{IY} \cdot \frac{\sin (90^{\circ} - \frac{\angle B}{2})}{\sin (90^{\circ} - \frac{\angle C}{2})} = \frac{XI}{IY} \cdot \frac{\cos \frac{\angle B}{2}}{\cos \frac{\angle C}{2}}$$ We have $BI = \frac{r}{\sin \frac{\angle B}{2}}$ and $CI = \frac{r}{\sin \frac{\angle C}{2}}$. So, $$\frac{NX}{NF} = \frac{\sin \frac{\angle B}{2} \cos \frac{\angle B}{2}}{\sin \frac{\angle C}{2} \cos \frac{\angle C}{2}} = \frac{\sin \angle B}{\sin \angle C} = \frac{b}{c} = \frac{AC}{AB}$$ as desired.

Let the midline $MI$ cut $EF$ at $X'$ since $AB\parallel MX'$ then $\hat{B}= \widehat{MX'B}+\widehat{X'BM}, \widehat{X'BA}=\widehat{BX'M}$ so $MX'B$ is isoceles which implies $\widehat{BX'C}= \frac{\pi}{2}$ so $X=X'$ likewise we prove $Y \in EF\cap MJ$ . $\frac{NX}{NY} =\frac{\sin \widehat{XMN}\cdot MX}{\sin\widehat{YMN} \cdot MY}=\frac{\sin \widehat{MAB} }{\sin\widehat{MAC}}=\frac{AC}{AB}$. R HAS

Here we go again. By Iran Lemma $X,Y \in EF$ also note that $\triangle IXY \sim \triangle ICB$ and bcs of meh config stuff $N = DI \cap \overline{EF}$ where $D$ is the other in touch point. $$\dfrac{NX}{NY} = \dfrac{IX \sin \angle XIN}{IY \sin \angle YIN} = \dfrac{IC \sin \angle BID}{IB \sin \angle CID} = \dfrac{\tfrac{IC}{ID} \cos B/2}{\tfrac{IB}{ID} \cos C/2} = \dfrac{\sin B/2 \cos B/2}{\sin C/2 \cos C/ 2} = \dfrac{\sin B }{\sin C} = \dfrac{AC}{AB}$$

Let $P=AI\cap EF$, $Q=AI\cap BC$ and $D$ be the tangency point on $BC$. From a well known lemma from EGMO we know $DI$ passes through $N$. Now delete $AM$ and consider $N$ as the intersection of $DI$ and $EF$. Notice that $P, D$ are the corresponding point of the similar triangles $\triangle DIY$ and $\triangle CIB$. So $G, N$ are also corresponding points. Hence $NX/NY=CQ/QB=AC/AB$.

Let $DEF$ be the intouch triangle. We have $X=BI\cap EF$, $Y=CI\cap EF$, and $N=DI\cap EF$. Since $XYBC$ is cyclic, we get $\triangle IXY\sim\triangle ICB$ so if the isogonal of $ID$ with respect to $IBC$ intersects $BC$ at $Z$, then $\frac{NX}{NY}=\frac{ZC}{ZB}$. However, by Incenter-Excenter, the isogonal of $ID$ is $AI$, so $\frac{ZC}{ZB}=\frac{AC}{AB}$.

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