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Source: 2022 Brazilian National Mathematical Olympiad - Problem 2

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238 posts

#1 h Nov 22, 2022, 1:10 AM • 4 Y

Y by VicKmath7, itslumi, Mango247, Rounak_iitr

Let $ABC$ be an acute triangle, with $AB<AC$ . Let $K$ be the midpoint of the arch $BC$ that does not contain $A$ and let $P$ be the midpoint of $BC$ . Let $I_B,I_C$ be the $B$ -excenter and $C$ -excenter of $ABC$ , respectively. Let $Q$ be the reflection of $K$ with respect to $A$ . Prove that the points $P,Q,I_B,I_C$ are concyclic.

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#2 h Nov 22, 2022, 5:35 AM

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Solution

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#3 h Nov 22, 2022, 6:56 AM • 3 Y

Y by BrodyGong1027, Ubfo, Rounak_iitr

Let $I_A$ be the $A-$ excenter, $K$ is mindpoint of the $II_a$ by the nine point circle, $\triangle I_BII_A\sim \triangle I_BBC$ and $\triangle I_CII_A \sim \triangle I_CBC \Rightarrow \angle I_BKA=\angle I_BPC$ and $\angle I_CKA=\angle I_CPB$ so $PQI_BI_C$ is cyclic.

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1385 posts

#4 h Nov 22, 2022, 9:59 AM

Y by

Nice config geo!
Let $D=AK \cap BC$ , $I_A$ be the $A$ -excenter, $M$ be the midpoint of the major arc $BC$ and $I_BI_C \cap BC=X$ . We have that $AMPD$ and $BCI_BI_C$ are cyclic, so spamming PoP gives $XD.XP=XA.XM=XB.XC=XI_C.XI_B$ , so $I_BI_CDP$ is cyclic. We want to prove that $Q$ lies on that circle. Indeed, we want $AI_C.AI_B=AQ.AD=AD.AK \iff$ $D$ is orthocenter for $\triangle I_BI_CK$ , which follows by Brokard for the cyclic quadrilateral $IBI_AC$ (since $K$ is its center).

This post has been edited 2 times. Last edited by VicKmath7, Nov 22, 2022, 10:00 AM

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#5 h Nov 25, 2022, 6:04 PM • 2 Y

Y by teomihai, HoripodoKrishno

Not that hard but yeah... Nice problem for a NUB like me.

Anyways moving onto the solution,

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -28, xmax = 6, ymin = -5, ymax = 19; /* image dimensions */ /* draw figures */ draw((-8.132112681822244,6.103664258932486)--(-8.99,-0.77), linewidth(0.4)); draw((-8.99,-0.77)--(-0.31,-0.77), linewidth(0.4)); draw((-0.31,-0.77)--(-8.132112681822244,6.103664258932486), linewidth(0.4)); draw(circle((-4.65,2.1787015619620003), 5.24694586416871), linewidth(0.4)); draw((-0.31,-0.77)--(-13.3200477963969,4.134066052134884), linewidth(0.4)); draw((-13.3200477963969,4.134066052134884)--(4.020047796396906,10.71722880012654), linewidth(0.4)); draw(circle((-5.0862639235495,8.574769577234072), 9.354947603412338), linewidth(0.4) + linetype("4 4")); draw((-8.99,-0.77)--(-13.3200477963969,4.134066052134884), linewidth(0.4)); draw((-0.31,-0.77)--(4.020047796396906,10.71722880012654), linewidth(0.4)); draw((-8.99,-0.77)--(-26.237391236422933,-0.77), linewidth(0.4)); draw((-26.237391236422933,-0.77)--(-13.3200477963969,4.134066052134884), linewidth(0.4)); draw((-11.614225363644488,15.275572820071684)--(-4.65,-3.0682443022067103), linewidth(0.4)); draw((-8.99,-0.77)--(4.020047796396906,10.71722880012654), linewidth(0.4)); /* dots and labels */ dot((-8.132112681822244,6.103664258932486),linewidth(2pt) + dotstyle); label("$A$", (-8.516662214016402,6.952946630403522), NE * labelscalefactor); dot((-8.99,-0.77),linewidth(2pt) + dotstyle); label("$B$", (-10.215226956958455,-2.259379272983231), NE * labelscalefactor); dot((-0.31,-0.77),linewidth(2pt) + dotstyle); label("$C$", (-0.02429676701291338,-0.9572815308149143), NE * labelscalefactor); dot((-4.65,-3.0682443022067103),linewidth(2pt) + dotstyle); label("$K$", (-4.669273367310101,-4.62494533814528), NE * labelscalefactor); dot((-4.65,-0.77),linewidth(2pt) + dotstyle); label("$P$", (-4.696326799454727,-2.2229913594429492), NE * labelscalefactor); dot((-13.3200477963969,4.134066052134884),linewidth(2pt) + dotstyle); label("$I_C$", (-15.751989828677138,4.626351465965678), NE * labelscalefactor); dot((4.020047796396906,10.71722880012654),linewidth(2pt) + dotstyle); label("$I_B$", (4.068985215404134,10.91917518067594), NE * labelscalefactor); dot((-11.614225363644488,15.275572820071684),linewidth(2pt) + dotstyle); label("$Q$", (-13.001730467854944,15.653525805985506), NE * labelscalefactor); dot((-5.5225278470990045,-0.77),linewidth(2pt) + dotstyle); label("$D$", (-6.941761667458694,-2.1959379272983231), NE * labelscalefactor); dot((-26.237391236422933,-0.77),linewidth(2pt) + dotstyle); label("$E$", (-27.09656861520515,-0.5244266165008968), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let $D = AK \cap BC$ , $E = I_B I_C$ . Also, trivial to notice that $I_BI_CBC$ is cyclic.

Now, $AQ \cdot AD = AK \cdot AD \stackrel{\dagger}{=} (\sqrt{bc})^2 = AB \cdot AC \stackrel{\ddagger}{=} AI_B \cdot AI_C \implies I_B I_C DQ\text{ is cyclic}.$
$\dagger$ follows straight from $\sqrt{bc}$ Inversion, and $\ddagger$ follows from here.

And, $EI_B \cdot EI_C = EB \cdot EC \stackrel{\star}{=} ED \cdot EP \implies I_B I_C PD \text{ is cyclic}.$
$\star$ is just EGMO Lemma 9.17 as it's well known $(E, D; B, C) = -1$ .

Both these together imply that $I_B I_C PDQ$ is cyclic. $\blacksquare$

$\textbf{Remark:}$ This might've been well known for $\sqrt{bc}$ Inversion that $I_B \xleftrightarrow{} I_C$ , and due to not being in touch with geo for so long I had forgotten the result and had to prove it myself

.

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2944 posts

#7 h Nov 25, 2022, 6:25 PM

Y by

Rafinha wrote:

Let $I_A$ be the $A-$ excenter, $K$ is mindpoint of the $II_a$ by the nine point circle, $\triangle I_BII_A\sim \triangle I_BBC$ and $\triangle I_CII_A \sim \triangle I_CBC \Rightarrow \angle I_BKA=\angle I_BPC$ and $\angle I_CKA=\angle I_CPB$ so $PQI_BI_C$ is cyclic.

nice solution!

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304 posts

#8 h Nov 25, 2022, 8:18 PM

Y by

My solution is also posted above I just wanted to post this since I think it is a very cool one!

2022 Brazilian National Mathematical Olympiad - Problem 2 wrote:

Let $ABC$ be an acute triangle, with $AB<AC$ . Let $K$ be the midpoint of the arch $BC$ that does not contain $A$ and let $P$ be the midpoint of $BC$ . Let $I_B,I_C$ be the $B$ -excenter and $C$ -excenter of $ABC$ , respectively. Let $Q$ be the reflection of $K$ with respect to $A$ . Prove that the points $P,Q,I_B,I_C$ are concyclic.

By some angle chasing we get:
$\angle I_AII_B=\angle BCI_B\land\angle II_AI_B=\angle CBIB_B\implies \Delta II_AI_B\cup\{ K\}\sim\Delta CBI_B\cup\{P\}\implies\angle IKI_B=\angle CPI_B$ By a similar argument, we can get $\angle IKI_C=\angle BPI_C$ .Thus, we have $\angle I_CQI_B=\angle I_CKI_B=\angle CPI_B+\angle BPI_C=180^\circ-\angle I_CPI_B\implies P,Q,I_B,I_C \text{ are concyclic.}\blacksquare$

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#9 h Nov 25, 2022, 11:24 PM

Y by

Straightforward by complex numbers

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#10 h Nov 26, 2022, 12:20 AM • 3 Y

Y by Mango247, Mango247, Mango247

Let $L$ be the midpoint of arc $BAC$ and $X=I_BI_C\cap BC$ . Let $K'$ be the reflection of $K$ over $L$ . Then, $\angle DAL=\angle DPL=90^\circ$ , so $ADLP$ is cyclic. Thus, by say Power of a Point at $K$ , $PDQK'$ is also cyclic. In addition, Power of a Point gives $XD\cdot XP=XA\cdot XL=XB\cdot XC=XI_C\cdot XI_B$ so $I_BI_CPD$ is cyclic.

Note $D$ is the orthocenter of $I_BI_CK$ . Note clearly $KA\perp I_BI_C$ (note $I_B,I_C,A,L$ are collinear) and $\measuredangle DI_CI_B=\measuredangle DPI_B=\measuredangle CPI_B$ . In addition, if $I_A$ is the $A$ -excenter, $II_AI_B\sim CBI_B$ , so the midpoints are corresponding. In particular, $\measuredangle CPI_B=\measuredangle I_BKI=90^\circ-\measuredangle I_CI_BK$ . Thus, $I_CD\perp I_BK$ and vice versa, so $D$ is the orthocenter. This implies that the reflection of $(I_BI_CK)$ is $(I_BI_CD)$ , so the reflection of $K$ over $I_BI_C$ lies on $(I_BI_CD)$ , as desired.

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6857 posts

#13 h Nov 26, 2022, 3:52 AM • 4 Y

Y by HamstPan38825, mod_x, Rounak_iitr, ehuseyinyigit

Solution from Twitch Solves ISL:

Let $I_A$ be the $A$ -excenter, and let $H = \overline{AIK} \cap \overline{BC}$ .
$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair A = (-3.81664,2.88806); pair B = (-3.5,-0.5); pair C = (1.5,-0.5); pair I = (-2.45079,0.65183); pair K = (-1.,-1.72347); pair P = (-1.,-0.5); pair Q = (-6.63328,7.49959); pair I_B = (3.85362,7.57292); pair I_C = (-5.85362,1.64391); pair I_A = (0.45079,-4.09877); pair H = (-1.74727,-0.5); pair T = (-9.36373,-0.5); import graph; size(9cm); pen zzttqq = rgb(0.6,0.2,0.); pen qqffff = rgb(0.,1.,1.); pen qqwuqq = rgb(0.,0.39215,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw(A--B--C--cycle, linewidth(0.6) + zzttqq); draw(I_A--I_B--I_C--cycle, linewidth(0.6) + qqffff); draw(A--B, linewidth(0.6) + zzttqq); draw(B--C, linewidth(0.6) + zzttqq); draw(C--A, linewidth(0.6) + zzttqq); draw(circle((-1.,1.44247), 3.16594), linewidth(0.6)); draw(circle((-1.37363,5.22015), 5.73234), linewidth(0.6)); draw(I_A--I_B, linewidth(0.6) + qqffff); draw(I_B--I_C, linewidth(0.6) + qqffff); draw(I_C--I_A, linewidth(0.6) + qqffff); draw(Q--I_A, linewidth(0.6)); draw(B--I_B, linewidth(0.6) + blue); draw(C--I_C, linewidth(0.6) + blue); draw(I_B--T, linewidth(0.6) + qqwuqq); draw(T--B, linewidth(0.6) + qqwuqq); dot("$A$", A, dir((-14.232, 52.186))); dot("$B$", B, dir((-41.152, -54.958))); dot("$C$", C, dir((13.913, -43.888))); dot("$I$", I, dir((5.740, 13.299))); dot("$K$", K, dir((6.148, 12.041))); dot("$P$", P, dir(110)); dot("$Q$", Q, dir((-50.427, 29.076))); dot("$I_B$", I_B, dir((6.270, 12.255))); dot("$I_C$", I_C, dir((-9.789, 50.090))); dot("$I_A$", I_A, dir((-12.421, -42.985))); dot("$H$", H, dir(225)); dot("$T$", T, dir((-60.449, -26.493))); [/asy]$

Then by Brokard's theorem on cyclic quadrilateral $BICI_A$ , it follows that $\triangle I_B I_C H$ is self-polar to this circle.
In particular $K$ is the orthocenter. Equivalently, $H$ is the orthocenter of $\triangle I_B K I_C$ .
Define $T = \overline{I_B I_C} \cap \overline{BC}$ . Then $(TH;BC) = -1$ .
I claim that $I_B I_C P H$ is cyclic. Indeed, $TH \cdot TP = TB \cdot TC = TI_B \cdot I_C$ .
The reflection of the orthocenter $K$ of $\triangle I_BI_CH$ over a side $\overline{I_B I_C}$ then coincides with $Q$ , which now lies on $(PHI_BI_C)$ .

This post has been edited 1 time. Last edited by v_Enhance, Nov 26, 2022, 3:53 AM

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#14 h Nov 27, 2022, 8:41 AM • 1 Y

Y by teomihai

rcorreaa wrote:

Let $ABC$ be an acute triangle, with $AB<AC$ . Let $K$ be the midpoint of the arch $BC$ that does not contain $A$ and let $P$ be the midpoint of $BC$ . Let $I_B,I_C$ be the $B$ -excenter and $C$ -excenter of $ABC$ , respectively. Let $Q$ be the reflection of $K$ with respect to $A$ . Prove that the points $P,Q,I_B,I_C$ are concyclic.

Can't believe no one has posted this sol yet.

We bary with respective triangle $ABC$ and set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ .

First, let's find $K$ . Note that $K$ lies on $AI$ where $I$ is the incenter so $K=(x:b:c)$ . Moreover, it lies on $(ABC)$ so
$a^2bc+b^2cx+c^2xb=0\implies x=-\frac{a^2}{b+c}\implies K=(-\frac{a^2}{b+c}:b:c).$ Now, let's find $Q$ . We know $K=(-\frac{a^2}{b+c}:b:c)$ and $A=(b+c-\frac{a^2}{b+c}:0:0)$ . Thus,
$Q=2A-K=(2(b+c)-\frac{a^2}{b+c}:-b:-c)=(2(b+c)^2-a^2:-b(b+c):-c(b+c)).$ And lastly, we know $P=(0:1:1),I_B=(a:-b:c),I_C=(a:b:-c)$ . Let's find the equation of $(PI_BI_C)$ .
$(x,y,z)\mapsto (0:1:1)\implies -a^2+2(v+w)=0\implies v+w=\frac{a^2}{2}$ $(x,y,z)\mapsto (a:-b:c)\implies a^2bc-b^2ca+c^2ab+(ua-vb+wc)(a-b+c)=0\implies abc+ua-vb+wc=0$ $(x,y,z)\mapsto (a:b:-c)\implies a^2bc+b^2ca-c^2ab+(ua+vb-wc)(a+b-c)=0\implies abc+ua+vb-wc=0$ The second plus the third equation gives $2abc+2ua=0\implies u=-bc$ . Plug this back to the second equation to get $vb=wc$ , so $w=\frac{vb}{c}$ . Plug this into the first equation, and we'll have $v+\frac{vb}{c}=\frac{a^2}{2}\implies v=\frac{a^2c}{2(b+c)}$ , so $w=\frac{a^2b}{2(b+c)}$ .
$(PI_BI_C): -a^2yz-b^2zx-c^2xy+\left(-bcx+\frac{a^2c}{2(b+c)}y+\frac{a^2b}{2(b+c)}z\right)(x+y+z)=0.$ Finally, we plug $(x,y,z)\mapsto (2(b+c)^2-a^2:-b(b+c):-c(b+c))=Q$ .
$-a^2yz-b^2zx-c^2xy=-a^2bc(b+c)^2+b^2c(b+c)(2(b+c)^2-a^2)+bc^2(b+c)(2(b+c)^2-a^2)=2bc(b+c)^2((b+c)^2-a^2)$ $-bcx+\frac{a^2c}{2(b+c)}y+\frac{a^2b}{2(b+c)}z=(-2bc(b+c)^2+a^2bc)-\frac{a^2bc}{2}-\frac{a^2bc}{2}=-2bc(b+c)^2$ $x+y+z=(2(b+c)^2-a^2)-b(b+c)-c(b+c)=(b+c)^2-a^2$ Hence, we're done.

This post has been edited 1 time. Last edited by Quidditch, Nov 27, 2022, 9:30 AM

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#15 h Mar 8, 2023, 3:00 AM • 1 Y

Y by samrocksnature

Rename $K$ to $M_A$ . Suppose $(ABC)$ is the unit circle, and let $A=x^2$ , etc., so $M_A=-yz$ , etc., and $I=-xy-yz-zx$ . By incenter-excenter we then have $I_B=-xy+xz+zy$ and $I_C=-xz+yx+yz$ . Furthermore, $P=\frac{y^2+z^2}{2}$ and $Q=2x^2+yz$ . We then wish to prove that
$\frac{-xz+yx+yz-\frac{y^2+z^2}{2}}{-xz+yx+yz-(2x^2+yz)} \div \frac{-xy+xz+zy-\frac{y^2+z^2}{2}}{-xy+xz+zy-(2x^2+zy)} \in \mathbb{R}.$ We can simplify the expression to
$\frac{2x(y-z)-(y-z)^2}{x(-z+y-2x)} \div \frac{-2x(y-z)-(y-z)^2}{x(-y+z-2x)}=\frac{2x-y+z}{-2x+y-z} \div \frac{-2x-y+z}{-2x-y+z}=-1,$ which of course is real, so we are done. $\blacksquare$

This post has been edited 1 time. Last edited by IAmTheHazard, Mar 8, 2023, 3:00 AM

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#16 h May 20, 2023, 7:33 PM • 1 Y

Y by teomihai

Beautiful problem! I liked the configuration
Let $M$ be the midpoint of $I_BI_C$ , then $M$ is on $(ABC)$ because it is nine-point circle of $I_AI_BI_C$
Let $K'$ be the reflection of $K$ with respect to $A$

Claim 1: $I_CQK'I_B$ is cyclic.
Proof: Notice $\angle I_CKI_B= \angle I_CQI_B$ because of the reflection and $KA \perp I_BI_C$ , then we observe $I_CKI_BK'$ is parallelogram as diagonals bisect each other, hence $\angle I_CKI_B= \angle I_CK'I_B$ , hence the claim.

Claim 2:The problem itself
Proof: Note that by Claim 1 it suffices to prove $I_CK'I_BP$ is cyclic, $\angle I_CK'P = \angle I_CI_BP$ $\angle K'KI_B=I_CI_BP$ $MP*MK = MI_B^2$ $MP^2+MP*PK=MI_B^2$ $MP^2+PC^2=MI_B^2$ $MC=MI_B$ which is true since $I_BCI_C=90^\circ$ and $M$ is midpoint.

To make more readable: here's diagram

Attachments:

This post has been edited 3 times. Last edited by math_comb01, May 20, 2023, 7:39 PM

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L567

#17 h May 21, 2023, 5:26 AM

Y by

Solved with mueller.25, starchan, Siddharth03, AdhityaMV

Let $AK$ intersect segment $BC$ at $R$ and let $I_BI_C$ intersect line $BC$ at $T$ . Let $N$ be the midpoint of major arc $BC$ . Then note that $(N,K;B,C) = -1$ so projecting through $A$ , we get that $(T,R;B,C) = -1$ as well. This means that since $P$ is the midpoint of $BC$ and $(BCI_BI_C)$ , we have $TR \cdot TP = TB \cdot TC = TI_B \cdot TI_C$ so points $P,R,I_B,I_C$ are concyclic.

Let $I$ be the incenter of $\triangle ABC$ . Note that $(A,R;I,I_A) = -1$ and $K$ is the midpoint of $II_A$ . So, $AQ \cdot AR = AK \cdot AR = AI \cdot AI_A = AI_B \cdot AI_C$ , implying that points $R,Q,I_B,I_C$ are concyclic.

Together, these imply that points $P,Q,I_B,I_C$ are concyclic, as desired. $\blacksquare$

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#18 h Oct 26, 2023, 1:51 PM • 1 Y

Y by L13832

IAmTheHazard wrote:

Rename $K$ to $M_A$ . Suppose $(ABC)$ is the unit circle, and let $A=x^2$ , etc., so $M_A=-yz$ , etc., and $I=-xy-yz-zx$ . By incenter-excenter we then have $I_B=-xy+xz+zy$ and $I_C=-xz+yx+yz$ . Furthermore, $P=\frac{y^2+z^2}{2}$ and $Q=2x^2+yz$ . We then wish to prove that
$\frac{-xz+yx+yz-\frac{y^2+z^2}{2}}{-xz+yx+yz-(2x^2+yz)} \div \frac{-xy+xz+zy-\frac{y^2+z^2}{2}}{-xy+xz+zy-(2x^2+zy)} \in \mathbb{R}.$ We can simplify the expression to
$\frac{2x(y-z)-(y-z)^2}{x(-z+y-2x)} \div \frac{-2x(y-z)-(y-z)^2}{x(-y+z-2x)}=\frac{2x-y+z}{-2x+y-z} \div \frac{-2x-y+z}{-2x-y+z}=-1,$ which of course is real, so we are done. $\blacksquare$

oh

Let $D$ and $E$ be the intersections of the interior and exterior $\angle A$ -bisectors with $\overline{BC}$ respectively. By $\sqrt{bc}$ inversion we have
$AI_B\cdot AI_C=AB\cdot AC=AD\cdot AK=AD\cdot AQ,$ hence $I_BI_CQD$ is cyclic. On the other hand, it is well-known (extraversion of incenter/excenter) that $I_BI_CBC$ is cyclic, so $EI_B\cdot EI_C=EB\cdot EC$ . Furthermore, since $(B,C;D,E)=-1$ , we have $EB\cdot EC=ED\cdot EP$ , hence $I_BI_CPD$ is cyclic by power of a point, so we're done. $\blacksquare$

This post has been edited 2 times. Last edited by IAmTheHazard, Oct 26, 2023, 1:54 PM

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#19 h May 27, 2024, 4:58 PM • 1 Y

Y by surpidism.

Let $H$ be the intersection of $AK$ and $BC$ . We first show that $QI_CHI_B$ is cyclic. First as $\triangle I_CAC \sim \triangle I_BAB$ , we see that

$\frac{I_CA}{AC} = \frac{AB}{I_BA} \implies I_CA \cdot I_BA = AB \cdot AC.$
Secondly, as $\triangle ABK \sim \triangle ACH$ , we again see

$\frac{AK}{AC} = \frac{AB}{AH} \implies AK \cdot AH = AB \cdot AC.$
So,

$QA \cdot AH = KA \cdot AH = BA \cdot AC = I_CA \cdot AI_B$
implying the claim. Now let $J$ be the midpoint of $I_BI_C$ . Note that $J$ lies on $(ABC)$ as it is the nine point circle of $\triangle I_AI_BI_C$ . Furthermore $J$ is the arc midpoint of arc $BC$ containing $A$ . That means $PHAJ$ is cyclic. Now, letting $I_CI_B$ intersect $BC$ at $X$ , we see

$HX \cdot XP = AX \cdot XJ = BX \cdot XC = I_CX \cdot XI_B$
finishing the problem.

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L13832

#20 h Jun 28, 2024, 4:09 AM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -5.803255101716253, xmax = 20.94508689023541, ymin = -2.1002703047388525, ymax = 22.319316398917017; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); draw((7.575733531948671,10.863916969703546)--(5,5)--(12,5)--cycle, linewidth(2) + zzttqq); /* draw figures */ draw((7.575733531948671,10.863916969703546)--(5,5), linewidth(2)); draw((5,5)--(12,5), linewidth(2)); draw((12,5)--(7.575733531948671,10.863916969703546), linewidth(2)); draw(circle((8.5,6.960275942712022), 4.011568492694037), linewidth(2)); draw((1.6247981902196311,10.169020646842817)--(8.970521844693746,-1.0807354791321093), linewidth(2)); draw((8.970521844693746,-1.0807354791321093)--(15.375201809780371,11.7746682239693), linewidth(2)); draw((1.6247981902196311,10.169020646842817)--(15.375201809780371,11.7746682239693), linewidth(2)); draw(circle((8.380234318788622,11.99749071039707), 6.998515561209342), linewidth(2)); draw((7.575733531948671,10.863916969703546)--(8.970521844693746,-1.0807354791321093), linewidth(2)); draw((1.6247981902196311,10.169020646842817)--(12,5), linewidth(2)); draw((5,5)--(15.375201809780371,11.7746682239693), linewidth(2)); draw((15.375201809780371,11.7746682239693)--(8.5,5), linewidth(2) + linetype("2 2") + qqwuqq); draw((8.5,2.9487074500179844)--(15.375201809780371,11.7746682239693), linewidth(2) + linetype("2 2") + zzttqq); draw((8.5,5)--(1.6247981902196311,10.169020646842817), linewidth(2) + linetype("2 2") + qqwuqq); draw((1.6247981902196311,10.169020646842817)--(8.5,2.9487074500179844), linewidth(2) + linetype("2 2") + zzttqq); /* dots and labels */ dot((7.575733531948671,10.863916969703546),dotstyle); label("$A$", (7.716463099778057,11.193041132615535), NE * labelscalefactor); dot((5,5),dotstyle); label("$B$", (5.128957223893979,5.338809088427837), NE * labelscalefactor); dot((12,5),dotstyle); label("$C$", (12.11522308878099,5.338809088427837), NE * labelscalefactor); dot((8.5,5),linewidth(4pt) + dotstyle); label("$P$", (8.622090156337485,5.274121441530735), NE * labelscalefactor); dot((8.5,2.9487074500179844),linewidth(4pt) + dotstyle); label("$K$", (8.622090156337485,3.2041167408234825), NE * labelscalefactor); dot((1.6247981902196311,10.169020646842817),linewidth(4pt) + dotstyle); label("$I_C$", (1.7651995852446765,10.416789369850315), NE * labelscalefactor); dot((8.970521844693746,-1.0807354791321093),linewidth(4pt) + dotstyle); label("$I_A$", (9.107247508065749,-0.8065173667968194), NE * labelscalefactor); dot((15.375201809780371,11.7746682239693),linewidth(4pt) + dotstyle); label("$I_B$", (15.511324550878843,12.033980542277858), NE * labelscalefactor); dot((6.651467063897343,18.779126489389107),linewidth(4pt) + dotstyle); label("$Q$", (6.7784922197700785,19.052590230613387), NE * labelscalefactor); dot((8.029478155306254,6.978150379168081),linewidth(4pt) + dotstyle); label("$I$", (8.169276628057771,7.247094671892335), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let $I_A$ be the $A-$ excenter. Note that $\angle I_CKI_B=\angle I_CQI_B=90^{\circ}$ this is because $Q$ is the reflection of $K$ over $A$ , so we have to show that $90^{\circ}=\angle I_CDI_B$ , so we need to show that $\angle I_BI_CD+\angle I_CI_BD=90^{\circ}$ . Now observe that $K$ is midpoint of the $II_a$ because the circumcircle of the orthic triangle is nine point circle of the reference triangle. Note that $II_CBA$ and $II_ABC$ are cyclic, this gives us $\triangle I_BI_AI\sim \triangle I_BCI$ and $\triangle I_CI_AI \sim \triangle I_CCB$ because of this similarity we have $\angle I_BPC=\angle I_BKA$ and $\angle I_CPB=\angle I_CKA$ so $\angle I_BI_CD+\angle I_CI_BD=90^{\circ}$ so $P,Q,I_B,I_C$ are concyclic.

This post has been edited 1 time. Last edited by L13832, Jun 28, 2024, 5:10 AM

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#21 h Jun 28, 2024, 5:13 AM

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Cute!

rcorreaa wrote:

Let $ABC$ be an acute triangle, with $AB<AC$ . Let $K$ be the midpoint of the arch $BC$ that does not contain $A$ and let $P$ be the midpoint of $BC$ . Let $I_B,I_C$ be the $B$ -excenter and $C$ -excenter of $ABC$ , respectively. Let $Q$ be the reflection of $K$ with respect to $A$ . Prove that the points $P,Q,I_B,I_C$ are concyclic.

Since $KA \perp I_BI_C$ and $KQ$ bisects $I_BI_C$ , we just need to prove $Q$ is the $K$ -HM point in $\triangle KI_BI_C$ . This is equivalent to showing that $L\overset{\text{def}}{:=} KA \cap BC$ is the orthocentre of $\triangle KI_BI_C$ , as $LQ \perp KQ$ . Finally, this last part follows as $L$ lies on ray $AK$ and $AL \cdot AK = AB \cdot AC = AI_B \cdot AI_C$ , as desired.

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#22 h Oct 8, 2024, 6:51 PM • 2 Y

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If we rephrase the problem according to $\triangle I_AI_BI_C$ where $I_A$ is the $A-$ excenter, we get the following problem.
New Problem Statement: $ABC$ is a triangle with altitudes $AD,BE,CF$ and $K=AD\cap (DEF), \ M$ is the midpoint of $EF$ . Prove that reflection of $K$ with respect to $BC,B,C,M$ are concyclic.
Proof: Let $N$ be the midpoint of $BC$ and $Q,K'$ be the reflections of $K$ to $BC,N$ . $P=EF\cap AD,R=EF\cap BC$ . Note that $K,M,N$ are collinear. $NB.NC=NM.NK=NM.NK'$ hence $B,C,M,K'$ are concyclic. Also $RB.RC=RE.RF=RD.RN=RP.RM$ which implies $B,C,M,P$ are cyclic. Since $P,M,K,Q$ lie on the circle with diameter $PK',$ we get that $P,K'\in (BCM)$ and $Q\in (PMK)$ thus, $B,C,M,P,K',Q$ are concyclic as desired. $\blacksquare$

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#23 h Today at 4:30 AM

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One of my quickest headsolves.

If we let $D=\overline{AI} \cap \overline{BC}$ then see that $(I_BI_CQD)$ is cyclic since $AI_B \cdot AI_C=AD \cdot AK=AD \cdot AQ$ (first two quantities are both equal to $AB \cdot AC$ by say $\sqrt{bc}$ inversion).

Now let $X$ be $I_A$ -Ex point of $\triangle I_AI_BI_C$ And now $(XD;BC)=-1$ (by Cevs Menelaus or whatever) and hence by Mclaurin's we get $XI_C \cdot XI_B=XB \cdot XC=XD \cdot XP$ and done by PoP.

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