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Source: 2020 China Second Round (A) P1

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165 posts

#1 h Sep 13, 2020, 10:50 AM • 1 Y

Y by k12byda5h

In triangle $ABC,$ $AB=BC,$ and let $I$ be the incentre of $\triangle ABC.$ $M$ is the midpoint of segment $BI.$ $P$ lies on segment $AC,$ such that $AP=3PC.$ $H$ lies on line $PI,$ such that $MH\perp PH.$ $Q$ is the midpoint of the arc $AB$ of the circumcircle of $\triangle ABC$ . Prove that $BH\perp QH.$

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618 posts

#2 h Sep 13, 2020, 11:10 AM • 1 Y

Y by Muaaz.SY

Extend $CI,PI$ to intersect the line parallel to $AC$ (through $B$ ) at $X$ and $Y$ respectively. Now look at triangle $IBX$ . Obviously it's circumcenter is $Q$ and thus we have an easier problem to solve:
Let $ABC$ be a right angled triangle $\angle C = 90^{\circ}$ . Let $M,N,P$ be the midpoints of $AB,BC,CA$ . Let $H$ be the foot from $P$ to $AN$ . We need to prove $\angle MHC=90^{\circ}$ , but this is easy (for example we could prove $H$ belongs to the circle around $MNPC$ )

This post has been edited 2 times. Last edited by Steve12345, Sep 13, 2020, 11:18 AM

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#4 h May 15, 2021, 5:26 PM • 1 Y

Y by Professor33

Let $N$ be the midpoint of $AC$ . Since $AP = 3PC$ , $P$ must be the midpoint of $CN$ . From $BA = BC$ , $B, I, N$ are collinear and $\angle INC = 90^{\circ}$ . Since $Q$ is the arc midpoint of $\overarc{AB}$ , $Q, I, C$ are collinear. An angle chase gives us $\angle QIB = \angle IBC + \angle ICB = \angle ABI + \angle ACQ = \angle ABI + \angle ABQ = \angle QBI \iff QI = QB$ . Since $M$ is the midpoint of $BI$ , $\angle QMI = 90^{\circ} = \angle CNI$ , thus $\triangle QMI \sim \triangle CNI$ . Also, since $\angle MHI = \angle PNI = 90^{\circ}, \triangle MHI \sim \triangle PNI$ . From these, $\frac{HM}{HI} = \frac{NP}{NI} = \frac{1}{2} \cdot \frac{NC}{NI} = \frac{1}{2} \cdot \frac{QM}{MI} = \frac{QM}{BI} \iff \frac{HM}{QM} = \frac{HI}{BI}$ . Together with $\angle HMQ = 90^{\circ} - \angle IMH = \angle HIB$ , $\triangle HMQ \sim \triangle HIB$ and $\angle HQM = \angle HBI = \angle HBM$ , thus, $H, M, B, Q$ are cyclic. Therefore, $\angle QHB = \angle QMB = 90^{\circ}$ and $BH \perp QH$ .

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Afo

#5 h Aug 25, 2021, 6:07 AM • 2 Y

Y by JG666, Mango247

Solution.

https://i.imgur.com/qoanvii.png

Let $PI \cap QM = D, PI \cap (BMQ) = H',E$ . It's enough to show that $HM \perp HI$ or $EBMQ$ is a rectangle. Since $C,I,Q$ are collinear, $D$ is midpoint of $QM$ . By Incenter-Excenter lemma, $QB=QI$ so
$\angle EDQ = \angle IDM = \angle BDM \implies EBMQ\text{ is a rectangle}$

This post has been edited 3 times. Last edited by Afo, Aug 25, 2021, 6:11 AM

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#6 h Aug 25, 2021, 11:50 AM

Y by

My solution at https://stanfulger.blogspot.com/2021/08/2020-china-2nd-round.html

Best regards,
sunken rock

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579 posts

#7 h Aug 26, 2021, 8:24 PM

Y by

Let $\gamma$ be the line through $B$ parallel to $AC$ . Then, let $w_{QB}$ be the circle with diameter $QB$ . Then, let $X= \gamma\cap w_{QB}$ . Also, let $Y = CI\cap \gamma$ .
Note that by Fact 5, $\triangle QBI$ is isosceles. Thus, since $M$ is the midpoint, $M$ is the foot form $Q$ to $BI$ . Thus by extension $MQ\parallel AC\parallel \gamma$ since they're all perpendicular to $BI$ .

Claim: $X$ is the midpoint of $YB$ .
Proof:
Firstly, we angle chase to show that $\triangle QYB$ is isosceles.
$\angle QYB = \angle CYB = \angle ICA = \angle BCI = \frac12(BQ)=\angle BQY$ Since $\angle QXB = \angle QMB =90$ , $X$ is the foot of the altitude, so it's also the midpoint. $\square$ .

Now, $X,I,P$ are collinear. This is true because $X,P$ are both the midpoints of segments $YB, CE$ .

Now, note that $(XBMQ)$ and $(MHEP)$ are clearly both concyclic, where $E$ is the foot from $B$ to $AC$ . Then, we let $H'\neq M$ be the second intersection point of $(XBMQ)\cap (MHEP)$ .

Then, by Reim's, $ES$ and $EM\cap (XBMQ), SH'\cap (XBMQ)$ are parallel. Let $SH'\cap (XBMQ)=X'$ . This means that $ES\parallel BX'$ . But we also have $BX\parallel AC\parallel ES$ .

Thus, $X=X'$ . Therefore, $H'$ lies on the line $P,I,X$ . But since $H=(MHEP)\cap \overline{PIX}$ , we have that $H=H'$ , and $H$ is the second intersection and therefore $H\in (XBMQ)$ . Thus,
$\angle QHB = \angle QMB = 90$ and we're done $\blacksquare$ .

Remark

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128 posts

#8 h Jan 1, 2023, 1:06 PM

Y by

Here is our solution: https://calimath.org/pdf/CMC2020-1.pdf
And we uploaded the solution with motivation to: https://youtu.be/A5w90STbHec

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