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Calculate the distance AD
MTA_2024   6
N Yesterday at 10:17 PM by WheatNeat
A semi-circle is inscribed in a quadrilateral $ABCD$. The center $O$ of the semi-circle is the midpoint of segment $AD$. We have $AB=9$ and $CD=16$.
Calculate the distance $AD$.
6 replies
MTA_2024
Friday at 3:50 PM
WheatNeat
Yesterday at 10:17 PM
Question from Gazeta matematica
abcdefghijklmop   5
N Yesterday at 9:25 PM by abcdefghijklmop
Determine how many subsets formed by 7 elements which are in geometric progession are in the set
{1,2,....,2025}.
5 replies
abcdefghijklmop
Yesterday at 7:30 PM
abcdefghijklmop
Yesterday at 9:25 PM
Unknown triangle area
smartvong   2
N Yesterday at 8:09 PM by vanstraelen
The diagram shows a convex quadrilateral $ABCD$. The points $E$ and $F$ divide $AB$ into three equal parts while the points $G$ and $H$ divide $CD$ into three equal parts. The line segments $AH$ and $ED$ intersect at $I$. The line segments $CF$ and $BG$ intersect at $J$. Given that the areas of the triangles $AID$, $EHI$ and $FJG$ are $154$, $112$, and $99$ respectively, find the area of the triangle $BJC$.

IMAGE
2 replies
smartvong
May 8, 2025
vanstraelen
Yesterday at 8:09 PM
Range if \omega for No Inscribed Right Triangle y = \sin(\omega x)
ThisIsJoe   1
N Yesterday at 5:45 PM by Lankou
For a positive number \omega , determine the range of \omega for which the curve y = \sin(\omega x) has no inscribed right triangle.
Could someone help me figure out how to approach this?
1 reply
ThisIsJoe
May 8, 2025
Lankou
Yesterday at 5:45 PM
2012 preRMO p17, roots of equation x^3 + 3x + 5 = 0
parmenides51   11
N Yesterday at 3:29 PM by Pengu14
Let $x_1,x_2,x_3$ be the roots of the equation $x^3 + 3x + 5 = 0$. What is the value of the expression
$\left( x_1+\frac{1}{x_1} \right)\left( x_2+\frac{1}{x_2} \right)\left( x_3+\frac{1}{x_3} \right)$ ?
11 replies
parmenides51
Jun 17, 2019
Pengu14
Yesterday at 3:29 PM
Interesting question from Al-Khwarezmi olympiad 2024 P3, day1
Adventure1000   3
N Yesterday at 2:38 PM by sqing
Find all $x, y, z \in \left (0, \frac{1}{2}\right )$ such that
$$
\begin{cases}
(3 x^{2}+y^{2}) \sqrt{1-4 z^{2}} \geq z; \\
(3 y^{2}+z^{2}) \sqrt{1-4 x^{2}} \geq x; \\
(3 z^{2}+x^{2}) \sqrt{1-4 y^{2}} \geq y.
\end{cases}
$$Proposed by Ngo Van Trang, Vietnam
3 replies
Adventure1000
May 7, 2025
sqing
Yesterday at 2:38 PM
Malaysia MO IDM UiTM 2025
smartvong   1
N Yesterday at 2:20 PM by jasperE3
MO IDM UiTM 2025 (Category C)

Contest Description

Preliminary Round
Section A
1. Given that $2^a + 2^b = 2016$ such that $a, b \in \mathbb{N}$. Find the value of $a$ and $b$.

2. Find the value of $a, b$ and $c$ such that $$\frac{ab}{a + b} = 1, \frac{bc}{b + c} = 2, \frac{ca}{c + a} = 3.$$
3. If the value of $x + \dfrac{1}{x}$ is $\sqrt{3}$, then find the value of
$$x^{1000} + \frac{1}{x^{1000}}$$.

Section B
1. Let $\mathbb{Z}$ be the set of integers. Determine all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that for all integer $a, b$:
$$f(2a) + 2f(b) = f(f(a + b))$$
2. The side lengths $a, b, c$ of a triangle $\triangle ABC$ are positive integers. Let
$$T_n = (a + b + c)^{2n} - (a - b + c)^{2n} - (a + b - c)^{2n} - (a - b - c)^{2n}$$for any positive integer $n$.
If $\dfrac{T_2}{2T_1} = 2023$ and $a > b > c$, determine all possible perimeters of the triangle $\triangle ABC$.

Final Round
Section A
1. Given that the equation $x^2 + (b - 3)x - 2b^2 + 6b - 4 = 0$ has two roots, where one is twice of the other, find all possible values of $b$.

2. Let $$f(y) = \dfrac{y^2}{y^2 + 1}.$$Find the value of $$f\left(\frac{1}{2001}\right) + f\left(\frac{2}{2001}\right) + \cdots + f\left(\frac{2000}{2001}\right) + f\left(\frac{2001}{2001}\right) + f\left(\frac{2001}{2000}\right) + \cdots + f\left(\frac{2001}{2}\right) + f\left(\frac{2001}{1}\right).$$
3. Find the smallest four-digit positive integer $L$ such that $\sqrt{3\sqrt{L}}$ is an integer.

Section B
1. Given that $\tan A : \tan B : \tan C$ is $1 : 2 : 3$ in triangle $\triangle ABC$, find the ratio of the side length $AC$ to the side length $AB$.

2. Prove that $\cos{\frac{2\pi}{5}} + \cos{\frac{4\pi}{5}} = -\dfrac{1}{2}.$
1 reply
smartvong
Yesterday at 1:01 PM
jasperE3
Yesterday at 2:20 PM
Nice problem
gasgous   2
N Yesterday at 1:47 PM by vincentwant
Given that the product of three integers is $60$.What is the least possible positive sum of the three integers?
2 replies
gasgous
Yesterday at 1:30 PM
vincentwant
Yesterday at 1:47 PM
Angle Formed by Points on the Sides of a Triangle
xeroxia   1
N Yesterday at 1:28 PM by vanstraelen

In triangle $ABC$, points $D$, $E$, and $F$ lie on sides $BC$, $CA$, and $AB$, respectively, such that
$BD = 20$, $DC = 15$, $CE = 13$, $EA = 8$, $AF = 6$, $FB = 22$.

What is the measure of $\angle EDF$?


1 reply
xeroxia
Yesterday at 10:28 AM
vanstraelen
Yesterday at 1:28 PM
Inequalities
sqing   1
N Yesterday at 1:08 PM by sqing
Let $ a,b,c\geq 0 , (a+8)(b+c)=9.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{38}{23}$$Let $ a,b,c\geq 0 , (a+2)(b+c)=3.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2(2\sqrt{3}+1)}{5}$$
1 reply
sqing
Yesterday at 12:50 PM
sqing
Yesterday at 1:08 PM
The best computer problems of the year
NT_G   9
N Apr 10, 2025 by bin_sherlo
Source: https://t.me/NeuroGeometry
1. I is incenter of triangle $ABC$. $K$ is the midpoint of arc $BC$ not containing $A$. $L$ is the midpoint of arc $(BAC)$ $M$ is the midpoint of $IK$ and $N$ is the midpoint of $LM$.  $D$ is projection of $I$ onto $BC$. $S$ is the second intersection of $KD$ and $(ABC)$
Prove that $(ISD)$ is tangent to $(AMN)$.

2. Circle $w$ with center $I$ is incircle of triangle $ABC$. $D$ is projection of $I$ onto $BC$ Points $E, F$ on segments $BI$, $CI$ are following condition: $IE = IF$. $M$ is the midpoint of $EF$. $P$ is the second intersection of $(IMD)$ and $w$. $Q$ is the second intersection of $AP$ and $(IMD)$.
Prove, that $I,E,F,Q$ are concyclic.

I am grateful to Savva Chuev for making the diagrams.
9 replies
NT_G
Dec 31, 2023
bin_sherlo
Apr 10, 2025
The best computer problems of the year
G H J
Source: https://t.me/NeuroGeometry
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NT_G
45 posts
#1 • 2 Y
Y by GeoKing, SBYT
1. I is incenter of triangle $ABC$. $K$ is the midpoint of arc $BC$ not containing $A$. $L$ is the midpoint of arc $(BAC)$ $M$ is the midpoint of $IK$ and $N$ is the midpoint of $LM$.  $D$ is projection of $I$ onto $BC$. $S$ is the second intersection of $KD$ and $(ABC)$
Prove that $(ISD)$ is tangent to $(AMN)$.

2. Circle $w$ with center $I$ is incircle of triangle $ABC$. $D$ is projection of $I$ onto $BC$ Points $E, F$ on segments $BI$, $CI$ are following condition: $IE = IF$. $M$ is the midpoint of $EF$. $P$ is the second intersection of $(IMD)$ and $w$. $Q$ is the second intersection of $AP$ and $(IMD)$.
Prove, that $I,E,F,Q$ are concyclic.

I am grateful to Savva Chuev for making the diagrams.
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#2 • 1 Y
Y by GeoKing
P1 has a generalisation:
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SBYT
196 posts
#3 • 3 Y
Y by GeoKing, vuanhnshn, Om245
Proof of part 1

It is well know that $\angle ASI=\frac{\pi}{2}$ (Sharky Devil).Let $AS$ meets $BC$ at $E$,so $EI$ is the diameter of $\odot ISD$.
By $KI^2=KB^2=KD\cdot KS$, we can know that $KI$ is a tangent line of $\odot IDS$,so $AK\perp IE$.
Let $\odot ISD$ meets $\odot ABC$ at $S,F$,$\angle SFI=\angle SEI=\frac{\pi}{2}-\angle SAI=\frac{\pi}{2}-\angle SLK=\angle SKL=\angle SFL$,so $F,I,L$ are conlinear.
By $MF=MI=MK$, we can get $MF$ is another tangent line of $\odot IDS$.
It means that $FI$ is the polar of $M$ to $\odot ISD$ ,and $L$ lies on this line.
By $N$ is the midpoint of $ML$, we know $N$ lies on the radical axis of $\odot ISD$ and $\odot M$.
Let $NI$ meets $\odot ISD$ at $G$ again,then $NI\cdot NG=NM^2=NL^2=NA^2$.
So $A,N,M,G$ are concyclic.
Let $GP$ is a tangent line of $\odot ISD$,then $\angle IGP=\angle GIK=\angle AIN=\angle NAG$,so $IG$ tangents $\odot AMN$,too.$\Box$.
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Om245
164 posts
#4 • 2 Y
Y by ATGY, GeoKing
Solution of frist part

It well known that $S$ is Sharky-Devil point (as if $S$ is Sharky-Devil point then $SD$ is angle bisector so $S - D - K$)
$$\angle SIA = \angle AES = \angle AEF - \angle SEF$$
Spiral similarity at $S$ sends $\overline{EF}$ to $\overline{BC} \implies \angle SEF = \angle SCB$
so $\angle SIA = 90 - \frac{\angle A}{2} - \angle SCB$. From $S - D - K \implies \angle KSC = \angle KAC$

So $\angle SIA = 90 - \angle SDA$ from $\overline{ID} \perp \overline{BC}$ we get $\angle SIA = \angle SDI$ hence we get $\overline{AK}$ tangent to $(SID)$.

Observe $\angle LAM = 90$ and $N$ is midpoint of $LM$ we get $AN = NM$.

Let $Y =\overline{LI} \cap (ABC)$ from $\angle LYK = 90$. $Y$ also lie on circle $w$ with center $M$ with radius $MI$.
$\overline{MI}$ is tangent to $(SID)$ and $MI = MY$ so $MY$ also tangent to $(SID)$.

Let $ X = \overline{LI} \cap (SID)$ and $ P = \overline{MX} \cap (SID)$
$$(Y,I;X,P)\stackrel{I}{=}(L,M;N,\infty{\overline{LM}}) = -1 \implies \overline{IP} \parallel  \overline{LM}$$
$$\angle IXM = \angle PIM = \angle IMN \implies \angle IXM = \angle XMN$$
Now as $AN=NM$ we get $\angle IXM = \angle XMN = \angle MAN$ so $X,A,N,M$ cyclic points.

Now as tangent to $I$ and $N$ are parallel, Homothety at $X$ sends $(SID)$ to $(ANM)$ hence $X$ is tangent point of $(SID)$ and $(ANM)$


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NT_G
45 posts
#5 • 3 Y
Y by Grifon, SBYT, mathlove_13520
My first solution of P1 was similar to SBYT's. I found it accidently while Cartesian bashing the problem for the New Year stream on NeuroGeometry. Trick with proving tangency using radical axis was new for me...

My second solution:
Due to the fact that $ID \parallel KL$, $\angle IDS = \angle LKS$, therefore $LI$ passes through the second intersection of $(SID)$ and $(ABC)$. It means that $(SID)$ contains the point of tangency of circumcircle and mixtilinear circle of $\triangle ABC$.
Now we invert the problem with center in $A$. Problem turns into:
In triangle $ABC$ $I_a$ is the $A$ - excenter. $N$ is the projection of $I_a$ onto $BC$. $L, L_a$ are foots of bisectors of $\angle BAC$. K is a point on $AI_a$ such that $(I,I_a; K, A) = -1$. $A'$ is a point on $\Gamma = (AL_{a}K)$ suiting: $KA = KA'$. $\gamma$ is a circle passing through $N, I_a$ that is tangent to $AI_a$. We have to prove that $KA'$ is tangent to $\gamma$.
Let's spot that due to the fact that $\angle L_{a}AK = \frac{\pi}{2}$, $A'$ is the reflection of $A$ in $KL_a$.Therefore, we have to prove that line $O_{\Gamma}K$ passes through $O_{\gamma}$. $L_a, K, O_{\Gamma}$ are obviously collinear. Let $I_{a}'$ be the reflection of $I_{a}$ over $O_{\Gamma}$. $I_{a}'$ lies on $BC$. Then after spotting that $I_{a}I_{a}' \parallel AL_{a}$ and projecting $(I,I_a; K, A)$ from $L_a$ onto $I_{a}I_{a}' $ we get what we needed.
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SerdarBozdag
892 posts
#6 • 1 Y
Y by GeoKing
First problem. $G=AS \cap BC$ is on $(SID)$. $LI \cap ABC = J$. $A'$ is the antipode of $A$ and $S,I,A'$ are collinear. By radical axis theorem on $(ASFIE), (IJK), (ABC)$, $G$ is on $JK$. $H = IN \cap (SID)$. Let $I'$ be the reflection of $I$ across $N$.

$LIMI'$ is parallelogram so $LI' = MI = MK$ and $I'M \parallel JL \implies LJMI'$ is cyclic.

$\angle I'HJ + \angle JMI' = \angle ISJ + \angle JMI + \angle IMI' = \angle JAA' +\angle JMI + \angle JIM = \angle JAA' + 90 + \angle JKA =  \angle JAA' + 90^{\circ} + \angle JA'A = 180^{\circ} \implies H \in (LJMI')$.

$NA^2 = NM^2 = NM \cdot NL = NI' \cdot NH = NI \cdot NH \implies HMNA $ is cyclic. If tangents to $(SID)$ at $I$ and $H$ intersect at $Q$. $\angle NHQ = \angle HIQ = \angle HAN \implies HQ$ is tangent to $(HMNA)$.
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SerdarBozdag
892 posts
#7 • 1 Y
Y by GeoKing
Second problem. Let $G= EF \cap BC$. $(IMDG) \cap (I) = P$ , $(IMDG) \cap (IEF) = Q$, $(IMDG) \cap AI = L$, $PM \cap GL = S$, $N$ is the antipode of $I$ in $(IEF)$ and $AI \cap (I) = J$. I will prove that $A \in PQ$.

$\angle IQG = 90^{\circ} = \angle IQN \implies N \in QG$.

$\angle EID = 90^{\circ} - \frac{B}{2} = \angle JIF \implies J $ is the reflection of $D$ across $AI$. $\angle IJM = \angle IDM = \angle IPM = \angle IJP \implies J \in PM$. $\angle LGD = \angle DIJ = 2 \cdot \angle DIM = 2 \cdot \angle DGM \implies ML = MD \implies M$ is the center of $(SLJD)$.

$N,S,A$ are collinear $\iff$ (by Menelaus)
$$\frac{NM}{NI} \cdot \frac{IA}{AJ} \cdot \frac{JS}{SM} = 1$$This is true because $\frac{NM}{NI} = \cos^2 (\angle IFE) = \frac{1+\cos (90^{\circ} - \angle A/2) }{2}= \frac{JA}{IA} \cdot \frac{JS}{SM}$.

Applying Pascal on $PQG  LIM$ shows that $PQ \cap LI \in SN \implies A=PQ \cap LI$ as desired.
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NT_G
45 posts
#9 • 1 Y
Y by GeoKing
P2: (R.Prozorov's solution)
Let $R$ be the second intersection of $(IMD)$, $BC$. Obviously, $R$ lies on $EF$, and $IR$ is diameter of $(IMD)$.
$\angle IQD = \angle PQI$. So, using DIT for $ABDC$, we get that $\angle BQI = \angle CQI$.
$K = IQ \cap BC$. $IQ \perp QR$, therefore $(R,K; B,C) = -1 \Rightarrow $ $CE$, $BF$, $IQ$ are concurent. Using isogonal theorem for $\angle BQC$ and points $E$, $F$, we prove that $\angle EQI = \angle IQF$. Therefore, due to the equality: $IE = IF$, we get what we needed.
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SBYT
196 posts
#10 • 1 Y
Y by GeoKing
Another solution which is similar to NT_G's:
We get $(R,K;B,C)=-1$ the same way,$IK$ meets $EF$ at $N$.$(R,K;B,C)=-1\implies(IR,IK;IB,IC)=-1\implies(R,N;E,F)=-1\implies(QR,QN;QE,QF)=-1$.
Due to $IQ\perp RQ$,$IQ$ is the angle bisector of $\angle EQF$,and $IE=IF$,so $I,E,Q,F$ are concyclic.$\Box$
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bin_sherlo
721 posts
#11 • 1 Y
Y by MS_asdfgzxcvb
I 'll present a solution to the first problem.
Let $AS\cap BC=R,MR\cap (ABC)=T$. Let $R,W'$ be the feet of the perpendicular from $P,N$ to $IN,PI$ respectively. Let $W,I'$ be the reflections of $I$ over $W',N$. Note that $P,T,I,R,S,D$ are concyclic.
Claim: $A,M,N,R$ are concyclic.
Proof: $A,P,K,W$ are concyclic because $IP.IW=IL.IT=IA.IK$ so $A,P,M,W'$ are concyclic. $IA.IM=IP.IW'=IN.IR$ which gives the result.
Let $PR\cap (AMNR)=V$ which is the antipode of $N$. Since $\measuredangle LAM=90$, we have $NA=NM$ and $NV$ is diameter thus, $NV\perp AI\perp IP$. Hence $(RPI)$ and $(RNV)$ are tangent to each other as desired.$\blacksquare$
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