Points P,I,Q are collinear iff ABC is right

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Points P,I,Q are collinear iff ABC is right

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Source: ELMO 2014 Shortlist G13, by David Stoner

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#1 h Jul 24, 2014, 1:36 PM • 3 Y

Y by HamstPan38825, Adventure10, Rounak_iitr

Let $ABC$ be a nondegenerate acute triangle with circumcircle $\omega$ and let its incircle $\gamma$ touch $AB, AC, BC$ at $X, Y, Z$ respectively. Let $XY$ hit arcs $AB, AC$ of $\omega$ at $M, N$ respectively, and let $P \neq X, Q \neq Y$ be the points on $\gamma$ such that $MP=MX, NQ=NY$ . If $I$ is the center of $\gamma$ , prove that $P, I, Q$ are collinear if and only if $\angle BAC=90^\circ$ .

Proposed by David Stoner

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#2 h Jul 25, 2014, 12:27 PM • 2 Y

Y by Adventure10, Mango247

I actually now don't mind this problem.

Forward: $\angle BAC = 90 \implies MA = MI \implies M = CI \cap \odot ABC \implies PX \perp CI \implies 2(\angle (AB, PX) + \angle (AC, QY)) = 2(90-b-c/2+90-c-b/2) = A = 90$ as desired.

Backward: Invert about incircle. Note then $IXP \equiv IYQ$ , so as $IX=r=IY$ , the moving over the circle $\odot IXY$ to the left with angle equivalent to a chord with length $r$ take $IXP \mapsto IYQ \implies PQ=r$ . But, $NP = NQ = r/2$ where $N$ is the ninepoint centre of $XYZ$ , so $NPQ$ degenerates, so $\angle XIY=90$ as desired.

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#3 h Jul 27, 2014, 10:34 AM • 1 Y

Y by Adventure10

the main part is to prove that when $P,I,Q$ are collinear then $\angle A=90$

and my idea was same with @IDmasterz

invert on $I$ with ratio $r^2$ then the incircle will go to the incircle now assume that $M \rightarrow M' , N \rightarrow N'$ now we now that $M',N'$ are on the $\bigcirc IXY$ and on the nine point circle of the $XYZ$ now it's known that $M',N'$ are both inside or both outside the triangle
(it is serbia 2013 national p3 that they are reflected from the angle A bisector ) now assume that they are both inside we have that $\angle MIN= \angle M'IN'=180-A/2 \Longrightarrow A/2= \angle M'XN' \leq \angle ZXK=A/2$ where $K$ is the foot of $X$ on $ZY$ so we have that $M'=L$
and $N'=K$ so now $\angle XKY= \angle XIY =90$ and we are done

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#4 h Jul 30, 2014, 4:30 AM • 2 Y

Y by Adventure10, Mango247

Here's a solution without inversion. I will only work on the backward direction.

$P,Q$ are the reflections of $X,Y$ over $MI,IN$ , hence $P,I,Q$ are collinear $\Rightarrow \angle MIP+\angle NIQ+\angle MIN=180$ $\Rightarrow \angle MIX+\angle NIY+\angle MIN=$ $\angle A-(\angle XMI+\angle YNI)+\angle MIN$ $=\angle A-180-2\angle MIN=180$ $\Rightarrow 180-\angle MIN=\frac {\angle A}{2}=\angle YXI,\angle XYI$ $\Rightarrow \triangle IXM\sim \triangle NYI\sim \triangle NIM$ .

Let $IX,IY\cap (IMN)=F,E$ resp., since $FX*IX=MX*XN=AX*XB$ , therefore $F,E\in (ABI),(ACI)$ resp. Since $MIF=\angle MNI$ by the similarity, therefore $MF=MI$ . Let $K$ denote the center of $(ABI)$ , which is the midpoint of arc $AB$ that doesn't contain $C$ , therefore $MK\perp FI\Rightarrow MK\parallel AB\Rightarrow M\equiv K\Rightarrow MA=MI$ , similarly $AN=NI$ , hence line $MXYN$ perpendicularly bisects $AI \Rightarrow AXIY$ is a square $\Rightarrow \angle A=90$ .

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#5 h Apr 12, 2015, 5:48 PM • 2 Y

Y by Adventure10, Mango247

Here is a weird solution... Observe the weird $P,Q$ condition is equivalent to $\angle MIN=180-(\angle A/2)$ by anglechasing. Notice that this is equivalent to $\angle MIX = \angle MNI$ which is equivalent to $MI^2=MX*MN$ but $MI^2-r^2=MX*MY$ (where $r$ is inradius) by power of a point, so it is equivalent to $\boxed{r^2=MX*YN}$ .

First I prove that when $\angle A=90$ it works. To see this, observe $XY$ is the perpendicular bisector of $AI$ and so $MA=MI$ but it is well known this implies $M$ is the midpoint of $AB$ so $MIC$ are collinear, also $NIB$ . And so $\angle MIN= \angle BIC =135$ so we are done in this case

Now assume we have $\angle A < 90$ that satisfies the solution (this is enough since $ABC$ is acute). Let $D = XY \cap IA$ and choose $D'$ on segment $AD$ such that $AD'=D'I$ , which is possible because $\angle A <90$ . Construct $M',N'$ on the circle $\omega$ such that $M'N' \perp AI$ at $D'$ . Construct $\gamma'$ centered at $I$ with radius $r'=\sqrt{IA*ID'}>\sqrt{IA*ID}=r$ and clearly this circle contains $\gamma$ . Finally, construct $B',C' \in \omega$ such that $AB',AC'$ are tangent to $\gamma$ . Clearly $B'$ and $C'$ lie on the minor arcs $AB, AC$ respectively, because $\gamma'$ contains $\gamma$ . Let $X',Y'$ be the tangency of $AB',AC'$ with $\gamma '$ , and $X'',Y'' = AB \cap M'N', AC \cap M'N'$ respectively. Notice, as I proved above, that $AB'C'$ is a right triangle so it satisfies $r'^2=M'X'*N'Y'$ .

It is easy to see $r^2=MX*NY > M'X'' * N'Y'' > M'X'*N'Y' = r'^2 > r^2$ , a contradiction. So we are done.

Alternatively, consider the circumcircle of $IMN$ and let it cut $IY$ at $Z$ (forget about the old $Z$ ). Then $YZ*YI=YM*YN=YA*YC$ by power of a point so $AIZC$ is cyclic, centered at the midpoint of arc $AC$ , $N_0$ . Then $\angle IZN=\angle NMI = \angle ZIN$ and so $NI=NZ$ and $N_0I=N_0Z$ so $NN_0$ is perpendicular to $IZ$ , which is perpendicular to $AC$ , which is tangent to the tangent to $\omega$ at $N_0$ , so $N_0=N$ and we finish easily.

This post has been edited 3 times. Last edited by JuanOrtiz, Apr 12, 2015, 6:05 PM

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#6 h Jul 5, 2015, 3:16 AM • 2 Y

Y by Adventure10, Mango247

After some basic angle chasing, we find that $P,I,Q$ are collinear iff $\triangle NYI \sim \triangle IXM$ .
This is equivalent to $\frac{NY}{YI} = \frac{IX}{XM} \iff xy=r^2$ , where $x=MX, y=NY$ . Thus we must show that $xy=r^2$ iff $\angle BAC=90$ .

Firstly we assume $xy=r^2$ . By Power of a Point,
$x(XY+y)=(s-a)(s-b)\implies x=\frac{(s-a)(s-b)-r^2}{XY} = \frac{(s-a)(s-b)-\frac{(s-a)(s-b)(s-c)}{s}}{XY} = \frac{(s-a)(s-b)c}{sXY}$ .
We have $XY=2(s-a)\sin\alpha = 2(s-a)\sqrt{\frac{(s-b)(s-c)}{bc}}$ , so
$x=\frac{c}{2s}\sqrt{\frac{bc(s-b)}{s-c}}$ , and similarly, $y=\frac{b}{2s}\sqrt{\frac{bc(s-c)}{s-b}}$ .
Thus $r^2=xy=\frac{b^2c^2}{4s^2}\implies \frac{bc}{2s}=r=\frac{abc}{4Rs}=\frac{bc}{2s} \frac{a}{2R} \implies \frac{a}{2R}=1 \implies \sin\angle A=1 \implies \angle BAC=90$ .

The other direction is basically the same but I'll write it out separately to be rigorous. Assume $\angle BAC=90$ . We note that the pair $(x,y)$ obtained above satisfies $xy=r^2$ , and thus $x(XY+y)=(s-a)(s-b)$ and $y(XY+x)=(s-a)(s-c)$ .
Hence if we let $M'X, N'Y$ be the lengths $x,y$ respectively, $BM'AN'$ and $CN'AM'$ are cyclic, where $M',F,E,N'$ are collinear in that order. Thus $BM'AN'C$ is cyclic and $M'=M, N'=N$ . So $MX\cdot NY=r^2$ , as desired.

This post has been edited 2 times. Last edited by leminscate, Jul 5, 2015, 3:29 AM
Reason: fixed latex

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#7 h Jun 28, 2019, 6:11 AM • 1 Y

Y by Adventure10

XmL wrote:

Here's a solution without inversion. I will only work on the backward direction.

$P,Q$ are the reflections of $X,Y$ over $MI,IN$ , hence $P,I,Q$ are collinear $\Rightarrow \angle MIP+\angle NIQ+\angle MIN=180$ $\Rightarrow \angle MIX+\angle NIY+\angle MIN=$ $\angle A-(\angle XMI+\angle YNI)+\angle MIN$ $=\angle A-180-2\angle MIN=180$ $\Rightarrow 180-\angle MIN=\frac {\angle A}{2}=\angle YXI,\angle XYI$ $\Rightarrow \triangle IXM\sim \triangle NYI\sim \triangle NIM$ .

Let $IX,IY\cap (IMN)=F,E$ resp., since $FX*IX=MX*XN=AX*XB$ , therefore $F,E\in (ABI),(ACI)$ resp. Since $MIF=\angle MNI$ by the similarity, therefore $MF=MI$ . Let $K$ denote the center of $(ABI)$ , which is the midpoint of arc $AB$ that doesn't contain $C$ , therefore $MK\perp FI\Rightarrow MK\parallel AB\Rightarrow M\equiv K\Rightarrow MA=MI$ , similarly $AN=NI$ , hence line $MXYN$ perpendicularly bisects $AI \Rightarrow AXIY$ is a square $\Rightarrow \angle A=90$ .

perfect

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#9 h Apr 30, 2020, 9:13 AM

Y by

Notice that if $C,I,M$ where collinear, similarly $B,I,N$ have to be collinear leading to that both $M,N$ lie on the perpendicular bisector of segment $AI$ , Thus $X,Y$ also do!! and that establishes the fact $\angle{BAC} = 90$ .
for the other direction here's two solutions:
the first solution
So we have to prove that $C,I,M$ are collinear. First of all, let $M'=(CI) \cap (ABC)$ and $S=(MIN) \cap (AIB)$ .
Direct application of radical axis on $(MIN),(AIBS),(ABC)$ gives $S \in IX$ . Simple angle chasing gives that:
$\angle{MNS} = \angle{MIS} = \angle{MIX} = \angle{MIP} = \frac{1}{2} \angle{XIP} = \frac{1}{2} \angle{YNQ} = \angle{INQ} = \angle{MNI}$ Hence $MI=MS$ , thus $M$ lies on the perpendicular bisector of segment $IS$ namely let it be $(\ell)$ , but $M'$ is the center of $(ASBI)$ so $M'$ also does. Therefore $M,M' = (ABC) \cap (\ell)$ , and since $(\ell),AB$ are parallel this forces $M'=M$ , as required.
$\blacksquare$
the second solution

claim(1): $NQIX$ is cyclic
proof:
$\angle IQN=\angle IYN =180 -\angle BIA=180 -\angle IXY$
$\blacksquare$
now do an inversion wrt the incircle
$N^*$ is on the ninepoint circle of $\triangle ABC$ and $X,N^*,Q$ are collinear and $N^*IXY$ are concyclic
claim(2): $N*$ is the foot of the altitude from $X$ to $ZY$
proof:
$\angle MXP=\angle MPX =\angle YPX=\angle XZY =90-\angle ZXQ \implies XQ \perp ZY \implies N$ is foor of the altitude $XQ$
$\blacksquare$
then $\angle XAY = \angle XIY =\angle XN^*Y=90$

and we win

This post has been edited 2 times. Last edited by Ali3085, Apr 30, 2020, 9:24 AM

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