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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
Social Club with 2k+1 Members
v_Enhance   24
N 5 minutes ago by mathwiz_1207
Source: USA December TST for IMO 2013, Problem 1
A social club has $2k+1$ members, each of whom is fluent in the same $k$ languages. Any pair of members always talk to each other in only one language. Suppose that there were no three members such that they use only one language among them. Let $A$ be the number of three-member subsets such that the three distinct pairs among them use different languages. Find the maximum possible value of $A$.
24 replies
v_Enhance
Jul 30, 2013
mathwiz_1207
5 minutes ago
A strong inequality problem
hn111009   1
N 10 minutes ago by Tung-CHL
Source: Somewhere
Let $a,b,c$ be the positive number satisfied $a^2+b^2+c^2=3.$ Find the minimum of $$P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{3abc}{2(ab+bc+ca)}.$$
1 reply
hn111009
Today at 2:02 AM
Tung-CHL
10 minutes ago
Altitude configuration with two touching circles
Tintarn   3
N 11 minutes ago by NumberzAndStuff
Source: Austrian MO 2024, Final Round P2
Let $ABC$ be an acute triangle with $AB>AC$. Let $D,E,F$ denote the feet of its altitudes on $BC,AC$ and $AB$, respectively. Let $S$ denote the intersection of lines $EF$ and $BC$. Prove that the circumcircles $k_1$ and $k_2$ of the two triangles $AEF$ and $DES$ touch in $E$.

(Karl Czakler)
3 replies
Tintarn
Jun 1, 2024
NumberzAndStuff
11 minutes ago
IMO Shortlist 2009 - Problem G3
April   48
N 18 minutes ago by Markas
Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram.
Prove that $GR=GS$.

Proposed by Hossein Karke Abadi, Iran
48 replies
April
Jul 5, 2010
Markas
18 minutes ago
Show that (DEN) passes through the midpoint of BC
v_Enhance   24
N 19 minutes ago by Markas
Source: Sharygin First Round 2013, Problem 21
Chords $BC$ and $DE$ of circle $\omega$ meet at point $A$. The line through $D$ parallel to $BC$ meets $\omega$ again at $F$, and $FA$ meets $\omega$ again at $T$. Let $M = ET \cap BC$ and let $N$ be the reflection of $A$ over $M$. Show that $(DEN)$ passes through the midpoint of $BC$.
24 replies
v_Enhance
Apr 7, 2013
Markas
19 minutes ago
2020 EGMO P5: P is the incentre of CDE
alifenix-   49
N 20 minutes ago by Markas
Source: 2020 EGMO P5
Consider the triangle $ABC$ with $\angle BCA > 90^{\circ}$. The circumcircle $\Gamma$ of $ABC$ has radius $R$. There is a point $P$ in the interior of the line segment $AB$ such that $PB = PC$ and the length of $PA$ is $R$. The perpendicular bisector of $PB$ intersects $\Gamma$ at the points $D$ and $E$.

Prove $P$ is the incentre of triangle $CDE$.
49 replies
alifenix-
Apr 18, 2020
Markas
20 minutes ago
The reflection of AD intersect (ABC) lies on (AEF)
alifenix-   61
N 20 minutes ago by Markas
Source: USA TST for EGMO 2020, Problem 4
Let $ABC$ be a triangle. Distinct points $D$, $E$, $F$ lie on sides $BC$, $AC$, and $AB$, respectively, such that quadrilaterals $ABDE$ and $ACDF$ are cyclic. Line $AD$ meets the circumcircle of $\triangle ABC$ again at $P$. Let $Q$ denote the reflection of $P$ across $BC$. Show that $Q$ lies on the circumcircle of $\triangle AEF$.

Proposed by Ankan Bhattacharya
61 replies
alifenix-
Jan 27, 2020
Markas
20 minutes ago
JBMO 2013 Problem 2
Igor   44
N 21 minutes ago by Markas
Source: Proposed by Macedonia
Let $ABC$ be an acute-angled triangle with $AB<AC$ and let $O$ be the centre of its circumcircle $\omega$. Let $D$ be a point on the line segment $BC$ such that $\angle BAD = \angle CAO$. Let $E$ be the second point of intersection of $\omega$ and the line $AD$. If $M$, $N$ and $P$ are the midpoints of the line segments $BE$, $OD$ and $AC$, respectively, show that the points $M$, $N$ and $P$ are collinear.
44 replies
Igor
Jun 23, 2013
Markas
21 minutes ago
Problem 1: Triangle triviality
ZetaX   134
N 21 minutes ago by Markas
Source: IMO 2006, 1. day
Let $ABC$ be triangle with incenter $I$. A point $P$ in the interior of the triangle satisfies \[\angle PBA+\angle PCA = \angle PBC+\angle PCB.\] Show that $AP \geq AI$, and that equality holds if and only if $P=I$.
134 replies
ZetaX
Jul 12, 2006
Markas
21 minutes ago
Incenter and midpoint geom
sarjinius   90
N 22 minutes ago by Markas
Source: 2024 IMO Problem 4
Let $ABC$ be a triangle with $AB < AC < BC$. Let the incenter and incircle of triangle $ABC$ be $I$ and $\omega$, respectively. Let $X$ be the point on line $BC$ different from $C$ such that the line through $X$ parallel to $AC$ is tangent to $\omega$. Similarly, let $Y$ be the point on line $BC$ different from $B$ such that the line through $Y$ parallel to $AB$ is tangent to $\omega$. Let $AI$ intersect the circumcircle of triangle $ABC$ at $P \ne A$. Let $K$ and $L$ be the midpoints of $AC$ and $AB$, respectively.
Prove that $\angle KIL + \angle YPX = 180^{\circ}$.

Proposed by Dominik Burek, Poland
90 replies
sarjinius
Jul 17, 2024
Markas
22 minutes ago
Rest in peace, Geometry!
mathisreaI   85
N 23 minutes ago by Markas
Source: IMO 2022 Problem 4
Let $ABCDE$ be a convex pentagon such that $BC=DE$. Assume that there is a point $T$ inside $ABCDE$ with $TB=TD,TC=TE$ and $\angle ABT = \angle TEA$. Let line $AB$ intersect lines $CD$ and $CT$ at points $P$ and $Q$, respectively. Assume that the points $P,B,A,Q$ occur on their line in that order. Let line $AE$ intersect $CD$ and $DT$ at points $R$ and $S$, respectively. Assume that the points $R,E,A,S$ occur on their line in that order. Prove that the points $P,S,Q,R$ lie on a circle.
85 replies
mathisreaI
Jul 13, 2022
Markas
23 minutes ago
Prove DK and BC are perpendicular.
yunxiu   62
N 23 minutes ago by Markas
Source: 2012 European Girls’ Mathematical Olympiad P1
Let $ABC$ be a triangle with circumcentre $O$. The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$. (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.)
Let $K$ be the circumcentre of triangle $AFE$. Prove that the lines $DK$ and $BC$ are perpendicular.

Netherlands (Merlijn Staps)
62 replies
yunxiu
Apr 13, 2012
Markas
23 minutes ago
AT // BC wanted
parmenides51   104
N 24 minutes ago by Markas
Source: IMO 2019 SL G1
Let $ABC$ be a triangle. Circle $\Gamma$ passes through $A$, meets segments $AB$ and $AC$ again at points $D$ and $E$ respectively, and intersects segment $BC$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $BDF$ at $F$ and the tangent to circle $CEG$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $AT$ is parallel to $BC$.

(Nigeria)
104 replies
parmenides51
Sep 22, 2020
Markas
24 minutes ago
IMO Shortlist 2010 - Problem G1
Amir Hossein   133
N 25 minutes ago by Markas
Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$

Proposed by Christopher Bradley, United Kingdom
133 replies
Amir Hossein
Jul 17, 2011
Markas
25 minutes ago
Easy Geometry
TheOverlord   33
N Apr 21, 2025 by math.mh
Source: Iran TST 2015, exam 1, day 1 problem 2
$I_b$ is the $B$-excenter of the triangle $ABC$ and $\omega$ is the circumcircle of this triangle. $M$ is the middle of arc $BC$ of $\omega$ which doesn't contain $A$. $MI_b$ meets $\omega$ at $T\not =M$. Prove that
$$ TB\cdot TC=TI_b^2.$$
33 replies
TheOverlord
May 10, 2015
math.mh
Apr 21, 2025
Easy Geometry
G H J
G H BBookmark kLocked kLocked NReply
Source: Iran TST 2015, exam 1, day 1 problem 2
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TheOverlord
97 posts
#1 • 2 Y
Y by Dadgarnia, Adventure10
$I_b$ is the $B$-excenter of the triangle $ABC$ and $\omega$ is the circumcircle of this triangle. $M$ is the middle of arc $BC$ of $\omega$ which doesn't contain $A$. $MI_b$ meets $\omega$ at $T\not =M$. Prove that
$$ TB\cdot TC=TI_b^2.$$
This post has been edited 2 times. Last edited by djmathman, Sep 10, 2015, 3:59 AM
Reason: latex/formatting
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Luis González
4148 posts
#2 • 6 Y
Y by phantranhuongth, AlastorMoody, lahmacun, Eliot, Adventure10, Mango247
Let $I$ and $I_a,I_c$ be the incenter and the excenters of $\triangle ABC$ againts $A,C.$ Since $\omega$ is 9-point circle of $\triangle I_aI_bI_c,$ then $\omega$ cuts $II_a,I_aI_b$ at their midpoints $M,N$ $\Longrightarrow$ $MN \parallel BI_b$ $\Longrightarrow$ $\angle BI_bT=\angle TMN=\angle TCI_b.$ Since $\angle BTI_b=\angle CTI_b$ ($TM$ bisects $\angle BTC$), then $\triangle TBI_b \sim \triangle TI_bC$ $\Longrightarrow$ $\tfrac{TB}{TI_b}=\tfrac{TI_b}{TC}$ $\Longrightarrow$ $TB \cdot TC={TI_b}^2.$
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aditya21
717 posts
#3 • 2 Y
Y by Adventure10, Mango247
nice and easy!
my solution = as $ABMT,ATCM$ are concylic hence
$\angle MTB=\angle MTC=\frac{\angle A}{2}$ which is due to $AM$ bisecting $\angle BAC$
and thus $\angle CTI_b=\angle I_bTC=180-\frac{\angle A}{2}$
also now let $\angle TCI_b=x$ than $\angle TCB=90+\frac{\angle C}{2}-x$
and thus $\angle CBT=180-\angle TCB=90-\frac{\angle C}{2}+x$
and so $\angle TBI_b=\frac{\angle B}{2}-\angle CBT=\frac{\angle A}{2}-x=\angle CI_bT$
and thus $\triangle TBI_b\sim \triangle TI_bC \Longrightarrow TB.TC={TI_b}^2$

so we are done :D
This post has been edited 2 times. Last edited by aditya21, May 19, 2015, 6:56 AM
Reason: e
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tranquanghuy7198
253 posts
#4 • 2 Y
Y by AlastorMoody, Adventure10
My solution:
Let $D$ be the midpoint of arc $BC$ containing $A$
$\Rightarrow$ $D$ is the circumcenter of $\triangle{I_bBC}$ and $MB, MC$ are tangent to $(I_bBC)$
$\Rightarrow$ $I_bM$ is the symmedian of $\triangle{I_bBC}$
Moreover, $T$ lies on the symmedian such that $TI_b$ bisects $\angle{BTC}$ $\Rightarrow$ $\triangle{TBI_b}$$\sim$$\triangle{TI_bC}$ and the conclusion follows
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colinhy
751 posts
#5 • 1 Y
Y by Adventure10
Here's a solution:
Observe that $\angle BTC = \angle A$ and since $MI_B$ is an angle bisector of $\angle BTC$, $\angle BTI_B = \angle CTI_B = 180 - \frac{1}{2}\angle A$. Now, let $D$ be the second intersection of the circumcircle $\tau$ of $CTI_B$ and $BC$, so we have $\angle CDI_B = \frac{1}{2} \angle A$, which means that $\triangle BCI_B \sim \triangle BI_BD$ and $BI_B$ is tangent to $\tau$, so $\angle TCI_B = \angle TI_BB$, so by AA symmetry, $\triangle BTI_B \sim \triangle I_BTC$, and the result follows.

Here's a slightly more difficult problem based on the above:
Let $I$ be the incenter of $\triangle ABC$. Show that it is possible to construct a right triangle with side lengths $IM, MT, TI_B$.
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infiniteturtle
1131 posts
#6 • 2 Y
Y by Adventure10, Mango247
Extravert (mostly for convenience) to get
Quote:
$ABC$ is a triangle with circumcircle $w$. Let $I$ be the incenter, $M$ be the midpoint of arc $BC$ not including $A$, and let $N$ be the midpoint of arc $BAC$. If $NI\cap w=T$, show that $TI^2=TB\cdot TC.$

The problem now falls easily after drawing $(BIC)$: Let $IT\cap (BIC)=X,  CT\cap (BIC)=Y$. It's trivial by angle chasing that $BT=TY$ and that $IT=TX$, now PoP kills it.
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Dukejukem
695 posts
#7 • 1 Y
Y by Adventure10
Let $N$ be the midpoint of arc $\widehat{BAC}$ of $\omega$ and let $I_c$ be the $C$-excenter. It is well-known that $B, C, I_b, I_c$ are inscribed in the circle $\Gamma$ of diameter $\overline{I_bI_c}$ with center $N.$ Moreover, note that $M$ is the midpoint of arc $\widehat{BC}$, implying that $TM$ and $TN$ are the internal and external bisectors of $\angle BTC$, respectively. Therefore, the inversion with power $TB \cdot TC$ combined with a reflection in $TM$ swaps $B$ and $C$ fixes lines $TM$ and $TN.$ It follows that the center of $\Gamma'$ (the image of $\Gamma$) is a point on line $TN.$ However, the center of $\Gamma'$ must also lie on the perpendicular bisector of $\overline{BC}$, implying that the center is $N$ itself. Therefore, $\Gamma$ is fixed under this inversion, and hence $I_b$ is fixed as well. Thus, $TI_b^2 = TB \cdot TC$ as desired. $\square$
This post has been edited 1 time. Last edited by Dukejukem, Sep 10, 2015, 9:54 PM
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hayoola
123 posts
#8 • 2 Y
Y by Adventure10, Mango247
the interestion between IbC and w is thw midpoint of arc ABC
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soroush.MG
158 posts
#9 • 2 Y
Y by Adventure10, Mango247
Let $D$ be the midpoint of arc $BAC$. And $c$ is a circle with centre $D$ and radius $DB$. $c \cap I_bT=E$.
As $ATM=90$ and $AI_b=AE$ so $TE=TI_b$ and it's enough to show that $TE^2=TB.TC$ We know that $MC$ and $MB$ are tangent to $c$. So we can simply show that $\triangle TCE \sim \triangle TEB$ and $TE^2=TB.TC$.
This post has been edited 1 time. Last edited by soroush.MG, Feb 19, 2017, 8:30 PM
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tenplusten
1000 posts
#10 • 1 Y
Y by Adventure10
An easy barycentric problem.
We have $I_b=(a:-b:c)$ and $M=(-a^2:b(b+c):c(b+c))$.It's easy to find the coordinates of $T$ by intersecting $MI_b$ and $\omega$.So we finish problem using Distance formula.
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artsolver
139 posts
#11 • 1 Y
Y by Adventure10
Murad.Aghazade wrote:
An easy barycentric problem.
We have $I_b=(a:-b:c)$ and $M=(-a^2:b(b+c):c(b+c))$.It's easy to find the coordinates of $T$ by intersecting $MI_b$ and $\omega$.So we finish problem using Distance formula.

Well, if you are about to bash, then bash to the end, since it is known that you can always finish problem that way, no matter how you start, but if you are doing it synthetic then it's nice to write just sketch of the proof...
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Anar24
475 posts
#12 • 3 Y
Y by tenplusten, Adventure10, Mango247
artsolver wrote:
Murad.Aghazade wrote:
An easy barycentric problem.
We have $I_b=(a:-b:c)$ and $M=(-a^2:b(b+c):c(b+c))$.It's easy to find the coordinates of $T$ by intersecting $MI_b$ and $\omega$.So we finish problem using Distance formula.

Well, if you are about to bash, then bash to the end, since it is known that you can always finish problem that way, no matter how you start, but if you are doing it synthetic then it's nice to write just sketch of the proof...

Tenplusten gave the sketch of bary solution,and it is good idea to allow others to finish by themselves.For that reason it is not correct to deny others for skipping the long calculations.
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WolfusA
1900 posts
#13 • 2 Y
Y by Adventure10, Mango247
In math publications leaving long calculations is acceptable, but you have to give final result or conclusion following from it.
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Taha1381
816 posts
#14 • 3 Y
Y by AlastorMoody, Adventure10, Mango247
It is just sum of some lemmas:).First note that $MC,MB$ are both tangent to circumcircle of $I_bBC$ by angle chasing.Also $BMCT$ is cyclic so $T$ is the midpoint of $I_b$ symmedian in triangle $BCI_B$ and so two triangles $TI_bC$ is similar to $TBI_b$ from which the relation follows.
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FISHMJ25
293 posts
#16 • 2 Y
Y by Adventure10, Mango247
Here is complex bash solution. $\omega$ is unit circle, $a=x^2$ $b=y^2$ $ c=z^2$. So $m=-yz$ and $Ib=xy+yz-xz$. We now want $t$. But its easy just solve $$\frac{m-t}{\bar m -\bar t }=\frac{m-Ib}{\bar m -\bar Ib }$$.We get $$t=\frac{x(xy+2yz-xz)}{z+2x-y}$$And now we need to prove $$|(t-y^2)(t-z^2)|=|Ib-t|^2$$. After substituting $t$ and $Ib$ we see that both sides are equal to $$\frac{(z+x)^2(x-y)^2(x-z)^2}{(z+2x-y)^2}$$and we are done.
Calculations are not bad can be done under 20 minutes.
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ComiCabE
25 posts
#17 • 1 Y
Y by Adventure10
Here is a different approach.

Denote by $N$ the midpoint of $\widehat{BCA}$. With elementary angle chasing we may show that $MN \parallel BI_B$. This implies $\angle TCI_B =\angle TMN =\angle TI_BB$. This combined with $\angle MTC =\angle MTB$ gives $\triangle I_BTC \sim \triangle I_BBT$, and the desired equality follows $\square$
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Krits
147 posts
#18
Y by
observe that $TM$ bisects $\angle CTB$ so $\angle CTI_B=\angle BTI_B$ let $(BCI_B)$ intersect $MT$ at $U$ so $\angle CUB=180-\frac{\angle A}{2}$ combining with $\angle CI_B U=\angle CBU$ we are done because we have $\triangle I_B TC \sim \triangle TBI_B$
This post has been edited 1 time. Last edited by Krits, Apr 11, 2020, 9:43 AM
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mathaddiction
308 posts
#19
Y by
We will show that $T$ is the $I_b-$dumpty point of $\triangle BI_bC$ which implies the result since the dumpty point is the center of spiral similarity sending $\overline{CI_b}$ to $\overline{I_bC}$.
Notice that the dumpty point is the intersection of the $I_b-$symmedian and $(BOC)$, where $O$ is the circumcenter of $\triangle BI_bC$. Now by Ceva's theorem
$$\frac{\sin\angle BI_bM}{\sin\angle CI_bT}\cdot\frac{\sin\angle I_bCM}{\sin\angle MCB}\cdot\frac{\sin\angle MBC}{\sin\angle MBI_b}=1$$Since $\angle MBC=\angle MCB$,
$$\frac{\sin\angle BI_bM}{\sin\angle CI_bT}=\frac{\sin\angle MBI_b}{\sin\angle I_bCM}=\frac{\cos\frac{C}{2}}{\sin\frac{B}{2}}=\frac{\sin\angle I_bCB}{\sin\angle I_bBC}$$which implies that $MT$ is a symmedian. Moreover,
$$\angle BTC=\angle A=2\angle IAC=2\angle II_bC=\angle BOC$$Hence $BI_BOC$ is cyclic which completes the proof.
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Gaussian_cyber
162 posts
#20 • 2 Y
Y by amar_04, Eliot
Storage
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jayme
9792 posts
#21
Y by
Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Le%20produit%20AB.AC.pdf p. 35-38.

Sincerely
Jean-Louis
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TheUltimate123
1740 posts
#22 • 1 Y
Y by snakeaid
Solved with nukelauncher.

First observe \(\measuredangle BTI_B=\measuredangle I_BTC\) since \(\overline{TM}\) bisects \(\angle BTC\).

[asy]         size(5cm); defaultpen(fontsize(10pt));

pair B,A,C,I,M,NN,IB,T,X;         B=dir(110);         A=dir(210);         C=dir(330);         I=incenter(A,B,C);         M=extension( (B+C)/2,origin,A,I);         NN=extension( (C+A)/2,origin,B,I);         IB=2NN-I;         T=2*foot(origin,M,IB)-M;         X=2*foot(origin,IB,C)-C;

draw(circumcircle(A,C,IB),gray);         draw(M--A,gray);         draw(B--IB,dashed);         draw(C--IB,dashed);         draw(unitcircle);         draw(A--B--C--cycle);         draw(M--IB);

dot("\(A\)",A,A);         dot("\(B\)",B,B);         dot("\(C\)",C,C);         dot("\(M\)",M,M);         dot("\(N\)",NN,SW);         dot("\(I_B\)",IB,S);         dot("\(T\)",T,SE);         dot("\(I\)",I,NW);     [/asy]

In addition, \[\measuredangle I_BTC=\measuredangle MTC=\measuredangle IAC=\measuredangle BI_BC=\measuredangle BI_BT+\measuredangle TI_BC,\]thus \(\measuredangle BI_BT=\measuredangle I_BTC+\measuredangle CI_BT=\measuredangle I_BCT\), so \[\triangle TBI_B\sim\triangle TI_BC.\]It follows that \(TB\cdot TC=TI_B^2\), as needed
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snakeaid
125 posts
#23
Y by
I really need to learn directed angles
This post has been edited 1 time. Last edited by snakeaid, Oct 19, 2020, 6:37 PM
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rafaello
1079 posts
#25
Y by
We claim that $\triangle BTI_B\sim\triangle I_bTC$.
Obviously, we have that$$\angle BTM=\angle BAM=\angle MAC=\angle MTC\implies \angle I_bTB=\angle CTI_b.$$We obtain that $\angle I_bCT=\angle I_bCA-\angle TCA$ and
$$ \angle BI_bT=\angle AII_b-\angle AMI_b=180^\circ-\angle AIB -\angle TCA=90^\circ-0.5\angle ACB-\angle TCA=\angle I_bCT-\angle TCA,$$hence we conclude that $\angle I_bCT=\angle BI_bT$.
$TB\cdot TC=TI_b^2$ follows directly from $\triangle BTI_B\sim\triangle I_bTC$.
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AMN300
563 posts
#26
Y by
We claim triangles $CTI_B, I_BTB$ are similar, which clearly implies the result. Relabel $M \rightarrow M_A$ and let $M_B$ be the arc midpoint of minor arc $BC$. First remark $\angle CTM_A = \angle BTM_A = A/2$ so $\angle CTI_B = \angle BTI_B$. Next let $\angle M_B BT = x$. Straightforward angle chasing yields $\angle ACI_B = 90$, so $\angle TCI_B = 90-\frac{C}{2}-\angle ACT = 90-\frac{C}{2}-\frac{A}{2}-x = \frac{B}{2}-x$. But $\angle TI_B B = \frac{2(B/2)-2x}{2}=\frac{B}{2}-x$ by standard angle facts. So $\angle TCI_B = \angle TI_B B$, so we're done.
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hakN
429 posts
#27
Y by
It is easy to see that $\angle BTI_B = \angle CTI_B$. Let $N$ be the midpoint of the arc $\widehat{AC}$ not containing $B$. Becuase $NIMC$ is a kite, we get that $NM\perp IC$ and we know that $I_BC\perp IC \implies NM\parallel I_BC$. So we get $\angle NBT = \angle NMT = \angle TI_BC
\implies \triangle BTI_B \sim \triangle I_BTC \implies TB\cdot TC = TI_B^2$.
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nixon0630
31 posts
#28 • 1 Y
Y by javohirsultanov20
It's enough to show, that: $$\triangle BTI_B\sim\triangle CTI_B$$Claim. $MT$ bisects $\angle CTB$.
Proof.
Claim. $\angle TCI_B = \angle TI_BB$.
Proof.
Therefore $\triangle BTI_B\sim\triangle CTI_B$ by $ASA$. $\blacksquare$
Attachments:
This post has been edited 2 times. Last edited by nixon0630, Oct 19, 2021, 5:42 AM
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trying_to_solve_br
191 posts
#29
Y by
Nice and cute, notice first that, as $M$ is the midpoint of the arc, then $\angle BTI_b=\angle CTI_b$ but notice we want $TI_b/TC=TB/TI_b$ and thus we just need one more angle to prove $BTI_b \sim I_bTC$. We'll show $\angle I_bBT=\angle TI_bC$, but $\angle I_bBT=B/2-\angle TBC$ and as $\angle BTI_b=90+B/2+C/2=\angle TBC+\angle TI_bC+\angle BCI_b=\angle TBC+\angle TI_bC+90+C/2$ and thus $\angle TBC \angle TI_bC=B/2=\angle TBI_b+\angle TBC$, proving our claim.
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JAnatolGT_00
559 posts
#30
Y by
Note that $\angle CTI_b=$ $\angle I_bTB=$ $\pi$ $-\frac{\angle BAC}{2}=$ $\pi$ $-\angle BI_bC$ $\implies$ $\angle TBI_b=$ $\angle TI_bC.$
Hence $\triangle BTI_b\sim \triangle I_bTC\implies |TB|\cdot |TC|=|TI_b|^2$ $\blacksquare$
This post has been edited 2 times. Last edited by JAnatolGT_00, Oct 22, 2021, 10:28 AM
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Mahdi_Mashayekhi
695 posts
#31
Y by
We will prove BTIb and IbTC are similar and it will follow TB . TC = TIb^2.
∠BTIb = 180 - ∠MTB = 180 - ∠CTM = ∠IbTC. Let ∠TBIb = x. ∠ACT = ∠B/2 + x and ∠ACIb = 90 - ∠C/2 so ∠TCIb = 90 - ∠C/2 - ∠B/2 - x.
∠CIbT = ∠A/2 - ∠TCIb = ∠A/2 - 90 + ∠C/2 + ∠B/2 + x = x = ∠TBIb so BTIb and IbTC are similar.
we're Done.
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Project_Donkey_into_M4
162 posts
#34 • 1 Y
Y by Mango247
Overcomplicated headsolve
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David_Kim_0202
384 posts
#35
Y by
Too easy for Iran TST

1. think about using simillar triangle theory
2. just use angle moving for the evidemce of proof.
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math_comb01
662 posts
#36
Y by
Easy problem
Claim 1: $MB$ and $MC$ are tangent to $(BI_BC)$
Proof
Now this combined with $\measuredangle BTI_B = \measuredangle CTI_B$ gives that $T$ is $I_B-$ dumpty point so this implies the conclusion as
$\triangle BTI_B \sim \triangle I_BTC$
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Cali.Math
128 posts
#37
Y by
We uploaded our solution https://calimath.org/pdf/IranTST2015-2.pdf on youtube https://youtu.be/GClFR42aUoU.
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math.mh
1 post
#38
Y by
this problem was proposed by Ali Zamani
This post has been edited 1 time. Last edited by math.mh, Apr 21, 2025, 1:40 PM
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