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An upper bound for Iran TST 1996
Nguyenhuyen_AG   0
34 minutes ago
Let $a, \ b, \ c$ be the side lengths of a triangle. Prove that
\[\frac{ab+bc+ca}{(a+b)^2} + \frac{ab+bc+ca}{(b+c)^2} + \frac{ab+bc+ca}{(c+a)^2} \leqslant \frac{85}{36}.\]
0 replies
Nguyenhuyen_AG
34 minutes ago
0 replies
Sharygin 2025 CR P2
Gengar_in_Galar   5
N 36 minutes ago by NicoN9
Source: Sharygin 2025
Four points on the plane are not concyclic, and any three of them are not collinear. Prove that there exists a point $Z$ such that the reflection of each of these four points about $Z$ lies on the circle passing through three remaining points.
Proposed by:A Kuznetsov
5 replies
Gengar_in_Galar
Mar 10, 2025
NicoN9
36 minutes ago
Kinda lookimg Like AM-GM
Atillaa   1
N an hour ago by Natrium
Show that for all positive real numbers \( a, b, c \), the following inequality always holds:
\[
\frac{ab}{b+1} + \frac{bc}{c+1} + \frac{ca}{a+1} \geq \frac{3abc}{1 + abc}
\]
1 reply
Atillaa
2 hours ago
Natrium
an hour ago
best source for inequalitys
Namisgood   1
N an hour ago by Jackson0423
I need some help do I am beginner and have completed Number theory and almost all of algebra (except inequalitys) can anybody suggest a book or resource from where I can study inequalitys
1 reply
Namisgood
an hour ago
Jackson0423
an hour ago
Find the Maximum
Jackson0423   0
an hour ago
Source: Own.
Let \( ABC \) be a triangle with \( AB \leq AC \) and \( \angle BAC = 60^\circ \).
A point \( X \) inside triangle \( ABC \) satisfies the following conditions:
\[
XA^2 + BC^2 \leq XC^2 + AB^2 \leq XB^2 + AC^2.
\]Find the maximum value of \( m \) such that
\[
\frac{XA}{AC} \geq m.
\]
0 replies
Jackson0423
an hour ago
0 replies
Geometry with parallel lines.
falantrng   33
N an hour ago by L13832
Source: RMM 2018,D1 P1
Let $ABCD$ be a cyclic quadrilateral an let $P$ be a point on the side $AB.$ The diagonals $AC$ meets the segments $DP$ at $Q.$ The line through $P$ parallel to $CD$ mmets the extension of the side $CB$ beyond $B$ at $K.$ The line through $Q$ parallel to $BD$ meets the extension of the side $CB$ beyond $B$ at $L.$ Prove that the circumcircles of the triangles $BKP$ and $CLQ$ are tangent .
33 replies
falantrng
Feb 24, 2018
L13832
an hour ago
Something nice
KhuongTrang   25
N 2 hours ago by KhuongTrang
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
25 replies
KhuongTrang
Nov 1, 2023
KhuongTrang
2 hours ago
Lord of the files
AndreiVila   2
N 2 hours ago by Rohit-2006
Source: Mathematical Minds 2024 P3
On the screen of a computer there is an $2^n\times 2^n$ board. On each cell of the main diagonal there is a file. At each step, we may select some files and move them to the left, on their respective rows, by the same distance. What is the minimum number of necessary moves in order to put all files on the first column?

Proposed by Vlad Spătaru
2 replies
AndreiVila
Sep 29, 2024
Rohit-2006
2 hours ago
Inspired by old results
sqing   6
N 2 hours ago by sqing
Source: Own
Let $ a,b\geq 0 $ and $  a^2+ab+b^2=2$ . Prove that
$$ (a+b-ab)\left( \frac{1}{a+1} + \frac{1}{b+1}\right)\leq 2 $$$$ (a+b-ab)\left( \frac{a}{b+1} + \frac{2b}{a+2}\right)\leq 2 $$$$ (a+b-ab)\left( \frac{a}{b+1} + \frac{2b}{a+1}\right)\leq 4$$Let $ a,b  $ be reals such that $  a^2+b^2=2$ . Prove that
$$ (a+b)\left( \frac{1}{a+1} + \frac{1}{b+1}\right)= 2 $$$$ (a+b)\left( \frac{a}{b+1} + \frac{b}{a+1}\right)=2 $$
6 replies
sqing
Today at 2:42 AM
sqing
2 hours ago
An upper bound for Viet Nam TST 2005
Nguyenhuyen_AG   0
2 hours ago
Let $a, \ b, \ c$ be the side lengths of a triangle. Prove that
\[\frac{a^3}{(a+b)^3}+\frac{b^3}{(b+c)^3}+\frac{c^3}{(c+a)^3} \leqslant \frac{9}{8}.\]Viet Nam TST 2005
0 replies
Nguyenhuyen_AG
2 hours ago
0 replies
Geometry :3c
popop614   4
N Apr 4, 2025 by goaoat
Source: MINE :<
Quadrilateral $ABCD$ has an incenter $I$ Suppose $AB > BC$. Let $M$ be the midpoint of $AC$. Suppose that $MI \perp BI$. $DI$ meets $(BDM)$ again at point $T$. Let points $P$ and $Q$ be such that $T$ is the midpoint of $MP$ and $I$ is the midpoint of $MQ$. Point $S$ lies on the plane such that $AMSQ$ is a parallelogram, and suppose the angle bisectors of $MCQ$ and $MSQ$ concur on $IM$.

The angle bisectors of $\angle PAQ$ and $\angle PCQ$ meet $PQ$ at $X$ and $Y$. Prove that $PX = QY$.
4 replies
popop614
Apr 3, 2025
goaoat
Apr 4, 2025
Geometry :3c
G H J
G H BBookmark kLocked kLocked NReply
Source: MINE :<
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popop614
270 posts
#1 • 3 Y
Y by sixoneeight, OronSH, goaoat
Quadrilateral $ABCD$ has an incenter $I$ Suppose $AB > BC$. Let $M$ be the midpoint of $AC$. Suppose that $MI \perp BI$. $DI$ meets $(BDM)$ again at point $T$. Let points $P$ and $Q$ be such that $T$ is the midpoint of $MP$ and $I$ is the midpoint of $MQ$. Point $S$ lies on the plane such that $AMSQ$ is a parallelogram, and suppose the angle bisectors of $MCQ$ and $MSQ$ concur on $IM$.

The angle bisectors of $\angle PAQ$ and $\angle PCQ$ meet $PQ$ at $X$ and $Y$. Prove that $PX = QY$.
This post has been edited 1 time. Last edited by popop614, Apr 3, 2025, 12:42 AM
Reason: asfdasdf
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sixoneeight
1138 posts
#2
Y by
popop614 wrote:
$MI \perp BI$. $DI$
mibidi
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Ianis
402 posts
#3
Y by
sixoneeight wrote:
popop614 wrote:
$MI \perp BI$. $DI$
mibidi

skibidi?
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ItzsleepyXD
95 posts
#4
Y by
Ianis wrote:
sixoneeight wrote:
popop614 wrote:
$MI \perp BI$. $DI$
mibidi

skibidi?

dop dop yes yes
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goaoat
94 posts
#5 • 2 Y
Y by mpcnotnpc, sixoneeight
sogood :love:

$\textbf{Dealing with S:}$
First, let's eliminate $S$. Note that the angle bisector concurrence is the same as $$AQ/AM=MS/SQ=MC/CQ=AM/CQ \implies AM^2=AQ \cdot CQ$$Furthermore, the $PX=QY$ relation is the same as $AQ/AP=CP/CQ$ by angle bisector theorem, which we can simplify to $AM^2=AP \cdot CP$ using the previous equation. Thus, it suffices to prove this last relation.

$\textbf{Dealing with Q:}$
The equality $AM^2=AM \cdot AC=AQ \cdot CQ$ implies $Q$ lies on the lemniscate of Bernoulli with foci $A$ and $C$, $\mathcal{L}$. Thus, the inverse of $Q$ with respect to $(AC)$ lies on the rectangular hyperbola with foci $A$ and $C$, $\mathcal{H}$.

Now, the pedal curve of $\mathcal{H}$ is the image of $\mathcal{L}$ after a $\times 1/2$ homothety at $M$ . This gives that the perpendicular bisector of $MQ$, or $BI$, is tangent to $\mathcal{H}$. However, since we have $\angle ABI = \angle CBI$, by the optical property, the unique point on $BI$ that satisfies this is its tangency with $\mathcal{H}$. Consequently, $B$ lies on $\mathcal{H}$.

Now, $B$ lying on $\mathcal{H}$ is equivalent to $|AB-BC|=\frac{2+\sqrt{2}}{4} AC$, and this can be rearranged with the median length formula to $AB \cdot BC = BM^2$.

Let $Q'$ be the harmonic conjugate of $B$ with respect to $(BAC)$. Then, $$BC/BQ'=BM/AB \implies AB \cdot BC = BM \cdot BQ'=BM^2$$Thus, $BQ'=BM$. Because $Q'$ also lies on the $B$-symmedian, we conclude $Q=Q'$ so $Q$ lies on $(BAC)$.

$\textbf{Dealing with D:}$
Furthermore, $ABCD$ is tangential so $AB+CD=AD+BC$ or $AB-BC=AD-CD$, and this implies $D$ lies on $\mathcal{H}$.

Claim: $\angle MBD = 90^\circ$

Proof: Let the reflection of $B$ over $M$ be $B'$. Note $B'$ lies on $\mathcal{H}$ as $M$ is its center. Let $N$ be the midpoint of $BD$. Realize $MN$ is the Newton line of $ABCD$ so $I$ lies on it. So, $BI \perp (B'D \parallel MN)$. Since triangle $B'BD$ lies on $\mathcal{H}$, its altitude $BI$ should intersect $\mathcal{H}$ at its orthocenter, but $BI$ is tangent to $\mathcal{H}$. Ergo, $B$ is the orthocenter of triangle $B'BD$, proving the claim.

$\textbf{Dealing with P:}$
The previous claim gives $\angle MTD = 90^\circ$, so $P$ is the point on the $D$-symmedian of $DAC$ satisfying $DM=DP$.

Using the same phantom point argument we did with $Q$, we can show that $P$ is the harmonic conjugate of $D$ with respect to $(DAC)$.

Claim: The inverse of $P$ with respect to $(AC)$ is $D'$, the reflection of $D$ over $AC$.

Proof: It's easy to show that the inverse of $(DAC)$ with respect to $(AC)$ is $(D'AC)$, the reflection of $(DAC)$ over $M$. Let $D''$ be the reflection of $D$ over the perpendicular bisector of $AC$. Note $D''$ is the intersection of $PD'$ with $(DAC)$ by some projective, so the inverse of $P$ is the reflection of $D''$ over $M$ which is $D'$ (it is not the reflection of $P$ over $M$ as they lie on different sides of $M$).

$\textbf{Finish:}$
Finally, $D'$ lies on $\mathcal{H}$ as $\mathcal{H}$ is symmetric about $AC$, so $P$ lies on $\mathcal{L}$. This means $AP \cdot CP = AM \cdot CM = AM^2$, which is what we needed to prove, so we are done.
This post has been edited 3 times. Last edited by goaoat, Apr 4, 2025, 8:14 PM
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