Denote a = BC, b = AB = AC the triangle sides, r, R the triangle inradius and circumradius and s, $\triangle$ the triangle semiperimeter and area. Let S, T be the tangency points of the incircle (I) with the sides AB, AC. Let K, L be the midpoints of of BC, PQ and M the midpoint of the arc BC of the circumcircle (O) opposite to the vertex A. The isosceles triangles $\triangle AST \sim \triangle APQ \sim \triangle ABC$ are all centrally similar with the homothety center A. The homothety coefficient for the triangles $\triangle AST \sim \triangle ABC$ is $h_{13} = \frac{AS}{AB} = \frac{s - a}{b}$. The homothety coefficient for the triangles $\triangle AST \sim \triangle APQ$ is $h_{12} = \frac{AS}{AP}$. The point K is the tangency point of the incircle (I) with the side BC, while the point M is the tangency point of the given circle with the triangle circumcircle (O). It follows that the triangles $\triangle ASK \sim \triangle APM$ are also centrally similar with the homothety center A and the same homothety coefficient as the triangles $\triangle AST \sim \triangle APQ$: $h_{12} = \frac{AS}{AP} = \frac{AK}{AM} = \frac{h}{2R}$, where h = AK is the A-altitude of the triangle $\triangle ABC$. The homothety coefficient of the triangles $\triangle APQ \sim \triangle ABC$ is then $h_{23} = \frac{h_{12}}{h_{13}} = \frac{s - a}{b} \cdot \frac{2R}{h}$ Denote h' = AL the A-altitude of the triangle $\triangle APQ$. Then $\frac{h'}{h} = h_{23} = \frac{s - a}{b} \cdot \frac{2R}{h}$ $KL = h - h' = h - 2R\ \frac{s - a}{b}$ Substituting $h = \frac{2 \triangle}{a}$, $2R = \frac{ab^2}{2 \triangle}$ and $s - a = \frac{\triangle^2}{s(s - b)^2}$, we get $KL = \frac{2 \triangle}{a} - \frac{ab^2}{2 \triangle} \cdot \frac{\triangle^2}{bs(s - b)^2} =$ (substituting 2s = 2b + a and $s - b = \frac a 2$ for an isosceles triangle) $= \frac{\triangle}{s}\left(\frac{2s}{a} - \frac{ab}{2(s - b)^2}\right) = \frac{\triangle}{s} \left(\frac{2b}{a} + 1 - \frac{2b}{a}\right) = \frac{\triangle}{s} = r$ which means that the point $L \equiv I$ is identical with the incenter of the triangle $\triangle ABC$.

I note: the circumcircle $C(O,R)$, the incircle $C(I,r)$ and the circle $w=C(I_1,r_1)$ interior tangent to the circumcircle in the point $T$ and tangent to the sides $AB,AC$ in the points $P,Q$ respectively. From the a wellknown property, the points $U\in CI\cap TP$, $V\in BI\cap TQ$ belong to the circumcircle. Thus, the quadrilaterals $AUPI$, $AVQI$ are cyclic and $UA\perp UT$, $VA\perp VT\Longrightarrow PI\perp AI,\ QI\perp AI\Longrightarrow I\in PQ$. Remark. The property $I\in PQ$ is true in any triangle $ABC$. See http://www.mathlinks.ro/Forum/viewtopic.php?t=46418

A generalization of this problem: "The circle $w=C(I_1,r_1)$ is internally tangent (in the point $T$) with the circumcircle of the triangle $ABC$ and is also tangent to the sides $[AB],[AC]$ in the points $P,Q$. Prove that the incenter $I$ of the triangle $ABC$ belongs to the segment $[PQ]$ and $AP=\frac{bc}{s},\ \ \frac{r_1}{r}=\frac{bc}{s(s-a)}.$ My solution. From a wellknown property, the points $M\in BI\cap TQ$, $N\in CI\cap TP$ belong to the circumcircle of the triangle $ABC$. From the Pascal's theorem applied to the inscribed hexagon $BACNTM$ results that the points $P\in BA\cap NT$, $Q\in AC\cap TM$, $I\in CN\cap MB$ are collineary, i.e. $I\in PQ$ (The other mentioned relations are easily solved). See http://www.mathlinks.ro/Forum/viewtopic.php?t=46418

A well-known consequence of the Pascal theorem.

Hm... Can you give more details, Erken?

See Virgil Nicula's post above.

there is yet another proof of the generalization were you construc the circumference and then prove it is unique and that the theorem holds, it is not hard but I am to lazy to write it now I'll just write the beggining

Let $ R$ be the tangency of circles and $ S$ be the center of inner circle. $ \angle PSA = 2k$ and $ \angle PRS = k$. $ \angle ABR = 90$. The circumcenter and $ S$ lie on $ AR$. $ AR \perp PQ$ at $ M$ and $ AR \perp BC$ at $ N$ . $ PMRB$ is cyclic. $ \angle PBM = \angle PRS = k$. $ PBNS$ is cyclic. $ \angle PBN = \angle PSA = 2k$. So $ \angle MBN = k$. $ BM$ is angle bisector.

Dear Mathlinkers, for an extension of the initial problem, see http://perso.orange.fr/jl.ayme/ vol. 4 A new mixtilinear incircle adventure I. Sincerely jean-Louis

Virgil Nicula wrote: A generalization of this problem: "The circle $ w = C(I_1,r_1)$ is internally tangent (in the point $ T$) with the circumcircle of the triangle $ ABC$ and is also tangent to the sides $ [AB],[AC]$ in the points $ P,Q$. Prove that the incenter $ I$ of the triangle $ ABC$ belongs to the segment $ [PQ]$ Let $ B'$ and $ C'$ be points in $ AC$ and $ AB$, respectively, such that $ AB = AB'$ and $ AC = AC'$. Let $ I_a$ be the center of the excircle of $ ABC$ tangent to $ AB$ and $ AC$ at $ D$ and $ E$ respectively ($ I_a$ lies in the bisector of $ \angle A$). We will show that the midpoint $ M$ of $ PQ$ is $ I$. Consider the inversion with center $ A$ and radius $ AB \times AC$. It maps $ B$ and $ C$ into $ C'$ and $ B'$ respectively. Then, the circumcircle of $ ABC$ becomes line $ B'C'$. Therefore, $ w$ is mapped into a circle tangent to a circle tangent to $ AB$, $ AC$ and $ B'C'$. Notice $ B'C'$ is tangent to both the incircle and $ (I_a)$ (because it is the reflection of $ BC$ over the bisector of $ \angle A$). Then $ w$ becomes $ (I_a)$ (it is easy to see that it cannot be the inverse of the incircle). Now, if we call $ M'$ the inverse of $ M$, we have $ \angle ADM' = \angle AMP = 90^o$. As $ M'$ is on the bisector of $ \angle A$, we get that $ M' = I_a$. This implies that $ AM \times AI_a = AB \times BC$. We will show that $ AI \times AI_a = AB \times BC$. Consider the circle $ K$ of diameter $ II_a$. It is known that $ \angle ICI_a = \angle IBI_a = 90^o$. Then both $ B$ and $ C$ belong to $ K$. By simmetry, we also get that $ B'$ and $ C'$ are in $ K$. Then, we know that $ AI \times AI_a$ is the power of $ A$ with respect to $ K$. Then $ AI \times AI_a = AB \times AC' = AB \times AC$. As we knew that $ AM \times AI_a = AB \times AC$, this implies that $ I = M$ (because both points are on the bisector of $ \angle A$, inside $ ABC$), then $ I \in PQ$, as we wanted to prove. Virgil Nicula wrote: $ AP = \frac {bc}{s},\ \ \frac {r_1}{r} = \frac {bc}{s(s - a)}.$ The first equality follows, because $ D$ is the inverse of $ P$ and $ AD = s$ is known. For the second one, an homotethy with center in $ A$ and ratio $ \frac {AP}{s-a}$ maps the incircle into $ w$. The ratio must be equal to $ \frac {r_1}r$ and the result follows using the first equality.

Let $\Gamma$ denote the circumcircle of $ABC$ and $\Omega$ denote the internally tangent circle through $P$ and $Q$. Let $M$ denote the point of tangency between $\Gamma$ and $\Omega$, $I$ denote the midpoint of $PQ$ and let $O$ denote the center of $\Omega$. Let $\angle{ABC}=a$. First note that $A$, $M$, $I$ and $O$ are collinear and the entire figure is symmetrical around line $AM$. Therefore $I$ lies on the angle bisector of $\angle{BAC}$. We now have that $\angle{POQ}=180-\angle{BAC}=2a$. Hence we have that $\angle{PMQ}=a$ and by symmetry, $\angle{PMI}=a/2$. Now note that $\angle{PIM}=\angle{PBM}=90$. Hence $PIMB$ is cyclic. Therefore $\angle{PBI}=\angle{PMI}=a/2$. Hence $I$ is also on the angle bisector of $\angle{ABC}$ and $I$ is the incenter of the triangle.

The 'standard' proof uses Casey's for the 'circles' $A, \;B, \;C$ and $\Omega$, all internally tangent to $\Gamma$, followed by the reciprocal of the Transversal Theorem (Cristea's). Best regards, sunken rock

Could you explain what you mean in detail? I do not know what Cristea's theorem is... and how casey's theorem is used here (i.e. what selection of +/- do we use).

To Altheman: For Cristea's theorem see here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=323170&p=1741338&hilit=cristea#p1741338 Regarding the use of Casey's: We have the 'circles' $A,B, \Omega, C$, all internally tangent to $\Gamma$, hence the following equality holds: $AP\cdot BC=AB\cdot CQ+AC\cdot BP$ $(\;1\; )$. If $BC=a$, etc, $a+b+c=2s$ and $AP=AQ=m$, from $(1)$ we get $a\cdot m=c(b-m)+b(c-m)$, or $m=\frac{bc}{s}$ $(\; 2 \;)$ If $\{ D \} \equiv AI \cap BC$, then: $\frac{DI}{AI}=\frac{a}{b+c} \; ( \; 3 \; )$, $CD=\frac{ab}{b+c} \; ( \; 4 \; )$ and $BD=\frac{ac}{b+c} \; ( \; 5 \; )$. Next, as per Cristea's, we have to prove: $CD\cdot \frac{BP}{PA}+BD\cdot \frac{CQ}{AQ}=BC\cdot \frac{DI}{AI}$ and, with the relations 2-5: $\frac{ab}{b+c} \cdot \frac{c-\frac{bc}{s}}{\frac{bc}{s}}+\frac{ac}{b+c}\cdot \frac{b-\frac{bc}{s}}{\frac{bc}{s}}=a \cdot \frac{a}{b+c}$, or $2s-b-c=a$, which is true. Best regards, sunken rock

This solution is motivated from snowEverywhere. Let do as usual notations midpoint of $PQ$ be $I.$ and let the angle bisector of $\angle A$ hits the $\odot ABC$ at $M$.note that $MA$ contains $I$.If you rotate the figure about $A$ with angle $90^{\circ}$ then $C \mapsto B$.so by the symmetry we can say that $AM$ is the diameter of the $\odot ABC$. our motive is to show that, $P,I,M,B$ is cyclic.$M$ is the midpoint of the arc $PMQ \implies$ $PM$ bisects $\angle BPQ$.From above we can infer $\angle ABM =90^{\circ}$ and $\angle PIM =90^{\circ}$.so concyclic proved. it implies that, $\angle IBM =\angle IPM$ and $\angle MPB=\angle BIM \implies \angle IBM =\angle BIM \implies MI=MB$ so it follows that $I$ is the incenter.

Let $O$ be the center of the circle. Let the circle be tangent to the circumcircle of ∆$ABC$ at $D$. Let $I $be the midpoint of $PQ$. Then $A, I, O, D$ are collinear by symmetry. Consider the $homothety$ with center $A$ that sends ∆$ABC$ to ∆$AB’C’$ such that$ D$ is on $B’C’$. Thus, $k=AB’/AB$. As right triangles $AIP, ADB’, ABD, APO$ are similar, we have $AI /AO = (AI / AP)(AP / AO) = (AD /AB’)(AB /AD) = AB/AB’=1/k$.Hence the homothety sends $I$ to $O$. Then $O$ being the incenter of ∆$AB’C’$ implies$ I$ is the incenter of ∆$ABC$. QED

This is true in general. Use $\sqrt[2]{bc} $ inversion. The mixitilinear Incircle becomes the excircle and the incenter the excenter. Now, it's trivial thereafter. Sorry, if this is resembling some previously posted solution.

By incenter/excenter lemma, it suffices to show $JM=JB$. Let $\angle MOP=2x$. Then $\angle MJP=x$ and $\angle JAB=90-2x \to \angle BAJ=2x$. Thus $\angle BJP=\angle MJP$. Since $\angle JBP=\angle JMP=90$, $BPMJ$ is a kite, done.

$\text{Midpoint of}$ $PQ=M$; $\text{line parallel to BC through}$ $I \cap AB=S$; $\text{line parallel to BC through}$ $I \cap AC=R$ Its apparent that $\triangle ASI \cong \triangle ARI \implies \angle AIS=90$ but $\angle AMP=90$ too and $PQ||SR$ so a contradiction thus $M=I$ (hopefully not a wrong solution)

Is not this true for all triangles? how AB=AC is used?

howcanicreateacccount wrote: Is not this true for all triangles? how AB=AC is used? yeah it is true for all triangles. It is something called Mixtillinear Incircles. See Evan Chen's handout on it (or EGMO chapter 4).

This fact is actually true for all triangles. First I claim $I$ actually lies on $\overline{PQ}$. Consider a $\sqrt{bc}$-inversion which swaps this circle $\omega$ with the $A$-excircle. The conclusion is equivalent to showing $E, F, A, I_A$ cyclic, where $E, F$ are the excentral touchpoints, which is obvious. Now $AK=AL$ which implies the result.

Label the touch points to $AB$, $AC$, and $(ABC)$ as $K$, $L$, and $T$, and the midpoints of arcs $AB$ and $AC$ as $X$ and $Y$. Homothety tells us $TKX$ and $TLY$ collinear. Pascal on $BACXTY$ then tells us that $KIL$ collinear, and noting $\triangle AKI \cong \triangle ALI$ implies $IK = IL$. $\blacksquare$