The time is now - Spring classes are filling up!

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k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
There is still time to train for the November 6th and November 12th AMC 10A/12A and AMC 10B/12B, respectively! Enroll in our weekend seminars to be held on November 2nd and 3rd (listed below) and you will learn problem strategies, test taking techniques, and be able to take a full practice test! Note that the “B” seminars will have different material from the “A” seminars which were held in October.

[list][*]Special AMC 10 Problem Seminar B
[*]Special AMC 12 Problem Seminar B[/list]
For those who want to take a free practice test before the AMC 10/12 competitions, you can simulate a real competition experience by following this link. As you assess your performance on these exams, be sure to gather data!

[list][*]Which problems did you get right?
[list][*]Was the topic a strength (e.g. number theory, geometry, counting/probability, algebra)?
[*]How did you prepare?
[*]What was your confidence level?[/list]
[*]Which problems did you get wrong?
[list][list][*]Did you make an arithmetic error?
[*]Did you misread the problem?
[*]Did you have the foundational knowledge for the problem?
[*]Which topics require more fluency through practice (e.g. number theory, geometry, counting/probability, algebra)?
[*]Did you run out of time?[/list][/list]
Once you have analyzed the results with the above questions, you will have a plan of attack for future contests! BEST OF LUCK to all competitors at this year’s AMC 10 and AMC 12!

Did you know that the day after both the AMC 10A/12A and AMC 10B/12B you can join a free math jam where our AoPS team will go over the most interesting problems? Find the schedule below under “Mark your calendars”.

Mark your calendars for these upcoming free math jams!
[list][*]November 20th: Amherst College Info Session, 7:30 pm ET: Matt McGann, Dean of Admission and Financial Aid at Amherst College, and Nathan Pflueger, math professor at Amherst College, will host an info session exploring both Amherst College specifically and liberal arts colleges generally. Topics include opportunities in math, the admission process, and financial aid for both US and international students.
[*]November 7th: 2024 AMC 10/12 A Discussion, Thursday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 A, administered November 6. We will discuss some of the most interesting problems from each test!
[*]November 13th: 2024 AMC 10/12 B Discussion, Wednesday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 B, administered November 12. We will discuss some of the most interesting problems from each test![/list]
AoPS Spring classes are open for enrollment. Get a jump on the New Year and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile!

Whether you've taken our classes at AoPS Online or AoPS Academies or reached the top echelons of our competition training with our Worldwide Online Olympiad Training (WOOT) program, you can now add your AoPS experience to the education section on your LinkedIn profile.

Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

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0 replies
jlacosta
Nov 1, 2024
0 replies
infinite sequence
gnoka   1
N 8 minutes ago by BR1F1SZ
Source: 46th International Tournament of Towns, Senior A-Level P4, Fall 2024
Does there exist an infinite sequence of real numbers ${a}_{1},{a}_{2},{a}_{3},\ldots$ such that ${a}_{1} = 1$ and for all positive integers $k$ we have the equality

$$
{a}_{k} = {a}_{2k} + {a}_{3k} + {a}_{4k} + \ldots ?
$$
Ilya Lobatsky
1 reply
gnoka
18 minutes ago
BR1F1SZ
8 minutes ago
five integer division
gnoka   1
N 11 minutes ago by LLL2019
Source: 46th International Tournament of Towns, Senior O-Level P3, Fall 2024
3. There are five positive integers written in a row. Each one except for the first one is the minimal positive integer that is not a divisor of the previous one. Can all these five numbers be distinct?

Boris Frenkin
1 reply
+1 w
gnoka
38 minutes ago
LLL2019
11 minutes ago
Thanks u!
Ruji2018252   1
N 15 minutes ago by tobiSALT
$a,b,c\in [0,2]$ and $a+b+c=3$. Find min and max
$$P=\dfrac{6-a^2-b^2-c^2}{a+1}$$
1 reply
Ruji2018252
an hour ago
tobiSALT
15 minutes ago
Several napkins of equal size
gnoka   0
15 minutes ago
Source: 46th International Tournament of Towns, Senior A-Level P7, Fall 2024
Several napkins of equal size and of shape of a unit disc were placed on a table (with overlappings). Is it always possible to hammer several point-sized nails so that all the napkins will be thus attached to the table with the same number of nails? (The nails cannot be hammered into the borders of the discs).

Vladimir Dolnikov, Pavel Kozhevnikov
0 replies
gnoka
15 minutes ago
0 replies
No more topics!
Isotomic conjugates [intersections of median & incircle]
Arne   113
N Today at 3:14 AM by Saucepan_man02
Source: belgian IMO preparation; IMO Shortlist 2005 geometry problem G6
Let $ABC$ be a triangle, and $M$ the midpoint of its side $BC$. Let $\gamma$ be the incircle of triangle $ABC$. The median $AM$ of triangle $ABC$ intersects the incircle $\gamma$ at two points $K$ and $L$. Let the lines passing through $K$ and $L$, parallel to $BC$, intersect the incircle $\gamma$ again in two points $X$ and $Y$. Let the lines $AX$ and $AY$ intersect $BC$ again at the points $P$ and $Q$. Prove that $BP = CQ$.
113 replies
Arne
Mar 25, 2006
Saucepan_man02
Today at 3:14 AM
Isotomic conjugates [intersections of median & incircle]
G H J
Source: belgian IMO preparation; IMO Shortlist 2005 geometry problem G6
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Arne
3660 posts
#1 • 8 Y
Y by Aref, rashah76, samrocksnature, Adventure10, jhu08, mathematicsy, MathLuis, Mango247
Let $ABC$ be a triangle, and $M$ the midpoint of its side $BC$. Let $\gamma$ be the incircle of triangle $ABC$. The median $AM$ of triangle $ABC$ intersects the incircle $\gamma$ at two points $K$ and $L$. Let the lines passing through $K$ and $L$, parallel to $BC$, intersect the incircle $\gamma$ again in two points $X$ and $Y$. Let the lines $AX$ and $AY$ intersect $BC$ again at the points $P$ and $Q$. Prove that $BP = CQ$.
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Arne
3660 posts
#2 • 4 Y
Y by samrocksnature, Adventure10, jhu08, Mango247
Hint: consider the circle passing through $M$ and $P$ which is tangent to $AB$ and $AC$.
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fagot
37 posts
#3 • 7 Y
Y by govind7701, myh2910, samrocksnature, Adventure10, jhu08, guptaamitu1, Mango247
Let $\ell: A\in\ell$ is the line, parallel $BC$. Consider the projective transformation, which maps $\ell$ in
the infinite line and preserve insircle of $ABC$. Then ratios of distances on the any line, parallel $\ell$, are preserved. Then the problem is trivial: the points $P'$ and $Q'$ in which maps the points $P$ and $Q$ are symmetric
of the center of incircle.
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Virgil Nicula
7054 posts
#4 • 10 Y
Y by Epistle, samrocksnature, Adventure10, Adventure10, jhu08, truongphatt2668, Mango247, bin_sherlo, and 2 other users
PP.Let $ABC$ be a triangle, and $M$ the midpoint of its side $BC$. Let $\gamma$ be the incircle of triangle $ABC$. The median $AM$ of triangle $ABC$ intersects the incircle $\gamma$ at two points $K$ and $L$. Let the lines passing through $K$ and $L$, parallel to $BC$, intersect the incircle $\gamma$ again in two points $X$ and $Y$. Let the lines $AX$ and $AY$ intersect $BC$ again at the points $P$ and $Q$. Prove that $BP = CQ$.

Lemma. Let $ABC$ be a triangle for which $b\ne c$. Denote: the middlepoint $M$ of the side $[BC]$; the points $D,E,F$ where the incircle $C(I,r)$ touches the side-lines $BC,CA,AB$ respectively; the intersections $K,L$ between the incircle and the median $AM$; the intersection $S\in EF\cap AM$. Then $I\in SD$ and the division $(A,K,S,L)$ is harmonically.

Proof. $\frac{FB}{FA}\cdot MC+\frac{EC}{EA}\cdot MB=\frac{SM}{SA}\cdot BC\Longrightarrow\frac{SM}{SA}=\frac{a}{b+c-a}\ \ (1)$. Therefore, $T\in BC,\ AT\perp BC\Longrightarrow $ $MD=\frac 12\left| b-c\right|\ ,$ $DT=$ $\frac{(p-a)|b-c|}{a},\ \frac{DM}{DT}=\frac{a}{b+c-a}\ \ (2)$. From the relations $(1)$ and $(2)$ results that $\frac{SM}{SA}=\frac{DM}{DT}$, i.e. $SD\parallel AT$ what means $I\in SD$. Immediately we have and the second part of the conclusion (the line $EF$ is the polar of the point $A$ w.r.t. the incircle).

Proof of the proposed problem. Define the point $S\in EF\cap AM$ . Observe that $KX\parallel LY\iff$ $KXLY$ is an isosceles trapezoid $\iff$ $S\in XY\cap KL\implies$ $\frac {AK}{AL}=\frac {SK}{SL}=\frac {KX}{YL}\implies$ $\boxed{\frac {AK}{AL}=\frac {KX}{YL}}\ (1)$ .Therefore, $\left\{\begin{array}{cccc}
\frac{AM}{AK} & = & \frac{MP}{KX} & (2)\\\\
\frac{AL}{AM} & = & \frac{LY}{MQ} & (3)\end{array}\right\|$ . From the product of the relations $(1)$ , $(2)$ and $(3)$ we obtain $MP=MQ$, i.e. $BP=CQ\ .$
This post has been edited 8 times. Last edited by Virgil Nicula, Feb 29, 2016, 8:52 AM
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Arne
3660 posts
#5 • 3 Y
Y by samrocksnature, Adventure10, jhu08
Very nice!
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Virgil Nicula
7054 posts
#6 • 4 Y
Y by samrocksnature, Adventure10, jhu08, ehuseyinyigit
Thanks, Arne ! I and you must at first prove that the points $A,K,Z,L$ form a harmonical division. This property is one among the first problems in which the my mathe-teacher applied the harmonical division (approx. in 1958 !)
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Davron
484 posts
#7 • 4 Y
Y by samrocksnature, Adventure10, jhu08, Mango247
What does harmonic division mean ? Virigil please can you send a link or something like that ...

Davron
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Virgil Nicula
7054 posts
#8 • 3 Y
Y by samrocksnature, Adventure10, Mango247
Davron and Arne, see that (in the previous message) I proved the mentioned classical harmonical division in a lemma.
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nsato
15646 posts
#9 • 6 Y
Y by samrocksnature, Adventure10, Flash_Sloth, CyclicISLscelesTrapezoid, and 2 other users
A less sophisticated solution, using Arne's hint:

Click to reveal hidden text
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Arne
3660 posts
#10 • 4 Y
Y by samrocksnature, Adventure10, Mango247, CyclicISLscelesTrapezoid
My solution is different:
Click to reveal hidden text
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Little Gauss
200 posts
#11 • 3 Y
Y by samrocksnature, Adventure10, ehuseyinyigit
Generalization of this problem :

Theorem
Let $ABC$ be a triangle, and $P, Q$ is the point on $BC$ such that $BP=CQ$. Let $w$ be the incircle of triangle ABC. The line $AP$ intersects the incircle $w$ at $R$ (closer to A) and line $AQ$ intersects the incircle $w$ at $S$ (farrer to A).
Let the lines passing through $R$ and $S$, parallel to BC, intersect the incircle w again in two points $X$ and $Y$. Let the lines $AX$ and $AY$ intersect $BC$ again at the points $Z$ and $W$.
than $BZ=CW$

it is modification of India 1998 and easy to prove. we can use trigonometry or power of point.

Sorry to my bad english.
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Summerburn
83 posts
#12 • 5 Y
Y by samrocksnature, Adventure10, Mango247, and 2 other users
Virgil Nicula wrote:
$\frac{FB}{FA}\cdot MC+\frac{EC}{EA}\cdot DB=\frac{SM}{SA}\cdot BC$.

Well, why is this?
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Summerburn
83 posts
#13 • 4 Y
Y by samrocksnature, Adventure10, and 2 other users
Hmm... I still don't understand. What theorem did you use here?
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April
1270 posts
#14 • 5 Y
Y by samrocksnature, Adventure10, Mango247, and 2 other users
Quote:
Lemma. Let be given a triangle $ABC$ and a point $M$ lies on segment $BC.$ A line cuts $AB,\,AC,\,AM$ at $B',\,C',\,M'$ respectively. We have:
\[BC\cdot\frac{AM}{AM'}=BM\cdot\frac{AC}{AC'}+CM\cdot\frac{AB}{AB'}\]
Proof. We have:

\begin{eqnarray*}\frac{AB'\cdot AC'}{AB\cdot AC}&=&\frac{S(AB'C')}{S(ABC)}\\&=&\frac{S(AB'M')+S(AC'M')}{S(ABC)}\\&=&\frac{S(AB'M')}{\frac{BC}{BM}\cdot S(ABM)}+\frac{S(AC'M')}{\frac{BC}{CM}\cdot S(ACM)}\\ &=&\frac{BM}{BC}\cdot\frac{AB'\cdot AM'}{AB\cdot AM}+\frac{CM}{BC}\cdot\frac{AC'\cdot AM'}{AC\cdot AM}\end{eqnarray*}

$\Longrightarrow BC\cdot\frac{AM}{AM'}=BM\cdot\frac{AC}{AC'}+CM\cdot\frac{AB}{AB'}$
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Summerburn
83 posts
#15 • 6 Y
Y by samrocksnature, Adventure10, Mango247, and 3 other users
Thanks. But also, can you tell why $A,K,S,L$ form a harmonic division (in Virgil's lemma)...
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