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!

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0 replies
jlacosta
Nov 1, 2024
0 replies
infinite sequence
gnoka   1
N 9 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
9 minutes ago
five integer division
gnoka   1
N 12 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
12 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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awesomeming327.
1584 posts
#103
Y by
Let $DEF$ be the contact triangle, let $G$ be the point on $EF$ such that $IG\perp BC$. Let the parallel line to $BC$ from $G$ intersect $AC$ and $AB$ at $J$ and $H$. Let $AX$ intersect $YL$ at $Z$. We have $\angle IGJ=90^\circ$ so $IGJF$ and $IGEH$ are cyclic. This means that $\angle FJI=\angle FGI=\angle IHE$ so $AJIH$ is cyclic. Since $AI$ is angle bisector, $IJ=IH$ which implies $GJ=GH$ so $A,G,M$ collinear. Thus, $X,G,Y$ collinear.

Now, we have
\[-1=(A,G;K,L)\stackrel{X}{=}(Z,Y;L,P_{\infty})\stackrel{A}{=}(Q,P;M,P_{\infty})\]and we're done.
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Markas
105 posts
#104
Y by
If we want to show that BP = CQ, we only need to prove M is the midpoint of PQ, so if we show that $(P,Q;M,P_\infty) = -1$ we are ready. Let $KL \cap FE = N$. Since AF and AE are tangents, we have that $(A,N;K,L) = -1$. By lemma 4.17, $FE \cap KL \cap DI = N$. Since $KX \parallel YL$, KXLY is a trapezoid and it is cyclic $\Rightarrow$ KXLY is an isosceles trapezoid. From $N \in DI$, $PI \perp KX$ and $PI \perp YL$ $\Rightarrow$ $N \in XY$. Now let $AX \cap YL = Z$. Now projecting we get $(P,Q;M,P_\infty)\stackrel{A}{=}(Y,Z;L,P_\infty)\stackrel{X}{=}(N,A;L,K) = -1$ $\Rightarrow$ we showed that $(P,Q;M,P_\infty) = -1$, which is what we wanted to prove $\Rightarrow$ we are ready.
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SenorSloth
37 posts
#105 • 1 Y
Y by bjump
Let $D$, $E$, $F$ be the tangency points of $\gamma$ to $BC$, $AC$, and $AB$, respectively. It is well-known that $AM$, $EF$, and the diameter of $\gamma$ through $D$ concur at a point $Z$. Then it is clear that $(AZ; KL)$ is a harmonic bundle. Then we reflect over line $DX$ and get that $(A'Z; XY)$ is a harmonic bundle. We finish by projecting through $A$ onto line $BC$. Since $AA' \parallel BC$, we have that $(\infty M; PQ)$ is a harmonic bundle. This implies that $M$ is the midpoint of $PQ$ which means we must have $BP = CQ$, as desired.
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EeEeRUT
27 posts
#106
Y by
Denote $B_1 \neq X$ and $C_1 \neq Y$ the intersection of $\omega$ with $AX$ and $AY$,respectively and the line parallel to $BC$through $Z$ intersects $\omega$ at $P_1$ and $Q_1$, where $P_1$ and $B$ is on the same side of $DI$. Also let $M$ be midpoint of $BC$

Note that, $BP = CQ$ is equivalent to $M$ is midpoint $PQ$

It is well known that $DI,EF$ and $AL$ is concurrented, say at $Z$.

From La Haire theorem, we get that $B_1C_1, XY$ and $EF$ concurrent at $Z$.

From Butterfly theorem, we get that $P_1Z = Q_1Z$.

Since $P_1Q_1 \parallel BC$ and $A,Z,M$ collinear, we can conclude that $M$ is midpoint of $PQ$, as desired.
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Ianis
330 posts
#107
Y by
Let $DEF$ be the intouch triangle, let $EF\cap AM=G$ and let $AX\cap LY=N$.
It's well known that $DI$, $AM$ and $EF$ are concurrent, so $D,I,G$ are collinear. It's clear that $DI$ is the perpendicular bisector of $XK$ and $YL$, so $X,Y,G$ are collinear. Then\begin{align*}\{P,Q;M,\infty \} & \underset{A}{=}\{N,Y;L,\infty \} \\
& \underset{X}{=}\{A,G;L,K\} \\
& \underset{E}{=}\{E,F;L,K\} \\
& =-1,
\end{align*}so $M$ is the midpoint of $PQ$, which implies the desired result.
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EpicBird08
1678 posts
#108
Y by
oops me when i learn ddit

Claim: Let the tangents to $X_1$ and $Y_1$ to $\gamma$ meet at point $Z.$ Then $AZ \parallel BC.$
Proof: Let $I$ be the center of $\gamma,$ and let $D$ be the tangency point of $\gamma$ with $BC.$ By reflecting across $ID,$ it suffices to show that if the tangents to $\gamma$ at $X$ and $Y$ meet at point $W,$ then $AW \parallel BC.$ Let $S = EF \cap AM;$ then it is well-known that $DS \perp BC,$ but by polars $DS \perp AW$ as well. Thus $AW \parallel BC,$ as claimed.

Now, applying DDIT, if $\gamma$ touches $AC,AB$ at points $E,F,$ respectively, there exists an involution swapping pairs $(AE,AF), (AZ, AZ), (AX_1, AX_2).$ Projecting onto $BC,$ we see that there exists an involution swapping $(C,B), (\infty, \infty),$ and $(P, Q).$ However, this is just the involution reflecting across the perpendicular bisector of $BC,$ and so $BP = CQ,$ as desired.
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MagicalToaster53
152 posts
#109
Y by
It suffices to show that $M$ bisects segment $\overline{QP}$. Let $Z = KX \cap AQ$, and $D, E, F$ denote the contact points of $\gamma$ with sides $\overline{BC}, \overline{CA}, \overline{AB}$, respectively. Before proceeding we introduce a lemma:

Lemma: Let $\triangle ABC$ be a triangle with incenter $I$ and contact points $D, E, F$ on $\overline{BC}, \overline{CA},$ and $\overline{AB}$, respectively. If $M$ is the midpoint of $BC$, then $DI, AM,$ and $EF$ concur.
Proof

Now using this lemma in our context of $\overline{EF}, \overline{KL}, \overline{DI}$, we subsequently find that $X, G, Y$ are collinear, where $G$ is the concurrency point of $\overline{EF}, \overline{KL}, \overline{DI}$. Hence \[(Q, M; P, P_{\infty}) \overset{A}{\underset{KX}{\longrightarrow}} (Z, K; X, P_{\infty}) \overset{Y}{\underset{AM}{\longrightarrow}} (A, K; G, L) = -1, \]where the final equality follows from the lemma. $\blacksquare$
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bjump
887 posts
#110
Y by
Let $J=KL \cap XY$, let $N$ be the point on $XY$ such that $AN \parallel BC$, let $D$, $E$ and $F$ be the intouch points on $BC$, $AC$, and $AB$ respectively. Now $(PQ ; M \infty) \stackrel{A}= (YX; JN)  \stackrel{\text{rflct}}= (LK; JA)$. By EGMO lemma 4.17 $FJE$ are collinear so then $(LK;JA) \stackrel{\gamma}=(LK; EF)=-1$. Therefore $-1=(PQ; M \infty)$ and we can conclude $M$ is the midpoint of $PQ$ by midpoints and parallel lines.
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bebebe
718 posts
#111
Y by
Let $P_{\infty}$ be point at infinity on line $BC,$ and $D,E,F$ be the tangency points of the incircle. We have $$-1= (B,C;M,P_{\infty}) \overset{A}{=} (F,E;EF\cap AM, EF \cap AP_{\infty}).$$Let $T=EF\cap AM$.

\textit{Claim:} $EF, AM=KL, XY$, and $DI$ are concurrent at $T$.

\textit{Proof:} $EF, AM, DI$ are concurrent follows from homothethy and simson lines. It can alternativey be proven using homothethy and harmonic bundles. Since $XY$ is the reflection of $KL$ across $DI,$ we know that it is also concurrent.


\textit{Claim:} $XE \cap FY, A, P_{\infty}$ are collinear.

\textit{Proof:} It's easy to see $polar(A)=EF$ and $polar(P_{\infty}) = DI.$ By Brocard, $T \in polar(XE \cap FY).$ Since $DI \cap EF=T$, by duality of poles and polars (concurrency and collinearity), we are done.


Taking perspecivity at $XE \cap FY \cap AP_{\infty},$ we get $$-1=(F,E;T, EF \cap AP_{\infty})=(Y, X; T, XY \cap AP_{\infty}) \overset{A}{=} (Q, P; M, P_{\infty}),$$which implies $M$ is the midpoint of $PQ,$ and the desired follows.
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joshualiu315
2430 posts
#112 • 1 Y
Y by dolphinday
Let $\triangle DEF$ be the intouch triangle of $\triangle ABC$, and denote $R$ as the concurrency of $\overline{AM}$, $\overline{EF}$, and $\overline{DI}$. Moreover, let $S$ be the point on $\overline{XY}$ such that $\overline{AS} \parallel \overline{BC}$. We have

\[-1 = (A,R;G,L)  \overset{\infty}{=} (S,R;X,Y) \overset{A}{=} (\infty,M,P,Q),\]
so $M$ is the midpoint of $\overline{PQ}$. This is equivalent to the desired conclusion, so we are done. $\square$
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Eka01
139 posts
#113 • 1 Y
Y by Sammy27
Let $\Delta DEF$ be the inotuch triangle of $\Delta ABC$, Let $DI \cap EF=G$. It is well known that $\overline{A-K-G-L-M}$ are collinear. Since $KY || XL || BC$, this implies $KXLY$ is an isoceles trapezoid and that $DI$ is the perpendicular bisector of $KY$ and $LX$.Since $G$ is the intersection of the perpendicular bisector of the parallel sides with one of the diagonals of a cyclic isoceles trapezoid, it follows that the other diagonal must pass through it as well, hence $\overline{X-G-Y}$ are collinear. Let $P_ \infty$ be the point at infinity along $BC$ and let the line through $A$ parallel to $BC$ intersect $XY$ at $T$. Also notice that $(K,L;E,F) \stackrel{E/F}{=} (K,L;A,G) =-1$.
Now,
$$(P,Q;M,P_\infty) \stackrel{A}{=} (X,Y;G,T) \stackrel{Reflection \ over \ DI}{=} (L,K;G,A)=-1$$It follows that $M$ is the midpoint of $PQ$ hence we are done.
Attachments:
This post has been edited 3 times. Last edited by Eka01, Aug 27, 2024, 8:15 AM
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cj13609517288
1736 posts
#114 • 1 Y
Y by OronSH
I've given up on this problem twice so I kind of remember the solution but like whatever lol.

Define $D,E,F$ as the intouch points. Construct $A'$, the reflection of $A$ over $ID$. Construct $Z=KL\cap XY$. Therefore, it also lies on $ID$, so by the ``an incircle concurrency'' lemma, it also lies on $EF$. Then
\[-1=(EF;KL)\stackrel{F}{=}(ZA;KL)=(ZA';XY)\stackrel{A}{=}(M\infty_{BC};PQ),\]as desired.
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cj13609517288
1736 posts
#115 • 1 Y
Y by dolphinday
Apply DDIT on $KXLY$ through $A$, with helper line $\ell=\overline{BC}$. Then there exists an involution on $\ell$ swapping $Q$ and $P$, $M$ and $M$, and $\infty$ with $A(KY\cap XL)\cap \ell$.

Note that $Z=KY\cap XL$ is the Miquel point of $KXLY$, so $D=KL\cap XY$ is the inverse of $Z$ wrt the incircle. We want to show that $AZ\parallel BC$, which is equivalent to $A$ being on the polar of $D$. This is equivalent to $D$ being on the polar of $A$, which is the $A$-intouch chord. This is exactly the "an incircle concurrency" lemma. $\blacksquare$
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CyclicISLscelesTrapezoid
354 posts
#117 • 1 Y
Y by OronSH
What's projective?

By the two homotheties at $A$ mapping either $\overline{KX}$ or $\overline{LY}$ to $\overline{BC}$, there exist two circles through $M$ tangent to $\overline{AB}$ and $\overline{AC}$ intersecting $\overline{BMC}$ again at $P$ and $Q$. Let one of these circles be $\omega$, and suppose it is tangent to $\overline{AB}$ at $D_1$ and $\overline{AC}$ at $E_1$. We will resolve the case when $D_1$ is on segment $AB$; other cases are analogous.

Let $P'$ be the reflection of $P$ over $M$. We will construct a circle through $M$ and $P'$ tangent to $\overline{AB}$ and $\overline{AC}$. Choose $D_2$ such that $BD_2=CE_1$ and $B$ lies on segment $AD_2$, and choose $E_2$ such that $CE_2=BD_1$ and $C$ lies on segment $AE_2$. Notice that
  • since $BD_2^2=CE_1^2=CM \cdot CP=BM \cdot BP'$, there exists a circle $\omega_1$ through $M$ and $P'$ tangent to $\overline{AB}$ at $D_2$
  • since $CE_2^2=BD_1^2=BM \cdot BP=CM \cdot CP'$, there exists a circle $\omega_2$ through $M$ and $P'$ tangent to $\overline{AC}$ at $E_2$
  • since $AD_1+D_1B+BD_2=AE_1+E_1C+CE_2$, there exists a circle $\omega_3$ tangent to $\overline{AB}$ at $D_2$ and $\overline{AC}$ at $E_2$.
If any two of $\omega_1$, $\omega_2$, and $\omega_3$ coincide, then they all coincide and we are done. Otherwise, radical axis on the three circles implies lines $AB$, $AC$, and $BC$ concur, a contradiction. $\blacksquare$
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Saucepan_man02
1129 posts
#118
Y by
Let $D, E, F$ denote the point of tangency of $\gamma$ with sides $BC, CA, AB$ respectively.

Lemma: Let $I$ denote the incenter of $\triangle ABC$, with $M$ being the midpoint of $AB$. Then: $AM, EF, DI$ are concurrent.
Proof: Let $X = DI \cap EF$. Let $B'C'$ be a line passing though $X$ with $B'C' \parallel BC$. Notice that, the pedal triangle of point $I$ onto $\triangle AB'C'$ is degenerate which implies $I$ lies on $\triangle AB'C'$. Since $AI$ is the angle bisector of $\angle B'AC'$, and $X$ is the foot of perpendicular of $I$ onto $B'C'$, we have: $X$ to be the midpoint of $BC$. Taking homothety gives us the result.

Note that, $KXLP$ is an isosceles trapezoid.
Due to the above lemma, let $R$ denote the concurrency point of lines $AM, EF, DI$. Note that: $R$ lies on $XP$.
Let $A'$ be a point such that: $AA' \parallel BC$ with $A', X, P$ collinear. Notice that, as we get $(A, A'), (K, X), (R, R), (P, L)$ are symmetric with-respect to line $DR$: $$-1=(AR;KL) = (A'R; XP).$$Taking perspective into line $BC$ from point $A$ gives: $$-1 = (A'R; XP) = (\infty M; QP).$$Therefore $M$ is the midpoint of $PQ$ and we are done.
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