Prove that a line touches a circle

Question

Let $I$ be the incenter of $\triangle ABC$. The circle passing through $I$ and centered at $A$ meets the circumcircle of $\triangle ABC$ at points $M$ and $N$. Prove that the line $MN$ touches the incircle of $\triangle ABC$.

I appreciate any help.

Answer 1

Let $AI=a$. Let $O$ be the circumcenter of $ABC$.

Let $H$ be intersection of $AO$ and $MN$. Look at the triangle AMO. MH is its altitude, so $$\frac{AH}{HO}=\frac{AM \cos \angle MAO}{OM \cos \angle AOM} = \frac{AM (\frac{AM^2 +AO^2 - OM^2}{2AM \cdot AO})}{OM \frac{AO^2+OM^2-AM^2}{2OA \cdot OM}} = \frac{AM^2+AO^2-OM^2}{AO^2+OM^2-AM^2}$$

Now, substitute $OA=OM=R$ and $AM=AI=a$.

$$\frac{AH}{HO} = \frac{AM^2+AO^2-OM^2}{AO^2+OM^2-AM^2}= \frac{a^2}{2R^2 -a^2}$$.

From that $$ AH = a^2 \frac{R}{(2R^2 -a^2) +a^2}=\frac{a^2}{2R} $$

Now let $MH\cap AI = S$ and $\phi=\angle IAO$ and let $T$ be the projection of $I$ onto line $MH$

By Euler's formula, $OI^2=R^2-2Rr$

So, $$\cos \phi = \frac{a^2 + R^2 - (R^2-2Rr)}{2aR}=\frac{a^2 +2Rr}{2aR}$$

$$ AS= \frac{AH}{\cos \phi}=\frac{a^2}{2R} \cdot \frac{2aR}{a^2 + 2Rr}=\frac{a^3}{a^2 +2Rr} $$

$$IS=a-AS=a-\frac{a^3}{a^2 + 2Rr} = \frac{2Rra}{a^2 + 2Rr}$$

And finally, from trapezium ITAH

$$ IT = \frac{IS}{SA} AH = \frac{2Rra}{a^3} \frac{a^2}{2R}=r$$

Answer 2

Now, there is a more geometric solution, requiring instead such heavy machinery as inversion and the following fact about mixtilinear circles.

Verrier's lemma If mixtilinear circle touches the sides of triangle at points P and Q, then I (the incenter of the triangle) lies on PQ.

(The lemma is also proven by inversion, by the way - centered at $A$ with radius $\sqrt{AB\cdot AC}$)

Now, the solution. Do an inversion centered at $A$ with radius $AI$. Since $M'=M, N'=N$, the line $MN$ maps to the circumcircle of $ABC$. It remains to prove that the image of the incircle is tangent to the circumcircle. Now, let $P$ and $Q$ be points of tangency of the incircle with $AB$ and $AC$ respectively. $A, P, Q, I$ lie on a circle, which maps to a line under our inversion. Since $AP=AQ$, we have $AP'=AQ'$ and since $I'=I$, we have that $P', Q'$ and $I$ lie on a line. The image of the incircle is tangent to $AB$ and $AC$ at $P'$ and $Q'$, so by Verrier lemma, it is the mixtilinear circle, so it touches the circumcircle.

Answer 3

Let $AI$ intersect the circumcircle of $ABC$ at $D$, the line $BC$ at $X$, the line $MN$ at $Y$. Let $MN$ intersect $BC$ at $Z$. It is enough to prove that $I$ lies on the bisector of $XZY$ because then the incircle of $ABC$, which is tangent to $BC$, is tangent to $MN$ as well.

Using the trefoil lemma we get that $DI=DB=DC$. Hence the circumcircle of $BIC$ is tangent to the circumcircle of $MIN$ at $I$.

Since $Z$ lies on $MN$ and on $BC$, $Z$ is the radical center of the circumcircles of $ABC$, $MIN$, $BIC$. Hence $Z$ lies on the radical axis of $MIN$ and $BIC$, i.e. on the line perpendicular to $AI$ through $I$. So, $ZI \perp AI$.

Now, angle chasing reveals that $\angle YXZ = \angle ZYX$. Indeed, $$\angle ZYX = \angle MDY + \angle YMD = \angle AMN + \angle YMD = \angle AMD$$ and $$\angle YXZ = \angle XDB + \angle DBX = \angle AMB + \angle BMD = \angle AMD.$$

This shows that $$\angle XZI = 90^\circ - \angle YXZ = 90^\circ - \angle ZYX = \angle IZY.$$ Hence $I$ lies on the bisector of $XZY$, as desired.