Very easy to get the idea but very difficult to write, i am bit unsure abput this writing, really bad and dirty

Attachments:
Sharygin_P10.pdf (196kb)

Pls check if this works. Here i present some lemmas i've quite frequently used in the proof.All these follow trivially from direct observation. In the proof we call the links which are not bad links as good links. $(1)$ Let $A_i,A_j,A_k$ be three points such that $\omega$ is the $A_i$ excircle of $\Delta A_iA_jA_k$ then $A_iA_j,A_i,A_k$ are bad links and $A_jA_k$ is a good link (this is just simple observation ) $(2)$Let $A_i,A_j,A_k$ be three points such that $\omega$ is the incircle of $\Delta A_iA_jA_k$ then $A_iA_j,A_i,A_k,A_jA_k$ are all good links (simple and trivial observation ) $(3)$If three collinear point $A_i,A_j,A_k$ are such that $A_iA_j$ is a good link and $A_iA_k$ is a good link then $A_jA_k$ is a bad link. ( proof follows directly from length chasing) $(4)$If three collinear point $A_i,A_j,A_k$ are such that $A_iA_j$ is a good link and $A_iA_k$ is a bad link then $A_jA_k$ is a good link. ( proof follows directly from length chasing) $(5)$If three collinear point $A_i,A_j,A_k$ are such that $A_iA_j$ is a bad link and $A_iA_k$ is a bad link then $A_jA_k$ is a bad link. ( proof again follows directly from length chasing) Proof: We'll set up a strong induction. Firstly we'll prove the base case when $n=2$. We form two subcases for the base case. Case 1: $A_1A_2$ is a bad link. Notice that $A_2A_3$ is not a bad link then $\omega$ is the $A_1$-excircle for $\Delta A_1A_2A_3 \implies A_1A_3$ is a bad link so the number of bad links is $2$ which is even. When $A_2A_3$ is a bad link , notice that in this case since both $A_2A_3$ and $A_2A_1$ are bad links we have that $\omega$ is the $A_2$-excircle for $\Delta A_1A_2A_3 \implies A_1A_3$ is not a bad link so total number of bad links is again $2$ which is even. Done with this case.$\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(5cm); 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 = -2.219040614352925, xmax = 24.431762479043854, ymin = -12.604196636201385, ymax = 4.99571683521713; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw(circle((9.54,-2.34), 6), linewidth(2) + rvwvcq); draw((xmin, -0.6568643668752885*xmin-3.2521592629754266)--(xmax, -0.6568643668752885*xmax-3.2521592629754266), linewidth(2) + dtsfsf); /* line */ draw((xmin, 0.6413869112453995*xmin-1.330747371356809)--(xmax, 0.6413869112453995*xmax-1.330747371356809), linewidth(2) + dtsfsf); /* line */ draw((xmin, -2.756905126843203*xmin + 6.36488299081697)--(xmax, -2.756905126843203*xmax + 6.36488299081697), linewidth(2) + dtsfsf); /* line */ /* dots and labels */ dot((9.54,-2.34),dotstyle); label("$A$", (9.60076361473149,-2.1688497106700533), NE * labelscalefactor); dot((-1.48,-2.28),dotstyle); label("$A_1$", (-1.4056719485154912,-2.099626845492399), NE * labelscalefactor); dot((2.264558276899077,0.12171066719869393),dotstyle); label("$A_2$", (2.332362771077823,0.2885620031366618), NE * labelscalefactor); dot((4.579455045405561,-6.260240102009595),linewidth(4pt) + dotstyle); label("$A_3$", (4.6513287545292314,-6.114553025796328), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Case 2: When $A_1A_2$ is not a bad link. Notice that in this case either $\omega$ will be the incircle of $\Delta A_1A_2A_3$ or $\omega$ will be the $A_3$-excicle of $\Delta A_1A_2A_3$ since $A_3A_2$ and $A_3A_1$ are tangents and $A_1A_2$ is not a bad link. In the case when it is incircle we have that the number of bad links is $0$ which is even and in the case of the excircle no. of bad links is $2$ which is again even. Hence we are done with the base case. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(5cm); 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 = -11.868472810071914, xmax = 28.14006192168888, ymin = -11.766292683061389, ymax = 14.654927980705356; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw(circle((9.54,-2.34), 6), linewidth(2) + rvwvcq); draw((xmin, -0.6568643668752885*xmin-3.2521592629754266)--(xmax, -0.6568643668752885*xmax-3.2521592629754266), linewidth(2) + dtsfsf); /* line */ draw((xmin, 0.6413869112453995*xmin-1.330747371356809)--(xmax, 0.6413869112453995*xmax-1.330747371356809), linewidth(2) + dtsfsf); /* line */ draw((xmin, 0.6413869112454009*xmin-1.33074737135682)--(xmax, 0.6413869112454009*xmax-1.33074737135682), linewidth(2) + dtsfsf); /* line */ draw((xmin, 0.07740374668491358*xmin + 2.9395154348612955)--(xmax, 0.07740374668491358*xmax + 2.9395154348612955), linewidth(2) + dtsfsf); /* line */ /* dots and labels */ dot((9.54,-2.34),dotstyle); label("$A$", (9.642609500251421,-2.075913816212818), NE * labelscalefactor); dot((7.571613967494814,3.5255867243972148),dotstyle); label("$A_2$", (7.668162331671019,3.795468553513125), NE * labelscalefactor); dot((-1.48,-2.28),linewidth(4pt) + dotstyle); label("$A_1$", (-1.372727334986615,-2.075913816212818), NE * labelscalefactor); dot((-8.432443930889926,2.2868126808999545),linewidth(4pt) + dotstyle); label("$A_3$", (-8.335251561033298,2.4964901531312793), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Now we'll set up a strong induction. Assume that our assertion that no. of bad links is even holds till some $n$. We claim that it even holds for $n+1$ points. Consider any arbittary configuration of $n+1$ points. Let $A_{n-2}A_{n-1} \cap l=A_{n}'$ where $l$ is the second tangent for $A_1$ apart from $A_1A_2$. Now observe that $A_1A_2A_3....A_{n-1}A_n'$ is a broken line that satisfies the problem statement and is a configuration of $n$ points. By induction hypothesis there are an even no. of bad links in this broken line. Now we form cases. Case 1: $A_{n-1}A_n'$ is a good link and $A_n'A_1$ is a bad link. We form subcases of this case. Subcase 1: When $A_{n-1}A_n$ is a good link. Notice that $A_{n-1}A_n$ is good and $A_{n-1}A_n'$ is good (assumed in case 1) $\implies A_nA_n'$ is a bad link $\implies \omega $ is an excircle in $\Delta A_nA_n'A_{n+1}$. Now if $A_{n-1}A_n > A_{n-1}A_n' \implies \omega$ is the $A_n$ -excircle of $\Delta A_nA_n'A_{n+1}$ and if $A_{n-1}A_n < A_{n-1}A_n' \implies \omega$ is the $A_n'$ -excircle $\Delta A_nA_n'A_{n+1}$. Now note that to obtain the broken line $A_1A_2.......A_{n-1}A_nA_{n+1}$ from the broken line $A_1A_2........A_{n-1}A_n'$ we need to remove the edges $A_{n-1}A_n'$ and $A_n'A_1$ from $A_1A_2........A_{n-1}A_n'$ and add the edges $A_{n-1}A_n , A_{n}A_{n+1} , A_{n+1}A_1$ . Note that since by the assumption in our claim we have that $A_n'A_1$ is a bad link and $A_{n-1}A_n'$ is a good link so after their removal the parity of the bad links is odd since their number decreases by $1$ and initially they were even by induction hypothesis. Now if $\omega$ is the $A_n$ -excircle of $\Delta A_nA_n'A_{n+1}$ we have that $A_nA_{n+1}$ is a bad link and that $A_n'A_{n+1}$ is a good link and now by assumption in subcase 1 since $A_n'A_1$ is a bad link $\implies A_1A_{n+1} $ is a good link. So after adding the edges $A_{n-1}A_n , A_{n}A_{n+1} , A_{n+1}A_1$ we get that only $A_{n}A_{n+1}$ is a bad link so we get that there is an increase of $1$ in the no. of bad links. Initially after the removal the parity of bad links was odd and since $\text{odd}+1=\text{even}$ we have that the parity in this case after adding the edges is even. Done. Now when $\omega$ is the $A_{n}'$ excircle of $\Delta A_nA_n'A_{n+1}$. In this case note that since again $A_n'A_1$ is a bad link and $A_{n-1}A_n'$ is a good link so after the removal the parity would become odd. Now note that in this case $A_nA_{n+1}$ is a good link and $A_n'A_{n+1}$ is a bad link and since $A_n'A_1$ is a bad link $\implies A_{n+1}A_1$ is a bad link. Now note that among $A_{n-1}A_n , A_{n}A_{n+1} , A_{n+1}A_1$ only $A_{n+1}A_1$ is a bad link so after adding the links the number of bad links increases by $1$. Since after the removal the parity of the bad links was odd and as $\text{odd}+1=\text{even}$ we get that in this case also after adding the edges the final parity is even. So in subcase 1 the final parity will be even. Done. $\square$ Subcase 2: When $A_{n-1}A_n$ is a bad link. Note that after the removal as proved in subcase 1 parity is odd. Now note that since $A_{n-1}A_n$ is a bad link as assumed in subcase 2 and since $A_{n-1}A_n'$ is a good link as assumed in case 1 $\implies A_nA_n'$ is a good link $\implies \omega$ is either the incircle for $\Delta A_nA_n'A_{n+1}$ or its the $A_{n+1}$ excircle for $\Delta A_nA_n'A_{n+1}$.In both the cases either $A_nA_{n+1}$ and $A_{n+1}A_1$ are both good or both bad links (good in the case of incircle and bad in the case of excircle since its the $A_{n+1}$ excircle.) So we see that among the links $A_{n-1}A_n , A_{n}A_{n+1} , A_{n+1}A_1$ we have that the only bad link is $A_{n-1}A_n$ (assumed in subcase 2) and so after adding these links the number of bad links increases by $1$ and since after the removal parity was odd and $\text{odd}+1=\text{even}$ we have that in the end there are any even number of bad links in subcase 2 . Hence we have proved case 1. $\blacksquare$ Case 2: $A_{n-1}A_n'$ is a bad link and $A_n'A_1$ is a good link. The proof for this case is infact similiar to the proof of case 1. Case 3: When both the links $A_{n-1}A_n'$ and $A_n'A_1$ are good. We again form subcases. Subcase 1: When $A_{n-1}A_n$ is a good link. Note that by assumption in case 3 since $A_{n-1}A_n'$ is a good and by assumption in subcase 1 since $A_{n-1}A_n$ is a good link $\implies A_nA_n'$ is a bad link. So we have that $\omega$ is an excircle for $\Delta A_nA_n'A_{n+1}$ and precicely as also stated in case 1 its either the $A_n$-excircle or the $A_n'$-excircle. If its the $A_n$-excircle $\implies A_n'A_{n+1}$ is a good link. And by assumption in claim 1 we have $A_n'A_1$ is a good link $\implies A_{n+1}A_1$ is a bad link. And also since $\omega$ is the $A_n$ excircle $\implies A_nA_{n+1}$ is a bad link. Now note that since both the links $A_{n-1}A_n'$ and $A_n'A_1$ are good so after their removal the parity of the link remains unchanged or it remains even. Now note that among the links $A_{n-1}A_n , A_{n}A_{n+1} , A_{n+1}A_1$ only $A_{n}A_{n+1} , A_{n+1}A_1$ are bad links so after adding the edges the number of bad links increases by $2$ and since $\text{even}+2=\text{even}$ we get that in the final the parity of the bad links remains even in subcase 1. Done. Now when its the $A_n'$-excircle. Note that in this case $A_nA_{n+1}$ is a good link and $A_n'A_{n+1}$ is a bad link and since by our assumption in case 3 we have $A_n'A_1$ is a good link $\implies A_nA_{n+1}$ is a good link. Now note that after the removal of the links $A_{n-1}A_n'$ and $A_n'A_1$ the parity remains unchanged that is remains even(as stated earlier).Now among $A_{n-1}A_n , A_{n}A_{n+1} , A_{n+1}A_1$ we see that no link is bad and so the no. of links remains the same and so the parity too remains the same i.e. even. Done with subcase 1.$\square$ Subcase 2: When $A_{n-1}A_n$ is a bad link. By assumption of subcase 2 we have that $A_{n-1}A_n$ is a bad link and by our assumption in case 3 we have that $A_{n-1}A_n'$ are good link. $\implies A_nA_n'$ is a good link $\implies \omega$ is either the incircle for $\Delta A_nA_n'A_{n+1}$ or its the $A_{n+1}$-excircle for $\Delta A_nA_n'A_{n+1}$. If its the incircle then $A_nA_{n+1}$ is a good link and $A_n'A_n$ is a good link and as $A_n'A_1$ is a good link as assumed in case 3 $\implies A_1A_{n+1}$ is a bad link. Now after removing the links $A_{n-1}A_n'$ and $A_n'A_1$ as also stated in subcase 1 the parity will not change and will remian even. And among $A_{n-1}A_n , A_{n}A_{n+1} , A_{n+1}A_1$ we see that only $A_{n-1}A_n, A_{n+1}A_1$ are bad links we get that after adding these links the number of bad links increases by $2$ and since $\text{even}+2=\text{even}$ . Now when $\omega$ is the $A_{n+1}$ excircle then we get that $A_{n+1}A_n'$ is a bad link and since by our assumption in case 3 we have $A_n'A_1$ is a good link $\implies A_1A_{n+1}$ is a good link. Also since $\omega$ is $A_{n+1}$-excircle $\implies A_{n+1}A_n$ is a bad link. Now after the removal of the links $A_{n-1}A_n'$ and $A_n'A_1$ we get that the parity remains unchanged i.e. even. Now among the links $A_{n-1}A_n , A_{n}A_{n+1} , A_{n+1}A_1$ we have that $A_{n-1}A_n, A_{n+1}A_1$ are the only bad links. So after adding these bad links we see that the parity still doesnt change since $\text{even}+2=\text{even}$ and so we get that final parity of bad links remains even in subcase 2. So we may conclude that the final parity of bad links remains even in case 3 as well.Done. $\blacksquare$ Case 4: When both the links $A_{n-1}A_n'$ and $A_n'A_1$ are bad. The proof is quite similiar to case 3. Hence we have proved for all possible cases and so we are done with the problem.$\blacksquare$.

Let line $A_iA_{i+1}$ touch $\omega$ at $T_i$. ($A_{n+1}=A_1$). This implies that the pole of $T_iT_{i-1}$wrt $\omega$ is $A_i$. For two points $X,Y$ on a circle , call $\widehat{XY}$ clockwise if the shortest path from $X$ to $Y$ on the circle is clockwise .Call it anticlockwise otherwise. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(3cm); 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 = -12.692277073170734, xmax = 13.493088780487813, ymin = -7.664935609756093, ymax = 7.89604; /* image dimensions */ /* draw figures */ draw(circle((-2.2,-0.31), 5.209836849652779), linewidth(0.8)); /* dots and labels */ dot((-1.88,4.89),linewidth(3pt) + dotstyle); label("$X$", (-1.7888624390243881,5.019942439024386), NE * labelscalefactor); dot((2.3674997701548017,2.1960618207929006),dotstyle); label("$Y$", (2.4608936585365893,2.4014058536585345), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] In the figure given above, $\widehat{XY}$ is clockwise but $\widehat{YX}$ is anticlockwise. Key Claim: The link $A_iA_{i+1}$ is bad if and only if $\widehat{T_{i-1}T_i}$ and $\widehat{T_iT_{i+1}}$ are of different types i.e one of them is clockwise and the other is anticlockwise. Proof: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(17cm); 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 = -11.707359535417035, xmax = 13.125168483813328, ymin = -8.073397046214067, ymax = 6.683638047181003; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); /* draw figures */ draw(circle((-2.2,-0.31), 5.209836849652779), linewidth(0.6)); draw((xmin, -0.061538461538461604*xmin + 4.774307692307693)--(xmax, -0.061538461538461604*xmax + 4.774307692307693), linewidth(0.6)); /* line */ draw((-9.438926109466331,5.355164683659467)--(-7.17066979684774,-1.8703979526728578), linewidth(0.6)); draw((-4.8278157584167705,5.071404046671803)--(-6.497436233075794,2.6352405373156422), linewidth(0.6)); draw((-2.2,-0.31)--(-1.88,4.89), linewidth(0.6)); draw((-7.17066979684774,-1.8703979526728578)--(-1.88,4.89), linewidth(0.6) + dtsfsf); draw((-6.497436233075794,2.6352405373156422)--(-1.88,4.89), linewidth(0.6) + rvwvcq); /* dots and labels */ dot((-2.2,-0.31),linewidth(3pt) + dotstyle); label("$O$", (-2.1203753575010507,-0.1961934860155951), NE * labelscalefactor); dot((-1.88,4.89),linewidth(3pt) + dotstyle); label("$T_i$", (-1.7947028588881937,5.014566491790113), NE * labelscalefactor); dot((-6.497436233075794,2.6352405373156422),linewidth(3pt) + dotstyle); label("$T_{i+1}$", (-6.5169540887746225,2.0870593159074377), NE * labelscalefactor); dot((-7.17066979684774,-1.8703979526728578),linewidth(3pt) + dotstyle); label("$T_{i-1}$", (-8.217871427409534,-1.9466831660597002), NE * labelscalefactor); dot((-9.438926109466331,5.355164683659467),linewidth(3pt) + dotstyle); label("$A_i$", (-9.366588451637123,5.482720708546094), NE * labelscalefactor); dot((-4.8278157584167705,5.071404046671803),linewidth(3pt) + dotstyle); label("$A_{i+1}$", (-4.746109877567212,5.197757272259845), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We add the centre $O$ of $\omega$ and the line $OT_i$. By definition, link $A_iA_{i+1}$ is bad iff $A_i$ and $A_{i+1}$ lie on the tangent to $\omega$ at $T_i$ such that they are on the same side of $T_i$. If this were true, it is easy to see that the points $T_{i-1}$ and $T_{i+1}$(which are on the circle $\omega$) will also lie on the same side of the line $OT_i$. Now, if $T_{i-1}$ and $T_{i+1}$ lie on the same side of $OT_i$, $\widehat{T_iT_{i+1}}$ and $\widehat{T_iT_{i-1}}$ are of the same type. As we have shown earlier $\widehat{T_iT_{i-1}}$ and $\widehat{T_{i-1}T_i}$ are of different types. Therefore, $\widehat{T_{i-1}T_i}$ and $\widehat{T_iT_{i+1}}$ are of different types iff the link is bad. $\hfill \square$ Main Proof: Colour chord $T_iT_{i+1}$ red if $\widehat{T_iT_{i+1}}$ is clockwise and colour it blue otherwise. Using the claim we proved earlier , we can rephrase the problem in terms of the points $T_j$. Quote: $T_1T_2 \dots T_n$ is a polygon with $n$ sides. All sides of the polygon are coloured either red or blue. A vertex $T_j$ is called bad if the two sides passing through the vertex are of different colours and called good otherwise. Prove that the number of bad vertices is even. Suppose that initially all sides of the polygons are coloured red. Then the number of bad links is $0$ which is even. Now, we colour some sides of the polygon blue one by one. We show that the parity of bad vertices does not change. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -9.204062561017535, xmax = 30.720918194763026, ymin = -10.852935714631785, ymax = 12.872974980401723; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); /* draw figures */ draw((-7.485300993062626,0.33768866375949286)--(-4.199573975009941,3.4446006059997845), linewidth(0.8) + dtsfsf); draw((-4.199573975009941,3.4446006059997845)--(-1.0256063793399997,4.1375090247727995), linewidth(0.8) + dtsfsf); draw((-1.0256063793399997,4.1375090247727995)--(5.411736349906079,0.0024103966112599203), linewidth(0.8) + rvwvcq); draw((6.816775808101945,2.6983524629077325)--(9.734745035673525,2.7638032510319626), linewidth(0.6)); draw((9.13703329288613,3.3588104157976932)--(9.734745035673525,2.7638032510319626), linewidth(0.6)); draw((9.734745035673525,2.7638032510319626)--(9.164313327367411,2.142594333076444), linewidth(0.6)); label("bad",(-1.6772219267310355,5.215232769866771),SE*labelscalefactor); label("good",(-5.015212121066787,4.560724888624467),SE*labelscalefactor); draw((11.707464244674087,0.40459984273583804)--(14.35822116370542,3.5135122786367807), linewidth(1) + dtsfsf); draw((14.35822116370542,3.5135122786367807)--(17.336232023357905,4.6589010708108125), linewidth(1) + rvwvcq); draw((17.336232023357905,4.6589010708108125)--(25.124875810141326,-0.44626040287915675), linewidth(1) + rvwvcq); label("good",(-5.015212121066787,4.560724888624467),SE*labelscalefactor); label("bad",(13.540086312152539,4.560724888624467),SE*labelscalefactor); label("good",(16.976252688674638,5.96791683329542),SE*labelscalefactor); /* dots and labels */ dot((-1.0256063793399997,4.1375090247727995),linewidth(3pt) + dotstyle); dot((-7.485300993062626,0.33768866375949286),linewidth(3pt) + dotstyle); dot((-4.199573975009941,3.4446006059997845),linewidth(3pt) + dotstyle); dot((5.411736349906079,0.0024103966112599203),linewidth(3pt) + dotstyle); dot((11.707464244674087,0.40459984273583804),linewidth(3pt) + dotstyle); dot((14.35822116370542,3.5135122786367807),linewidth(3pt) + dotstyle); dot((17.336232023357905,4.6589010708108125),linewidth(3pt) + dotstyle); dot((25.124875810141326,-0.44626040287915675),linewidth(3pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Consider the following cases: Side $T_{i}T_{i+1}$ is coloured blue (assuming that this side was red earlier) and $T_{i-1}T_{i}$ and $T_{i+1}T_{i+2}$ are both red. Then the number of bad vertices increases by $2$ since the vertices $T_i$ and $T_{i+1}$ become bad. Side $T_{i}T_{i+1}$ is coloured blue and $T_{i-1}T_{i}$ and $T_{i+1}T_{i+2}$ are both blue. Then the number of bad vertices decreases by $2$ since the vertices $T_{i}$ and $T_{i+1}$ which were both bad earlier become good. Side $T_{i}T_{i+1}$ is coloured blue and one of the adjacent sides is blue and the other is red.(shown in the figure) Assume without loss of generality that $T_{i-1}T_{i}$ is red and $T_{i+1}T_{i+2}$ is blue. The number of bad links does not change since the vertex $T_{i+1}$ which was previously bad becomes good, and the vertex $T_{i}$ which was previously good becomes bad. Therefore, the parity of bad vertices is invariant. We showed in the key claim that there was a bijection between the number of bad links and bad vertices. Hence, the number of bad links is always even. $\blacksquare$