Let $ I$ be incenter of triangle $ ABC$, $ M$ be midpoint of side $ BC$, and $ T$ be the intersection point of $ IM$ with incircle, in such a way that $ I$ is between $ M$ and $ T$. Prove that $ \angle BIM-\angle CIM=\frac{3}2(\angle B-\angle C)$, if and only if $ AT\perp BC$.
Problem
Source: Iranian National Olympiad (3rd Round) 2007
Tags: geometry, incenter, circumcircle, trigonometry, geometry proposed
lomos_lupin
03.09.2007 03:34
when $ AT\perp BC$ then $ T$ is on the nine-point circle.
Assume $ AT\perp BC$ and incircle meets $ BC$ at $ E$. Now note that $ \angle BIM-\angle BIC= 2\angle EIM-\frac{1}{2}(\angle C-\angle B)$ So we should prove $ \angle EIM=\angle B-\angle C$.
Now note that if we invert to $ M$ with respect to the incircle and since the invertion preserves the angles, nine-point circle changes to the tangent line that is perpendicular to $ IM$ at it's other intersection with incircle beside of $ T$ which means that $ T$ was initially on the
nine-point circle.
We have $ \angle EIM=\angle H_{a}TM$ where $ H_{a}$ is the feet of the altitude from $ A$. So it comes down to showing that arc $ H_{a}M$ is twice the $ \angle B-\angle C$. Let $ N$ be the mid point of $ AC$ then note that $ \angle H_{a}NC=180-2\angle C$ since $ H_{a}N$ is median in the right triangle $ AH_{a}C$ and $ \angle MNC=A$. hence $ \angle H_{a}NM=\angle B-\angle C$. Since $ N$ is on the nine-point circle $ \angle H_{a}NM=\angle H_{a}TM$ which finishes the proof.
Virgil Nicula
03.09.2007 09:18
Prpblem. $ w = C(I,r)$ is the incircle of $ \triangle ABC$. Denote the points $ \{\begin{array}{c}M\in (BC)\ ,\ MB = MC\\ \ D\in BC\ ,\ AD\perp BC\\ \ T\in MI\cap w\end{array}$. Prove that $ T\in AD\Longleftrightarrow$ $ m(\widehat{BIM}-m(\widehat{CIM}) =\frac{3}{2}\cdot (B-C)$. Proof. Suppose w.l.o.g. that $ b > c$. Calculate easily $ ME =\frac{b-c}{2}$ and $ MD =\frac{b^{2}-c^{2}}{2a}$. Denote the points $ \{\begin{array}{c}E\in BC\cap w\\ \ S\in MI\cap AD\end{array}$ and $ AS = x$. Observe that $ IE\parallel AD$ in the $ D$- right triangle $ MDA$. Thus, $ IE\parallel AD\Longleftrightarrow$ $ \frac{IE}{SD}=\frac{ME}{MD}$ $ \Longleftrightarrow$ $ \frac{r}{h_{a}-x}=\frac{a}{b+c}$ $ \Longleftrightarrow$ $ ax = ah_{a}-(b+c)r$ $ \Longleftrightarrow$ $ ax = 2pr-(b+c)r$ $ \Longleftrightarrow$ $ x = r$, i.e. $ \boxed{\ AS = r\ }$. Therefore, $ \boxed{\ T\in AD\Longleftrightarrow T\equiv S\ }$. ============================================================================================================== Since $ \triangle ATI$ is $ A$- isosceles and $ m(\widehat{TAI}) =\frac{B-C}{2}$, obtain $ m(\widehat{DTM}) = B-C$. Since $ IE\parallel TD$, obtain $ T\in AD$ $ \Longleftrightarrow$ $ \phi\equiv\boxed{m(\widehat{EIM}) = B-C}$. Since $ \{\begin{array}{c}m(\widehat{BIM}) =(90^{\circ}-\frac{B}{2})+\phi\\ \ m(\widehat{CIM}) =(90^{\circ}-\frac{C}{2})-\phi\end{array}\|$ $ \implies$ $ m(\widehat{BIM})-m(\widehat{CIM}) =$ $ 2\phi-\frac{B-C}{2}=\frac{3}{2}\cdot (B-C)$, obtain $ m(\widehat{BIM})-m(\widehat{CIM}) =\frac{3}{2}\cdot (B-C)$. Remark. $ T\in AD\Longleftrightarrow$ the circumcenter $ O$ of $ \triangle ABC$ belongs to the $ A$- Nagel's line, i.e. $ O\in AF$, where $ M\in (EF)$ and $ MF = ME$ $ \Longleftrightarrow$ $ OM = r$ $ \Longleftrightarrow$ $ r = R\cos A$ (the our triangle is acute !) $ \Longleftrightarrow$ $ AT = OM$ $ \Longleftrightarrow$ the point $ T$ is middle of the segment $ AH$, where $ H$ is the orthocenter of $ \triangle ABC$ $ \Longleftrightarrow$ the point $ T$ belongs to the Euler's circle of $ \triangle ABC$ $ \Longleftrightarrow$ $ \cos B+\cos C = 1$ $ \Longleftrightarrow$ $ \frac{r}{R}+\frac{a}{b+c}= 1$. Commentary. The remarkable relation $ \boxed{\ AS = r\ }\ \ (*)$ is well-known. From this moment the our nice problem became more easily ! By the way, I don't like the right hand of the equivalence from the conclusion, is even nastily. In the my opinion, $ m(\widehat{EIM}) = B-C$ is more nicely. In fact, this problem is the property $ (*)$ somehow "painted".
MysticTerminator
06.10.2008 03:38
I think this is closely related to the solution given. Denote by D the intersection of AI and the circumcircle; it is not difficult to see that D is the center of the circle passing through B, I, C, and the A-excenter. In particular, we will use that DM is perpendicular to BC. The angle condition is easily seen to be equivalent to DIM = IDM, so we wish to show that AT is perpendicular to BC if and only if MID is isosceles. This follows easily once we note that IT / IA = DM / DI, so both conditions are in fact equivalent to the similarity of the triangles AIT and DIM.
nikolapavlovic
08.09.2016 11:21
Let $D_{1}$ be the touch point if the $A$-excircle,let $AD_{1}\cup \odot I={Z}$,$\odot I$ touches $BC$ in $D$. $IM$ is the $I$ midline in $\triangle DZD_{1}$ and hence $IM||D_{1}Z$.$ID\perp BC$,$AT\perp BC$ and hence $AT||IZ$ $\implies$ $AZIT$ is a parallelogram but as $IT=IZ$ its a rhombus and hence $AD_{1}$ is the isogonal to $AT$ $\implies$ $AD_{1}$ passes thru the center of $\odot ABC$. $$\angle CIM=\angle \frac{\pi}{2}-\frac{\gamma}{2}+ \angle DIM=\angle \frac{\pi}{2}-\frac{\gamma}{2}+\angle HAO$$$$=\angle \frac{\pi}{2}-\frac{\gamma}{2}+\frac{\pi}{2}-\beta-\frac{\pi}{2}+\gamma=\frac{\pi}{2}+\frac{\gamma}{2}-\beta$$Analogously we have $\angle BIM=\frac{\pi}{2}-\gamma+\frac{\beta}{2}$ so we are done.$\clubsuit$