Estonian Mathematical Olympiad

Let $X$ and $Y$ be the reflections of points $D$ and $E$, respectively, across the point $I$ (Fig. 32). Then $\angle XBI=\angle IBD=\angle IBA=\angle CBI$, implying that $X$ lies on the line $BC$. As $\angle DIB=90^{\circ }$, points $D$, $I$ and $X$ lie on a line, i.e., $X$ is the point of intersection of lines $ID$ and $BC$. Analogously, we see that $Y$ is the point of intersection of lines $IE$ and $BC$. Hence $\angle EID=\angle YIX$, which along with the equalities $IX=ID$ and $IY=IE$ shows that triangles $DEI$ and $XYI$ are equal.

Thus also the altitudes drawn from the vertex $I$ in triangles $DEI$ and $XYI$ are equal. The altitude drawn from the vertex $I$ of the triangle $XYI$ equals the inradius of the triangle $ABC$. Hence the same holds for the triangle $DEI$, i.e., the incircle of the triangle $ABC$ passes through the foot of the altitude drawn from the vertex $I$ of the triangle $DEI$. The line $DE$ is perpendicular to this altitude which is the inradius of the triangle $ABC$; consequently, the line $DE$ is tangent to the incircle of the triangle $ABC$.

Fig. 32

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Alternative solution.

Let $U$ and $V$ be the reflections of points $B$ and $C$, respectively, across the point $I$ (Fig. 33). Then $IU=IB$ and $IV=IC$, implying that $BC\parallel UV$. As the line $BC$ is tangent to the incircle of the triangle $ABC$ whose centre $I$ is the centre of reflection, the line $UV$ is also tangent to the incircle of the triangle $ABC$ by symmetry.

Now let the line $UV$ intersect the side $AB$ and $AC$ at points $D^{\prime }$ and $E^{\prime }$, respectively (Fig. 34). Then $\angle D^{\prime }UB=\angle CBU=\angle CBI=\angle IBA=\angle UB{D}^{\prime }$, implying $D^{\prime }B=D^{\prime }U$. As $D^{\prime }I$ is the median drawn from the vertex angle of the isosceles triangle $D^{\prime }BU$, it must also be its altitude. This implies $D^{\prime }I\perp BI$ yielding $D^{\prime }=D$.

Analogously, we get $E^{\prime }=E$. Hence $DE=D^{\prime }{E}^{\prime }=UV$. Altogether, we have shown that the line $DE$ is tangent to the incircle of the triangle $ABC$.

Fig. 33 Fig. 34

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Alternative solution.

Denote $\angle BAI=\angle IAC=\alpha$, as well as $\angle CBI=\angle IBA=\beta$ and $\angle ACI=\angle ICB=\gamma$. Let $K$ be the circumcentre of the triangle $DEI$ (Fig. 35). By construction, $\alpha +\beta +\gamma =\frac{180^{\circ }}{2}=90^{\circ }$.

From the triangle $BCI$, we get $$\angle BIC=180^{\circ }-(\beta +\gamma )=180^{\circ }-(90^{\circ }-\alpha )=90^{\circ }+\alpha ,$$ hence $\angle EID=360^{\circ }-90^{\circ }-90^{\circ }-(90^{\circ }+\alpha )=90^{\circ }-\alpha$. From the circum-circle of the triangle $DEI$, we now get $$\angle EKD=2(90^{\circ }-\alpha )=180^{\circ }-2\alpha =180^{\circ }-\angle DAE.$$ Thus the quadrilateral $ADKE$ is cyclic. Its chords $DK$ and $EK$ are equal, implying that the corresponding inscribed angles are also equal, i.e., $\angle DAK=\angle KAE$. Hence $K$ lies on the bisector of the angle $BAC$, i.e., on the line $AI$.

Fig. 35 Fig. 36

From the triangle $ABI$, we get $$\angle AIB=180^{\circ }-(\alpha +\beta )=180^{\circ }-(90^{\circ }-\gamma )=90^{\circ }+\gamma ,$$ hence $\angle KID=\angle AID=(90^{\circ }+\gamma )-90^{\circ }=\gamma$. Therefore also $\angle IDK=\gamma$ as $DK=IK$. On the other hand, $\angle KDE=\frac{180^{\circ }-(180^{\circ }-2\alpha )}{2}=\alpha$, implying $\angle IDE=\gamma +\alpha =90^{\circ }-\beta$. Since also $\angle BDI=90^{\circ }-\beta$, the line $DI$ is the external bisector of the angle $ADE$, whereas $I$ is the point of intersection of this external bisector and the internal bisector of $DAE$. This means that $I$ is the excentre of the triangle $ADE$. As the incentre of the triangle $ABC$ is $I$ and the incircle of $ABC$ is tangent to the prolongation of the side $AD$ of the triangle $ADE$, these circles must coincide. Hence $DE$ is tangent to the incircle of the triangle $ABC$.

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Alternative solution.

Let $r$ be the inradius of the triangle $ABC$. We show that the point $D$ lies at distance $2r$ from the side $BC$.

To this end, let $D^{\prime }$ be the projection of the point $D$ on the side $BC$ (Fig. 36) and let $\angle CBI=\angle IBA=\beta$. Then $BI=\frac{r}{\sin \beta }$, $DB=\frac{BI}{\cos \beta }=\frac{r}{\sin \beta \cos \beta }$ and $D{D}^{\prime }=DB\sin 2\beta =\frac{r}{\sin \beta \cos \beta }\cdot 2\sin \beta \cos \beta =2r$.

Analogously, we can show that the point $E$ lies at distance $2r$ from the side $BC$. Thus the line $DE$ is parallel to the side $BC$ and at distance $2r$ from it. As the line $BC$ is tangent to the incircle of the triangle $ABC$, also the line $DE$ is tangent to this circle.