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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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0 replies
jlacosta
May 1, 2025
0 replies
Easy functional equation
fattypiggy123   15
N 8 minutes ago by ariopro1387
Source: Singapore Mathematical Olympiad 2014 Problem 2
Find all functions from the reals to the reals satisfying
\[f(xf(y) + x) = xy + f(x)\]
15 replies
+1 w
fattypiggy123
Jul 5, 2014
ariopro1387
8 minutes ago
Iran TST Starter
M11100111001Y1R   5
N 9 minutes ago by DeathIsAwe
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[
a_n < \frac{1}{1404}
\]
5 replies
M11100111001Y1R
May 27, 2025
DeathIsAwe
9 minutes ago
My journey to IMO
MTA_2024   2
N 23 minutes ago by Royal_mhyasd
Note to moderators: I had no idea if this is the ideal forum for this or not, feel free to move it wherever you want ;)

Hi everyone,
I am a random 14 years old 9th grader, national olympiad winner, and silver medalist in the francophone olympiad of maths (junior section) Click here to see the test in itself.
While on paper, this might seem like a solid background (and tbh it kinda is); but I only have one problem rn: an extreme lack of preparation (You'll understand very soon just keep reading :D ).
You see, when the francophone olympiad, the national olympiad and the international kangaroo ended (and they where in the span of 4 days!!!) I've told myself :"aight, enough math, take a break till summer" (and btw, summer starts rh in July and ends in October) and from then I didn't seriously study maths.
That was until yesterday, (see, none of our senior's year students could go because the bachelor's degree exam and the IMO's dates coincide). So they replaced them with us, junior students. And suddenly, with no previous warning, I found myself at the very bottom of the IMO list of participants. And it's been months since I last "seriously" studied maths.
I'm really looking forward to this incredible journey, and potentially winning a medal :laugh: . But regardless of my results I know it'll be a fantastic journey with this very large and kind community.
Any advices or help is more than welcome <3 .Thank yall for helping me reach and surpass a ton of my goals.
Sincerely.
2 replies
+1 w
MTA_2024
29 minutes ago
Royal_mhyasd
23 minutes ago
Very odd geo
Royal_mhyasd   1
N 34 minutes ago by Royal_mhyasd
Source: own (i think)
Let $\triangle ABC$ be an acute triangle with $AC>AB>BC$ and let $H$ be its orthocenter. Let $P$ be a point on the perpendicular bisector of $AH$ such that $\angle APH=2(\angle ABC - \angle ACB)$ and $P$ and $C$ are on different sides of $AB$, $Q$ a point on the perpendicular bisector of $BH$ such that $\angle BQH = 2(\angle ACB-\angle BAC)$ and $R$ a point on the perpendicular bisector of $CH$ such that $\angle CRH=2(\angle ABC - \angle BAC)$ and $Q,R$ lie on the opposite side of $BC$ w.r.t $A$. Prove that $P,Q$ and $R$ are collinear.
1 reply
Royal_mhyasd
37 minutes ago
Royal_mhyasd
34 minutes ago
No more topics!
Find the smallest positive real number
April   5
N Feb 23, 2024 by HamstPan38825
Source: IMO Shortlist 2007, G6 / USA TST 2008, Day 2, Problem 6
Determine the smallest positive real number $ k$ with the following property. Let $ ABCD$ be a convex quadrilateral, and let points $ A_1$, $ B_1$, $ C_1$, and $ D_1$ lie on sides $ AB$, $ BC$, $ CD$, and $ DA$, respectively. Consider the areas of triangles $ AA_1D_1$, $ BB_1A_1$, $ CC_1B_1$ and $ DD_1C_1$; let $ S$ be the sum of the two smallest ones, and let $ S_1$ be the area of quadrilateral $ A_1B_1C_1D_1$. Then we always have $ kS_1\ge S$.

Author: Zuming Feng and Oleg Golberg, USA
5 replies
April
Jul 13, 2008
HamstPan38825
Feb 23, 2024
Find the smallest positive real number
G H J
Source: IMO Shortlist 2007, G6 / USA TST 2008, Day 2, Problem 6
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April
1270 posts
#1 • 4 Y
Y by Davi-8191, Adventure10, Mango247, and 1 other user
Determine the smallest positive real number $ k$ with the following property. Let $ ABCD$ be a convex quadrilateral, and let points $ A_1$, $ B_1$, $ C_1$, and $ D_1$ lie on sides $ AB$, $ BC$, $ CD$, and $ DA$, respectively. Consider the areas of triangles $ AA_1D_1$, $ BB_1A_1$, $ CC_1B_1$ and $ DD_1C_1$; let $ S$ be the sum of the two smallest ones, and let $ S_1$ be the area of quadrilateral $ A_1B_1C_1D_1$. Then we always have $ kS_1\ge S$.

Author: Zuming Feng and Oleg Golberg, USA
This post has been edited 1 time. Last edited by levans, Jul 10, 2016, 5:34 PM
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Vrajitoarea Andrei
81 posts
#2 • 1 Y
Y by Adventure10
This is from IMO Shorlist 2007,problem G6
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math154
4302 posts
#3 • 4 Y
Y by Mediocrity, 277546, Adventure10, Mango247
The answer is $k=1$. We can get $\frac{S}{S_1}$ arbitrarily close to $1$ by taking $AA_1BB_1CC_1DD_1$ arbitrarily close to a medial triangle configuration (e.g. approaching $D_1=A=A_1$).

Lemma. Let $\triangle{XYZ}$ be a triangle and $X_1,Y_1,Z_1$ points on sides $YZ$, $ZX$, $XY$. Then $[X_1Y_1Z_1] \ge \min_{\text{cyc}}([XY_1Z_1])$.

Proof. For convenience, take an affine transformation sending $\triangle{X_1Y_1Z_1}$ to an equilateral triangle, and construct its antimedial triangle $\triangle{X_2Y_2Z_2}$. Let $O$ be the circumcenter of $\triangle{X_1Y_1Z_1}$.

Suppose the ``border areas'' $[XY_1Z_1],\ldots$ are all strictly greater than $[X_1Y_1Z_1]$. Then $d(X,Y_1Z_1) > d(X_1,Y_1Z_1)$, so $X$ cannot lie inside $(OY_1X_2Z_1)$. Since $Y_1Z_1$ separates $X,X_1$, we must have $\angle{Y_1XZ_1}<60^\circ$. By symmetry, we obtain $\angle{X}+\angle{Y}+\angle{Z} < 3\cdot60^\circ=180^\circ$, contradiction.$\blacksquare$

We will now prove $S_1 \ge S$. Inspired by the degenerate case, we define $T = A_1C_1\cap B_1D_1$, $f(A) = [AA_1D_1]$, $g(A) = [TA_1D_1]$, and $h(A) = \max([B_1A_1D_1],[C_1A_1D_1])$; similarly define $f(X),g(X)$ for $X=B,C,D$.

Case 1: There exists $X\in\{A,B,C,D\}$ such that $f(X) \le h(X)$. WLOG $f(A) \le h(A)$ and $[B_1A_1D_1] \le [C_1A_1D_1]$. Then
\begin{align*}
S_1-S &\ge [C_1A_1D_1]+[A_1B_1C_1]-f(A)-\min(f(B),f(C),f(D)) \\
&\ge [A_1B_1C_1]-\min(f(B),f(C),f(D)).
\end{align*}But $A_1BCDD_1$ is convex, so letting $B' = CB\cap D_1A_1$ and $D' = CD\cap A_1D_1$, we find $f(B) \le [B'A_1B_1]$, $f(C) = [CB_1C_1]$, $f(D) \le [D'C_1D_1] \le [D'C_1A_1]$. The lemma now immediately gives $S_1-S \ge0$, as desired.

Case 2: $f(X)>h(X)>g(X)$ for every $X\in\{A,B,C,D\}$, so $d(A,A_1D_1)>d(T,A_1D_1)$, etc. Taking an affine transformation if necessary, WLOG assume $\angle{A_1TB_1}=90^\circ$. For convenience, we write $T=(0,0)$, $A_1=(-a,0)$, $B_1=(0,b)$, $C_1=(c,0)$, $D_1=(0,-d)$ for positive reals $a,b,c,d$.

Then in particular, either $x(A)<-a$ or $y(A)<-d$ must hold (analogous statements hold for $B,C,D$). WLOG suppose $y(A)<-d$. Then we easily deduce
\begin{align*}
y(D)>-d &\implies x(D)>c>x(C) \\
&\implies y(C)>b>y(B)\implies x(B)<-a<x(A).
\end{align*}A simple ratio calculation (note $A$ is bounded by $C_1D_1$, $D_1A_1$, and $x=-a$) yields
\[ [C_1A_1D_1] \le h(A) < f(A) \le \frac{a}{c}[A_1D_1C_1], \]whence $a>c$. But similarly, we deduce $b>d$, $c>a$, and $d>b$, contradiction.

Comment. There are certainly other ways to prove the lemma. Most obviously, we can directly use barycentric coordinates or standard area calculations. If we want to avoid both affine transformations and calculation, we can compare $\triangle{X_1Y_1Z_1}$ with the medial triangle $\triangle{X'Y'Z'}$ of $\triangle{XYZ}$.

Comment. The key is to look at degenerate cases (and realize the answer is $k=1$). "Simple case analysis" is one way to motivate this, but smoothing provides a bit more insight, both for why the degenerate case should be optimal and how to go about proving $S_1\ge S$. Indeed, suppose we have a configuration with $S/S_1$ maximal (or at least ``close'' to maximal). Then we can smooth to the case $f(A)\le f(B)=f(D)\le f(C)$ (otherwise, consider rotating $BC,CD$ about $B_1,C_1$ to increase $S$ while fixing $S_1$). Now if $f(A)<f(B)=f(D)$ and $B_1$ is closer to $AB$ than $D_1$ (or $C_1$ is closer to $AD$ than $A_1$), we can slide $A_1$ away from $A$ a bit (resp. $D_1$ away from $A$) to decrease $S_1$ while increasing $S$.

Hence an ``optimal'' configuration (with $S/S_1$ maximal) should have $d(D_1,AB) \le d(B_1,AB)$ (and $d(A_1,AD) \le d(C_1,AD)$). While we can't do much more obvious smoothing, this at least suggests we consider the case when at least one of $A_1,D_1$ coincides with $A$ (or just $f(A)$ is really small).
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JuanOrtiz
366 posts
#4 • 4 Y
Y by stroller, Modesti, Adventure10, Mango247
Lemma: in a triangle ABC with X,Y,Z on sides BC,CA,AB then (XYZ) $\ge$ (AYZ),(BXZ) or (CXY)
Proof: WLOG ABC is quadrilateral, now just computations.

The proof k=1 is easy with this result.WLOG rays AB and DC intersect at P, then apply lemma to triangle APD and we get S1 > (A1C1D1) $\ge$ (AA1D1), (DDAC1) or (PA1C1). In the last case, (PA1C1) > S so we are done. Otherwise, WLOG (A1C1D1) $ \ge$ (DC1D1) and we easily get (A1B1C1) > (A1BB1) or (B1CC1), so we get S1 > S.

For equality let D=A=D1, and A1, B1, C1 be midpoints of ABC.
This post has been edited 4 times. Last edited by JuanOrtiz, Apr 17, 2015, 5:53 PM
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awesomeming327.
1739 posts
#5
Y by
We claim that the answer is $1$. First, we'll show that $S_1=S$ can be achieved. Let $ABC$ be a triangle, and $D$ be infinitely close to the midpoint of $AC$. Then, we can let $A_1B_1C_1D_1$ essentially be equivalent to the medial triangle of $\triangle ABCD$, giving an answer of $1$.

Now, we show that $S_1\ge S$ is true. We claim that for any triangle $XYZ$ and $X_1$ on $YZ$, $Y_1$ on $ZX$, and $Z_1$ on $XY$, \[[X_1Y_1Z_1]\ge \text{min}([XY_1Z_1],[YZ_1X_1],[ZX_1Y_1])\]Assume otherwise, then apply the affine transform that will take $[X_1Y_1Z_1]$ to an equilateral triangle. Let $X'$, $Y'$ and $Z'$ be such that $\triangle X_1Y_1Z_1$ is the medial triangle of $\triangle X'Y'Z'$. Evidently, $X'$ is closer to $Y_1Z_1$ than $X$. Since $X'Y_1Z_1$ is equilateral, that means $X$ lies outside $(X'Y_1Z_1)$ so $\angle YXZ=\angle Y_1XZ_1>60^\circ$. Similarly, $\angle XZY,\angle ZYX>60^\circ$, contradiction because the sum of angles in a triangle adds to $180^\circ$. This proves our claim.

Now, let $E$ be $AB$ intersection with $CD$. Without loss of generality, $BC$ is closer to $E$ than $AD$. Let $T_1=[AA_1D_1]$, $T_2=[BB_1A_1]$, $T_3=[CC_1B_1]$, $T_4=[DD_1C_1]$ then by our claim,
\begin{align*}
S_1 &= [A_1C_1D_1] + [A_1B_1C_1] \\
&= \text{min}(T_1,T_2+T_3,T_4)+\text{min}(T_3,T_1+T_4,T_2) \\
&\ge S
\end{align*}as desired.
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HamstPan38825
8868 posts
#6
Y by
The answer is $k=1$. For a construction, consider the degenerate case where $ABCD$ is a triangle $ABC$ with $D$ the midpoint of $\overline{BC}$, where $A_1B_1C_1D_1$ are the midpoints of their respective sides (hence forming the medial triangle of $ABC$.)

To show that this is minimal, consider the following:

Claim: For any triangle $DEF$ inscribed on the sides of a triangle $ABC$, $[DEF] > \operatorname{min}([AEF], [BDF], [CDE])$.

Proof. Take an affine transform sending $DEF$ to an equilateral triangle. Assume for the sake of contradiction that the result is not true; then, let $XYZ$ be the image of $DEF$ upon a homothety of ratio $4$ at its center. It is equivalent to show that there cannot exist points $A, B, C$ outside $XYZ$ such that $DEF$ is inscribed in $ABC$.

However, notice that for any point $A$ outside $XYZ$, we always have $\angle EAF < 60^\circ$ (as the circle for which $\angle EAF = 60^\circ$ is tangent to triangle $XYZ$). Hence the angles of $ABC$ add up to less than $180^\circ$, contradiction. $\blacksquare$

Now WLOG $BC < AD$ and construct the point $E = \overline{AB} \cap \overline{CD}$. The lemma implies that $[B_1A_1C_1] < \operatorname{min}([BA_1B_1], [CB_1C_1])$ and $[A_1D_1C_1] < \operatorname{min}([DC_1D_1], [AA_1D_1])$, and summing yields the result.
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