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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Something nice
KhuongTrang   33
N 2 minutes ago by NguyenVanHoa29
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
33 replies
+1 w
KhuongTrang
Nov 1, 2023
NguyenVanHoa29
2 minutes ago
Geometry hard
Lukariman   2
N 2 minutes ago by Primeniyazidayi
Given triangle ABC inscribed in circle (O). The bisector of angle A intersects (O) at D. Let M, N be the midpoints of AB, AC respectively. OD intersects BC at P and AD intersects MN at S. The circle circumscribed around triangle MPS intersects BC at Q different from P. Prove that QA is tangent to (O).
2 replies
Lukariman
43 minutes ago
Primeniyazidayi
2 minutes ago
help!!!!!!!!!!!!
Cobedangiu   3
N 3 minutes ago by sqing
help
3 replies
Cobedangiu
Mar 23, 2025
sqing
3 minutes ago
Inequality
nguyentlauv   1
N 4 minutes ago by NguyenVanHoa29
Source: Own
Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3$ and $k\ge 0$, prove that
$$\frac{\sqrt{a+1}}{b+c+k}+\frac{\sqrt{b+1}}{c+a+k}+\frac{\sqrt{c+1}}{a+b+k} \geq \frac{3\sqrt{2}}{k+2}.$$
1 reply
nguyentlauv
Yesterday at 12:19 PM
NguyenVanHoa29
4 minutes ago
Great similarity
steven_zhang123   3
N 17 minutes ago by Lil_flip38
Source: a friend
As shown in the figure, there are two points $D$ and $E$ outside triangle $ABC$ such that $\angle DAB = \angle CAE$ and $\angle ABD + \angle ACE = 180^{\circ}$. Connect $BE$ and $DC$, which intersect at point $O$. Let $AO$ intersect $BC$ at point $F$. Prove that $\angle ACE = \angle AFC$.
3 replies
steven_zhang123
43 minutes ago
Lil_flip38
17 minutes ago
AD=BE implies ABC right
v_Enhance   117
N 2 hours ago by cj13609517288
Source: European Girl's MO 2013, Problem 1
The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$ so that $CD = BC$. The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$. Prove that, if $AD=BE$, then the triangle $ABC$ is right-angled.
117 replies
v_Enhance
Apr 10, 2013
cj13609517288
2 hours ago
Geometry
gggzul   6
N 3 hours ago by Captainscrubz
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
6 replies
gggzul
Yesterday at 8:22 AM
Captainscrubz
3 hours ago
Geometry
Lukariman   5
N 3 hours ago by Lukariman
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
5 replies
Lukariman
Yesterday at 12:43 PM
Lukariman
3 hours ago
Aime type Geo
ehuseyinyigit   4
N 5 hours ago by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
4 replies
ehuseyinyigit
Monday at 9:04 PM
ehuseyinyigit
5 hours ago
n variables with n-gon sides
mihaig   1
N 5 hours ago by mihaig
Source: Own
Let $n\geq3$ and let $a_1,a_2,\ldots, a_n\geq0$ be reals such that $\sum_{i=1}^{n}{\frac{1}{2a_i+n-2}}=1.$
Prove
$$\frac{24}{(n-1)(n-2)}\cdot\sum_{1\leq i<j<k\leq n}{a_ia_ja_k}\geq3\sum_{i=1}^{n}{a_i}+n.$$
1 reply
mihaig
Apr 25, 2025
mihaig
5 hours ago
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   3
N Today at 5:48 AM by EeEeRUT
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
3 replies
BR1F1SZ
Monday at 9:45 PM
EeEeRUT
Today at 5:48 AM
area of O_1O_2O_3O_4 <=1, incenters of right triangles outside a square
parmenides51   2
N Today at 4:35 AM by Solilin
Source: Thailand Mathematical Olympiad 2012 p4
Let $ABCD$ be a unit square. Points $E, F, G, H$ are chosen outside $ABCD$ so that $\angle AEB =\angle BF C = \angle CGD = \angle DHA = 90^o$ . Let $O_1, O_2, O_3, O_4$, respectively, be the incenters of $\vartriangle ABE, \vartriangle BCF, \vartriangle CDG, \vartriangle DAH$. Show that the area of $O_1O_2O_3O_4$ is at most $1$.
2 replies
parmenides51
Aug 17, 2020
Solilin
Today at 4:35 AM
Geo metry
TUAN2k8   3
N Today at 4:34 AM by TUAN2k8
Help me plss!
Given an acute triangle $ABC$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively. Lines $BD$ and $CE$ intersect at point $F$. The circumcircles of triangles $BDF$ and $CEF$ intersect at a second point $P$. The circumcircles of triangles $ABC$ and $ADE$ intersect at a second point $Q$. Point $K$ lies on segment $AP$ such that $KQ \perp AQ$. Prove that triangles $\triangle BKD$ and $\triangle CKE$ are similar.
3 replies
TUAN2k8
Yesterday at 10:33 AM
TUAN2k8
Today at 4:34 AM
China South East Mathematical Olympiad 2013 problem 2
s372102   3
N Today at 2:11 AM by AGCN
$\triangle ABC$, $AB>AC$. the incircle $I$ of $\triangle ABC$ meet $BC$ at point $D$, $AD$ meet $I$ again at $E$. $EP$ is a tangent of $I$, and $EP$ meet the extension line of $BC$ at $P$. $CF\parallel PE$, $CF\cap AD=F$. the line $BF$ meet $I$ at $M,N$, point $M$ is on the line segment $BF$, the line segment $PM$ meet $I$ again at $Q$. Show that $\angle ENP=\angle ENQ$
3 replies
s372102
Aug 10, 2013
AGCN
Today at 2:11 AM
Equality with Fermat Point
nsato   13
N Apr 6, 2025 by Nari_Tom
Source: 2012 Baltic Way, Problem 11
Let $ABC$ be a triangle with $\angle A = 60^\circ$. The point $T$ lies inside the triangle in such a way that $\angle ATB = \angle BTC = \angle CTA = 120^\circ$. Let $M$ be the midpoint of $BC$. Prove that $TA + TB + TC = 2AM$.
13 replies
nsato
Nov 22, 2012
Nari_Tom
Apr 6, 2025
Equality with Fermat Point
G H J
G H BBookmark kLocked kLocked NReply
Source: 2012 Baltic Way, Problem 11
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nsato
15654 posts
#1 • 4 Y
Y by soheil74, dmusurmonov, Adventure10, Mango247
Let $ABC$ be a triangle with $\angle A = 60^\circ$. The point $T$ lies inside the triangle in such a way that $\angle ATB = \angle BTC = \angle CTA = 120^\circ$. Let $M$ be the midpoint of $BC$. Prove that $TA + TB + TC = 2AM$.
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Nguyenhuyhoang
207 posts
#2 • 2 Y
Y by Adventure10, Mango247
Construct equilateral triangle $BCU$ outside triangle $ABC$, $AU$ intersects $(O)$ at $I$, we easily have $A,T,U$ are collinear and $B,C,U,T$ are concyclic, this leads to $TB+TC=TU \Rightarrow TA+TB+TC=AU$.
Construct parallelogram $ABNC$, now we only have to prove that $AU=AN$. Notice that $UB, UC$ are tangents of $(O)$ at $B,C$, so we have $ABIC$ is a harmonic quadrilateral and $AI$ is the symmedian of triangle $ABC$. We have $\widehat{BAI}=\widehat{CAM}$ and after several angle calculations, we have $\widehat{AUN}=\widehat{ANU}$, hence proved
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sunken rock
4392 posts
#3 • 2 Y
Y by Adventure10, Mango247
As before, construct the parallelogram $ABNC$; additionally, construct the equilateral triangle $\Delta ABP$, $C$ and $P$ on different sides of $AB$.
We see that a $60^\circ$ rotation about $B$ will map $A$ to $P$ and $U$ to $C$, hence $AU=PC$ (Torricelli problem).
On the other side we see that $\Delta PAC\equiv\Delta NCA$ (s.a.s.), and $AN=PC$, hence $AU=PC=AN$, done.

Best regards,
sunken rock
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vslmat
154 posts
#4 • 3 Y
Y by soheil74, Adventure10, Mango247
Another proof:

Let $O$ be the circumcenter of $ABC$. Is is obvious that $T$ must lie on the circumcircle of $BOC, AT$ meets this circle again at $S$. Then $\Delta BSC$ is equilateral. If we choose a point $F$ on $AS$ so that $BF = BT$ then $\Delta BTF$ is also equilateral. But then it is easy to see that $\Delta BTC\cong\Delta BFS$, hence $TC = FS$. Thus $TA + TB + TC = AS$ and to complete the proof it remains to show that $AS = 2.AM$
Notice that $SB, SC$ are in fact tangents to the circumcircle of $ABC$ and $AS$ is the A-symmedian, then $\angle BAS = \angle MAC$. By sinus law in $AMC$: $AM/sinC = MC/sinMAC$ and in triangle $ABS$: $AS/sinC = BS/sinBAS = 2. MC/sin MAC$. Indeed, $AS = 2.AM$ q.e.d.

Note; In general, the relationship between A-symmedian $AS$ and median $AM$ is $AM = AS. cosA$
Attachments:
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vlwk
12 posts
#5 • 2 Y
Y by Adventure10, Mango247
Let $AB=c$, $BC=a$, $AC=b$, $AM=d$. By Stewart's Theorem we have \[AM^2=(2d)^2=2b^2+2c^2-a^2=2b^2+2c^2-(b^2+c^2-2bc\cos 60^{\circ})=b^2+c^2+bc.\]Hence it suffices to show $b^2+c^2+bc=(TA+TB+TC)^2$. Now Cosine Rule on $\triangle{ATB}$, $\triangle{BTC}$, $\triangle{CTA}$ yields
\begin{align}
c^2&=TA^2+TB^2-2TA \cdot TB\cos 120^{\circ} \nonumber \\
&=TA^2+TB^2+TA \cdot TB \\
b^2&=TA^2+TC^2+TA \cdot TC \\
b^2+c^2-bc&=b^2+c^2-2bc\cos 60^{\circ} \nonumber \\
&=a^2 \nonumber \\
&=TB^2+TC^2+TB \cdot TC
\end{align}Now $(1)+(2)-(3)$ gives $bc=2TA^2+TA \cdot TB + TA \cdot TC - TB \cdot TC$. Denote this as $(4)$, then $(1)+(2)+(4)$ gives $b^2+c^2+bc=4TA^2+TB^2+TC^2+2TA\cdot TB+2TA\cdot TC-TB\cdot TC$. It suffices to show this is equivalent to $(TA+TB+TC)^2=TA^2+TB^2+TC^2+2TA\cdot TB+2TA\cdot TC+2TB\cdot TC \iff TA^2=TB\cdot TC$.

To prove this, extend $AT$ to $D$ such that $AT=TD$ and extend $BT$ to $E$ such that $TE=TC$. Then $\angle{CTA}=\angle{ATB}=\angle{DTE} \implies \triangle{DTE} \equiv \triangle{ATC}$. Therefore $\angle{ADE}=\angle{TDE}=\angle{TAC}=60^{\circ}-\angle{BAP}=\angle{ABP}=\angle{ABE} \implies ABDE$ is cyclic, so by Power of a Point $AP \cdot PD=BP \cdot PE \iff PA^2=PB\cdot PC$, as desired. Hence done.
This post has been edited 3 times. Last edited by vlwk, Jul 28, 2016, 10:15 AM
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KRIS17
134 posts
#6 • 1 Y
Y by Adventure10
The problem becomes trivial once you observe that $T$ is the circumcenter of $triangle\ ABC$ implying that $ABC$ is an equilateral triangle!

How come no one observed this? Am I missing something?
This post has been edited 1 time. Last edited by KRIS17, Aug 28, 2019, 2:32 PM
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Pluto1708
1107 posts
#7 • 1 Y
Y by Adventure10
KRIS17 wrote:
The problem becomes trivial once you observe that $T$ is the circumcenter of $triangle\ ABC$ implying that $ABC$ is an equilateral triangle!

How come no one observed this? Am I missing something?

Clearly $T$ is not the circumcenter of $ABC$ so $ABC$ is not equilateral
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KRIS17
134 posts
#8 • 1 Y
Y by Adventure10
Pluto1708 wrote:
KRIS17 wrote:
The problem becomes trivial once you observe that $T$ is the circumcenter of $triangle\ ABC$ implying that $ABC$ is an equilateral triangle!

How come no one observed this? Am I missing something?

Clearly $T$ is not the circumcenter of $ABC$ so $ABC$ is not equilateral

It is given that $\angle BTC = 2 * \angle BAC$ (120 = 2*60)
So why can't we use Central angle theorem?
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LKira
252 posts
#9 • 2 Y
Y by Adventure10, Mango247
KRIS17 wrote:
Pluto1708 wrote:
KRIS17 wrote:
The problem becomes trivial once you observe that $T$ is the circumcenter of $triangle\ ABC$ implying that $ABC$ is an equilateral triangle!

How come no one observed this? Am I missing something?

Clearly $T$ is not the circumcenter of $ABC$ so $ABC$ is not equilateral

It is given that $\angle BTC = 2 * \angle BAC$ (120 = 2*60)
So why can't we use Central angle theorem?
$T$ lies on circumcircle of $BOC,$ not circumcenter
Look at the figure at post 4
This post has been edited 1 time. Last edited by LKira, Aug 28, 2019, 4:27 PM
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KRIS17
134 posts
#10 • 2 Y
Y by Adventure10, Mango247
LKira wrote:
KRIS17 wrote:
Pluto1708 wrote:
Clearly $T$ is not the circumcenter of $ABC$ so $ABC$ is not equilateral

It is given that $\angle BTC = 2 * \angle BAC$ (120 = 2*60)
So why can't we use Central angle theorem?
$T$ lies on circumcircle of $BOC,$ not circumcenter
Look at the figure at post 4
True, but my point is that $O$ and $T$ happen to be one and the same as per the given inputs in the problem using central angle theorem on Point $T$ and vertex $A$.
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LKira
252 posts
#11 • 1 Y
Y by Adventure10
It is given that $\angle BTC = 2 * \angle BAC$ (120 = 2*60)
So why can't we use Central angle theorem?[/quote]
$T$ lies on circumcircle of $BOC,$ not circumcenter
Look at the figure at post 4[/quote]
True, but my point is that $O$ and $T$ happen to be one and the same as per the given inputs in the problem using central angle theorem on Point $T$ and vertex $A$.[/quote]

Did you look at the figure at post 4 ?
O and T coincide is just one small case, not the whole problem
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KRIS17
134 posts
#12 • 1 Y
Y by Adventure10
Even though most people have given the solution in the general case, I still believe that the problem indirectly asks about the special case where $T$ coincides with circumcenter (due to the inputs given in the problem).
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rafaello
1079 posts
#13
Y by
Let $X$ be the point on $AT$ such that $XBC$ is equilateral triangle. Let $A'$ be the reflection of $A$ over $M$.

By Ptolemy, $TX=TB+TC$. Hence, we need $AX=AA'$. Reflect diagram over $BC$, note that $X$ goes to $N$, the midpoint of arc $BAC$ and $A'O$ goes to $AK$, where $K$ is the midpoint of arc $BC$ as $\angle BAC=60^\circ$. Thus, $A'X\parallel AN\perp AK$. Also $K$ lies on the perpendicular bisector of $A'X$ as it is center of $(BTC)$. We conclude that $AX=AA'$.

[asy]import olympiad;
size(9cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta;

pair A,B,C,M,a,I,x,X,T,N,K,O; 
A=dir(120);B=dir(210);C=dir(330);M=midpoint(B--C);a=2M-A;path w=circumcircle(a,B,C);I=incenter(A,B,C);x=foot(a,A,I);X=2x-a;T=intersectionpoints(A--X,w)[0];N=2M-X;K=extension(X,N,A,I);O=(0,0);

draw(A--B--C--cycle,deep);draw(w,deep);draw(A--X,med);draw(A--a,med);draw(B--X--C,deep);draw(B--T,med);draw(C--T,med);draw(circumcircle(A,B,C),deep);draw(A--N,deep);

dot("$A$",A,dir(A)); 
dot("$B$",B,dir(B)); 
dot("$C$",C,dir(C)); 
dot("$M$",M,dir(M));
dot("$A'$",a,dir(a));
dot("$X$",X,dir(X));
dot("$T$",T,dir(T));
dot("$N$",N,dir(N));dot("$K$",K,dir(K));dot("$O$",O,N);
[/asy]
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Nari_Tom
117 posts
#14
Y by
I will provide nice lemma which technically solves the problem.

Lemma: Let $X$ be the point in circumcircle of equilateral triangle $ABC$. Let's assume $X$ lies on minor arc $BC$, Then we have $AX=BX+CX$.

Let's construct equilateral triangles $AZB$ and $AYC$ outside of the $ABC$. Let $X=ZB \cap YC$. Let $T'=ZC \cap BY$. It's easy to conclude that $T'=T$. Since $AZXC$ is a isosceles trapezoid, we have that $AT=ZC$. But $ZC=ZT+TC=TB+TA+TC$, by our lemma and we are done.
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