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Source: MEMO 2016 T6

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#1 h Aug 25, 2016, 2:27 PM • 3 Y

Y by doxuanlong15052000, Adventure10, Mango247

Let $ABC$ be a triangle for which $AB \neq AC$ . Points $K$ , $L$ , $M$ are the midpoints of the sides $BC$ , $CA$ , $AB$ .
The incircle of $ABC$ with center $I$ is tangent to $BC$ in $D$ . A line passing through the midpoint of $ID$ perpendicular to $IK$ meets the line $LM$ in $P$ .

Prove that $\angle PIA = 90 ^{\circ}$ .

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#2 h Aug 25, 2016, 4:28 PM • 3 Y

Y by Garfield, anhtaitran, Adventure10

Let $H$ be the orthocenter of $\triangle BIC$ , $S$ the midpoint of $IK$ and $S'$ its invers WRT the incircle,let $R,Q$ be the touch points of incircle with $AB,AC$ , $W$ the reflection of $I$ wrt $BC$

Lemma(well-known):The pole of $LM$ is $H$
The main idea will be to prove that polar of $P$ is perpendicular to $RQ$ .
Point $P$ lies on the polar of $H$ and hence its polar passes thru $H$ ,polar of $S'$ passes thru $P$ and hence the polar of $P$ is $HS'$ .The inversion with wrt $\odot I$ takes the midpoint of $ID$ to $W$ and hence $K$ is the midpoint of $IS'$ .By the last conclusion $S'$ is the reflection of $I$ over $K$ and so as $I$ is the orthocenter of $HBC$ we have that $HS'$ contains the center of $\odot HBC$ .Its well known that the feet of perpendicular form $C$ to $IB$ , $B$ to $IC$ lie on $PQ$ and let those be $R',Q'$ . $P'Q'$ and $BC$ are anti-parallel and hence as $HS',HI$ are isogonals we have $HS'\perp R'Q'\equiv RQ$ . $\clubsuit$

This post has been edited 2 times. Last edited by nikolapavlovic, Dec 20, 2016, 5:08 PM

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#3 h Aug 25, 2016, 8:53 PM • 7 Y

Y by rkm0959, Kezer, randomusername, Tawan, UK2019Project, GeoKing, Adventure10

Denote by $\omega$ the circumcircle of $\triangle BIC$ and let $\gamma$ be the circle with center $K$ passing through $D.$ Let $X$ be the midpoint of arc $\widehat{BC}$ on $\odot(ABC)$ and let $T$ be the midpoint of $\overline{ID}.$ By the Incenter/Excenter Lemma, $X$ is the center of $\omega.$

Claim: $LM$ is the radical axis of $\omega$ and $\gamma.$

Proof: Since $AX$ is the bisector of $\angle BAC$ , the reflection $B^*$ of $B$ in $AX$ is the second intersection of $AC$ with $\odot(BIC).$ Since $AB^* = AB$ , we have $\text{pow}(L, \omega) = LC \cdot LB^* = \left(\frac{b}{2}\right)\left(c - \frac{b}{2}\right) = \left(\frac{c}{2}\right)^2 - \left(\frac{b - c}{2}\right)^2 = KL^2 - KD^2 = \text{pow}(L, \gamma).$ Thus, $L$ lies on the radical axis of $\omega$ and $\gamma.$ Similarly $M$ lies on said radical axis, and the claim follows. $\blacksquare$
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Back to the proof at hand, notice that $T$ lies on the radical axis of $\gamma$ and the degenerate $\odot(I).$ Moreover, $TP \perp KI$ , so $TP$ is the radical axis of $\gamma$ and $\odot(I).$ By the claim, $P$ is the radical center of $\omega, \gamma, \odot(I)$ , so $P$ lies on the radical axis of $\omega$ and $\odot(I).$ Hence $PI$ is tangent to $\omega$ , which gives $PI \perp AI$ , as desired. $\square$

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#4 h Aug 25, 2016, 11:22 PM • 4 Y

Y by Ankoganit, GeoKing, Adventure10, Mango247

Let $\omega$ be the incircle, $H$ be the orthocenter of $\triangle BIC$ , and $G$ be the antipode of $H$ in $\odot(BHC)$ . It is easy to see that $G$ lies on the polar of $E$ with respect to $\omega$ , so by La Hire's Theorem, $P$ lies on the polar of $G$ with respect to $\omega$ as well. It is well-known that $LM$ is the polar of $H$ with respect to $\omega$ , so it follows that line $GH$ is the polar of $P$ with respect to $\omega$ . Finally, since line $HG$ is the reflection of line $AI$ about $K$ , it follows that they are parallel, so we're done.

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#6 h Sep 3, 2016, 2:40 AM • 3 Y

Y by GeoKing, Adventure10, Mango247

the incircle touch $AB$ and $AC$ at $F$ and $E$ rep, $BI \cap ED \equiv X, CI \cap FD \equiv Y, XY \equiv LM$ (well-known), $BI \cap FD \equiv X', CI \cap \equiv Y'$ is clear that $X'Y' \perp AI$ . Denote $R$ the midpoint of $ID$ and $Q \equiv IR \cap XY$ so $Q$ is midpoint of $YX \Rightarrow QR \perp X'Y'$ also $I$ is orthocenter of $\triangle QRP \Rightarrow IP \perp QR \Rightarrow X'Y' \parallel IP \Rightarrow AI \perp IP$

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#7 h Sep 3, 2016, 8:15 PM • 2 Y

Y by CeuAzul, Adventure10

Let $E$ and $F$ the points of tangency of the incircle with $AC$ and $AB$ , let $N$ the midpoint of $ID$ . Working in complex numbers where the incircle of $\triangle ABC$ is the unit circle $\Longrightarrow$ $i=0$ . Suppose that $d=1$ $\Longrightarrow$ $n=\frac{1}{2}$ . Finding $a$ , $b$ , $c$ , $m$ , $k$ , $l$ in funtion of $e$ and $f$ we get:
$a=\frac{2ef}{e+f},$ $b=\frac{2f}{1+f},$ $c=\frac{2e}{1+e},$ $m=\frac{f}{1+f}+\frac{ef}{e+f},$ $l=\frac{e}{1+e}+\frac{ef}{e+f},$ $k=\frac{e}{1+e}+\frac{f}{1+f}$ .
From $PN\perp KI$ we get $\frac{p-n}{\overline{p}-\overline{n}}=-\frac{k-i}{\overline{k}-\overline{i}}$ $\Longrightarrow$ $\frac{p-\frac{1}{2}}{\overline{p}-\frac{1}{2}}$ $=$ $-\frac{k}{\overline{k}}=-\frac{e+f+2ef}{2+e+f}...(1)$
From $M$ , $L$ , $P$ are collinear we get: $\frac{p-l}{\overline{p}-\overline{l}}=\frac{m-l}{\overline{m}-\overline{l}}=\frac{\frac{f}{1+f}-\frac{e}{1+e}}{\frac{1}{1+f}-\frac{1}{1+e}}=-1$ $\Longrightarrow$ $p-l=\overline{l}-\overline{p}...(2)$

By $(1)$ and $(2)$ we get: $p=\frac{ef(e+1)(f+1)}{(e+f)(ef-1)}$ and $\overline{p}=-\frac{(e+1)(f+1)}{(e+f)(ef-1)}$
$\Longrightarrow$ $AI\perp PI$ $\Longleftrightarrow$ $\frac{a-i}{\overline{a}-\overline{i}}=-\frac{p-i}{\overline{p}-\overline{i}}$ $\Longleftrightarrow$ $\frac{a}{\overline{a}}=-\frac{p}{\overline{p}}=ef$ $\Longleftrightarrow$ $\frac{a}{\overline{a}}=\frac{\frac{2ef}{e+f}}{\overline{\frac{2ef}{e+f}}}=ef$ which it is true.
Hence $\measuredangle PIA=90^{\circ}$ .

This post has been edited 3 times. Last edited by FabrizioFelen, Jun 26, 2017, 1:00 PM
Reason: Thanks CeuAzul

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#8 h Sep 4, 2016, 2:08 PM • 5 Y

Y by danepale, harapan57, UK2019Project, GeoKing, Adventure10

Let's recall the classic "midpoints of altitudes lemma" and its configuration: $S$ is the midpoint of the $A$ -altitude, and $E,F$ are the topmost points of the $A$ -excircle and incircle. It's known that $K$ is the midpoint of $\overline{DE}$ . By homothety, $A,F,E$ are collinear, and so after an affinity from line $BC$ with ratio $1/2$ , we get the line of $S,I,E$ .

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Let $T$ be the midpoint of $\overline{ID}$ . Since $\overline{IK}$ and $\overline{FE}$ are parallel, $TP$ hits $AE$ perpendicularly at $U$ . Clearly, $ASUP$ is cyclic with diameter $\overline{AP}$ . We need only show that $I$ also lies on its circle. (So, the crux of this proof is introducing new points to connect that pesky $P$ to the rest of the diagram, and then forgetting $P$ .)

But $ASIU$ being cyclic is equivalent by PoP to $EI\cdot ES=EU\cdot EA$ , but as $AS\parallel FI$ , we just want $EI^2=EU\cdot EF$ . This is but a simple computation: it's easy to check that if $DE=a$ , $DF=b$ , $EF=c$ , then both sides are equal to $c\left(\frac{a^2}{c}+\frac{b^2}{4c}\right)$ , so we are done. $\blacksquare$

This post has been edited 1 time. Last edited by randomusername, Sep 4, 2016, 2:11 PM

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#9 h Sep 5, 2016, 6:01 PM • 1 Y

Y by Adventure10

A rather straight forward solution with not much insight (and computation). Assume for the sake of convenience, $AB<AC$ .

We shall exploit the perpendicularity lemma. Let $T$ be the midpoint of $ID$ . Let the line through $T$ perpendicular to $IK$ meet the $A$ midline at $P$ . We show that $\angle PIA=90^{\circ}$ . Clearly, we have $KP^2-PI^2=KT^2-IT^2=TD^2+DK^2-IT^2=DK^2$ . It suffices to show that $PI^2=AP^2-AI^2$ which is equivalent to saying that $KP^2-AP^2=DK^2-AI^2$ . Let $Z$ be the midpoint of arc $BC$ of the circumcircle and let $ZK$ meet $LM$ at $W$ . Clearly, $KP^2-PZ^2=WK^2-WZ^2$ and so it suffices to showing that $AP^2-PZ^2=AI^2-DK^2+WK^2-WZ^2$ . Therefore, it suffices to showing that $AI^2-IZ^2=AI^2-DK^2+WK^2-WZ^2$ . This reduces to $DK^2=IZ^2+WK^2-WZ^2$ . Let $ZK$ meet the line through $A$ parallel to $BC$ at $X$ and the circumcircle of triangle $ABC$ at $Y$ . By the "shooting lemma", $ZI^2=ZK\cdot ZY$ and $WK^2-WZ^2=-ZM\cdot ZX$ . We get that the requested condition is equivalent to $DK^2=ZK\cdot XY$ .

Now comes the part with calculations. Let $BC=a,CA=b,AB=c$ and the usual notations. We get that $XY=\frac{b^2-c^2}{2a}\cdot \tan \left(\frac{B-C}{2}\right)$ and $ZK=\frac{a}{2}\cdot \tan \frac{A}{2}$ and $DK=\frac{b-c}{2}$ . Thus, we only need to show that $\tan (A/2)\cdot \tan \left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}$ . This is rather easy, considering that the RHS evaluates to $\begin{align*} \frac{b-c}{b+c}=\frac{\sin B-\sin C}{\sin B+\sin C}=\frac{2\sin (\frac{B-C}{2})\cos (\frac{B+C}{2})}{2\sin (\frac{B+C}{2})\cos (\frac{B-C}{2})}=\tan (A/2)\cdot \tan \left(\frac{B-C}{2}\right) \end{align*}$ which holds because of the standard sum-to-product trig formulas. The conclusion holds.

This post has been edited 1 time. Last edited by anantmudgal09, Sep 5, 2016, 6:10 PM

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#10 h Sep 7, 2016, 1:07 PM • 5 Y

Y by randomusername, AlastorMoody, GeoKing, Adventure10, Mango247

This problem was proposed by Burii.

Let me post yet another solution:
Let $X$ , $Y$ , $Z$ be reflections of $D$ about $I$ , $K$ , $P$ respectively.
$[asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(13.4cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 9.1, ymin = -2.86, ymax = 6.3; /* image dimensions */ /* draw figures */ draw(circle((-0.1,0.49), 2.53)); draw((-1.88,5.48)--(-3.06,-2.04)); draw((-3.06,-2.04)--(6.52,-2.04)); draw((6.52,-2.04)--(-1.88,5.48)); draw((-0.1,0.49)--(1.73,-2.04)); draw((3.56,-2.04)--(-1.88,5.48)); draw((-1.88,5.48)--(6.79,5.48)); draw((3.35,1.72)--(-0.1,0.49)); draw((-0.1,0.49)--(-1.88,5.48)); draw((-0.1,3.02)--(6.79,5.48)); draw((-0.1,-2.04)--(-0.1,5.48)); draw((-0.1,0.49)--(6.79,5.48)); draw((-0.1,-2.04)--(6.79,5.48)); draw((-0.1,-0.77)--(3.35,1.72)); draw((-2.47,1.72)--(3.35,1.72)); /* dots and labels */ dot((-1.88,5.48),dotstyle); label("$A$", (-1.8,5.6), NE * labelscalefactor); dot((-3.06,-2.04),dotstyle); label("$B$", (-3.36,-1.9), NE * labelscalefactor); dot((6.52,-2.04),dotstyle); label("$C$", (6.6,-1.92), NE * labelscalefactor); dot((-0.1,0.49),dotstyle); label("$I$", (-0.36,0.44), NE * labelscalefactor); dot((-0.1,-2.04),dotstyle); label("$D$", (-0.28,-2.4), NE * labelscalefactor); dot((-2.47,1.72),dotstyle); label("$M$", (-2.8,1.8), NE * labelscalefactor); dot((2.32,1.72),dotstyle); label("$L$", (2.4,1.84), NE * labelscalefactor); dot((1.73,-2.04),dotstyle); label("$K$", (1.82,-1.92), NE * labelscalefactor); dot((3.56,-2.04),dotstyle); label("$Y$", (3.64,-1.92), NE * labelscalefactor); dot((1.68,-9.56),dotstyle); label("$A'$", (-4.3,6.3), NE * labelscalefactor); dot((-0.1,3.02),dotstyle); label("$X$", (-0.02,3.14), NE * labelscalefactor); dot((3.35,1.72),dotstyle); label("$P$", (3.48,1.42), NE * labelscalefactor); dot((6.79,5.48),dotstyle); label("$Z$", (6.88,5.6), NE * labelscalefactor); dot((-0.1,5.48),dotstyle); label("$E$", (-0.02,5.6), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
We shall prove that $A, X, Y$ are collinear. Since $K$ is the midpoint of segments $BC$ , $DY$ , we have $BD = CY$ . Therefore $Y$ is the common point of segment $BC$ and the $A$ -excircle. Consider homothety centered at $A$ mapping incircle to $A$ -excircle. It's easy to see that this homothety maps $X$ to $Y$ . Therefore $A,X,Y$ are collinear.

Consider homothety with centre $D$ and ratio $2$ . We easily see that $AZ \parallel LM$ . Since $IX \perp BC \parallel LM$ , we have $IX \perp AZ.$
Moreover $IZ \parallel g \perp KI \parallel XY$ , where $g$ is the line perpendicular to $IK$ passing through the midpoint of $ID$ . Since $A,X,Y$ are collinear, we have $IZ \perp AX.$
Thus $X$ is the orthocentre of triangle $AIZ$ . Therefore $ZX \perp AI$ . Since $ZX \parallel PI$ , we conclude that $PI \perp AI.$ In other words, $\angle PIA = 90^\circ.$

This post has been edited 1 time. Last edited by timon92, Sep 7, 2016, 1:08 PM

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#11 h Sep 16, 2016, 9:50 AM • 2 Y

Y by Adventure10, Mango247

Let H be the orthocenter of BIC and K' be the reflection of I in K. Clearly by la Hire we get that HK' is the polar of P wrt the incircle. It suffices to show that HK' || AI. Note that I is the orthocenter of BIC and thus K' is the antipode of H in (BHC). By a straightforward angle chase the angle bw HK' and BC equals 90°+C/2-B/2 which is just that between AI and BC and we're done.

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#12 h Jul 8, 2018, 5:16 PM • 9 Y

Y by UK2019Project, SerdarBozdag, dgreenb801, Beginner2004, lazizbek42, DanDumitrescu, GeoKing, Adventure10, Mango247

My solution
Let $N,J$ be the midpoint of $AD,ID$ . It's well-known that $\overline{N,I,K}$ (Newton's theorem in triangle). It's easy to see $I$ is the orthocenter of $PNJ$ . So $IP \perp NJ$ and then $IP \perp IA$

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#13 h Jul 8, 2018, 10:27 PM • 1 Y

Y by Adventure10

i like this

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#14 h Jul 7, 2023, 1:51 PM

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设 ${DI}$ 交 $\triangle{ABC}$ 于另一点 ${T},$ $AT,BC$ 相交于点 $S.$ 由内切圆性质知 ${S}$ 为 $A-$ 旁切圆切点且 $KD=KS,$ 从而 $KI\| AS.$
设 $AS,NP$ 相交于点 $V,$ 则 $\angle AVP=\frac{\pi}2.$ 设过 ${A}$ 垂直于 $BC$ 垂线分别交 $LM,BC$ 于点 $Q,E,$ 则 ${Q}$ 为 ${AE}$ 中点 $.$
由 $TD\| AE,$ ${I}$ 为 $TD$ 中点 $,$ ${Q}$ 为 $AE$ 中点 $,$ 有 $S,I,Q$ 三点共线 $.$ 由 $\angle AQP=\angle AVP=\frac{\pi}2,$ $A,Q,V,P$ 四点共圆 $.$
由 $SV\cdot ST=TS^2-TV\cdot TS=(TD^2+DS^2)-TN\cdot TD=DS^2+DN\cdot DT=DS^2+DI^2=IS^2,$
可以得到 $\triangle SIV\sim\triangle STI.$ 从而 $\angle SIV=\angle STI=\angle VAQ,$ 即 $A,Q,I,V$ 四点共圆 $,$ 故 $A,Q,I,V,P$ 五点共圆 $,$
进而 $\angle AIP=\angle AQP=\frac{\pi}2.\blacksquare$

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