incircle with center I of triangle ABC touches the side BC

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incircle with center I of triangle ABC touches the side BC

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Source: Vietnam TST 2003 for the 44th IMO, problem 2

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orl

3647 posts

orl

#1 h Jun 26, 2005, 12:16 PM • 5 Y

Y by HWenslawski, Adventure10, jhu08, Miku_, ehuseyinyigit

Given a triangle $ABC$ . Let $O$ be the circumcenter of this triangle $ABC$ . Let $H$ , $K$ , $L$ be the feet of the altitudes of triangle $ABC$ from the vertices $A$ , $B$ , $C$ , respectively. Denote by $A_{0}$ , $B_{0}$ , $C_{0}$ the midpoints of these altitudes $AH$ , $BK$ , $CL$ , respectively. The incircle of triangle $ABC$ has center $I$ and touches the sides $BC$ , $CA$ , $AB$ at the points $D$ , $E$ , $F$ , respectively. Prove that the four lines $A_{0}D$ , $B_{0}E$ , $C_{0}F$ and $OI$ are concurrent. (When the point $O$ concides with $I$ , we consider the line $OI$ as an arbitrary line passing through $O$ .)

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mecrazywong

606 posts

mecrazywong

#2 h Jun 26, 2005, 1:28 PM • 2 Y

Y by Adventure10, jhu08

$A_0D\cap(I)$ is the tangency point of the circle passing through B,C tangent to the incircle.
Then we consider the Apollonian circles with the ratio s-a/s-b,etc. It's easy to prove they intersect at exactly 2 points.
The concurrency of $A_0D,B_0E,C_0F$ can be proved by radical axis.
The remaining is to show that I,O have the same power wrt the 3 Apollonian circles. For I it's easy; for O it's simple computation.

I know I have over-compressed the solution...

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darij grinberg

6555 posts

darij grinberg

#3 h Jun 27, 2005, 3:27 PM • 6 Y

Y by tobash_co, Adventure10, jhu08, Miku_, Mango247, and 1 other user

You can find an alternative solution in Hyacinthos message #9551:

[url=http://groups.yahoo.com/group/Hyacinthos/message/9551?expand=1]Hyacinthos message #9551[/url] (CORRECTED) wrote:

From: Darij Grinberg
Subject: Re: midpoints of AH, BK, CL

Dear Orlando,

In Hyacinthos message #9539, you wrote:

orl wrote:

Given a triangle $ABC$ inscribed in a circle with center $O$ . Let $H$ , $K$ , $L$ be the feet of the altitudes of triangle $ABC$ from the vertices $A$ , $B$ , $C$ , respectively. Denote by $A_{0}$ , $B_{0}$ , $C_{0}$ the midpoints of these altitudes $AH$ , $BK$ , $CL$ , respectively. The incircle of triangle $ABC$ has center $I$ and touches the sides $BC$ , $CA$ , $AB$ at the points $D$ , $E$ , $F$ , respectively. Prove that the four lines $A_{0}D$ , $B_{0}E$ , $C_{0}F$ and $OI$ are concurrent. (When the point $O$ concides with $I$ , we consider the line $OI$ as an arbitrary line passing through $O$ .)

The first three lines $A_{0}D$ , $B_{0}E$ , $C_{0}F$ pass through the excenters $I_{a}$ , $I_{b}$ , $I_{c}$ of triangle ABC (proof: see below). Hence, these three lines coincide with the lines $DI_{a}$ , $EI_{b}$ , $FI_{c}$ , respectively. But triangles DEF and $I_{a}I_{b}I_{c}$ are homothetic (since their corresponding sidelines are parallel, what can be easily seen); hence, the lines $DI_{a}$ , $EI_{b}$ , $FI_{c}$ concur at the homothetic center T of the two triangles DEF and $I_{a}I_{b}I_{c}$ .

Now, O is the nine-point center of triangle $I_{a}I_{b}I_{c}$ , and I is the orthocenter of triangle $I_{a}I_{b}I_{c}$ . Hence, the circumcenter of triangle $I_{a}I_{b}I_{c}$ is the reflection of the orthocenter in the nine-point center, i. e. the reflection of I in O.

Obviously, the circumcenter of triangle DEF is I.

Since the circumcenters of two homothetic triangles lie on one line with the homothetic center, it follows that the reflection of I in O and the point I lie on one line with the homothetic center T of triangles DEF and $I_{a}I_{b}I_{c}$ . In other words, the point T lies on the line OI.

Hence, altogether, the point T lies on the lines $DI_{a}$ , $EI_{b}$ , $FI_{c}$ (i. e., on the lines $A_{0}D$ , $B_{0}E$ , $C_{0}F$ ), and on the line OI. The proof is complete.

Two notes:

(1) In the above, I have left out the proof that the lines $A_{0}D$ , $B_{0}E$ , $C_{0}F$ pass through the points $I_{a}$ , $I_{b}$ , $I_{c}$ .
Here is this proof:

The "semiprojection" of a point on a line will mean the midpoint between the point and its orthogonal projection on the line. Of course, if three points are collinear, their semiprojections on any line are collinear, too (in fact, point $\mapsto$ its semiprojection on a given line is an affine mapping).

If D' is the point of tangency of the A-excircle of triangle ABC with the side BC, and D" is the point diametrically opposite to D' on the A-excircle, then the tangent to the A-excircle at D" is parallel to BC; hence, if this tangent meets AB and AC at B" and C", respectively, the triangles ABC and AB"C" are homothetic. The homothetic center of these two triangles is A, of course.

Now, the incircle of triangle ABC touches BC at D; the incircle of triangle AB"C" is the A-excircle of triangle ABC and touches B"C" at D". Hence, the points D and D" are collinear with the homothetic center A of the two triangles ABC and AB"C".

Now, since the points D, D" and A are collinear, the points D, $I_{a}$ , $A_{0}$ - being their semiprojections on the line BC - are collinear, too, i. e. the line $A_{0}D$ passes through $I_{a}$ . Similarly, the lines $B_{0}E$ and $C_{0}F$ pass through $I_{b}$ and $I_{c}$ , respectively, qed..

(2) The point T is the triangle center X(57) in Kimberling's ETC; in fact, T is the isogonal conjugate of the Mitten point X(9).

Sincerely,
Darij Grinberg

This post has been edited 4 times. Last edited by darij grinberg, Jul 29, 2007, 3:13 PM

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Remike

126 posts

Remike

#4 h Sep 11, 2005, 11:51 AM • 3 Y

Y by Adventure10, jhu08, Mango247

darij grinberg wrote:

But triangles A0B0C0 and IaIbIc are homothetic (since their corresponding sidelines are parallel, what can be easily seen);

What????

This is obviously an incorrect statement (a counterexample can be constructed very easyly). May be you meant that triangles $DEF$ and $I_AI_BI_C$ are homothetic...

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darij grinberg

6555 posts

darij grinberg

#5 h Sep 11, 2005, 12:18 PM • 3 Y

Y by Adventure10, jhu08, Mango247

Remike wrote:

What????

This is obviously an incorrect statement (a counterexample can be constructed very easyly). May be you meant that triangles $DEF$ and $I_AI_BI_C$ are homothetic...

Okay, I'm sorry, I confused some notations (mainly DEF and A0B0C0). Now it should be correct.

Darij

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Virgil Nicula

7054 posts

Virgil Nicula

#6 h Sep 16, 2005, 4:02 PM • 4 Y

Y by Adventure10, jhu08, Mango247, ehuseyinyigit

Remark. The exincircles with the centers $I_a,\ I_b,\ I_c$ of the $\triangle ABC$ touch the sides $BC,\ CA,\ AB$ in the points $M,\ N,\ P$ respectively. Then $I\in MA_0\cap NB_0\cap PC_0.$

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plane geometry

467 posts

plane geometry

#7 h Mar 19, 2009, 8:51 AM • 2 Y

Y by jhu08, Adventure10

Virgil Nicula wrote:

Remark. The exincircles with the centers $I_a,\ I_b,\ I_c$ of the $\triangle ABC$ touch the sides $BC,\ CA,\ AB$ in the points $M,\ N,\ P$ respectively. Then $I\in MA_0\cap NB_0\cap PC_0.$

dear Virgil Nicula,do you have a proof for this nice problem?

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Luis González

4151 posts

Luis González

#8 h Mar 19, 2009, 2:40 PM • 4 Y

Y by jhu08, Adventure10, Mango247, ehuseyinyigit

Virgil Nicula wrote:

Remark. The exincircles with the centers $I_a,\ I_b,\ I_c$ of the $\triangle ABC$ touch the sides $BC,\ CA,\ AB$ in the points $M,\ N,\ P$ respectively. Then $I\in MA_0\cap NB_0\cap PC_0.$

Basically, note that $\triangle AA_0I \sim \triangle I_aMI \Longrightarrow \frac{AA_0}{r_a}=\frac{IA}{II_a}$

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mathVNpro

469 posts

mathVNpro

#10 h Apr 21, 2009, 4:57 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Let me restatement this problem so that it can fits my solution

:
"let $ABC$ be the acute triangle with incircle $(I)$ . Denote $D$ , $E$ , $F$ the tagent points of $(I)$ with $BC$ , $CA$ , $AB$ , respectively. Let $A_1$ , $B_1$ , $C_1$ be the projections of $A$ , $B$ , $C$ onto $BC$ , $CA$ , $AB$ , respectively. Suppose $A_2$ , $B_2$ , $C_2$ are the midpoints of $AA_1$ , $BB_1$ , $CC_1$ , respectively.
Prove that: $DA_2$ , $EB_2$ , $FC_2$ , $OI$ are concurent, where $O$ is the circumcenter of triangle $ABC$ ."

Proof:
Let $X$ be the intersection of $EF$ with $BC$ . It is well- known that $(X,D,B,C) = - 1$ $(1)$ . Denote $X'$ the intersection of $AD$ with $EF$ . From $(1)$ , we also get that $(X,X',E,F) = - 1$ . Therefore $X$ is the pole of $AD$ wrt $(I)$ . Hence, $IX$ is perpendicular to $AD$ . It is easy to notice that triangle $IXD$ is similar to triangle $DAA_1$ . Now, let me call $A_3$ is the midpoint of $XD$ , from the notice above, we get, $IA_3$ is also perpendicular to $DA_2$ . Therefore, $A_3$ is the pole of $DA_2$ wrt $(I)$ . Define the same for $B_3$ , $C_3$ .
So now, in order to prove $DA_2$ , $EB_2$ , $FC_2$ are concurent, we need to prove that $A_3$ , $B_3$ , $C_3$ are collinear.
Indeed, because, $(X,D,B,C) = - 1$ , $A_3$ is the midpoint of $XD$ , we get, $A_3D^2 = \overline {A_3B}.\overline {A_3C}$ , which also means that the power of $A_3$ wrt $(I)$ and $(ABC)$ are equal. With the same argument for $B_3$ , $C_3$ . Hence, $(A_3B_3C_3)$ is the radical axis wrt $(I)$ and $(ABC)$ , further, $A_3$ , $B_3$ , $C_3$ are collinear. As the result, $DA_2$ , $EB_2$ , $FC_2$ are concurent.
Let $P$ be the intersection of these lines, we get that, $P$ is the pole of $(A_3B_3C_3)$ wrt $(I)$ . Therefore, $IP$ is perpendicular to $(A_3B_3C_3)$ . $OI$ is also perpendicular to $(A_3B_3C_3)$ (Because $(A_3B_3C_3)$ is the radical axis wrt $(I)$ and $(ABC)$ ), where $O$ is the circumcenter of triangle $ABC$ . Hence, $O,I,P$ are collinear. Which leads to the result that $DA_2$ , $EB_2$ , $FC_2$ , $OI$ are concurent.
Our proof is completed.

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v_Enhance

6882 posts

v_Enhance

#11 h Jul 30, 2012, 9:50 PM • 7 Y

Y by Durjoy1729, HamstPan38825, jhu08, Adventure10, Mango247, and 2 other users

Here is a solution combining projective geometry and homogeneous/barycentric coordinates. The calculations fit on two pages when I performed them by hand.

$[asy]import graph; size(12cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-10.98,xmax=12.96,ymin=-6.14,ymax=7.52; pen zzttqq=rgb(0.6,0.2,0), qqttff=rgb(0,0.2,1), zzzzzz=rgb(0.6,0.6,0.6), zzzzqq=rgb(0.6,0.6,0), qqzzzz=rgb(0,0.6,0.6), aqgreen=rgb(0,0.4,0.4); pair A=(-1.88,-3), B=(9.12,-3), C=(0.12,6), L=(0.12,-3), O=(3.62,0.5), I=(1.86,0), L_C=(-8.78,-3), C_0=(0.12,1.5), F=(1.86,-3), P=(0.81,-0.29), P_C=(1.11,-3), K_C=(2.74,-3); D(A--B--C--cycle,zzttqq); D(A--B,zzttqq); D(B--C,zzttqq); D(C--A,zzttqq); D(CR(I,3),linetype("4 4")+qqttff); D(C--L,zzzzzz); D(C_0--F,zzzzqq); D(O--L_C,linewidth(1.2)+green); D(C--P_C,qqzzzz); D(L_C--A); D(C--K_C,qqzzzz); D(F--I,linetype("0 3 4 3")+zzzzzz); D(CR(O,6.52),linewidth(1.3)+aqgreen); D(A); MP("A",(-1.81,-2.89),SW*lsf); D(B); MP("B",(9.2,-2.89),SE*lsf); D(C); MP("C",(0.19,6.12),NW*lsf); D(L); MP("L",(0.07,-3.52),NE*lsf); D(O); MP("O",(3.69,0.62),NE*lsf); D(I); MP("I",(1.95,0.13),NE*lsf); D(L_C); MP("L_C",(-8.98,-2.79),NE*lsf); D(C_0); MP("C_0",(0.19,1.62),NE*lsf); D(F); MP("F",(1.91,-3.48),NE*lsf); D(P); MP("P",(0.89,-0.17),NE*lsf); D(P_C); MP("P_C",(0.99,-3.42),NE*lsf); D(K_C); MP("K_C",(2.73,-3.44),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

Let $P = C_0F \cap IO$ , $L_C = IO \cap AB$ and $K_C = AI \cap AB$ . Define $P_C = AP \cap AB$ . Let $C_\infty$ be a point at infinity on line $AL$ . Our goal is to compute the barycentric coordinates of $P_C$ with respect to $\triangle ABC$ .

We know that $K_C = (a:b:0)$ and $F = (s-b: s-a: 0)$ . The coordinates of $L_C$ can be computed and simplified using the identities $aS_A - cS_C = s(c-a)(s-b)$ , $I=(a:b:c)$ and $O=(a^2S_A: b^2S_B: c^2S_C)$ as $\begin{align*} L_C &= \left( \det \left[ \begin{array}{cc} a^2S_A & c^2S_C \\ a & c \end{array}\right] : \det \left[ \begin{array}{cc} b^2S_B & c^2S_C \\ b & c \end{array} \right] : 0\right) \\ &= \left( a(aS_A - cS_C) : b(bS_B - cS_C) : 0 \right) \\ &= \left( a(a-c)(s-b) : b(b-c)(s-a) : 0 \right) \end{align*}$
Now, note that $(C,L; C_{\infty}, C_0) = -1$ is a harmonic bundle. Taking a perspective at $F$ with onto line $IO$ , we see that $(CF \cap IO, L_C; I, P) = -1$ is a harmonic bundle. Taking perspective at $C$ onto line $AB$ , we see that $(F, L_C; K_C, P_C) = -1$ . There's a lemma that states that if we have $K_C = F+L_C$ when adding componentwise, then $P_C = -F + L_C$ . So, we want to find a real $r$ such that $\frac{r(s-b) + a(a-c)(s-b)}{r(s-a) + b(b-c)(s-a)} = \frac{a}{b}$ because this will give us the coordinates of $P_C$ for free. Solving for $r$ (this is the hardest part), $\begin{align*} r( b(s-b) - a(s-a) ) &= ab( (b-c)(s-a) - (a-c)(s-b) ) \\ \implies r(a-b)(s-c) &= \frac{1}{2} ab( (b-c)(b+c-a) - (a-c)(a+c-b) ) \\ &= \frac{1}{2} ab( (b^2-c^2)-(a^2-c^2) + b(a-c) - a(b-c) ) \\ &= ab(b-a)(b+c-a) \\ \implies r &= -ab \end{align*}$
So what of $P_C$ ? We get $\frac{-r(s-b) + a(a-c)(s-b)}{-r(s-a) + b(b-c)(s-a) } = \frac{ab(s-b) + a(a-c)(s-b)}{ab(s-a) + b(b-c)(s-a) } = \frac{a(s-b)}{b(s-a)}.$ Hence, $P_C = (a(s-b) : b(s-a) : 0)$ .

If we define $P_A$ and $P_B$ analogously, we see that they concur at the point $P = (a(s-b)(s-c) : b(s-c)(s-a) : c(s-a)(s-b) ).$ It remains to show that this point actually lies on $IO$ , which is surprisingly easy, as we have $\det \left[ \begin{array}{ccc} a^2S_A & b^2S_B & c^2S_C \\ a & b & c \\ a(s-b)(s-c) & b(s-c)(s-a) & c(s-a)(s-b) \end{array} \right]$ which simplifies as $\sum_{\text{cyc}} a(s-b)(s-c) \left( bc(bS_B - cS_C) \right) = abc(s-a)(s-b)(s-c) \sum_{\text{cyc}} (b-c) = 0.$ This implies the points are collinear, so we're done.

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Virgil Nicula

7054 posts

Virgil Nicula

#12 h Jul 31, 2012, 2:58 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Quote:

Remark. The exincircles with the centers $I_a,\ I_b,\ I_c$ of the $\triangle ABC$ touch the sides $BC,\ CA,\ AB$ in the points $M,\ N,\ P$ respectively. Then $I\in MA_0\cap NB_0\cap PC_0.$

$I\in MA_0\iff$ $\frac {ID}{A_0H}=\frac{MD}{MH}\iff$ $\frac{r}{\frac {h_a}{2}}=\frac {|b-c|}{\frac {s|b-c|}{a}}\iff$ $\frac {2r}{h_a}=\frac as\iff$ $2sr=ah_a(=2S)$ O.K.

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thecmd999

2860 posts

thecmd999

#13 h Jan 24, 2014, 12:17 AM • 5 Y

Y by jhu08, guptaamitu1, Adventure10, Mango247, ehuseyinyigit

Solution

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jayme

9805 posts

jayme

#14 h Sep 18, 2015, 11:27 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Dear Mathlinkers,
also a link

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=497757

Sincerely
Jean-Louis

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Stranger8

238 posts

Stranger8

#15 h May 21, 2016, 1:07 PM • 3 Y

Y by jhu08, Adventure10, Mango247

I use Three line Coordinates maybe this way is similar to Evan chen's solution,but still....
I use $A$ $B$ $C$ to represent the $cosA$ $cosB$ $cosC$
because $A_0$ is the midpoint of the altitudes $AH$ ,then we can get $A_0$ $(1,C,B)$ ,and we can get $D$ $(0,1+C,1+B)$
then line $A_0D$ is $(C-B,-1-B,1+C)$ similarily we can get line $B_0E$ is $(1+A,A-C,-1-C)$ and line $C_0F$ is $(-1-C,1+B,B-A)$
notice that incenter $I$ is $(1,1,1)$ and circumcentre $O$ is $(A,B,C)$ then line $OI$ is $(C-B,A-C,B-A)$ the rest part is to calculate the Det which is easy only takes 5minutes

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navi_09220114

487 posts

navi_09220114

#16 h May 21, 2016, 2:12 PM • 4 Y

Y by k12byda5h, jhu08, Adventure10, Mango247

It is well known (and proveable using cross ratio) that $A_0D$ passes through the A-excenter, call it $I_a$ . Now we want to prove $DI_a, EI_b, F_c$ concur at $OI$ , where $I_a, I_b, I_c$ are the excenters. But just note that $I_bI_c\parallel EF$ and similarly to other two sides, so $\triangle I_aI_bI_c$ and $\triangle DEF$ are homothetic. So the three lines concur at the center of homothety, say $P$ . $P$ also lie on the line through circumcenter of the two triangles, which is line $IX_{40}$ , which coincide with $OI$ , because $O$ is midpoint of $IX_{40}$ . So $P, O, I$ colinear and we are done.

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Delray

348 posts

Delray

#18 h Apr 20, 2017, 3:30 PM • 3 Y

Y by jhu08, Adventure10, Mango247

Midpoints are nasty to work with, so we delete them from the problem by instead considering the lines $AI_A$ , $BI_B$ , and $CI_C$ , as the midpoint of the altitude from A, D and $I_A$ are collinear. Observing that the sides of $\triangle{DEF}$ are parallel to those of $\triangle{I_AI_BI_C}$ , we see that there is a homothety mapping to $I_A$ to $D$ , and so on for the other two vertices. This already implies that $A_0D$ , $B_0E$ and $C_0F$ are concurrent. Let this point of concurrency be $T$ .

Denote the circumcenters of $\triangle{DEF}$ and $\triangle{I_AI_BI_C}$ as $I$ and $O_{I}$ respectively. Let the incenter of $\triangle{I_AI_BI_C}$ , which is also the circumcenter of $\triangle{ABC}$ , be $O$ . $O_I$ , otherwise known as the Bevan Point of $\triangle{ABC}$ , is on line connecting containing the incenter and circumcenter of $\triangle{ABC}$ , implying that $O_{I}$ , $O$ and $I$ are collinear. Our homothety however, maps the circumcenter of $\triangle{I_AI_BI_C}$ to the circumcenter of $\triangle{DEF}$ , implying that $T$ , $O_I$ , and $I$ are collinear. It follows that $T$ , $I$ and $O$ are
collinear, implying that $IO$ also passes through $T$ as desired. $\square$

This post has been edited 1 time. Last edited by Delray, Apr 20, 2017, 3:31 PM

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trumpeter

3332 posts

trumpeter

#19 h May 21, 2017, 1:13 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Let $I_A$ be the $A$ -excenter, $I_B$ be the $B$ -excenter, and $I_C$ be the $C$ -excenter of $\triangle{ABC}$ . Let $\omega$ be the incircle of $\triangle{ABC}$ , $\Gamma$ be the circumcircle of $\triangle{ABC}$ , and $\Omega$ be the circumcircle of $\triangle{I_AI_BI_C}$ (centered at $P$ ). Let $Q$ be the exsimilicenter of $\Gamma$ and $\Omega$ . Note that $Q$ lies on $OP$ .

If $M_A$ is the midpoint of arc $BC$ of $\Gamma$ which does not contain $A$ , then the incenter-excenter lemma gives that $M_A$ is the midpoint of $II_A$ . Similar statements hold for $B$ and $C$ , so $\Gamma$ maps to $\Omega$ under a homothety centered at $I$ with ratio $2$ . In particular $P$ lies on $OI$ , so $Q$ lies on $OI$ . Furthermore, the tangent to $\Omega$ at $I_A$ is parallel to the tangent to $\Gamma$ at $M_A$ , which is parallel to $BC$ .

Now, let $A_1$ be the foot of $A$ onto $BC$ , $D_1$ be the tangency point of the $A$ -excircle of $\triangle{ABC}$ onto $BC$ , and $D_2$ be the antipode of $D_1$ with respect to the $A$ -excircle of $\triangle{ABC}$ . Then the homothety centered at $A$ sending $\omega$ to the $A$ -excircle of $\triangle{ABC}$ sends $D$ to $D_2$ , so $A$ , $D$ and $D_2$ are collinear. Now, consider the spiral similarity sending $AA_1$ to $D_2D_1$ . Because $D\in AD_2$ and $D\in A_1D_1$ , $D$ is the center of this. But this maps $A_0$ to $I_A$ , so $A_0$ , $D$ , and $I_A$ are collinear. Thus, we can instead consider $DI_A$ .

Now, I claim that $Q$ lies on $DI_A$ . Consider the homothety centered at $P$ sending $\omega$ to $\Omega$ . It sends $D$ to $I_A$ because these are the ``bottom-most'' points (their tangents are parallel to $BC$ ). Thus, $Q$ lies on $DI_A$ .

But then by symmetry, $Q$ lies on $EI_B$ and $FI_C$ also, so $A_0D$ , $B_0E$ , $C_0F$ concur at $Q$ on $OI$ .

This post has been edited 2 times. Last edited by Luis González, Apr 21, 2019, 3:25 AM
Reason: Unhiding solution

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Kagebaka

3001 posts

Kagebaka

#20 h Apr 19, 2019, 8:28 PM • 8 Y

Y by SD2014, AlastorMoody, lilavati_2005, Limerent, jhu08, Hermione.Potter, Adventure10, Mango247

I didn't see this solution while skimming so here's a cevian nest solution:

Solution

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khanhnx

1619 posts

khanhnx

#21 h Jun 10, 2019, 10:54 AM • 2 Y

Y by jhu08, Adventure10

Here is my solution for this problem
Solution
Let $X$ , $Y$ , $Z$ be second intersections of $DA_0$ , $EB_0$ , $FC_0$ with $(I)$ ; $D'$ is reflection of $D$ through $I$ ; $S$ $\equiv$ $AD$ $\cap$ $(I)$ $(S \not \equiv D)$
Since: $D(DSD'X) = D(AHD'A_0) = - 1$ then: $DXSD'$ is harmonic quadrilateral
But: $DFD'E$ is harmonic quadrilateral so: $EF$ , $XD'$ , $BC$ and tangent at $S$ of $(I)$ concurrent
Hence: $X(BCDD') = - 1$ or $XD$ is bisector of $\angle{BXC}$
Then: $(BCX)$ tangents $(I)$ at $X$
Similarly: $(CAY)$ tangents $(I)$ at $Y$ , $(ABZ)$ tangents $(I)$ at $Z$
Let $A_1$ , $A'$ be midpoint of $\stackrel\frown{BC}$ which not containing $A$ of $(O)$ , midpoint of $\stackrel\frown{BC}$ which not containing $X$ of $(BCX)$ ; similarly, we have: $B_1$ , $B'$ , $C_1$ , $C'$
We have: $IA'^2 - IC'^2 = \overline{A'D} . \overline{A'X} - \overline{C'F} . \overline{C'Z} = BA'^2 - BC'^2$
Then: $C'A'$ $\perp$ $IB$
But: $C_1A_1$ $\perp$ $IB$ so: $C_1A_1$ $\parallel$ $C'A'$
Similarly: $B_1C_1$ $\parallel$ $B'C'$ , $A_1B_1$ $\parallel$ $A'B'$
Hence: $O$ is circumcenter of $\triangle A'B'C'$
Let $J$ be external homothetic center of $(I)$ and $(A'B'C')$ then: $J$ ; $D$ ; $A'$ are collinear
But: $D$ , $X$ , $A'$ are collinear so: $D$ , $J$ , $X$ , $A'$ are collinear or $D$ , $J$ , $X$ are collinear
Similarly: $E$ , $J$ , $Y$ are are collinear; $F$ , $J$ , $Z$ are collinear
Hence: $DX$ , $EY$ , $FZ$ concurrent at a point $J$ lies on $IO$ or $DA_0$ , $EB_0$ , $FC_0$ concurrent at a point $J$ lies on $IO$

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AlastorMoody

2125 posts

AlastorMoody

#22 h Jul 12, 2019, 12:23 PM • 3 Y

Y by jhu08, Adventure10, Mango247

Lemma #2, Result #3, Result #4 (here)

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a1267ab

223 posts

a1267ab

#23 h Jul 27, 2019, 1:36 PM • 5 Y

Y by Kagebaka, AlastorMoody, jhu08, Adventure10, Mango247

Anyone want to try this generalization?

Let $ABC$ be a triangle with circumcenter $O$ . Let $D, E, F$ be the feet of the altitudes from $A, B, C$ respectively. Denote by $A_{0}$ , $B_{0}$ , $C_{0}$ the midpoints of $AD, BE, CF$ . Let $P$ be a point on the Darboux cubic of $ABC$ , and $X, Y, Z$ be the feet from $P$ to $BC, CA, AB$ . Prove that $A_0X, B_0Y, C_0Z, OP$ concur.

( $P$ being on the Darboux cubic just means that $AX, BY, CZ$ concur.)

This post has been edited 1 time. Last edited by a1267ab, Jul 27, 2019, 3:46 PM

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Dr_Vex

562 posts

Dr_Vex

#24 h Jun 7, 2020, 9:36 AM • 1 Y

Y by jhu08

See that ${ A_{0} –D –I _ {A}}$
We know that isotomic conjugate of Gergonne point is Nagel point, Due to which $\Delta ABC$ forms a cevian nest WRT its excentral triangle.
$I_{A}D,I_{B}E, I_{C}F$ concur
Now homothety centred at $X$ taking
$\Delta DEF \leftrightarrow \Delta I_{A} I_B I {C}$
finishes the question $\blacksquare$

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srijonrick

168 posts

srijonrick

#25 h Jun 21, 2020, 9:22 PM • 5 Y

Y by amar_04, Aritra12, abhradeep12, jhu08, kamatadu

orl wrote:

To prove that the four lines $A_{0}D$ , $B_{0}E$ , $C_{0}F$ and $OI$ are concurrent. First we will prove that the lines $A_{0}D$ , $B_{0}E$ , $C_{0}F$ are concurrent.

Recall that by Midpoint of Altitudes Lemma, the set of points $\{A_0, D, I_A \}$ , $\{B_0, E, I_B \}$ and $\{C_0, F, I_C \}$ are collinear $-(\bigstar)$ . Hence it suffices to show that the lines $\overline{I_AD}, \overline{I_BE}$ and $\overline{I_CF}$ are concurrent. Note that in $\triangle I_AI_BI_C$ , the lines $AI_A, BI_B$ and $CI_C$ are concurrent at $I$ since the points $\{A, I, I_A \}$ , $\{B, I, I_B \}$ and $\{C, I, I_C \}$ are collinear (by Fact5). Next note that the lines $AD, BE$ and $CF$ are concurrent at the "Gergonne Point". So, by the Cevian Nest (Theorem) we have the lines $\overline{I_AD}, \overline{I_BE}$ and $\overline{I_CF}$ concurrent, at some point. Let us denote that concurrency point by $X$ .

So, $(X=\overline{I_AD} \cap \overline{I_BE} \cap \overline{I_CF})$ . Next we will prove that the line $OI$ passes through $X$ as well.

Next recall that by Duality of Orthocenters and Excenters Lemma, $\triangle ABC$ is the orthic triangle of $\triangle I_AI_BI_C$ and $I$ is the orthcenter of $\triangle I_AI_BI_C$ . Also from $(\bigstar)$ we can conclude that there exists a homothety $\psi$ that maps $\triangle DEF$ to $\triangle I_AI_BI_C$ . And hence the point $I$ , which is the circumcenter of $\triangle DEF$ gets mapped to the circumcenter of $\triangle I_AI_BI_C$ (let us denote it by $O'$ ), and hence $X, I, O'$ are collinear $- (\star)$ . Next note that in $\triangle I_AI_BI_C$ , the points: $I$ (the orthocenter), $O'$ (the circumcenter) and the Nine-Point center must be collinear, as all of these lie on the Euler Line $- (\star\star)$ . Hence from $(\star)$ and $(\star\star)$ we can conclude that $X, I, O'$ and the Nine-Point center of $\triangle I_AI_BI_C$ are collinear. $-(\bigstar\bigstar)$

Now, note that in $\triangle I_AI_BI_C$ , $AI_A, BI_B$ and $CI_C$ are the altitudes, that is $A, B$ and $C$ are the feet of the altitudes, and the circumcircle of $\triangle ABC$ passes through these $3$ points only, hence $\odot(ABC)$ is the Nine-Point Circle of $\triangle I_AI_BI_C$ and thus the center of $\odot(ABC)$ , $O$ is the Nine-Point center of $\triangle I_AI_BI_C$ . Hence from $(\bigstar\bigstar)$ we can say that $X, I, O'$ and $O$ are collinear, and thus $OI$ passes through $X$ as well. $\quad \blacksquare$

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Afo

1002 posts

Afo

#27 h Aug 21, 2020, 4:45 AM • 2 Y

Y by jhu08, Mango247

The proof that $A_0D,B_0E,C_0F$ are concurrent is identical to the solutions above (That is, there is a homethethy between $DEF$ and $I_AI_BI_C$ ). Now, we "include" proving $OI$ . Let $P$ be the circumcircle of $I_AI_BI_C$ . It's enough to show that $I,O,P$ are collinear. Note that $I$ , $O$ , $P$ , is the orthocenter, nine-point center, and circumcenter respectively so they lie on the Euler's line of $\triangle I_AI_BI_C$ .
$[asy] import graph; size(31.668149578472285cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.4012066955558575,xmax=28.266942882916428,ymin=-7.3203210542584065,ymax=14.233001568944603; pen xdxdff=rgb(0.49019607843137253,0.49019607843137253,1.), qqccqq=rgb(0.,0.8,0.), ffttcc=rgb(1.,0.2,0.8), ffqqtt=rgb(1.,0.,0.2); pair A=(9.011309345848499,7.215059903683832), B=(7.060803027120685,-0.9462221037893297), C=(14.294050373775246,-1.2914678077732946), I=(9.97820131745357,1.2452595365401127), O=(10.848355810206165,2.462293101426641), F=(7.713890745265604,1.7864186429857236), D=(9.867207656966674,-1.0801728862885847), I_A=(11.232649004007717,-6.499969895476381), I_B=(20.375805631921757,9.05569770470921), I_C=(1.8067677366971093,6.0481844618646425), P=(11.71851037050289,3.6793262492811234); draw(A--B,linewidth(2.)); draw(B--C,linewidth(2.)); draw(C--A,linewidth(2.)); draw(circle((10.848355810206167,2.4622931014266416),5.095442276078402),linewidth(2.)+xdxdff); draw(circle(I,2.3280797979904406),linewidth(2.)+qqccqq); draw(D--(11.955936837608213,2.4734768455411915),linewidth(2.)); draw((11.955936837608213,2.4734768455411915)--F,linewidth(2.)); draw(F--D,linewidth(2.)); draw((xmin,1.3986407873809124*xmin-12.710659810748405)--(xmax,1.3986407873809124*xmax-12.710659810748405),linewidth(2.)+linetype("4 4")+ffttcc); draw(circle(P,10.190884714789368),linewidth(2.)+ffqqtt); draw(I_C--I_B,linewidth(2.)); draw(I_B--I_A,linewidth(2.)); draw(I_A--I_C,linewidth(2.)); dot(A,ds); label("$A$",(9.11120887133264,7.464808717394179),NE*lsf); dot(B,ds); label("$B$",(6.488846327373972,-1.2763997624679915),NE*lsf); dot(C,ds); label("$C$",(14.630657654331356,-1.3762992879521305),NE*lsf); dot(I,linewidth(4.pt)+ds); label("$I$",(10.085229244803001,1.445862306974799),NE*lsf); dot(O,linewidth(4.pt)+ds); label("$O$",(10.5847268722237,2.6696314941555026),NE*lsf); dot((11.955936837608213,2.4734768455411915),linewidth(4.pt)+ds); label("$E$",(12.05824487311476,2.6696314941555026),NE*lsf); dot(F,linewidth(4.pt)+ds); label("$F$",(7.23809276850502,1.420887425603764),NE*lsf); dot(D,linewidth(4.pt)+ds); label("$D$",(9.885430193834722,-1.5760983389204086),NE*lsf); dot(I_A,linewidth(4.pt)+ds); label("$I_A$",(11.45884772020992,-6.9456978336928845),NE*lsf); dot(I_B,linewidth(4.pt)+ds); label("$I_B$",(20.474779895153528,9.263000176108683),NE*lsf); dot(I_C,linewidth(4.pt)+ds); label("$I_C$",(1.3440207649407778,6.191089767471405),NE*lsf); dot(P,linewidth(4.pt)+ds); label("$P$",(11.583722127065096,3.9183755627072414),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

This post has been edited 2 times. Last edited by Afo, Aug 21, 2020, 5:51 AM

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meql

20 posts

meql

#28 h Aug 21, 2020, 11:10 PM • 4 Y

Y by amar_04, a1267ab, jhu08, Mango247

a1267ab wrote:

Let $P'$ be the reflection of $P$ over $O$ .

Claim: There exist points $A',B',C'$ on $PX,PY,PZ$ such that $A\in B'C', B\in C'A'$ , and $C\in A'B'$ .

Proof: Let $A'B'C'$ be the antipedal triangle of $P'$ . Since $P$ and $P'$ are reflections over $O$ , $P$ is the isogonal conjugate of $P'$ with respect to $A'B'C'$ . Thus, $A'P\perp BC$ , so $A'$ lies on $PX$ . Similarly, $B',C'$ lie on $PY,PZ$ respectively. $\square$

By cevian nest, $AA',BB',CC'$ concur at some point $Q$ . Projecting $(AD;A_0P_{\infty})$ through $X$ onto $AA'$ , we see that $Q$ lies on $A_0X$ . Similarly, $Q$ lies on $B_0Y$ and $C_0Z$ . But by Sondat's theorem, $P, Q$ , and $P'$ are collinear, so $Q$ lies on $OP$ . Thus, $A_0Z, B_0Y, C_0Z, OP$ concur ar $Q$ . $\square$

This post has been edited 1 time. Last edited by meql, Aug 21, 2020, 11:11 PM
Reason: added space for readability

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ike.chen

1162 posts

ike.chen

#30 h Jun 11, 2021, 7:37 PM • 2 Y

Y by Siddharth03, jhu08

Let $I_a$ , $I_b$ , $I_c$ be the $A$ -excenter, $B$ -excenter, and $C$ -excenter of $ABC$ respectively, and define $O_1$ as the circumcenter of $I_aI_bI_c$ . In addition, we label the feet of the altitudes from $A, B, C$ as $X, Y, Z$ respectively. The Midpoint of Altitudes Lemma implies $I_a \in A_0D$ , $I_b \in B_0E$ , and $I_c \in C_0F$ .

Claim: $DEF$ and $I_aI_bI_c$ are homothetic.

Proof. Properties of internal and external bisectors yield $CI \perp \overline{I_aCI_b}$ . Now, since $CD = CE$ and $\angle ECI = \angle ICD$ , we can conclude $IC \perp DE$ , so $\overline{I_aCI_b} \parallel DE$ . Analogous processes give $I_bI_c \parallel EF$ and $I_cI_a \parallel FD$ , as required. $\square$

Thus, $\overline{A_0DI_a}$ , $\overline{B_0EI_b}$ , and $\overline{C_0FI_c}$ concur at a homothetic center of $DEF$ and $I_aI_bI_c$ , which we denote with $J$ .

Claim: $OI$ is the Euler Line of $I_aI_bI_c$ .

Proof. It's well-known that $ABC$ is the orthic triangle of $I_aI_bI_c$ . Hence, $I$ is the orthocenter of $I_aI_bI_c$ .

Now, we also know $(ABC)$ is the Nine-Point Circle of $I_aI_bI_c$ , so $O$ is the Nine-Point Center of $I_aI_bI_c$ . The desired result follows readily. $\square$

Hence, we have $O_1 \in OI$ .

Claim: $J \in IO_1$ .

Proof. Because $I$ is the circumcenter of $DEF$ and $O_1$ is the circumcenter of $I_aI_bI_c$ , we know $IO_1$ passes through any of the homothetic centers between the two triangles. $\square$

Now, we can conclude $J \in \overline{OIO_1}$ , which finishes. $\blacksquare$

An Alternate Start: You can also use the Cevian Nest Lemma on $I_aI_bI_c$ and $ABC$ to show that $I_aD, I_bE, I_cF$ concur. In fact, this was how I initially proved the concurrency.

This post has been edited 1 time. Last edited by ike.chen, Aug 25, 2022, 5:21 AM
Reason: Overhaul and Reflection

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srisainandan6

2811 posts

srisainandan6

#31 h Jun 25, 2021, 3:44 PM • 1 Y

Y by jhu08

Denote $I_A,I_B,I_C$ as the excenters of $A,B,C$ respectively.

By Lemma 4.14 (EGMO), we have that $A_0,B_0,C_0$ lie on $I_AD,I_BE,I_CF$ respectively. Hence if suffices to show that $I_AD,I_BE,I_CF$ $,$ $OI$ are concurrent.

Claim: There exists a homotethy sending $\triangle I_AI_BI_C$ to $\triangle DEF$ .

Proof: It suffices to show $I_AI_B \parallel DE$ , and the rest of the corresponding sides follow with the same logic. We will show $\angle I_BCB = \angle EDB$ . Since $CD=CE$ , we have $\angle EDC = 90 - \frac{\angle C}{2}$ , so $\angle EDB = 90 + \frac{\angle C}{2}$ . It is easy to see that $\angle I_BCB = 90 + \frac{\angle C}{2}$ , so we are done.

All is left now is to show that $OI$ passes through $X$ . Observe that $I$ is the circumcenter of $\triangle DEF$ . Label $O'$ as the circumcenter of $\triangle I_AI_BI_C$ . If the center of homotethy is $X$ , then $O'I$ passes through $X$ .

It is well known that $O'$ lies on $OI$ , the Euler line, so $OI$ also passes through $X$ . $\blacksquare$

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hakN

429 posts

hakN

#32 h Jul 30, 2021, 11:52 PM • 1 Y

Y by jhu08

Let $I_A , I_B , I_C$ be the excenters corresponding to vertices $A,B,C$ respectively.
It is well known that $A_0 , D , I_A$ , $B_0 , E , I_B$ , $C_0 , F , I_C$ are collinear.
Also note that $\triangle DEF$ and $\triangle I_AI_BI_C$ are homothetic, thus $DI_A , EI_B , FI_C$ are concurrent at the homothety center say $P$ .
Let $K$ be the center of $(I_AI_BI_C)$ . Since $O$ is the 9-point center of $(I_AI_BI_C)$ and $I$ is the orthocenter of $\triangle I_AI_BI_C$ , we have $I,O,K$ are collinear. But, since $P$ is the center of homothety, we know $P,I,K$ are collinear. Thus, $P$ lies on $OI$ , as desired.

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IAmTheHazard

5005 posts

IAmTheHazard

#33 h Aug 3, 2021, 2:06 PM • 1 Y

Y by jhu08

Let $I_A,I_B,I_C$ denote the $A,B,C$ -excenters respectively, and rename $H,K,L$ as $X,Y,Z$ respectively.. It is well-known that

$A_0,D,I_A$ collinear,
$B_0,E,I_B$ collinear,
$C_0,F,I_C$ collinear,

so it suffices to show that $\overline{I_AD},\overline{I_BE},\overline{I_CF},\overline{OI}$ concur.
Since $\overline{AI_A},\overline{BI_B},\overline{CI_C}$ are concurrent at the orthocenter of $\triangle I_AI_BI_C$ and $\overline{AD},\overline{BE},\overline{CF}$ are concurrent at the Gergonne point of $\triangle ABC$ , by Cevian nest $\overline{I_AD},\overline{I_BE},\overline{I_CF}$ concur, say at point $P$ .
I now claim that $\overline{I_BI_C}$ and $\overline{EF}$ are parallel. It is well-known that $\overline{AI_A} \perp \overline{I_BI_C}$ and $A,I_A,I$ are collinear, so it suffices to show that $\overline{AI} \perp \overline{EF}$ . But this follows as $IE=IF$ and $AE=AF$ , so $AEIF$ is a kite.
Since $\overline{I_CI_A} \parallel \overline{FD}$ and $\overline{I_AI_B} \parallel \overline{DE}$ similarly, with the previous concurrency we find that $\triangle I_AI_BI_C$ and $\triangle DEF$ are homothetic. Now consider the homothety sending $\triangle DEF$ to $\triangle I_AI_BI_C$ . This sends the circumcenter of $\triangle DEF$ (which is $I$ ) to the circumcenter of $\triangle I_AI_BI_C$ . Since $\triangle ABC$ is the orthic triangle of $\triangle I_AI_BI_C$ , we find that $I$ (the orthocenter of $\triangle I_AI_BI_C$ ) is collinear with the circumcenter of $\triangle I_AI_BI_C$ and the circumcenter $O$ of $\triangle DEF$ , which is the 9-point center of $\triangle I_AI_BI_C$ , hence $P,O,I$ are collinear as desired. $\blacksquare$

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Mogmog8

1080 posts

Mogmog8

#34 h Sep 14, 2021, 12:24 AM • 2 Y

Y by centslordm, jhu08

Solution

This post has been edited 2 times. Last edited by Mogmog8, Sep 14, 2021, 3:59 AM

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AwesomeYRY

579 posts

AwesomeYRY

#35 h Mar 18, 2022, 2:19 AM

Y by

Claim: $\triangle I_A I_BI_C\sim \triangle DEF$ .
Proof: Note that the sides are parallel, for example $EF\parallel I_BI_C$ since both are perpendicular to $AI$ . Additionally,
$\angle EDF = 180 - \angle FDB - \angle EDC = 180 - (90 - \frac12 B) - (90-\frac12 C) = 180 - (90 + \frac12 A) = 180-\angle BIC = \angle BI_A C$ $\square$ .

Now, consider the homothety centered at $X$ that sends $\triangle DEF$ to $\triangle I_A I_B I_C$ . It clearly sends $D\to I_A, E\to I_B, F\to I_C$ , and $I$ goes to $O'$ , the circumcenter of $\triangle A'B'C'$ . Note that $I, O$ are the orthocenter and nine point center of $\triangle I_A I_B I_C$ . Thus, the circumcenter of $\triangle I_AI_BI_C$ lies on $IO$ , so $X,I,O_{(ABC)}$ are also collinear and we're done. $\blacksquare$ .

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pinkpig

3766 posts

pinkpig

#36 h Jul 13, 2022, 7:38 PM • 1 Y

Y by hungrypig

Partial Solution

Can I have a hint on how to prove that the point of concurrency lies on $OI?$ Thanks in advance!

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franzliszt

23531 posts

franzliszt

#37 h Jul 14, 2022, 2:21 AM • 1 Y

Y by Mango247

The results of this entire problem are quite well-known nowadays. Let $\triangle I_aI_bI_c$ be the excentral triangle of $\triangle ABC$ . Note that $A_0,D,I_a$ (and $B_0,E,I_b$ and $C_0,F,I_c$ ) are colinear by the well-known Midpoints of Altitudes Lemma (Lemma 4.14 in EGMO). Hence, it suffices to show that $DI_a,EI_b,FI_c$ concur on $\overline{OI}$ . However, these three lines meet at $X_{57}$ , the Isogonal Mittenpunkt point, which is known to lie on $\overline{OI}$ . Done. $\blacksquare$ .

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franzliszt

23531 posts

franzliszt

#38 h Jul 14, 2022, 3:09 PM

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But suppose that we didn't know what the Isogonal Mittenpunkt point was...

Set $\triangle ABC$ as the reference triangle with $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ . We also have $D=(0:s-c:s-b)$ and $I_a=(-a:b:c)$ . Foot of $A$ -altitude has coordinates $H=\left(0:a^2+b^2-c^2:a^2-b^2+c^2\right)$ . We can normalize by multiplying each component by $\frac{1}{2a^2}$ . Applying the midpoint formula on $\overline{AH}$ gives $A_0=\left(2a^2:a^2+b^2-c^2:a^2-b^2+c^2\right)$ .

We claim that $A_0,D,I_a$ are colinear. To prove this, we just need to verify $\begin{vmatrix}-a&b&c\\ 0&s-c&s-b\\ 2a^2&a^2+b^2-c^2&a^2-b^2+c^2\end{vmatrix}=0$ which is true upon expansion.

We can do similar things to show that $B_0,E,I_b$ and $C_0,F,I_c$ are colinear (just permute variables). Thus, it suffices to show that $DI_a,EI_b,FI_c$ concur on the line $OI$ .

The equation of line $DI_a$ is given by $\begin{vmatrix}x&y&z\\ 0&s-c&s-b\\ -a&b&c\end{vmatrix}=0 \iff (c-b)(-a+b+c)x+a(a-b+c)y-a(a+b-c)z=0.$ We can find the equtaions of lines $EI_b,FI_c$ by permuting variables.

To see that $DI_a,EI_b,FI_c$ concur, we need to check that $\begin{vmatrix}(c-b)(-a+b+c)&a(a-b+c)&-a(a+b-c)\\ -b(-a+b+c)&(a-c)(a-b+c)&b(a+b-c)\\ c(-a+b+c)&-c(a-b+c)&(b-a)(a+b-c)\end{vmatrix}=0$ however this is quite clear since $\begin{vmatrix}(c-b)(-a+b+c)&a(a-b+c)&-a(a+b-c)\\ -b(-a+b+c)&(a-c)(a-b+c)&b(a+b-c)\\ c(-a+b+c)&-c(a-b+c)&(b-a)(a+b-c)\end{vmatrix}=(a+b-c)(a-b+c)(-a+b+c)\begin{vmatrix}(c-b)&a&-a\\ -b&(a-c)&b\\ c&-c&(b-a)\end{vmatrix}=0.$
So the lines concur. Their concurrency point is the non-trivial solution to $\begin{bmatrix}(c-b)(-a+b+c)&a(a-b+c)&-a(a+b-c)\\ -b(-a+b+c)&(a-c)(a-b+c)&b(a+b-c)\\ c(-a+b+c)&-c(a-b+c)&(b-a)(a+b-c)\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$ which we can solve with Cramer's Rule to be $\left(\frac{a}{-a+b+c} ,\frac{b}{a-b+c},\frac{c}{a+b-c}\right).$
Finally, it suffices to check that this point lies on $OI$ . Since $I=(a:b:c)$ and $O=(a^2(b^2+c^2-a^2):b^2(c^2+a^2-b^2):c^2(a^2+b^2-c^2))$ , we just need to verify that $\begin{vmatrix}a&b&c\\ a^2(b^2+c^2-a^2)&b^2(c^2+a^2-b^2)&c^2(a^2+b^2-c^2)\\ \frac{a}{-a+b+c}&\frac{b}{a-b+c}&\frac{c}{a+b-c}\end{vmatrix}=0$ which (after factoring out $abc$ ) is true upon expansion.

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th1nq3r

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th1nq3r

#39 h Jul 20, 2022, 10:36 PM

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Solution

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YaoAOPS

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YaoAOPS

#40 h Jun 12, 2023, 8:54 PM

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Note that $A_0$ , $D$ , $I_A$ are collinear and so on. As such, it remains to show that the lines $I_AD$ , $I_BE$ , $I_CF$ are collinear. Since $\triangle I_AI_BI_C \sim \triangle DEF$ , that follows and the lines intersect at the homothety center.
Then, $I$ and $O$ are the orthocenter and nine-point center of $I_AI_BI_C$ respectively. As such, the circumcenter of $I_AI_BI_C$ lies on $IO$ . Since $I$ is the circumcenter of $DEF$ , it follows that $IO$ also is concurrent.

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eibc

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eibc

#41 h Feb 22, 2024, 7:54 PM

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Let $\triangle I_AI_BI_C$ be the excentral triangle. By the midpoints of altitudes lemma, we have $A_0, D$ , and $I_A$ collinear, and analogous results for the others. Therefore it now suffices to show that $DI_A, EI_B, FI_C, OI$ are collinear. But $\overline{DE} \parallel \overline{I_AI_B}$ as they're both perpendicular to $\overline{IC}$ , and similar for the other sides, so $DI_A, EI_B$ , and $FI_C$ concur at the center of homothety taking $\triangle DEF$ to $\triangle I_AI_BI_C$ . But it's well-known that the Euler line of both triangles is $OI$ , and as such the result follows.

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cursed_tangent1434

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cursed_tangent1434

#42 h Mar 10, 2024, 2:36 AM

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Solved with kingu. Very fun problem.

It is well known (Midpoint of the Altitudes) that $A_0D$ passes through $I_A$ and similarly for the other sides. Then, we can eliminate the points $A_0$ , $B_0$ and $C_0$ from the picture. It remains to show that $I_AD$ , $I_BE$ and $I_CF$ are concurrent on $IO$ . For this, note that since $\triangle DEF$ and $\triangle I_AI_BI_C$ are homothetic, it follows that $I_AD$ , $I_BE$ and $I_CF$ are concurrent, say at point $R$ . Then, note that since $\triangle M_AM_BM_C \sim \triangle DEF$ are homothetic, and the circumcenters of these two triangles are $O$ and $I$ respectively, their orthocenters must also be concurrent at this concurrency point. Clearly, $I$ is the orthocenter of $\triangle M_AM_BM_C$ ( $AI \perp M_BM_C$ etc. as mentioned before) and thus, the orthocenter of $\triangle DEF$ must also lie on $IO$ . Then, since $\triangle DEF$ and $\triangle I_AI_BI_C$ are homothetic, the line joining their orthocenters must be concurrent at $X$ as well. It is well known that the orthocenter of $\triangle I_AI_BI_C$ is $I$ . We previously showed that the orthocenter of $\triangle DEF$ lies on $IO$ from which it follows that $IO$ passes through $X$ , as the problem requires.

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blueberryfaygo_55

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blueberryfaygo_55

#43 h May 1, 2024, 2:52 AM • 1 Y

Y by megarnie

Lemma. In $\Delta ABC$ , let $D$ be the contact point of incircle $(I)$ with $BC$ , $I_a$ be the $A$ -excenter, and $M$ be the midpoint of altitude $AH$ $(H \in BC)$ . Then, $M, D, I_a$ are collinear.

Proof. Let $E$ be the point diametrically across $D$ in $(I)$ , $Y$ be the point diametrically across $X$ in $(I_a)$ , and $B', C'$ be the intersection of the line parallel to $BC$ through $E$ with $AB, AC$ , respectively. Then, $AB'C'$ and $ABC$ are homothetic triangles, and the homothety centered at $A$ that sends $\Delta AB'C'$ to $\Delta ABC$ sends $(I)$ , the $A$ -excircle of $AB'C'$ , to $I_a$ . This homothety thus sends $E$ to $X$ and $D$ to $Y$ , so $A,D,Y$ are collinear. Now, consider the homothety centered at $D$ that sends $H$ (the foot of the altitude from $A$ to $BC$ ) to $X$ ; since $AH \parallel XY$ , this homothety sends $A$ to $Y$ , which implies that the homothety sends $AH$ to $XY$ and the midpoint $M$ of $AH$ to the midpoint $I_a$ of $XY$ . Therefore, $M, D, I_a$ are collinear. $\blacksquare$

Letting $I_a, I_b, I_c$ be the $A$ -excenter, $B$ -excenter, and $C$ -excenter in $\Delta ABC$ and applying the lemma, $A_{0}D$ , $B_{0}E$ , and $C_{0}F$ are equivalent to $I_aD$ , $I_bE$ , and $I_cF$ , respectively. By the Duality of Orthocenters and Excenters, we now introduce a new, but equivalent configuration where $\Delta I_aI_bI_c$ becomes the reference triangle $\Delta ABC$ :

In triangle $ABC$ , let $AD, BE, CF$ be altitudes from $A,B,C$ to their opposite sides, respectively. Let $H$ be the orthocenter of $\Delta ABC$ and $X,Y,Z$ be the feet of altitudes from $H$ to $EF, DF, DE$ , respectively. In this new configuration, $I_aI_bI_c$ becomes $ABC$ , the original $ABC$ becomes $DEF$ , and $I$ , which was previously the incenter of $ABC$ , becomes $H$ , the center of $DEF$ . $D,E,F$ , which were originally the touchpoints of the incenter of $ABC$ , becomes $X,Y,Z$ . We first show that $AX, BY, CZ$ concur.

In $\Delta DEF$ , $H$ is the incenter, so $HF \perp XY$ . However, $HF \perp AB$ , so $XY \parallel AB$ . Similarly, $XZ \parallel AC$ and $YZ \parallel BC$ . Thus, $\Delta XYZ$ and $\Delta ABC$ are homothetic triangles, and $AX, BY, CZ$ concur at their homothetic center.

We now note that $O$ in our original configuration becomes the the circumcenter of $\Delta DEF$ , so $O$ must be $N_9$ , the center of the $9$ -point circle of $\Delta ABC$ . Therefore, $OI$ in our original configuration becomes $N_9H$ , the Euler line of $\Delta ABC$ . We need to show that $N_9H$ passes through the homothetic center of $\Delta XYZ$ and $\Delta ABC$ . However, this is clear because the same homothety that sends $\Delta XYZ$ to $\Delta ABC$ will send $H$ , the circumcenter of $\Delta XYZ$ to the circumcenter of $\Delta ABC$ , and we know that the circumcenter of $ABC$ is on the Euler line, so we are done. $\blacksquare$

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Eka01

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Eka01

#44 h Sep 5, 2024, 6:54 AM

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Note by midpoints of altitudes lemma, the midpoint of the $A$ altitude, $D$ and the $A$ excenter $I_a$ are collinear and analogous results for other vertices. Since the excentral triangle and the intouch triangle are homothetic, the lines joining corresponding vertices concur at the exsimilicenter. Note that $I$ is the orthocenter of the excentral triangle and $O$ is its nine point center. Since $I$ is mapped to the bevan point of $\Delta ABC$ which is well known to $OI$ , we are done.

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Ilikeminecraft

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Ilikeminecraft

#45 h Apr 28, 2025, 4:12 AM

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We begin by (re)naming a bunch of points
Let $A_0B_0C_0$ be the orthic triangle.
Let $M_A$ be the midpoint of $A_0, A.$ Define $M_B, M_C$ similarly.
Let $A_1B_1C_1$ be the intouch triangle.
Let $A_2$ denote the second intersection of $A_1, M_A$ with the incircle. 2002 ISLG 7 implies $A_2$ is the tangency of $\omega_a$ with the incircle.
Define $B_2, C_2$ similarly.
Let $A_3$ be the second intersection of $A_2A_1$ with $\omega_a.$ By curvilinear properties, $A_3$ is the midpoint of arc $BC.$ Define $B_3, C_3$ similarly.
Let $S_A$ denote the midpoint of arc $BC$ in $(ABC).$ Let $I_A$ denote the $A$ -excenter. By 2002 ISLG 7, it follows $M_A, A_1, A_3, I_A$ are collinear.
Claim: $A_3, B_3, C_3$ have circumcenter $O$
Proof: Take a half homothety centered at $I_A.$ This moves $I\to S_A, A_1\to A_3.$ Thus, $S_AA_3 = \frac r2,$ which is fixed. Thus, $OA_3 = OB_3 = OC_3 = R + \frac r2.$

Finally, note that $DEF$ and $I_AI_BI_C$ are homothetic due to obvious reasons. Thus, the triangle formed by the respective midpoints are also homothetic, which is $A_3B_3C_3.$ Thus, $A_3A_1, B_3B_1, C_3C_1$ are concurrent, and also the line connecting the circumcenters, which is $O, I.$

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