Right-angled triangle if circumcentre is on circle

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liberator

95 posts

liberator

#1 h Jan 4, 2016, 9:41 PM • 9 Y

Y by Davi-8191, anantmudgal09, HWenslawski, megarnie, son7, Adventure10, Funcshun840, PikaPika999, Rounak_iitr

Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$ . Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$ , respectively. Suppose that the circumcentre of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$ . Prove that triangle $ABC$ is right-angled.

Proposed by Alexander A. Polyansky, Russia

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huynguyen

535 posts

huynguyen

#2 h Jan 5, 2016, 5:42 AM • 6 Y

Y by mamavuabo, quangminhltv99, Tawan, son7, Adventure10, PikaPika999

Proving a geometry problem with "if, suppose" is not my strength, but let I try this hard and nice problem.

The center of $(A_1B_1C_1)$ lies on $(O)$ , which means it lies outside triangle $A_1B_1C_1$ .It implies that this triangle is obtuse, and WLOG, assumethag $\widehat{C_1A_1B_1}>90$ .
Let $(I)$ be the incircle of triangle $ABC$ , and its tangent points on $AB,AC$ are $F,E$ respectively.
Note that $CC_1,BB_1$ cut each other at the Nagel point, so $BC_1=AF=AE=B_1C,BA_1=AB_1,A_1C=AC_1$ .
It means if we let $S$ be the midpoint of arc $BAC$ ,then $S$ is indeed the spiral similarity mapping $B\rightarrow C,C_1\rightarrow B_1$ , which means $SC_1=SB_1,SC=SB$ .Combining with the constraint, it follows that $S$ is actually the center of $(A_1B_1C_1)$ .
Let $I_a,I_b,I_c$ be respectively the centers of $A$ -excircle, $B$ -excircle and $C$ -excircle.
Note that $I_aA_1,I_bB_1,I_cC_1$ are concurrent at a point $U$ .
Also note that $U$ is a intersection of $(BC_1A_1)$ and $(CB_1A_1)$ .
According to that, by angle chasing, we get:
$\widehat{BUC}=\widehat{BUA_1}+\widehat{CUA_1}=360-\widehat{BAC}-\widehat{C_1A_1B_1}= 180-\frac{\widehat{C_1SB_1}}{2}$ .
Combining with $SC_1=SB_1$ , we conclude that $S$ is also the center of $(BUC)$ .
Now by symmetry, $C_1U=BA_1=AB_1,B_1U=CA_1=AC_1$ .Hence $AC_1B_1U$ is a parallelogram.But note that $\widehat{AC_1U}=90$ so it is actually a rectangle, and the conclusion follows.

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liberator

95 posts

liberator

#4 h Jan 5, 2016, 3:33 PM • 11 Y

Y by quangminhltv99, rkm0959, A_Math_Lover, HolyMath, Aryan-23, mijail, son7, Adventure10, antimonio, endless_abyss, MS_asdfgzxcvb

$[asy] unitsize(2.5cm); void b() { pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); pen dd=linetype("4 8"); /* Define the excenter */ pair excenter(pair A=(0,0), pair B=(0,0), pair C=(0,0)) { return extension(A,bisectorpoint(C,A,B),B,rotate(90,B)*bisectorpoint(A,B,C)); } /* Draw points */ pair A=D("A",dir(115),N), B=D("B",(-1,0),W), C=D("C",(1,0),E), I=D(incenter(A,B,C)), Ia=D("I_a",excenter(A,B,C),S), Ib=D("I_b",excenter(B,C,A),E), Ic=D("I_c",excenter(C,A,B),W), A1=D("A_1",foot(Ia,B,C),SSW), B1=D("B_1",foot(Ib,C,A),E), C1=D("C_1",foot(Ic,A,B),N), Ma=D("M_a",midpoint(Ib--Ic),N), Mb=D("M_b",midpoint(Ic--Ia),SW), Mc=D("M_c",midpoint(Ia--Ib),SE), V=D("V",circumcenter(Ia,Ib,Ic),SE); /* Draw paths */ D(unitcircle,heavyblue); D(circumcircle(B,C,V),linetype("2 2")+rgb(0.6,0,1)); D(circumcircle(B,C1,V),linetype("2 2")+rgb(0.6,0,1)); D(A--B--C--cycle); D(Ia--Ib--Ic--cycle,gray(0.3)+linewidth(1)); D(A1--B1--C1--cycle); D(I--Ia,dd+red); D(I--Ib,dd+red); D(I--Ic,dd+red); D(V--Ia,dd+heavygreen); D(V--Ib,dd+heavygreen); D(V--Ic,dd+heavygreen); } b(); pathflag=false; b(); [/asy]$
Let the excenters opposite $A,B,C$ be $I_a,I_b,I_c$ . Let the midpoint of $\overline{I_bI_c}$ be $M_a$ , which lies on $(ABC)$ , the nine-point circle of $\triangle I_aI_bI_c$ ; analogously define $M_b,M_c$ .

$M_aB=M_aC$ and $BC_1=s-a=B_1C$ , so $\triangle M_aBC_1\cong\triangle M_aCB_1$ (SAS), thus $M_a$ is equidistant from $B_1,C_1$ , with analogous results for $M_b,M_c$ . It follows that the circumcentre of $\triangle A_1B_1C_1$ is one of $M_a,M_b,M_c$ ; WLOG, suppose it is $M_a$ .

By isogonal conjugacy, $I_aA_1,I_bB_1I_cC_1$ concur at the Bevan point $V$ of $\triangle ABC$ . $M_aM_b$ is the common perpendicular bisector of $\overline{C_1A_1}$ and $\overline{I_cC}$ , so $C_1A_1\parallel I_cC$ . $(A_1C_1M_b)$ is the circle on diameter $\overline{VB}$ , so by Reim's theorem, $V \in (I_bI_cBC)$ .

Hence $\angle I_cI_aI_b=\tfrac{1}{2}\angle I_cVI_b=45^{\circ}\implies\angle CAB=180^{\circ}-2\angle I_cI_aI_b=90^{\circ}$ , as required.

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rterte

209 posts

rterte

#5 h Feb 23, 2016, 1:45 PM • 5 Y

Y by rkm0959, son7, sayheykid, Adventure10, Mango247

Fairly easy for a #3. Anyway, here is my solution
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(4.22cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 2.24, xmax = 6.46, ymin = -0.88, ymax = 3.1; /* image dimensions */ /* draw figures */ draw((3.5,-0.32)--(3.5,2.62)); draw((3.5,-0.32)--(5.3,-0.32)); draw((3.5,2.62)--(5.3,-0.32)); draw(circle((4.4,1.15), 1.723629890666787)); draw((3.5,1.9736298906667877)--(4.102372300006038,1.6361252433234716)); draw((4.102372300006038,1.6361252433234716)--(4.653629890666787,-0.32)); draw((4.653629890666787,-0.32)--(3.5,1.9736298906667877)); draw((2.93,0.25)--(4.653629890666787,-0.32)); draw((2.93,0.25)--(3.5,1.9736298906667877)); draw((2.93,0.25)--(4.4,1.15)); draw((2.93,0.25)--(3.5,-0.32)); draw((2.93,0.25)--(5.3,-0.32)); draw((2.93,0.25)--(3.5,2.62)); draw((2.93,0.25)--(2.93,-0.32)); draw((2.93,-0.32)--(3.5,-0.32)); draw((2.93,0.25)--(3.5,0.25)); /* dots and labels */ dot((3.5,-0.32),linewidth(3.pt) + dotstyle); label("$A$", (3.22,-0.62), NE * labelscalefactor); dot((3.5,2.62),linewidth(3.pt) + dotstyle); label("$B$", (3.22,2.78), NE * labelscalefactor); dot((5.3,-0.32),linewidth(3.pt) + dotstyle); label("$C$", (5.42,-0.62), NE * labelscalefactor); dot((4.4,1.15),linewidth(3.pt) + dotstyle); label("$O$", (4.48,1.26), NE * labelscalefactor); dot((4.102372300006038,1.6361252433234716),linewidth(3.pt) + dotstyle); label("$A_1$", (4.18,1.76), NE * labelscalefactor); dot((4.653629890666787,-0.32),linewidth(3.pt) + dotstyle); label("$B_1$", (4.58,-0.78), NE * labelscalefactor); dot((3.5,1.9736298906667877),linewidth(3.pt) + dotstyle); label("$C_1$", (3.08,1.84), NE * labelscalefactor); dot((2.93,0.25),linewidth(3.pt) + dotstyle); label("$I$", (2.78,-0.08), NE * labelscalefactor); dot((3.5,0.25),linewidth(3.pt) + dotstyle); label("$H$", (3.58,0.36), NE * labelscalefactor); dot((2.93,-0.32),linewidth(3.pt) + dotstyle); label("$F$", (2.76,-0.64), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let $I$ be the circumcenter of $A_1B_1C_1$ . By the assumption, $I$ lies on $(O)$ , WLOG, assume $I$ is on the arc $BC$ that contains $A$ . We have $IB_1=IC_1$ and $CB_1=BC_1=p-a$ (set $BC=a,CA=b,AB=c,p=\frac{a+b+c}{2}$ ) Since $\angle{ABI}=\angle{ACI}$ hence $IB_1C=IC_1B$ (s.a.s), thus $IB=IC$ , meaning $I$ is the midpoint of the arc $BC$ that contains $A$ of $(O)$ . Using Stewart's theorem for $IBC$ w.r.t $IB_1$ , we have $IC^2\cdot BA_1+IB^2\cdot CA_1=BC(IA_1^2+BA_1\cdot CA_1)$ or $(BA_1+CA_1)IB^2=BC(IA-1^2+BA-1\cdot CA_1)$ (since $IB=IC$ ). Notice that $BA_1+CA_1=BC$ , we have $IB^2-IA_1^2=\frac{a^2-(b-c)^2}{4}\quad (1)$ Let $H,F$ be the perpendicular projection of $I$ on $AB$ and $AC$ . We have $IBH=ICF$ , thus $AF=AH$ , which gives us $BH=CF=\frac{b+c}{2}$ . On the other hand $\begin{align*} IB^2-IC_1^2&=HB^2-HC_1^2\\ &=\left(\frac{b+c}{2}\right)^2-\left(\frac{b+c}{2}-\frac{b+c-a}{2}\right)^2\quad (2) \end{align*}$ Since $IA_1=IC_1$ , from $(1)$ and $(2)$ , we have $\begin{align*} \frac{a^2-(b-c)^2}{4}&=\left(\frac{b+c}{2}\right)^2-\left(\frac{b+c}{2}-\frac{b+c-a}{2}\right)^2\\ \Leftrightarrow a^2&=b^2+c^2 \end{align*}$ Done.
You can have a look at here

This post has been edited 5 times. Last edited by rterte, Mar 4, 2016, 12:35 PM
Reason: "me" to "my". I blame typo

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rkm0959

1721 posts

rkm0959

#7 h Feb 25, 2016, 2:54 PM • 7 Y

Y by quangminhltv99, rterte, Tawan, Davi-8191, son7, Adventure10, Mango247

Set $\omega$ as the circumcircle of $\triangle ABC$ , $\omega'$ as the circumcircle of $\triangle A_1B_1C_1$ .
As the circumcircle of $\triangle A_1B_1C_1$ is outside $\triangle ABC$ , or outside $\triangle A_1B_1C_1$ , one of the angles is obtuse.
WLOG $\angle B_1A_1C_1 > 90$ . We will show $\angle A = 90$ .
Set $M_a$ as the midpoint of arc $BAC$ on $\omega$ . Set $I_b$ as the $B$ -excenter.

First, I claim that $M_a$ must be the center of $\omega'$ .
Trivially $M_aB=M_aC$ , $\angle C_1BM_a=\angle B_1CM_a$ and $BC_1=B_1C$ , so $\triangle M_aBC_1 \equiv \triangle M_aCB_1$ .
Now this gives $M_aB_1=M_aC_1$ . Meanwhile, the center of $\omega'$ clearly lies on arc $BAC$ .
Also, there are two intersections between $\omega$ and the perpendicular bisector of $B_1C_1$ .
However, only one of the two lies on arc $BAC$ . This forces $M_a$ to be the center of $\omega'$ .

Let us consider the intersections of $\omega'$ and the $B$ -excircle.
Since $AM_a$ is the external angle bisector of $A$ , we have $A, M_a, I_b$ colinear.
Now note that $B$ -excircle is symmetric wrt $\overline{AM_aI_b}$ .
One intersection of $\omega'$ and the $B$ -excircle is $B_1$ , so the other must be $B_1$ reflected over $AI_b$ .
This point is the tangency point of $BA$ and the $B$ -excircle. Call this point $B'_1$ .
By symmetry, note that the $A'_1$ lies on $\omega'$ , where $A'_1$ is $A_1$ reflected over the midpoint of $BC$ .

We are ready to finish. By Power of a Point with point $B$ and circle $\omega'$ , we have $BC_1 \cdot BB'_1 = BA'_1 \cdot BA_1 \implies s(s-a)=(s-b)(s-c) \implies a^2=b^2+c^2$ so $\angle BAC= 90$ as desired.

This post has been edited 1 time. Last edited by rkm0959, Feb 26, 2016, 3:16 AM

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PROF65

2016 posts

PROF65

#8 h Feb 25, 2016, 10:03 PM • 2 Y

Y by Tawan, Adventure10

Slightly different!
Let $M_1,M_2$ and $M_3$ the midpoints of the sides $I_bI_c,I_aI_c, I_aI_b$ resp. of the excentral triangle $I_aI_bI_c$ which are in the same time the midpoints of the arcs $\overarc{BAC},\overarc{ABC},\overarc{BCA}$ of the circumcircle of $ABC$
it's known $BC_1=B_1C$ then the similarity that send $B\to C_1,C\to B_1$ is rotation with center ,the second intersection of $\cal{C}$ $(ABC)$ and $\cal{C}$ $(AB_1C_1)$ but the center is on $\cal{C} (ABC)$ and on the bisector of $BC$ that it 's $M_1$ so $M_1B_1=M_1C_1$ similarly we get $M_2A_1=M_2C_1,M_3B_1=M_3A_1$ hence ,since the center of the circumcircle of $A_1B_1C_1$ is on the circumcircle of $ABC$ ,the bisectors of the sides of $A_1B_1C_1$ meet at $M_1,M_2$ or $M_3$ WLG suppose it 's $M_1$ then $M_1M_2 \perp A_1C_1$ but $M_1M_2\parallel I_aI_b$ thus $A_1C_1\perp I_aI_b$ similarly $A_1B_1\perp I_aI_c \implies \widehat{(A_1C1,A1B_1)}=\widehat{(I_aI_b,I_aI_c)}=\frac{1}{2}\widehat{(AC,AB)}$ thus $\widehat {(AB,AC)}=\widehat{(M_1B,M_1C)}=\widehat{(M_1C_1,M_1B_1)}=2\widehat{(A_1C_1,A_1B_1)}=\widehat{(AC,AB)}$ so $\widehat {(AB,AC)}=\frac{\pi}{2}$ .
R.HAS

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anantmudgal09

1979 posts

anantmudgal09

#9 h Apr 21, 2016, 8:56 PM • 3 Y

Y by Tawan, Adventure10, Mango247

Not hard for an IMO 3

Indeed, we see that if $M,N,P$ are the midpoints of arcs $BC,CA,AB$ respectively and $I_a,I_b,I_c$ are the corresponding excentres then,
A rotation about $M$ sends $B_1C$ to $C_1B$ and thus, points $A,M,B_1,C_1$ are concyclic. Similarly for the other two vertices. This gives that in fact $M$ lies on the perpendicular bisector of $B_1C_1$ and so one of since the circumcentre of $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$ , then it must actually be one of $M,N,P$ . Let it be $M$ say. We prove that $\angle A=90^{\circ}$

Indeed, we notice that $MN$ is the perpendicular bisector of $A_1C_1$ and $MP$ of $A_1B_1$ . Since, $MN \parallel I_aI_b$ , $A_1C_1 \perp I_aI_b$ and similarly $A_1B_1 \perp I_aI_c$ .

Now, this means that $\angle B_1A_1C_1=180^{\circ}-(90^{\circ}+\frac{\angle A}{2})$ and so $\angle B_1MC_1=180^{\circ}-\angle A=\angle B_1AC_1=\angle A$ and so, $\angle A=90^{\circ}$

Lol, IMO 3

QED

This post has been edited 2 times. Last edited by anantmudgal09, Apr 21, 2016, 8:56 PM
Reason: Typoes

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mathcool2009

352 posts

mathcool2009

#10 h Feb 2, 2017, 2:46 AM • 6 Y

Y by Kayak, e_plus_pi, Gems98, Ruy, son7, Adventure10

Terrible Solution (complex bash)

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Aiscrim

409 posts

Aiscrim

#12 h Jun 1, 2017, 2:07 PM • 3 Y

Y by son7, Adventure10, Mango247

Let $X,Y,Z$ be the intersection points of $(AB_1C_1),(BC_1A_1),(CA_1B_1)$ with $(ABC)$ . As $X$ is the center of the spiral similarity that takes $\overrightarrow{C_1B}$ to $\overrightarrow{B_1C}$ and $BC_1=CB_1$ , we get that $XB=TC$ and $XB_1=XC_1$ . Therefore, $X$ lies on the perpendicular bisector of $B_1C_1$ . The analogous relations are true for $Y$ and $Z$ .

As the circumcenter of $(A_1B_1C_1)$ is both on the perpendicular bisectors of $A_1B_1,B_1C_1,C_1A_1$ and on $(ABC)$ , it has to be one of the points $X,Y,Z$ . Suppose wlog that it is $X$ . In this case, $XY$ is the perpendicular bisector of $C_1A_1$ and $XZ$ is the perpendicular bisector of $A_1B_1$ so $\widehat{YAZ}=\widehat{YXZ}=\dfrac{\widehat{C_1XB_1}}{2}=\dfrac{\hat{A}}{2}$ . We are now done:
$\hat{A}=\widehat{BAC}=\widehat{YAB}+\widehat{YAZ}+\widehat{ZAC}=\left ( \hat{A}+\dfrac{\hat{B}}{2}-90^\circ \right ) +\dfrac{\hat{A}}{2}+\left ( \hat{A}+\dfrac{\hat{C}}{2}-90^\circ \right )$ which implies $\hat{A}=90^\circ$ .

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62861

3564 posts

62861

#13 h Jul 12, 2017, 10:10 PM • 5 Y

Y by HolyMath, KST2003, son7, Adventure10, Mango247

Solution

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H.HAFEZI2000

328 posts

H.HAFEZI2000

#14 h Jul 5, 2018, 2:20 PM • 2 Y

Y by son7, Adventure10

will use a mixture of angle chasing and complex numbers (which is my favorite

). first of all according to the problem $\triangle A_1B_1C_1$ has an angle which more than 90 and let $\angle A$ be the largest of all thereupon we can easily see that $B_1C_1$ would the largest side so $\angle B_1A_1C_1$ is the one which is more than 90. afterward we claim the mid point of arc $BAC$ (we call $N$ is the center of $\odot A_1B_1C_1$ . initially it's well-known that $N$ is the miquel point of $BCB_1C_1$ and $\overline{NB_1}=\overline{NC_1}$ , with the fact center of $\odot A_1B_1C_1$ is on the circumcircle of $\triangle ABC$ the cliam will be proved.
as mentioned before $N$ is the miquel point of $BCB_1C_1$ which implies $AB_1C_1N$ is concyclic and $\angle B_1NC_1=\angle A \implies$ $\angle B_1A_1C_1=180-\frac{A}{2}$ and this the beginning of complex. I just explain it briefly and would go into details. take $\odot ABC$ the unit circle.
we put $A=a^2, B=b^2$ and $C=c^2$ and then we can calculate $A_1=\frac {c^2+b^2+\sum ab-(\frac{(cb)(a+b+c)}{a})}{2}$
and then we have that $180- \angle AIB=\angle B_1A_1C_1 \implies \frac{\frac{a^2-I}{a^2-b^2}}{\frac{a_1-b_1}{a_1-c_1}}$ has to be real
$\frac{\frac{a^2-I}{a^2-b^2}}{\frac{a_1-b_1}{a_1-c_1}}=\frac{\frac{(a+b)(a+c)}{a^2-b^2}}{\frac{(b^2-a^2)(1-\frac{c(\sum a)}{ab}}{(c^2-a^2)(1+\frac{b(\sum a)}{ac}}}$ is real and I think the rest is straightforward to give us $b^2=c^2$

This post has been edited 4 times. Last edited by H.HAFEZI2000, Aug 31, 2018, 1:24 PM

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math_pi_rate

1218 posts

math_pi_rate

#15 h Jul 26, 2018, 11:38 AM • 2 Y

Y by Adventure10, Mango247

Restate the problem in terms of the excentral triangle as follows-

Restated problem wrote:

Let $O$ be the circumcenter and $\triangle DEF$ be the orthic triangle of $\triangle ABC$ . Let $AO \cap EF = A_1$ . Define $B_1$ and $C_1$ analogously. Suppose that the circumcenter of $\triangle A_1B_1C_1$ lie on $\odot DEF$ . Show that $\triangle DEF$ is right-angled.

My solution: Let $\triangle M_AM_BM_C$ be the medial triangle of $\triangle ABC$ . Note that $\measuredangle OM_AD = \measuredangle OB_1D = \measuredangle OC_1D = 90^{\circ} \Rightarrow M_A,D,B_1,C_1,O$ lie on a circle. All other such results hold cyclically.

Also, $M_A$ lies on $\odot (DEF)$ , as it is the nine point circle of $\triangle ABC$ . So $M_A$ is the center of the spiral similarity that takes $B_1F$ to $C_1E$ $\Rightarrow \triangle M_AC_1B_1 \sim \triangle M_AEF \Rightarrow$ As $M_AE = M_AF$ , we get that $M_AC_1 = M_AB_1$ . All other such results hold cyclically. As the circumcenter of $\triangle A_1B_1C_1$ lies on $\odot (M_AM_BM_C)$ , WLOG we can assume that $M_A$ is the circumcenter of $\triangle A_1B_1C_1$ .

Thus, $M_AM_B$ is the perpendicular bisector of $\overline{A_1C_1}$ . But $M_AM_B \perp CF$ $\Rightarrow A_1C_1 \parallel CF$ $\Rightarrow \measuredangle OCF = \measuredangle OC_1A_1 = \measuredangle OEA_1 = \measuredangle OEF$ $\Rightarrow O$ lies on $\odot (BCEF)$ . Hence, $\measuredangle EDF = \measuredangle C_1DB_1 = \measuredangle C_1OB_1 = \measuredangle COB = \measuredangle CEB = 90^{\circ}$ . $\blacksquare$

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v_Enhance

6872 posts

v_Enhance

#16 h Aug 30, 2018, 3:05 PM • 11 Y

Y by RAMUGAUSS, srijonrick, v4913, centslordm, suvamkonar, son7, hakN, PNT, Adventure10, Mango247, Funcshun840

We ignore for now the given condition and prove the following important lemma.

Lemma: Let $(AB_1C_1)$ meet $(ABC)$ again at $X$ . From $BC_1 = B_1C$ follows $XC_1 = XB_1$ , and $X$ is the midpoint of major arc $\widehat{BC}$ .

Proof. This follows from the fact that we have a spiral similarity $\triangle XBC_1 \sim \triangle XCB_1$ which must actually be a spiral congruence since $BC_1 = B_1C$ . $\blacksquare$

We define the arc midpoints $Y$ and $Z$ similarly, which lie on the perpendicular bisectors of $\overline{A_1 C_1}$ , $\overline{A_1 B_1}$ .

$[asy] pair A = dir(110); pair B = dir(180); pair C = dir(0); pair X = dir(90); pair Y = dir(235); pair Z = dir(325); pair A_1 = foot(Y+Z-X, B, C); pair B_1 = foot(Z+X-Y, C, A); pair C_1 = foot(X+Y-Z, A, B); draw(unitcircle, grey); draw(C_1--X--B_1, red); draw(A--B--C--cycle, grey); draw(A_1--B_1--C_1--cycle, heavycyan); draw(Y--X--Z, orange); draw(circumcircle(X, B_1, C_1), dotted+grey); draw(X--A_1, red+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(Z)); dot("$A_1$", A_1, dir(270)); dot("$B_1$", B_1, dir(B_1)); dot("$C_1$", C_1, dir(C_1)); /* TSQ Source: A = dir 110 B = dir 180 C = dir 0 X = dir 90 Y = dir 235 Z = dir 325 A_1 = foot Y+Z-X B C R270 B_1 = foot Z+X-Y C A C_1 = foot X+Y-Z A B unitcircle 0.1 grey / grey C_1--X--B_1 red A--B--C--cycle grey A_1--B_1--C_1--cycle 0.1 cyan / heavycyan Y--X--Z orange circumcircle X B_1 C_1 dotted grey X--A_1 red dashed */ [/asy]$

We now turn to the problem condition which asserts the circumcenter $W$ of $\triangle A_1B_1C_1$ lies on $(ABC)$ .

Claim: We may assume WLOG that $W = X$ .

Proof. This is just configuration issues since we already knew that the arc midpoints both lie on $(ABC)$ and the relevant perpendicular bisectors. Suppose (WLOG) that $\angle B_1 A_1 C_1 > 90^{\circ}$ (since $W$ lies $(ABC)$ and hence outside $\triangle ABC$ , hence outside $\triangle A_1 B_1 C_1$ ). Then $A$ and $X$ lie on the same side of line $\overline{B_1 C_1}$ , and since $W$ is supposed to lie both on $(ABC)$ and the perpendicular bisector of $\overline{B_1C_1}$ it follows $W = X$ . $\blacksquare$

Consequently, $\overline{XY}$ and $\overline{XZ}$ are exactly the perpendicular bisectors of $\overline{A_1 C_1}$ , $\overline{A_1 B_1}$ . The rest is angle chase, the fastest one is $\begin{align*} \angle A &= \angle C_1 X B_1 = \angle C_1 X A_1 + \angle A_1 X B_1 = 2 \angle YXA_1 + 2 \angle A_1 X Z \\ &= 2 \angle YXZ = 180^{\circ} - \angle A \end{align*}$ which solves the problem.

Remark: Angle chasing is also possible even without the points $Y$ and $Z$ , though it takes much longer. Introduce the Bevan point $V$ and use the fact that $VA_1B_1C$ is cyclic (with diameter $\overline{VC}$ ) and similarly $VA_1C_1B$ is cyclic; a calculation then gives $\angle CVB = 180^{\circ} - \frac{1}{2} \angle A$ . Thus $V$ lies on the circle with diameter $\overline{I_b I_c}$ .

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Blast_S1

356 posts

Blast_S1

#17 h Dec 9, 2018, 7:41 PM • 2 Y

Y by Adventure10, Mango247

WLOG let $\angle A$ be the largest angle in $\triangle ABC$ , and define $K_A$ as the midpoint of $\widehat{BAC}$ on $(ABC)$ . Furthermore, let $K_B$ and $M$ be the midpoints of $\widehat{ABC}$ and $\overline{A_1C_1}$ respectively.

Lemma: Suppose the circumcenter $K$ of $\triangle A_1B_1C_1$ lies on $(ABC)$ , then $K=K_A$ .

Proof: Already done with basically all the solutions above.
$[asy] size(9cm); defaultpen(fontsize(9pt)); pair A=(32.5,53.3), B=(0,0), C=(120,0), D=(80,0), E=(100.8,11.7), F=(11.7,19.2), K=(60,60), L=(28.8,-51.2), M=(45.9,9.6); dot(M); draw(A--B--C--cycle, linewidth(0.5)); draw(circle((60,0),60), linewidth(0.5)); draw(F--K--B, linewidth(0.5)); draw(E--K--C, linewidth(0.5)); draw(E--F, linewidth(0.4)+grey); draw(F--D, linewidth(0.4)+grey); draw(circumcircle(D,E,F), linewidth(0.8)+dashed); draw(F--L--A, linewidth(0.5)); draw(D--L--C, linewidth(0.5)); real xmin=-100, ymin=-110, xmax=200, ymax=80; clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); draw(K--L, linewidth(0.7)+dotted); dot("$A$", A, (0,1)); dot("$B$", B, (-1,0)); dot("$C$", C, (1,0)); dot("$A_1$", D, (0.5,-1)); dot("$B_1$", E, (0,-1)); dot("$C_1$", F, (-0.5,0)); dot("$K_A$", K, (0,1)); dot("$K_B$", L, (-0.5,-1)); [/asy]$
Using the above lemma, the condition we are trying to prove is equivalent to showing that $K_B$ , $M$ , and $K_A$ are collinear. We will proceed with barycentric coordinates wrt $\triangle ABC$ (and using the standard notation).

It is well known that $A_1=(0:s-b:s-c)$ and $C_1=(s-a:s-b:0)$ , which have coordinate sums of $c$ and $a$ respectively, so
$M=(a(s-a):(a+c)(s-b):c(s-c)).$ Now, since $K_A$ lies on the exterior bisector of angle $A$ , $K_A$ must be in the form of $(t:-b:c)$ for some value $t$ . Intersecting this with $(ABC)$ gives us
$a^2bc-b^2ct+c^2bt=0\implies t=\frac{a^2}{b-c},$ or $K_A=(a^2:b(c-b):c(b-c))$ (and $K_B=(a(c-a):b^2:c(a-c))$ by symmetry). Finally, the desired collinearity occurs iff
$0=ac\cdot\text{det}\left( \begin{array}{ccc} a & b(c-b) & b-c \\ c-a & b^2 & a-c \\ s-a & (a+c)(s-c) & s-c \end{array}\right),$ which can be shown to be equivalent to $b^2+c^2=a^2$ with a few minutes of omitted expansion and row/column operations.

This post has been edited 1 time. Last edited by Blast_S1, Dec 9, 2018, 8:02 PM

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Generic_Username

1088 posts

Generic_Username

#18 h May 29, 2019, 4:38 PM • 1 Y

Y by Adventure10

$[asy] import geometry; size(8cm,0); defaultpen(fontsize(9pt)); pair B = (-25,0), C = (25,0), N = (0,25), A = (-7,24); triangle t = triangle(A,B,C); drawline(t); circle c1 = excircle(t.BC); circle c2 = excircle(t.AC); circle c3 = excircle(t.AB); clipdraw(c1,bp+green); clipdraw(c2,bp+green); clipdraw(c3,bp+green); dot("$B$", B, (-1,0)); dot("$C$", C, (1,0)); dot("$N_A$", N, (0,1)); dot("$A$", A, (0,1)); draw(circumcircle(A,B,C)); draw(box((-60,-20), (60,50)), invisible); draw(extouch(t)); dot("$A_1$",extouch(t).A); dot("$B_1$",extouch(t).B); dot("$C_1$",extouch(t).C); point V = circumcenter(excenter(t.AB),excenter(t.BC),excenter(t.CA)); dot("$V$", V); point NN = (0,25); line l = perpendicular(NN, line(V,NN)); point VV = reflect(l)*V; dot("$V'$", VV); point BB = (-25,0); point CC = (25,0); point BBB = reflect(l)*BB; dot("$B'$", BBB); draw(circumcircle(BB,CC,V)); dot("$I_C$",excenter(t.AB)); dot("$I_B$",excenter(t.AC)); [/asy]$
Let $V$ be the circumcenter of $\triangle ABC$ and $N_A$ be the midpoint of major arc $BC$ in $\odot(ABC)$ . Then as $\triangle N_AC_1B\cong \triangle N_AB_1C$ by SAS, it follows that $N_AC_1=N_AB_1$ and so $N_A$ must be the circumcenter of $\triangle A_1B_1C_1$ (it lies on $\odot (ABC)$ and on the perpendicular bisector of $B_1C_1$ ).

Now since $A_1B_1C_1$ is the pedal triangle of $V$ WRT $\triangle ABC$ (isogonal lines in $I_AI_BI_C$ ,) we are motivated to reflect $V$ over $N_A$ to $V'$ . Then from homothety at $V$ , $V'$ is the circumcenter of the triangle formed by reflecting $V$ across the sidelines of $\triangle ABC$ , and so by a well known lemma, $V,V'$ are isogonal conjugates WRT $\triangle ABC$ .

So if $B'$ is the reflection of $B$ across $AN_A$ , $\angle V'BA=\angle VBC$ . But using the fact that $V',V$ are actually reflections across $AN_A$ ( $\angle AN_AV=90^{\circ}$ since $N_A$ is the midpoint of $I_BI_C$ ), we get
$\angle VBC=\angle V'BA=\angle VB'A=\angle VB'C$ so $B'BVC$ is cyclic. This means $V$ lies on the circle with diameter $I_BI_C$ , so $\angle C_1VB_1=\angle I_CVI_B=90^{\circ}$ . Now $AC_1VB_1$ has three right angles so it must be a rectangle, done.

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IAmTheHazard

5001 posts

IAmTheHazard

#75 h Dec 25, 2023, 4:36 PM

Y by

Solved with a small hint. Pure synthetic eludes me.

WLOG let $\angle A$ be the largest angle of $\triangle ABC$ and let $O$ be the center of $(A_1B_1C_1)$ .

Let $M$ be the midpoint of arc $BAC$ . Then $MB=MC$ , $BC_1=CB_1$ , and $\angle C_1BM=\angle B_1CM$ , so $\triangle MC_1B \cong \triangle MB_1C \implies MB_1=MC_1$ . Thus $M$ is the intersection between the perpendicular bisector of $\overline{B_1C_1}$ and $(ABC)$ that lies on the same side of $\overline{BC}$ as $A$ . $O$ is also an intersection between this perpendicular bisector and $(ABC)$ , and it's clear that $A$ and $O$ lie on the same side of $\overline{BC}$ , hence $O=M$ .

WLOG let $\angle B>\angle C$ $\angle OBA_1=90^\circ-\tfrac{1}{2}\angle A$ , so $\angle OBC_1=\tfrac{1}{2}\angle A+\angle B-90^\circ$ . Furthermore, $OB=\tfrac{a}{2\cos(90^\circ-\frac{1}{2}\angle A)}=\tfrac{a}{2\sin \frac{1}{2}\angle A}$ . Thus by the law of cosines, since $OC_1=OA_1$ we should have
$\begin{align*} \left(\frac{a}{2\sin \frac{1}{2}\angle A}\right)^2+(s-a)^2-2(s-a)\left(\frac{a}{2\sin \frac{1}{2}\angle A}\right)\cos\left(\frac{1}{2}\angle A+\angle B-90^\circ\right)&=\left(\frac{a}{2\sin \frac{1}{2}\angle A}\right)^2+(s-c)^2-2(s-c)\left(\frac{a}{2\sin \frac{1}{2}\angle A}\right)\cos\left(90^\circ-\frac{1}{2}\angle A\right)\\ (s-a)^2-a(s-a)\frac{\sin(\frac{1}{2}\angle A+\angle B)}{\sin \frac{1}{2}\angle A}&=(s-c)^2-a(s-c)=-(s-c)(s-b)\\ \frac{\sin(\frac{1}{2}\angle A+\angle B)}{\sin \frac{1}{2}\angle A}&=\frac{(s-a)^2+(s-b)(s-c)}{a(s-a)}. \end{align*}$ Let $D$ be the intersection of the (interior) $\angle BAC$ -bisector with $\overline{BC}$ . Then by law of sines $\tfrac{\sin(\frac{1}{2}\angle A+\angle B)}{\sin \frac{1}{2}\angle A}=\tfrac{AB}{BD}$ , and by angle bisector theorem this equals $\tfrac{AC}{CD}$ , hence it equals $\tfrac{b+c}{a}$ . Then
$\begin{align*} \frac{b+c}{a}&=\frac{(s-a)^2+(s-b)(s-c)}{a(s-a)}\\ 2(b+c)(b+c-a)&=(b+c-a)^2+(a-b+c)(a+b-c)\\ 2(b+c)^2-2a(b+c)&=(b+c)^2-2a(b+c)+a^2+a^2-(b-c)^2\\ (b+c)^2+(b-c)^2&=2a^2\\ b^2+c^2&=a^2, \end{align*}$ or $\angle BAC=90^\circ$ , as desired. $\blacksquare$

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joshualiu315

2513 posts

joshualiu315

#76 h Jan 10, 2024, 3:29 AM • 1 Y

Y by eg4334

WLOG let $\angle ABC$ be the largest angle in $\triangle ABC$ . Denote the midpoint of arc $ABC$ as $O_1$ .

It is clear that $O_1A = O_1C$ , $A_1C=AC_1$ , and $\angle A_1CO_1 = \angle C_1AO_1$ . This implies that $\triangle OAC_1 \cong \triangle O_1CA_1$ ; in particular, $O_1A_1=O_1C_1$ , making $O_1$ the desired circumcenter.

Define $O_2$ and $O_3$ similarly to $O_1$ except with arcs $BAC$ and $BCA$ instead. Note that $\triangle O_1B_1O_2 \cong \triangle O_1C_1O_2$ and $\triangle O_1A_1O_3 \cong \triangle O_1B_1O_3$ . Thus,

$\angle B = \angle A_1O_1C_1 = 2 \angle O_2O_1O_3.$
Convert this into arc lengths as follows:

Denote $m (XY)$ denote the measure of arc $XY$ .
The factor of $2$ caused by the inscribed angle theorem will be omitted.

$2 \angle O_2O_1O_3 = 2 m(O_1O_2) = 2( m(AC) - m(AO_2) - m(CO_3)).$
Realize that arc $AO_2$ has measure $\frac{\angle A}{2}$ because the angle bisector of $\angle BAC$ would intersect $(ABC)$ at the antipode of $O_2$ . Analogously, arc $CO_3$ has measure $\frac{\angle C}{2}$ . Plugging this in, we get

$2 \angle O_2O_1O_3 = 2 \left( \angle B - \frac{\angle A}{2} - \frac{\angle C}{2} \right) = \angle B$ $\implies \angle A + \angle C = \angle B.$
We are done. $\square$

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AngeloChu

470 posts

AngeloChu

#77 h Jan 19, 2024, 7:56 PM

Y by

assume $AB \geq BC \geq AB$ . let the excenters be $A_2, B_2,$ and $C_2$ respectively, and let the circumcenter of $A_1B_1C_1$ be $O$ , let $AA_2$ intersect $BC$ at $Y$ , let $AO$ intersect $CC_2$ at $Z$ ,
let $OD\perp B_1C_1$ , and $OE\perp B_1A_1$ . note that $AA_2,BB_2,$ and $CC_2$ intersect at $X$
cyclic quadrilaterals give that $B_2BC=B_2A_2X=A_1A_2B=C_1C_2B=AC_2X$ , $AB_2C=ABC_2=CBA_2$
we can then prove $OAB=OCB$ , $OBC=OAC=AC_2B-CC_2C_1-BAO$ , as well as $AYB=OZC=CA_2B$
$XCO=BB_2A_2$ , so $AOC=90$ and $ABC=90$

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OronSH

1728 posts

OronSH

#78 h Jan 31, 2024, 1:40 AM • 1 Y

Y by megarnie

solved with GrantStar (angle chasing too hard grantstar too orz)

Rename $A_1,B_1,C_1$ to $D,E,F.$ Notice that $BF=CE$ and symmetric. Now construct $A'$ to be the intersection of perpendicular bisectors of $EF,BC.$ We have $\triangle A'FB\cong\triangle A'EC$ directly congruent by SSS so $A'$ is the Miquel point of $BFEC,$ so it is on $(ABC).$ Thus it must be the midpoint of arc $BAC.$ Define $B',C'$ similarly. Let $A_1$ be the point on $(ABC)$ such that $A'A_1$ is the perpendicular bisector of $EF,$ and define $B_1,C_1$ similarly. If the circumcenter of $DEF$ lies on $(ABC)$ we must have that three of $A',B',C',A_1,B_1,C_1$ coincide. However $A',B',C'$ are distinct, and if $A_1,B_1,C_1$ coincide, it would imply that the circumcenter of $DEF$ lies inside $\triangle DEF,$ impossible. Thus we may assume that $A',B_1,C_1$ coincide. This implies that $A'B',A'C'$ are the perpendicular bisectors of $DF,DE$ respectively.

Let $A'D,A'E,A'F$ intersect $(ABC)$ again at $D',E',F'.$ Since $A'B'$ bisects $DF$ we have that $B'$ is the arc midpoint of $D'F',$ and since $AC,D'F'$ share an arc midpoint we have $AC\parallel D'F'.$ Similarly $AB\parallel D'E',$ so $\angle A=\angle E'D'F'=180^\circ-\angle EA'F.$ However from the spiral similarity we have $\angle EA'F=\angle BA'C=\angle A,$ so $\angle A=90^\circ$ as desired.

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shendrew7

793 posts

shendrew7

#79 h Feb 22, 2024, 5:53 AM

Y by

Suppose $P$ and $P'$ are the midpoints of arc $ACB$ and $AB$ , respectively. Note that $AB_1 = A_1B$ as well as $AP = BP$ and $AP' = BP'$ . SSA (in)congruency tells us at least one of
$\triangle PAB_1 \cong \triangle PBA_1, \quad \triangle P'AB_1 \cong \triangle P'BA_1$
must hold. We can suppose WLOG the first holds. Then our desired center is $P$ , so we can also consider the midpoints $M$ and $N$ of arcs $CAB$ and $ABC$ , respectively. Then $MP$ and $NP$ are the perpendicular bisectors of $B_1C_1$ and $C_1A_1$ , respectively, so
$\angle C = \angle APB = \angle A_1PB_1 = 2 \angle NPM = 180 - \angle C \implies \angle C = 90. \quad \blacksquare$

This post has been edited 1 time. Last edited by shendrew7, Feb 22, 2024, 5:53 AM

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Pyramix

419 posts

Pyramix

#80 h Apr 18, 2024, 3:32 AM

Y by

Similar ideas are used in GOTEEM 2020/5.

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Markas

105 posts

Markas

#81 h May 5, 2024, 12:50 PM

Y by

Let X be the arc midpoint of arc BAC. We have that $BC_1=CB_1$ from equal tangents, $XB=XC$ by X being the midpoint, also $\angle ACX =\angle ABX$ since ABCX is cyclic $\Rightarrow$ $\triangle XC_1B\cong\triangle XB_1C$ $\Rightarrow$ X is on the perpendicular bisector of $B_1C_1$ . Denote the arc midpoints of arc ABC and ACB, be Y and Z. Similarly they lie on the perpendicular bisectors of $A_1C_1,A_1B_1$ , respectively.
Assume WLOG that the circumcenter of $\triangle A_1B_1C_1$ is X (from the three arc midpoints). Now $\angle BAC = \angle BXC = \angle BXB_1 + \angle B_1XC = \angle BXB_1 + \angle BXC_1 = \angle C_1XB_1$ $\Rightarrow$ $\angle C_1AB_1 = \angle C_1XB_1$ $\Rightarrow$ $C_1AXB_1$ is cyclic. Now we have that $\angle YXZ = \angle YXA_1 + \angle A_1XZ = \frac{1}{2}\angle C_1XA_1 + \frac{1}{2}\angle A_1XB_1 = \frac{1}{2}\angle C_1XB_1 = \frac{1}{2}\angle C_1AB_1 = \frac{1}{2}\angle BAC$ $\Rightarrow$ $\angle YXZ = \frac{1}{2}\angle BAC$ . Also $\angle XYZ = 90 - \frac{1}{2}\angle BAC$ $\Rightarrow$

$\angle YXZ = \frac{1}{2}\angle BAC = 90 - \frac{1}{2}\angle BAC$ $\Rightarrow$ $\angle BAC = 90^{\circ}$ $\Rightarrow$

$\triangle ABC$ is right-angled as we wanted. We are ready.

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john0512

4178 posts

john0512

#82 h Jul 8, 2024, 6:17 AM • 2 Y

Y by centslordm, OronSH

wow, there was a mop class (sparrow v cannon) where we were building up to this problem, but it was kind of ruined as we were never able to get to it, possibly due to too much stuff happening during class
this is pretty nice though

Let $S$ denote the circumcenter of $\triangle A_0B_0C_0$ . The perpendicular bisector of $B_0C_0$ meets $(ABC)$ at two points. We claim that one of them is the midpoint of arc $BAC$ (this is the "sparrow lemma"). Note that $BC_0=CB_0$ . If $J_a$ is the midpoint of arc $BAC$ , then $J_aB=J_aC$ , and $\angle C_0BJ_a=\angle B_0CJ_a$ as well, so $\triangle J_aC_0B$ and $\triangle J_aB_0C$ are congruent and thus $J_aC_0=J_aB_0$ , showing the claim.

Call the second such point the $A$ -buh point (since it's annoying) and denote it by $X_a$ . We have that $X_a$ is second intersection of $(ABC)$ with the perpendicular to $B_0C_0$ through $J_a$ . By the given condition, $S$ is either $J_a$ or $X_a$ , and similarly either $J_b$ or $X_b$ and either $J_c$ or $X_c$ . Clearly, $X_a$ , $X_b$ , and $X_c$ do not coincide, so $S$ must be some arc midpoint, WLOG $J_a$ .

Importantly, this means that $J_a$ is also both $X_b$ and $X_c$ , which means that $J_aJ_b$ is perpendicular to $A_0C_0$ . Since $J_aJ_b$ is the midline in the excentral triangle, $J_aJ_b$ is parallel to $I_aI_b$ , which is in turn perpendicular to the internal $C$ -bisector. Therefore, the internal $C$ -bisector is parallel to $A_0C_0$ .

Let $CI$ meet $AB$ at $F$ . Then, $\frac{BA_0}{BC}=\frac{BC_0}{BF}$ $\frac{s-c}{a}=\frac{s-a}{\frac{ac}{a+b}}$ $(s-c)c=(s-a)(a+b)$ $(a+b-c)c=(-a+b+c)(a+b)$ $a^2=b^2+c^2$ and we are done.

Alternative angle chase (after getting that $A_0C_0$ and $A_0B_0$ are parallel to the internal bisectors): we know $AJ_aB_0C_0$ cyclic so $\angle C_0AB_0=\angle C_0J_aB_0=2(180-\angle C_0A_0B_0)=2(180-\angle BIC)=2(90-\angle A/2)$ so $\angle A=90$ .

This post has been edited 2 times. Last edited by john0512, Jul 8, 2024, 6:35 AM

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Ywgh1

139 posts

Ywgh1

#83 h Aug 6, 2024, 7:24 AM

Y by

Nice one, but quite easy for P3.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(11.651746085090494cm); 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 = -9.548320615797111, xmax = 11.103425469293384, ymin = -3.6265775477501117, ymax = 5.545945680581804; /* image dimensions */ pen qqffff = rgb(0.,1.,1.); pen zzttqq = rgb(0.6,0.2,0.); pen zzffzz = rgb(0.6,1.,0.6); pen ffccww = rgb(1.,0.8,0.4); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen afeeee = rgb(0.6862745098039216,0.9333333333333333,0.9333333333333333); pen ffefdv = rgb(1.,0.9372549019607843,0.8352941176470589); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); pen ffzztt = rgb(1.,0.6,0.2); /* draw figures */ draw(circle((0.39170126705322017,1.2593588720578868), 1.857541583711824), linewidth(1.) + qqffff); draw((-1.465734736439853,1.2791636226574639)--(-0.17979008689644715,3.0268028931175093), linewidth(1.) + zzttqq); draw((2.2491372705462935,1.2395541214583097)--(-0.17979008689644715,3.0268028931175093), linewidth(1.) + zzttqq); draw((-1.465734736439853,1.2791636226574639)--(2.2491372705462935,1.2395541214583097), linewidth(1.) + zzttqq); draw((-1.465734736439853,1.2791636226574639)--(3.00856072208783,3.5120521747993507), linewidth(1.) + zzffzz); draw((2.2491372705462935,1.2395541214583097)--(-2.1855486867822345,2.7215375763025706), linewidth(1.) + zzffzz); draw((-0.17979008689644715,3.0268028931175093)--(0.7671538157020328,-3.195131835870218), linewidth(1.) + zzffzz); draw((0.7671538157020328,-3.195131835870218)--(0.8146013590071536,1.2548497364354834), linewidth(1.) + ffccww); draw((0.8067633169011869,0.5197401711159279)--(3.00856072208783,3.5120521747993507), linewidth(1.) + ffccww); draw((0.8067633169011869,0.5197401711159279)--(-2.1855486867822345,2.7215375763025706), linewidth(1.) + ffccww); draw((0.7671538157020328,-3.195131835870218)--(3.00856072208783,3.5120521747993507), linewidth(1.) + qqwuqq); draw((-2.1855486867822345,2.7215375763025706)--(3.00856072208783,3.5120521747993507), linewidth(1.) + qqwuqq); draw((0.7671538157020328,-3.195131835870218)--(-2.1855486867822345,2.7215375763025706), linewidth(1.) + qqwuqq); draw(circle((0.8067633169011869,0.5197401711159279), 3.7150831674236473), linewidth(1.) + afeeee); draw((0.8146013590071536,1.2548497364354834)--(-1.030036728431871,1.871290934767146), linewidth(1.) + afeeee); draw((0.8067633169011869,0.5197401711159279)--(-1.465734736439853,1.2791636226574639), linewidth(1.) + afeeee); draw((0.8067633169011869,0.5197401711159279)--(2.2491372705462935,1.2395541214583097), linewidth(1.) + afeeee); draw((0.8146013590071536,1.2548497364354834)--(1.6570099584366114,1.6752521294662914), linewidth(1.) + afeeee); draw((0.8146013590071536,1.2548497364354834)--(0.4115060176527976,3.116794875550961), linewidth(1.) + ffefdv); draw((0.4115060176527976,3.116794875550961)--(-1.030036728431871,1.871290934767146), linewidth(1.) + ffefdv); draw((0.4115060176527976,3.116794875550961)--(1.6570099584366114,1.6752521294662914), linewidth(1.) + ffefdv); draw(circle((1.5279502937237401,0.8796471462871188), 0.806004768171585), linewidth(1.) + dotted + eqeqeq); draw(circle((-0.32948570976933295,0.899451896886696), 1.1980162124518678), linewidth(1.) + dotted + eqeqeq); draw(circle((0.3134866150023695,1.7732715321167192), 1.3470941977637165), linewidth(1.) + dotted + cqcqcq); draw(circle((0.4115060176527973,3.11679487555096), 2.6269605003572587), linewidth(1.) + dotted + ffzztt); draw((-1.030036728431871,1.871290934767146)--(1.6570099584366114,1.6752521294662914), linewidth(1.) + afeeee); /* dots and labels */ dot((-1.465734736439853,1.2791636226574639),dotstyle); label("$B$", (-1.4070278095853517,1.421023992101179), NE * labelscalefactor); dot((2.2491372705462935,1.2395541214583097),linewidth(4.pt) + dotstyle); label("$C$", (2.2972604172409987,1.3531798853827477), NE * labelscalefactor); dot((-0.17979008689644715,3.0268028931175093),dotstyle); label("$A$", (-0.13155860327884275,3.157833124093021), NE * labelscalefactor); dot((-2.1855486867822345,2.7215375763025706),linewidth(4.pt) + dotstyle); label("$Ic$", (-2.1261753408007236,2.8321814118445503), NE * labelscalefactor); dot((3.00856072208783,3.5120521747993507),linewidth(4.pt) + dotstyle); label("$Ib$", (3.0571144124874294,3.619173049778354), NE * labelscalefactor); dot((0.7671538157020328,-3.195131835870218),linewidth(4.pt) + dotstyle); label("$Ia$", (0.8182588907791958,-3.083824694002661), NE * labelscalefactor); dot((0.8067633169011869,0.5197401711159279),linewidth(4.pt) + dotstyle); label("$V$", (0.8589653548102546,0.6340323541673756), NE * labelscalefactor); dot((-1.030036728431871,1.871290934767146),linewidth(4.pt) + dotstyle); label("$C_{1}$", (-0.9728255265873912,1.9773456671923157), NE * labelscalefactor); dot((1.6570099584366114,1.6752521294662914),linewidth(4.pt) + dotstyle); label("$B_{1}$", (1.7138010994624893,1.787382168380708), NE * labelscalefactor); dot((0.8146013590071536,1.2548497364354834),linewidth(4.pt) + dotstyle); label("$A_{1}$", (0.8725341761539409,1.366748706726434), NE * labelscalefactor); dot((-0.023360782794746484,1.998977572999846),linewidth(4.pt) + dotstyle); label("$I$", (0.031267252845392446,2.1130338806291786), NE * labelscalefactor); dot((0.4115060176527976,3.116794875550961),linewidth(4.pt) + dotstyle); label("$M_{1}$", (0.46546953584335293,3.2256772308114523), NE * labelscalefactor); dot((-0.7091974355401003,-0.2367971297838247),linewidth(4.pt) + dotstyle); label("$M_{3}$", (-0.660742635682607,-0.12582164107905533), NE * labelscalefactor); dot((1.8878572688949316,0.15846016946456634),linewidth(4.pt) + dotstyle); label("$M_{2}$", (1.9444710623051558,0.26767417788784637), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
The problem can be divided into two claims:

Claim 1: Proving that $O$ the Center of $\triangle A_1B_1C_1$ is the midpoint of $I_cI_b$ .

Claim 2: Showing that $V$ lies on $(I_cBCI_b)$

The finishing is an easy angle chase.

This post has been edited 8 times. Last edited by Ywgh1, Aug 6, 2024, 8:49 AM

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Martin2001

132 posts

Martin2001

#84 h Aug 21, 2024, 9:49 PM

Y by

This problem was mostly just me relearning some $9$ point circle properties. We look at this problem from the excentral triangle, let them be $I_A, I_B, I_C.$ Let $M_A$ be the midpoint of $I_BI_C,$ define $M_B,M_C$ similarly. WLOG let $M_A$ be the center of $(A_1B_1C_1)$ by spiral. Then note that by using the previous spiral on $M_B,$ we have that $\overline{M_AM_B}$ is the perpendicular bisector of both of $I_CC, C_1A_1,$ so $I_CC\| C_1A_1.$ As $V$ lies on $(BC_1A_1)$ because it is the antipode of $B,$ by Reim we see that $(I_CBVCI_B)$ is cyclic. Therefore we get that $\angle I_CI_AI_B=45\deg,$ so we're done $.\blacksquare$

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Eka01

204 posts

Eka01

#85 h Aug 30, 2024, 4:53 PM • 1 Y

Y by Sammy27

sketch

Attachments:

This post has been edited 2 times. Last edited by Eka01, Aug 30, 2024, 4:56 PM

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TestX01

339 posts

TestX01

#86 h Sep 22, 2024, 10:51 AM

Y by

Edit: Line $2$ and $7$ are meant to say "perpendicular bisector of $B_1C_1$ ".

This post has been edited 1 time. Last edited by TestX01, Sep 22, 2024, 11:23 AM

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cursed_tangent1434

583 posts

cursed_tangent1434

#87 h Dec 21, 2024, 12:39 PM

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Posting a solution from old, for storage. Let $C_3$ be the point where the $C-$ excircle is tangent to $\overline{AC}$ . Consider the case when $O$ lies inside the angle $C_3ACB$ . We first note the following claim.

Claim : $O$ lies on the circle $(AB_1C_1)$ .

Proof : Let $O'=(AB_1C_1) \cap (ABC)$ . Then,
$\angle BC_1O' = 180 - \angle O'C_1A = 180 - \angle O'B_1A = \angle OB_1C$ and
$\angle O'BC_1 = \angle O'BA = \angle O'CA = \angle O'CB_1$ Let $B_2,C_2$ be the points where the incircle touches $AC$ and $AB$ respectively.

Now, we have the following key lemma. Let $ABC$ be a triangle. Suppose its incircle and $A$ -excircle are tangent to $BC$ at $X$ and $D$ respectively. Then, $BX = CD$ and $BD = CX$ .

Then,
$B_1C = AB_2 = AC_2 = BC_1$ Thus, clearly $\triangle O'C_1B \cong \triangle O'B_1C$ Thus, $O'C_1=O'B_1$ which means $O'$ lies on the perpendicular bisector of $B_1C_1$ and also lies on $(ABC)$ but then, this means that $O'=O$ .

Now, since $\triangle O'C_1B \cong \triangle O'B_1C$ we have
$\angle C_1OB_1 = \angle C_1OC + \angle COB_1 = \angle C_1OC + \angle BOC_1 = \angle BOC = \angle A$ We then also have,
$\angle OAC_3 = \angle OBC = \angle OCB = \angle OAB = \angle OAC_1$ Then, note that $AC_3=AC_1$ which gives us that $\triangle OC_3A \cong \triangle OAC_1$ . Thus,
$OC_3=OC_1=OB_1$ Thus, $C_3$ also lies on $(A_1B_1C_1)$ . Thus,
$\begin{align*} 360 - 2 \angle C_1A_1B_1 &= \angle BAC\\ 360 - 2(180 - \angle C_1C_3A) &= \angle A\\ 360 - 180 - \angle A &= \angle A\\ 2\angle A &= 180 \\ \angle A &= 90^\circ \end{align*}$ The other cases are entirely similar (with $\angle B$ and $\angle C$ being $90^\circ$ ). Thus, $\triangle ABC$ is right angled as claimed and we are done.

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Ilikeminecraft

343 posts

Ilikeminecraft

#88 h Mar 15, 2025, 10:25 PM • 1 Y

Y by PikaPika999

show that the circumcenter must be one of the arc midpoints by using miquel's theorem.
rest is angle chase, which is easy

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numbertheory97

42 posts

numbertheory97

#89 h Apr 6, 2025, 4:04 PM • 1 Y

Y by PikaPika999

$[asy] size(9cm); pair IA = dir(110); pair IB = dir(225); pair IC = dir(315); pair A = foot(IA, IB, IC); pair B = foot(IB, IC, IA); pair C = foot(IC, IA, IB); pair A1 = foot(IA, B, C); pair B1 = foot(IB, C, A); pair C1 = foot(IC, A, B); pair D = (IB+IC)/2; pair E = (IC+IA)/2; pair F = (IA+IB)/2; pair X = 2*foot(C, IB, IC) - C; pair Y = 2*foot(B, IB, IC) - B; draw(IA--IB--IC--cycle); draw(IA--A, paleblue); draw(IB--B, paleblue); draw(IC--C, paleblue); draw(B--C, heavymagenta); draw(B--X, heavymagenta); draw(C--Y, heavymagenta); draw(circumcircle(IB, B, C), heavyblue); draw(circumcircle(A, B, C), dotted); draw(circumcircle(A1, B1, C1), dotted); draw(IA--A1, purple); draw(IB--B1, purple); draw(IC--C1, purple); dot("$I_A$", IA, dir(110)); dot("$I_B$", IB, dir(180)); dot("$I_C$", IC, dir(0)); dot("$A$", A, dir(270)); dot("$B$", B, dir(60)); dot("$C$", C, dir(150)); dot("$A_1$", A1, dir(150)); dot("$B_1$", B1, dir(45)); dot("$C_1$", C1, dir(270)); dot("$D$", D, dir(270)); dot("$E$", E, dir(30)); dot("$F$", F, dir(150)); dot("$X$", X, dir(210)); dot("$Y$", Y, dir(300)); [/asy]$

Very beautiful configuration!

We reframe the problem as follows: let $I_A, I_B, I_C$ be the excenters opposite $A, B, C$ , respectively, and let $D, E, F$ denote the midpoints of sides $I_BI_C$ , $I_CI_A$ , $I_AI_B$ . Then $A, B, C$ are the feet of the perpendiculars from $I_A, I_B, I_C$ to the sides of $\triangle I_AI_BI_C$ , and $A_1, B_1, C_1$ are the feet of the perpendiculars from $I_A, I_B, I_C$ to the sides of $\triangle ABC$ .

Note that $\triangle A_1B_1C_1$ is obtuse since the circumcenter of $\triangle A_1B_1C_1$ lies outside the triangle; WLOG $\angle B_1A_1C_1$ is the obtuse angle. We claim that the circumcenter coincides with $D$ .

Claim: $DB_1 = DC_1$ .

Proof. Denote $a = BC$ , $b = CA$ , $c = AB$ , $s = (a + b + c)/2$ . We show $B_1$ and $C_1$ have the same power with respect to $\omega = (I_BI_CBC)$ , which will prove the claim since $D$ is the center of $\omega$ .

Let $X, Y$ denote the reflections of $C, B$ over $\overline{I_BI_C}$ . Since $\angle BAI_C = \angle CAI_B$ , we have $B, A, X$ collinear and $C, A, Y$ collinear. This implies $\text{Pow}_\omega(C_1) = BC_1 \cdot C_1X = BC_1(C_1A + AC) = (s - a)(s - b + b) = s(s - a),$ and we also have $\text{Pow}_\omega(B_1) = s(s - a)$ since this quantity is symmetric in $b$ and $c$ . $\square$

Notice that $A_1$ and $D$ lie on opposite sides of $\overline{B_1C_1}$ , so $D$ is indeed the circumcenter. We now compute the power of $A_1$ : $\text{Pow}_\omega(A_1) = BA_1 \cdot A_1C = (s - b)(s - c),$ which must equal $s(s - a)$ from before because $\omega$ and $(A_1B_1C_1)$ are concentric. Expanding gives
$\begin{align*} &s(s - a) = (s - b)(s - c) \\ &\implies as = bs + cs - bc \\ &\implies (b + c + a)(b + c - a) = 2bc \\ &\implies (b + c)^2 - a^2 = 2bc \\ &\implies a^2 = b^2 + c^2, \end{align*}$ so $\angle BAC = 90^\circ$ as desired. $\square$

This post has been edited 3 times. Last edited by numbertheory97, Apr 6, 2025, 4:07 PM

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