## sábado, 9 de marzo de 2019

### Parallel Lines Associated with a Mixtilinear Incircle

Let $I$, $O$ and $\tau$ be the incenter, circumcenter and circumcircle of triangle $ABC$. The circle $\omega$ is tangent to lines $AB$ and $BC$, and touches internally $\tau$ at $T$. The tangents to $\tau$ at $T$ and $B$ intersect at $P$. Prove that $IP\parallel{AC}$.

Proof. It suffices to show that $\angle{\frac{ACB}{2}}=\angle{CIP}$. We know from lemma 2 in a previous problem that $\triangle{BPI}$ is isosceles with $\angle{IBP}=\angle{BIP}=\frac{\angle{ABC}}{2}+\angle{BAC}$. We have
$$\angle{CIP}=\angle{BIC}-\angle{BIP}=180^\circ-\frac{\angle{ABC}}{2}-\frac{\angle{ACB}}{2}-\frac{\angle{ABC}}{2}-\angle{BAC}.$$
But $180^\circ-\angle{ABC}-\angle{BAC}=\angle{ACB}$. Hence, $\angle{CIP}=\frac{\angle{ACB}}{2}$.

$\square$

## viernes, 8 de marzo de 2019

### TST Peru, 2019

Let $I$, $O$ and $\tau$ be the incenter, circumcenter and circumcircle of triangle $ABC$. The line $BI$ intersects $\tau$ again at $M$. The circle $\omega$ is tangent to lines $AB$ and $BC$, and touches internally $\tau$ at $T$. The tangents to $\tau$ at $T$ and $B$ intersect at $P$. The lines $PI$ and $TM$ intersect at $Q$. Prove that the lines $QB$ and $MO$ intersect at $\tau$.

Solution.

Lemma 1. $\angle{MBT}=\angle{OTI}$.

Proof. By properties of angles in a circle, $\angle{BOT}=2\angle{BAC}+2\angle{TBC}$. As $\triangle{OTB}$ is isosceles, $\angle{OTB}=90^\circ-\angle{BAC}-\angle{TBC}$. Moreover,

$$\angle{OTC}=90^\circ-\angle{BAC}-\angle{TBC}+\angle{BAC}=90^\circ-\angle{TBC}.$$

$\angle{ATI}=\angle{CTI}$ (This is a well-known property of mixtilinear incircle. See $[1]$ and $[2]$)   and  $\angle{ATC}=180^\circ-\angle{ABC}$, then,  $\angle{ATI}=\angle{CTI}=90^\circ-\frac{\angle{ABC}}{2}$. It follows that

$$\angle{OTI}=\angle{OTC}-\angle{CTI}=90^\circ-\angle{TBC}-90^\circ+\frac{\angle{ABC}}{2}=\frac{\angle{ABC}}{2}-\angle{TBC}.$$

But $\angle{MBT}=\frac{\angle{ABC}}{2}-\angle{TBC}$, therefore, $\angle{MBT}=\angle{OTI}$.

$\square$

Lemma 2. $BP=PT=IP$.

Proof. We already know that $\angle{OBT}=\angle{OTB}=90^\circ-\angle{BAC}-\angle{TBC}$. But $\angle{BAC}+\angle{TBC}=\angle{TBP}$, then, $\angle{OBT}=\angle{OTB}=90^\circ-\angle{TBP}$. Let $R$ be the circumradius, then, by the Law of Sines,
$$\frac{BT}{\sin{2\angle{TBP}}}=\frac{R}{\sin{(90^\circ-\angle{TBP)}}}$$
$$BT=2R\sin{\angle{TBP}}$$
Focusing on quadrilateral $OBIT$, $\angle{BIT}=360^\circ-\angle{BOT}-\angle{OBI}-\angle{OTI}$. Let $B'$ be the orthogonal projection of $B$ onto $AC$. We have $\angle{B'BC}=90^\circ-\angle{ACB}$. Since $O$ is the isogonal conjugate of the orthocenter, it follows $\angle{OBI}=\frac{\angle{ABC}}{2}+\angle{ACB}-90^\circ$. Moreover, we know from lemma 1 that $\angle{OTI}=\frac{\angle{ABC}}{2}-\angle{TBC}$. Then,
$$\angle{BIT}=360^\circ-(2\angle{BAC}+2\angle{TBC})-(\frac{\angle{ABC}}{2}+\angle{ACB}-90^\circ)-(\frac{\angle{ABC}}{2}-\angle{TBC}).$$
But
$$\angle{ABC}+\angle{BAC}+\angle{ACB}=180^\circ,$$
then,
$$\angle{BIT}=270^\circ-\angle{BAC}-\angle{TBC}.$$
Now, focusing on $\triangle{BIT}$, we have
$$\angle{BIT}=360^\circ-(270^\circ-\angle{BAC}-\angle{TBC})=90^\circ+\angle{BAC}+\angle{TBC}=90^\circ+\angle{TBP}$$
and
$$\angle{ITB}=180^\circ-\angle{BIT}-\angle{MBT}=90^\circ-\angle{BAC}-\frac{\angle{ABC}}{2}.$$
As $\frac{\angle{ABC}}{2}+\frac{\angle{ACB}}{2}+\frac{\angle{BAC}}{2}=90^\circ$, it follows $\angle{ITB}=\frac{\angle{ACB}}{2}-\frac{\angle{BAC}}{2}$. Now, by the Law of Sines,
$$\frac{BI}{\sin{(\frac{\angle{ACB}}{2}-\frac{\angle{BAC}}{2})}}=\frac{2R\sin{\angle{TBP}}}{\sin{(90^\circ+\angle{TBP})}}=2R\tan{\angle{TBP}}$$
$$BI=2R\tan{\angle{TBP}}\sin{(\frac{\angle{ACB}}{2}-\frac{\angle{BAC}}{2})}$$
As $\angle{BPT}=180^\circ-2\angle{TBP}$, again, by the Law of Sines,
$$\frac{BP}{\sin{\angle{TBP}}}=\frac{2R\sin{\angle{TBP}}}{\sin{(180^\circ-2\angle{TBP)}}}$$
$$BP=R\tan{\angle{TBP}}$$
Finally, by the Law of Cosines,
$$IP^2=4R^2\tan^2{\angle{TBP}}\sin^2{(\frac{\angle{ACB}}{2}-\frac{\angle{BAC}}{2})}+R^2\tan^2{\angle{TBP}}$$
$$-4R^2\tan^2{\angle{TBP}}\sin{(\frac{\angle{ACB}}{2}-\frac{\angle{BAC}}{2})}\cos{(\frac{\angle{ABC}}{2}+\angle{BAC}})$$
But, as $\frac{\angle{ABC}}{2}=90^\circ-\frac{\angle{BAC}}{2}-\frac{\angle{ACB}}{2}$, it follows
$$\cos{(\frac{\angle{ABC}}{2}+\angle{BAC}})=\cos{(90^\circ+\frac{\angle{BAC}}{2}-\frac{\angle{ACB}}{2}})=\sin{(\frac{\angle{ACB}}{2}-\frac{\angle{BAC}}{2})}$$
Thus,
$$IP^2=R^2\tan^2{\angle{TBP}}$$
$$IP=R\tan{\angle{TBP}}$$
Therefore, $BP=PT=IP$.

$\square$
Remark. $IP\parallel{AC}$. A proof can be found here.

Back to the main problem.

Let $R$ be the second intersection of $QB$ with $\tau$. It suffices to show that $MR$ is the diameter of $\tau$. By property of angles in a circle, $\angle{PTQ}=\frac{\angle{ABC}}{2}-\angle{TBC}$. We now from lemma 2 that $\angle{BIT}=90^\circ+\angle{BAC}+\angle{TBC}$ and $\triangle{BIP}$, $\triangle{ITP}$ are isosceles. It follows

$$\angle{TIP}=90^\circ+\angle{BAC}+\angle{TBC}-\frac{\angle{ABC}}{2}-\angle{BAC}=90^\circ+\angle{TBC}-\frac{\angle{ABC}}{2}.$$

Consequently, $\angle{TPQ}=180^\circ+2\angle{TBC}-\angle{ABC}$. Hence, $\angle{PQT}=\frac{\angle{ABC}}{2}-\angle{TBC}$, meaning $\triangle{PQT}$ is isosceles with $PQ=PB$. We deduce that $\angle{PBQ}=90^\circ-\angle{BAC}-\frac{\angle{ABC}}{2}$. Notice that

$$\angle{MBC}+\angle{CBP}+\angle{PBQ}=\frac{\angle{ABC}}{2}+\angle{BAC}+\angle{PBQ}=90^\circ.$$

Therefore, $MB\perp{BR}$.

$\square$

References.
$[1]$ Evan Chen, Euclidean Geometry in Mathematical Olympiads.
$[2]$ Arseniy Akopyan, Geometry in Figures.