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Ovchinnikov Denis

#1 h Sep 10, 2010, 4:14 PM • 2 Y

Y by Adventure10, Mango247

Quadrilateral $ABCD$ is inscribed into circle $\omega$ , $AC$ intersect $BD$ in point $K$ . Points $M_1$ , $M_2$ , $M_3$ , $M_4$ -midpoints of arcs $AB$ , $BC$ , $CD$ , and $DA$ respectively. Points $I_1$ , $I_2$ , $I_3$ , $I_4$ -incenters of triangles $ABK$ , $BCK$ , $CDK$ , and $DAK$ respectively. Prove that lines $M_1I_1$ , $M_2I_2$ , $M_3I_3$ , and $M_4I_4$ all intersect in one point.

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#2 h May 29, 2011, 1:42 AM • 4 Y

Y by anantmudgal09, Adventure10, and 2 other users

Obviously $M_1M_3\cap M_2M_4$ , $I_1I_3\perp I_2I_4$ , and $K=I_1I_3\cap I_2I_4$ . Define $4\alpha=\angle{AOB}$ , etc. such that $\alpha+\beta+\gamma+\delta=90^\circ$ .

First we show that $M_1M_3\|I_1I_3$ . Indeed, since $I_1KI_3$ bisects $\angle{AKB}=\angle{CKD}$ ,
$\angle{AOI_1}-\angle{DOI_3}=\angle{AOI_1}+\angle{COI_3}-4\gamma=2\alpha-2\gamma=\angle{AOM_1}-\angle{COM_3},$ so $\angle{I_1OM_1}=\angle{I_3OM_3}$ . Similarly, $M_2M_4\|I_2I_4$ .

Let $T=M_1M_3\cap M_2M_4$ , $R=M_1M_3\cap I_2I_4$ , and $S=M_2M_4\cap I_1I_3$ . Note that $M_1I_2=M_1I_4$ , $I_2R=I_4R$ , etc. We have
$\frac{M_1R}{M_3R}=\frac{[M_1I_2I_4]}{[M_3I_2I_4]}=\frac{M_1I_4}{M_3I_4}\frac{M_1I_2}{M_3I_2}\frac{\sin2\gamma}{\sin2\alpha}=\frac{\tan\alpha}{\tan\gamma},\; \frac{M_2S}{M_4S}=\frac{\tan\beta}{\tan\delta}.$ Also,
$\frac{M_1T}{M_3T}=\frac{M_1M_2}{M_3M_2}\frac{M_1M_4}{M_3M_4}=\tan(\alpha+\beta)\tan(\alpha+\delta),\; \frac{M_2T}{M_4T}=\tan(\beta+\alpha)\tan(\beta+\gamma)$ and
$M_1M_3=2R\sin(\alpha+2\beta+\gamma),\; M_2M_4=2R\sin(\beta+2\gamma+\delta)$ while
$\frac{I_1K}{I_3K}=\frac{AB}{CD}=\frac{\sin2\alpha}{\sin2\gamma},\;\frac{I_2K}{I_4K}=\frac{\sin2\beta}{\sin2\delta}$ and
$I_1I_3=\frac{2M_2M_4\sin\beta\sin\delta}{\sin(\beta+\delta)},\; I_2I_4=\frac{2M_1M_3\sin\alpha\sin\gamma}{\sin(\alpha+\gamma)}.$ Now let $X=M_1I_1\cap M_3I_3$ and $Y=M_2I_2\cap M_4I_4$ , so it suffices to show that $d(X,I_1I_3)=d(Y,I_1I_3)$ (so $d(X,I_2I_4)=d(Y,I_2I_4)$ follows by symmetry).

All segments will be directed from now on. By easy ratios with similar triangles, we can compute
$\begin{align*} d(X,I_1I_3)=\frac{I_1I_3\cdot ST}{M_1M_3-I_1I_3} &= M_2M_4\cdot\frac{\frac{M_2T}{M_2M_4}-\frac{M_2S}{M_2M_4}}{\frac{M_1M_3}{I_1I_3}-1} \\ &= M_2M_4\cdot\frac{\frac{\tan(\beta+\alpha)\tan(\beta+\gamma)}{\tan(\beta+\alpha)\tan(\beta+\gamma)+1}-\frac{\tan\beta}{\tan\beta+\tan\delta}}{\frac{\sin(\alpha+2\beta+\gamma)\sin(\beta+\delta)}{2\sin\beta\sin\delta\sin(\beta+2\gamma+\delta)}-1} \\ &= M_2M_4\frac{\sin(\alpha+\beta)\sin(\beta+\gamma)\sin(\beta+\delta)-\sin\beta\cos\delta\cos(\alpha-\gamma)}{\cos(\beta-\delta)\sin(\beta+\delta)-2\sin\beta\sin\delta\cos(\alpha-\gamma)}\frac{2\sin\beta\sin\delta}{\sin(\beta+\delta)} \\ &= M_2M_4\frac{[2\sin(\alpha+\beta)\sin(\beta+\gamma)]\cos(\alpha+\gamma)-[2\sin\beta\sin(\alpha+\beta+\gamma)]\cos(\alpha-\gamma)}{\cos(\beta-\delta)\cos(\alpha+\gamma)-2\sin\beta\sin\delta\cos(\alpha-\gamma)}\frac{\sin\beta\sin\delta}{\sin(\beta+\delta)} \\ &= M_2M_4\frac{-\cos(\alpha+2\beta+\gamma)\cos(\alpha+\gamma)+\cos(\alpha+2\beta+\gamma)\cos(\alpha-\gamma)}{\cos(\beta-\delta)\cos(\alpha+\gamma)-2\sin\beta\sin\delta\cos(\alpha-\gamma)}\frac{\sin\beta\sin\delta}{\sin(\beta+\delta)} \\ &= M_2M_4\frac{-2\sin(\beta-\delta)\sin\alpha\sin\gamma}{\cos(\beta-\delta)\cos(\alpha+\gamma)-2\sin\beta\sin\delta\cos(\alpha-\gamma)}\frac{\sin\beta\sin\delta}{\sin(\beta+\delta)} \\ \end{align*}$ and
$\begin{align*} \frac{d(Y,I_1I_3)-I_2K}{d(Y,I_1I_3)-M_2S}=\frac{I_2I_4}{M_2M_4}\implies d(Y,I_1I_3) &= \frac{M_2M_4\cdot I_2K-I_2I_4\cdot M_2S}{M_2M_4-I_2I_4} \\ &= M_2M_4\cdot\frac{\frac{I_2K}{I_2I_4}-\frac{M_2S}{M_2M_4}}{\frac{M_2M_4}{I_2I_4}-1} \\ &= M_2M_4\cdot\frac{\frac{\sin2\beta}{\sin2\beta+\sin2\delta}-\frac{\tan\beta}{\tan\beta+\tan\delta}}{\frac{\sin(\beta+2\gamma+\delta)\sin(\alpha+\gamma)}{2\sin\alpha\sin\gamma\sin(\alpha+2\beta+\gamma)}-1} \\ &= M_2M_4\frac{\cos\beta-\cos\delta\cos(\beta-\delta)}{\cos(\alpha-\gamma)\sin(\alpha+\gamma)-2\sin\alpha\sin\gamma\cos(\beta-\delta)}\frac{2\sin\alpha\sin\beta\sin\gamma}{\sin(\beta+\delta)} \\ &= M_2M_4\frac{-\sin\delta\sin(\beta-\delta)}{\cos(\alpha-\gamma)\cos(\beta+\delta)-2\sin\alpha\sin\gamma\cos(\beta-\delta)}\frac{2\sin\alpha\sin\beta\sin\gamma}{\sin(\beta+\delta)} \\ \end{align*}$ Thus it remains to show that
$\begin{align*} \cos(\beta-\delta)\cos(\alpha+\gamma)-2\sin\beta\sin\delta\cos(\alpha-\gamma) &= \cos(\alpha-\gamma)\cos(\beta+\delta)-2\sin\alpha\sin\gamma\cos(\beta-\delta) \\ \Longleftrightarrow \cos(\beta-\delta)\cos(\alpha-\gamma) &= \cos(\alpha-\gamma)\cos(\beta-\delta), \end{align*}$ as desired.

Edit: We can also solve this problem with complex numbers. Note that $m_1^2=ab$ , etc., and $abcd=-m_1m_2m_3m_4$ , so there exist complex numbers $p,q,r,s$ such that
$(a,b,c,d,m_1,m_2,m_3,m_4)=(p^2,q^2,r^2,s^2,-pq,qr,rs,sp).$ Then $I_1=AM_2\cap BM_4$ , etc., so
$i_1=\frac{pr(q^2+sp)-qs(p^2+qr)}{pr-qs},\; x_1=\frac{pr(q+s)-qs(p+r)}{p+r-q-s},$ etc. where $\{M_i,X_i\}=M_iX_i\cap(ABCD)$ for all $i$ . Then it just remains to show that $m_1x_1\cap m_3x_3=m_2x_2\cap m_4x_4$ , where to cycle through the variables (for ease of computation) we set
$(p,q,r,s)\mapsto(-q,r,s,p)\mapsto(-r,s,p,-q)\mapsto(-s,p,-q,-r)\mapsto\cdots.$

This post has been edited 3 times. Last edited by math154, Mar 30, 2013, 2:04 PM

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#3 h May 29, 2011, 2:05 AM • 2 Y

Y by Adventure10, Mango247

See here also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=406516

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#4 h Feb 4, 2013, 4:39 PM • 2 Y

Y by Adventure10, Mango247

First note that $I_1=AM_2 \cap BM_4$ , etc. It is also easy to show that $I_2I_4$ passes through $K$ and is parallel to $M_2M_4$ . Now construct $P,Q$ on the circle such that $AP \parallel M_1M_4$ , $BQ \parallel M_1M_2$ . Let $X=AP \cap BQ, Y=M_4P \cap M_2Q$ .

We have $M_4P=AM_1=BM_1=M_2Q$ , so $PQ \parallel M_2M_4$ . Also $DM_4=AM_4=PM_1$ , so $M_4P \parallel M_1D$ , and similarly $M_2Q \parallel M_1C$ . So $\triangle XPQ$ and $\triangle M_1M_4M_2$ have sides that are pairwise parallel, so $M_4P,M_2Q,XM_1$ are concurrent, or $Y,X,M_1$ are collinear. Applying Pascal's theorem to hexagon $M_4PAM_2QB$ yields that $Y,X,I_1$ are also collinear, so $Y,M_1,I_1$ are collinear. But $\triangle YM_4M_2$ and $\triangle M_1I_4I_2$ also have sides that are pairwise parallel, so $YM_1,M_4I_4,M_2I_2$ are concurrent.

Attachments:

sharygin 2.pdf (600kb)

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#5 h Mar 30, 2017, 6:37 PM • 1 Y

Y by Adventure10

Let $X_1$ be the intersection of the external bisectors of $\angle DAB$ and $\angle ABC$ . Define $X_2, X_3, X_4$ cyclically. Note that the corresponding sides of $M_1M_2M_3M_4$ and $X_1X_2X_3X_4$ are parallel. Since $X_1X_3$ is the bisectors of the angle between $AD$ and $BC$ we get $\angle (AD, X_1X_3)=\frac{\angle ABC-\angle BCD}{2}=\frac{\angle CBD-\angle ACB}{2}=\angle (AD, M_1M_3),$ so we get $M_1M_3 \parallel X_1X_3$ and similarly $M_2M_4 \parallel X_2X_4$ . Hence, they are homothetic and the lines $M_iX_i$ are concurrent at their center of homothety.

I claim that $M_1, I_1, X_1$ are collinear. Indeed, let $T=NM_4\cap LM_2$ where $N, L$ are the midpoint of arcs $DAB, ABC$ respectively. Pascal's Theorem for the cyclic hexagon $(NAM_2LBM_4)$ yields $T, X_1, I_1$ are collinear. Since $\triangle X_1NL$ and $\triangle M_1M_4M_2$ are homothetic, we conclude that $T, X_1, M_1$ are collinear. Evidently, the claim holds and the conclusion is established. $\square$

EDIT: Thanks @tastymath, I think I have corrected it now!

This post has been edited 1 time. Last edited by anantmudgal09, Mar 31, 2017, 10:40 AM

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#6 h Mar 31, 2017, 4:48 AM • 4 Y

Y by anantmudgal09, guptaamitu1, Adventure10, Mango247

regarding the above solution

Now, a fast solution:

Note that $I_1 = AM_2\cap BM_4$ and similarly for $I_2,I_3,I_4$ . It suffices to show that $M_1I_1,M_2I_2,M_3I_3$ concur; then by similar reasoning we will get that $M_1I_1,M_2I_2,M_4I_4$ concur, which will finish the problem. From basic angle-chasing we can get that $I_1I_3||M_1M_3$ , and similarly $I_2I_4||M_2M_4$ . It's also clear that $I_1I_3\perp I_2I_4,M_1M_3\perp M_2M_4$ .

It's enough to show that triangles $M_1I_2M_3,I_1M_2I_3$ are perspective, and by Desargues', if $J_1 = AM_2\cap CM_1, J_2=BM_3\cap DM_2$ , then it suffices to show $J_1,J_2, \infty_{I_1I_3} = I_1I_3\cap M_1M_3$ are collinear, i.e. $J_1J_2||I_1I_3 \iff J_1J_2\perp M_2M_4$ . But $J_1,J_2$ are the incenters of $\triangle ABC,\triangle DBC$ , so by Fact 5 we know $M_2J_1=M_2J_2$ , so since $\triangle M_2J_1J_2$ is isosceles with axis of symmetry $M_2M_4$ , we are done.

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jayme

#7 h Mar 31, 2017, 5:43 AM • 3 Y

Y by Googlu15, Adventure10, Mango247

Dear Mathlinkers,

see : http://jl.ayme.pagesperso-orange.fr/Docs/Forme.pdf p. 5-11.

Sincerely
Jean-Louis

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#8 h Mar 31, 2017, 10:41 AM • 2 Y

Y by Adventure10, Mango247

I have edited my solution @tastymath. I think it's correct now

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#9 h Nov 20, 2017, 7:46 AM • 3 Y

Y by buratinogigle, Adventure10, Mango247

Simple proof :

Let $AD\cap BC =P, AB\cap CD = Q$ and $J_1, J_3$ be the incenter and $P$ - exceneter of $PAB$ and $PCD$ respectively. Similarly construct $J_2, J_4$ . Then it's well known that $I_k, J_k, M_k$ are collinear (this fact was on some regional level of Russian Olympiad and also can be proved with mixtilinear circles facts). Now note that $J_1J_2J_3J_4$ and $M_1M_2M_3M_4$ are homothetic, so $I_kM_k$ are concurrent at their homothety center. Done

This post has been edited 1 time. Last edited by Mosquitall, Nov 20, 2017, 8:20 AM

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#10 h Dec 6, 2019, 7:59 AM • 2 Y

Y by Adventure10, Mango247

so beautifull =)

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#11 h Sep 15, 2023, 12:38 PM • 1 Y

Y by mxlcv

nice problem
solution

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#12 h Nov 4, 2023, 1:55 AM • 2 Y

Y by PRMOisTheHardestExam, mathuz

Solution from Twitch Solves ISL:

In fact, letting $O$ denote the center of the cyclic quadrilateral, we show the four lines concur on line $OK$ . For this, it suffices to compute the coordinates of $\overline{OK} \cap \overline{M_1I_1}$ and appeal to a suitable symmetry.
Let's set up a complex number system with $A=a^2$ , $B=b^2$ , $C=c^2$ , $D=d^2$ , such that $\begin{align*} K &= \frac{a^2c^2(b^2+d^2)-b^2d^2(a^2+c^2)}{a^2c^2-b^2d^2} \\ \overline K &= \frac{a^2+c^2-(d^2+b^2)}{a^2c^2-b^2d^2} \\ M_1 &= -ab \\ M_3 &= + cd \\ M_2 &= -bc \\ M_4 &= -da. \end{align*}$ Then we can compute $I_1$ as $\begin{align*} I_1 &= \overline{AM_2} \cap \overline{BM_4} = \frac{a^2(-bc)(b^2-ad) - b^2(-ad)(a^2-bc)}{a^2(-bc)-b^2(-ad)} \\ &= \frac{bd(a^2-bc)-ac(b^2-ad)}{bd-ac}. \\ \end{align*}$ Hence $\begin{align*} M_1 - I_1 &= -ab - \frac{bd(a^2-bc)-ac(b^2-ad)}{bd-ac} \\ &= \frac{-ab(bd-ac) - bd(a^2-bc) + ac(b^2-ad)}{bd-ac} \\ &= \frac{(a+b)(abc-abd-acd+bcd)}{bd-ac} \end{align*}$ Now, note that $\begin{align*} \overline{M_1I_1} \cap \overline{KO} &= \frac {(\overline{M_1}I_1-M_1\overline{I_1})(K-O) - (\overline{K}O-K\overline{O})(M_1-I_1)} {(\overline{M_1}-\overline{I_1})(K-O)-(\overline{K}-\overline{O})(M_1-I_1)} \\ &= K \cdot \frac{\overline{M_1}I_1-M_1\overline{I_1}}{(\overline{M_1}-\overline{I_1})(K)-(\overline{K})(M_1-I_1)} \\ &= K \cdot \frac{ -\frac{1}{ab} \cdot \frac{bd(a^2-bc)-ac(b^2-ad)}{bd-ac} +ab \cdot \frac{ ac \cdot \left( \frac{1}{a^2}-\frac{1}{bc} \right) - bd \cdot \left( \frac{1}{b^2}-\frac{1}{ad} \right) }{ac-bd} }{ \frac{(1/a+1/b) \cdot (a+d-b-c)}{ac-bd} \cdot K - \frac{(a+b)(abc-abd-acd+bcd)}{bd-ac} \cdot \overline K } \\ &= \frac{K}{a+b} \cdot \frac{ \frac{bd(a^2-bc)-ac(b^2-ad)}{ab} +ab \cdot \frac{ ac \cdot \left( \frac{1}{a^2}-\frac{1}{bc} \right) - bd \cdot \left( \frac{1}{b^2}-\frac{1}{ad} \right) }{1} }{\frac{(a+d-b-c)}{ab} \cdot K + \frac{abc-abd-acd+bcd}{1} \cdot \overline K}. \end{align*}$ We focus on the latter fraction. The numerator is short enough to be computed first: $\begin{align*} \frac{bd(a^2-bc)-ac(b^2-ad)}{ab} & + ab \cdot \frac{ ac \cdot \left( \frac{1}{a^2}-\frac{1}{bc} \right) - bd \cdot \left( \frac{1}{b^2}-\frac{1}{ad} \right) }{1} &= -\frac{(a-b)(a+b)(ab-cd)}{ab}. \end{align*}$ The denominator is more painful, but because we know the numerator we know some of the factors that must appear, and divide those out: $\begin{align*} &\frac{(a+d-b-c)}{ab} \cdot K + \frac{abc-abd-acd+bcd}{1} \cdot \overline K \\ &= \frac{(a-b)(ab-cd) \left[ a^2bc-a^2bd-a^2cd+ab^2c-ab^2d+abc^2-abd^2+ac^2d-acd^2+b^2cd+bc^2d-bcd^2 \right] }{ab(a^2c^2-b^2d^2)} \end{align*}$ Thus, simplifying gives $\begin{align*} &\overline{M_1I_1} \cap \overline{KO} \\ &= K \cdot \frac{a^2c^2-b^2d^2}{ a^2bc-a^2bd-a^2cd+ab^2c-ab^2d+abc^2-abd^2+ac^2d-acd^2+b^2cd+bc^2d-bcd^2 }. \end{align*}$ The symmetry of this expression is enough to conclude.

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#13 h Nov 21, 2023, 1:19 PM

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Let $I_B$ and $I_C$ be the incenters of $\Delta ACD$ and $\Delta ABD$ respectively. Note that $I_4=AI_BM_3 \cap DI_CM_1$ , and $BI_1I_CM_4$ , $CI_3I_BM_4$ are straight lines. Since $I_1I_3$ bisects $\angle CKD$ , one can show $I_1I_3 \parallel M_1M_3$ by angle chasing. On the other hand, $AI_CI_BD$ is cyclic with circumcenter $M_2$ , so by Reim's we get $I_BI_C \parallel M_1M_3$ . Hence $I_BI_C \parallel M_1M_3 \parallel I_1I_3$ , so by Desargue's Theorem on $\Delta I_1M_1I_C$ and $\Delta I_3M_3I_B$ we get that $M_1I_1 \cap M_3I_3$ , $I_4$ and $M_4$ are collinear, so $M_1I_1$ , $M_3I_3$ and $M_4I_4$ concur. Similarly, $M_1I_1$ , $M_3I_3$ and $M_2I_2$ concur and we are done.

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#14 h Feb 13, 2024, 5:24 PM

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Let $X_1$ be the intersection of the external bisectors of $\angle DAB$ and $\angle ABC$ . Define $X_2, X_3, X_4$ cyclically. Note that the corresponding sides of $M_1M_2M_3M_4$ and $X_1X_2X_3X_4$ are parallel. Now clearly $M_1-I_1-X_1$ follows pascal's.
Remark: Interesting Config! Other properties: $I_1I_2 \parallel M_3M_4$ and $I_1I_3 \parallel M_1M_3$ , and $K-O-E$ where $E$ is the concurrence point and $O$ is center. The first 2 can be used in orignal soln maybe replacing pascal.

This post has been edited 6 times. Last edited by math_comb01, Feb 13, 2024, 5:58 PM

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#15 h Mar 13, 2024, 4:27 AM

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why did I do this

We employ complex numbers with $\omega$ the unit circle, noting that for $A = a^2$ and similar, the arc midpoints $\widehat{AB}, \widehat{BC}, \widehat{CD}, \widehat{DA}$ containing no other vertices of the quadrilateral are given by $-ab, -b, -cd, +da$ . So we may compute
$\begin{align*} i_1 &= \frac{-a^2bc(b^2+ad)-ab^2d(a^2-bc)}{-a^2bc-ab^2d} = \frac{ac(b^2+ad)+bd(a^2-bc)}{ac+bd} \\ k &= \frac{a^2c^2(b^2+d^2)-b^2d^2(a^2+c^2)}{a^2c^2-b^2d^2}. \end{align*}$ I claim that the concurrency point lies on the line $\overline{OK}$ . In particular, it will suffice to show that $\overline{I_1M_1}$ intersects $\overline{OK}$ at a point cyclic with respect to $a, b, c, d$ . To do this, we will compute a few things first. To begin, we have
$\begin{align*} \overline m_1 i_1 - m_1 \overline i_1 &= -\frac{ac(b^2+ad)+bd(a^2-bc)}{ab(ac+bd)} + \frac{ab\left[\frac 1{ac}\left(\frac 1{b^2}+\frac 1{ad}\right)+\frac 1{bd}\left(\frac 1{a^2}-\frac 1{bc}\right)\right]}{\frac 1{ac}+\frac 1{bd}} \\ &= \frac{(b-a)(a+b)(ab+cd)}{ab(ac+bd)}. \end{align*}$ Furthermore, we have
$\begin{align*} \overline k(m_1-i_1) &= \frac{\frac 1{a^2c^2}\left(\frac 1{b^2}+\frac 1{d^2}\right)-\frac 1{b^2d^2}\left(\frac 1{a^2}+\frac 1{c^2}\right)}{\frac 1{a^2c^2}-\frac 1{b^2d^2}} \left(-ab-\frac{ac(b^2+ad)+bd(a^2-bc)}{ac+bd}\right) \\ &= -\frac{(a+b)(abc+abd+acd-bcd)(a^2-b^2+c^2-d^2)}{(ac-bd)(ac+bd)^2}. \end{align*}$ Also,
$\begin{align*} k(\overline m_1-\overline i_1) &= \frac{a^2c^2(b^2+d^2)-b^2d^2(a^2+c^2)}{a^2c^2-b^2d^2} \left(-\frac 1{ab}-\frac{\frac 1{ac}\left(\frac 1{b^2}+\frac 1{ad}\right)+\frac 1{bd}\left(\frac 1{a^2}-\frac 1{bc}\right)}{\frac 1{ac}+\frac 1{bd}}\right) \\ &= \frac{(a^2b^2c^2-a^2b^2d^2+a^2c^2d^2-b^2c^2d^2)(a+b)(a-b-c-d)}{ab(ac-bd)(ac+bd)^2}. \end{align*}$ So by the complex intersection formula, we want the point
$\begin{align*} x &= k \cdot \frac{\overline m_1 i_1-m_1 \overline i_1}{k(\overline m_1-\overline i_1) - \overline k(m_1-i_1)} \\ &= \frac{\frac{(b-a)(a+b)(ab+cd)}{ab(ac+bd)}}{\frac{(a^2b^2c^2-a^2b^2d^2+a^2c^2d^2-b^2c^2d^2)(a+b)(a-b-c-d)}{ab(ac-bd)(ac+bd)^2}+\frac{(a+b)(abc+abd+acd-bcd)(a^2-b^2+c^2-d^2)}{(ac-bd)(ac+bd)^2}} \\ &= \frac{\frac{(b-a)(a+b)(ab+cd)}{ab(ac+bd)}}{\frac{(a-b)(a+b)(ab+cd)(a^2bc+a^2bd+a^2cd+ab^2c+ab^2d+abc^2-abd^2-ac^2d-acd^2-b^2cd-bc^2d-bcd^2)}{ab(ac-bd)(ac+bd)^2}} \\ &= \frac{(ac-bd)(ac+bd)}{a^2bc+a^2bd+a^2cd+ab^2c+ab^2d+abc^2-abd^2-ac^2d-acd^2-b^2cd-bc^2d-bcd^2}. \end{align*}$ This is cyclic in $a, b, c, d$ , which concludes the proof.

This post has been edited 2 times. Last edited by HamstPan38825, Mar 13, 2024, 4:27 AM

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#16 h Dec 18, 2024, 3:59 AM • 3 Y

Y by eg4334, anantmudgal09, GeoKing

We claim $M_1I_1,M_2I_2,OK$ concur, where $O$ is the center of $\omega$ . This implies the result.

Fix $A,B,C$ and move $M_4$ with degree $2$ , such that $M_3$ also moves with degree $2$ and $D$ has degree $4$ . Then $I_1,I_2$ have degree $1$ and $K$ has degree $2$ , so the concurrence is degree $4$ .

We check $6$ cases. If $A=M_4=D$ then $I_1=A$ and $I_2$ is the incenter of $ABC$ so the lines concur at $A$ . Similarly if $C=M_3=D$ then they concur at $C$ .

If $M_4=M_2$ then $ABCD$ is an isosceles trapezoid so $M_2\in OK$ . Additionally $I_1=M_2$ so the lines concur at $M_2$ . Similarly if $M_3=M_1$ then they concur at $M_1$ .

If $M_4$ is the antipode of $M_2$ then $ABCD$ is again an isosceles trapezoid and $M_2,M_4,O,K,I_2,I_4$ are collinear, so $M_2I_2$ and $OK$ coincide. Similarly if $M_3$ is the antipode of $M_1$ then $M_1I_1$ and $OK$ coincide.

This is six cases so we are done.

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