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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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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
Easy Diophantne
anantmudgal09   20
N 15 minutes ago by Adywastaken
Source: India Practice TST 2017 D1 P2
Find all positive integers $p,q,r,s>1$ such that $$p!+q!+r!=2^s.$$
20 replies
anantmudgal09
Dec 9, 2017
Adywastaken
15 minutes ago
Converse of a classic orthocenter problem
spartacle   43
N 24 minutes ago by ihategeo_1969
Source: USA TSTST 2020 Problem 6
Let $A$, $B$, $C$, $D$ be four points such that no three are collinear and $D$ is not the orthocenter of $ABC$. Let $P$, $Q$, $R$ be the orthocenters of $\triangle BCD$, $\triangle CAD$, $\triangle ABD$, respectively. Suppose that the lines $AP$, $BQ$, $CR$ are pairwise distinct and are concurrent. Show that the four points $A$, $B$, $C$, $D$ lie on a circle.

Andrew Gu
43 replies
spartacle
Dec 14, 2020
ihategeo_1969
24 minutes ago
Symmetric points part 2
CyclicISLscelesTrapezoid   22
N 25 minutes ago by ihategeo_1969
Source: USA TSTST 2022/6
Let $O$ and $H$ be the circumcenter and orthocenter, respectively, of an acute scalene triangle $ABC$. The perpendicular bisector of $\overline{AH}$ intersects $\overline{AB}$ and $\overline{AC}$ at $X_A$ and $Y_A$ respectively. Let $K_A$ denote the intersection of the circumcircles of triangles $OX_AY_A$ and $BOC$ other than $O$.

Define $K_B$ and $K_C$ analogously by repeating this construction two more times. Prove that $K_A$, $K_B$, $K_C$, and $O$ are concyclic.

Hongzhou Lin
22 replies
CyclicISLscelesTrapezoid
Jun 27, 2022
ihategeo_1969
25 minutes ago
Periodicity of factorials
Cats_on_a_computer   0
43 minutes ago
Source: Thrill and challenge of pre-college mathematics
Let a_k denote the first non zero digit of the decimal representation of k!. Does the sequence a_1, a_2, a_3, … eventually become periodic?
0 replies
Cats_on_a_computer
43 minutes ago
0 replies
Cyclic Quad. and Intersections
Thelink_20   11
N an hour ago by americancheeseburger4281
Source: My Problem
Let $ABCD$ be a quadrilateral inscribed in a circle $\Gamma$. Let $AC\cap BD=E$, $AB\cap CD=F$, $(AEF)\cap\Gamma=X$, $(BEF)\cap\Gamma=Y$, $(CEF)\cap\Gamma=Z$, $(DEF)\cap\Gamma=W$, $XZ\cap YW=M$, $XY\cap ZW=N$. Prove that $MN$ lies over $EF$.
11 replies
Thelink_20
Oct 29, 2024
americancheeseburger4281
an hour ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   15
N an hour ago by math90
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
15 replies
OgnjenTesic
May 22, 2025
math90
an hour ago
Easy Number Theory
math_comb01   39
N an hour ago by Adywastaken
Source: INMO 2024/3
Let $p$ be an odd prime and $a,b,c$ be integers so that the integers $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$are divisible by $p$.
Prove that $p$ divides each of $a,b,c$.
$\quad$
Proposed by Navilarekallu Tejaswi
39 replies
math_comb01
Jan 21, 2024
Adywastaken
an hour ago
Painting Beads on Necklace
amuthup   46
N an hour ago by quantam13
Source: 2021 ISL C2
Let $n\ge 3$ be a fixed integer. There are $m\ge n+1$ beads on a circular necklace. You wish to paint the beads using $n$ colors, such that among any $n+1$ consecutive beads every color appears at least once. Find the largest value of $m$ for which this task is $\emph{not}$ possible.

Carl Schildkraut, USA
46 replies
amuthup
Jul 12, 2022
quantam13
an hour ago
Iran geometry
Dadgarnia   38
N 2 hours ago by cursed_tangent1434
Source: Iranian TST 2018, first exam day 2, problem 4
Let $ABC$ be a triangle ($\angle A\neq 90^\circ$). $BE,CF$ are the altitudes of the triangle. The bisector of $\angle A$ intersects $EF,BC$ at $M,N$. Let $P$ be a point such that $MP\perp EF$ and $NP\perp BC$. Prove that $AP$ passes through the midpoint of $BC$.

Proposed by Iman Maghsoudi, Hooman Fattahi
38 replies
Dadgarnia
Apr 8, 2018
cursed_tangent1434
2 hours ago
hard problem (to me)
kjhgyuio   2
N 2 hours ago by kjhgyuio
........
2 replies
kjhgyuio
Apr 19, 2025
kjhgyuio
2 hours ago
PE is bisector of BPC
goldeneagle   44
N 2 hours ago by cursed_tangent1434
Source: Iran TST 2012 -first day- problem 2
Consider $\omega$ is circumcircle of an acute triangle $ABC$. $D$ is midpoint of arc $BAC$ and $I$ is incenter of triangle $ABC$. Let $DI$ intersect $BC$ in $E$ and $\omega$ for second time in $F$. Let $P$ be a point on line $AF$ such that $PE$ is parallel to $AI$. Prove that $PE$ is bisector of angle $BPC$.

Proposed by Mr.Etesami
44 replies
goldeneagle
Apr 23, 2012
cursed_tangent1434
2 hours ago
find question
mathematical-forest   9
N 2 hours ago by JARP091
Are there any contest questions that seem simple but are actually difficult? :-D
9 replies
mathematical-forest
May 29, 2025
JARP091
2 hours ago
Interesting inequality
sqing   1
N 2 hours ago by Zok_G8D
Source: Own
Let $  a, b,c>0,b+c\geq 3a$. Prove that
$$ \sqrt{\frac{a}{b+c-a}}-\frac{ 2a^2-b^2-c^2}{(a+b)(a+c)}\geq \frac{2}{5}+\frac{1}{\sqrt 2}$$$$ \frac{3}{2}\sqrt{\frac{a}{b+c-a}}-\frac{ 2a^2-b^2-c^2}{(a+b)(a+c)}\geq \frac{2}{5}+\frac{3}{2\sqrt 2}$$
1 reply
sqing
Yesterday at 2:49 AM
Zok_G8D
2 hours ago
Basic ideas in junior diophantine equations
Maths_VC   6
N 2 hours ago by Adywastaken
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
6 replies
Maths_VC
May 27, 2025
Adywastaken
2 hours 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
4149 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
9801 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
698 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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827681
163 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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