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Cyclic quadrilateral and a tangent
a_507_bc   6
N 16 minutes ago by User141208
Source: IMOC 2023 G6
Triangle $ABC$ has circumcenter $O$. $D$ is the foot from $A$ to $BC$, and $P$ is apoint on $AD$. The feet from $P$ to $CA, AB$ are $E, F$, respectively, and the foot from $D$ to $EF$ is $T$. $AO$ meets $(ABC)$ again at $A'$. $A'D$ meets $(ABC)$ again at $R$. If $Q$ is a point on $AO$ satisfying $\angle ABP = \angle QBC$, prove that $D, P, T, R$ lie on acircle and $DQ$ is tangent to it.
6 replies
+1 w
a_507_bc
Sep 9, 2023
User141208
16 minutes ago
Geometry
blug   0
18 minutes ago
Source: own
Let $H$ be the orthocenter of triangle $ABC$. Let $p, q$ be angle bisectors of $AHB$ and $AHC$ respectively. We denote $K=p\cap AB, L=p\cap AC, M=q\cap AC, N=q\cap AB$. Circumcircles of $NKH$ and $MHL$ intersect at $P\ne H$. Prove that
$$\angle BAC=\angle PKL+\angle PMN.$$
0 replies
blug
18 minutes ago
0 replies
XZ passes through the midpoint of BK, isosceles, KX = CX, angle bisector
parmenides51   5
N an hour ago by Kyj9981
Source: 1st Girls in Mathematics Tournament 2019 p5 (Brazil) / Torneio Meninas na Matematica (TM^2 )
Let $ABC$ be an isosceles triangle with $AB = AC$. Let $X$ and $K$ points over $AC$ and $AB$, respectively, such that $KX = CX$. Bisector of $\angle AKX$ intersects line $BC$ at $Z$. Show that $XZ$ passes through the midpoint of $BK$.
5 replies
parmenides51
May 25, 2020
Kyj9981
an hour ago
Solution needed ASAP
UglyScientist   7
N an hour ago by Captainscrubz
$ABC$ is acute triangle. $H$ is orthocenter, $M$ is the midpoint of $BC$, $L$ is the midpoint of smaller arc $BC$. Point $K$ is on $AH$ such that, $MK$ is perpendicular to $AL$. Prove that: $HMLK$ is paralelogram(Synthetic sol needed).
7 replies
UglyScientist
4 hours ago
Captainscrubz
an hour ago
Diophantine Equation with prime numbers and bonus conditions
p.lazarov06   10
N an hour ago by mathbetter
Source: 2023 Bulgaria JBMO TST Problem 3
Find all natural numbers $a$, $b$, $c$ and prime numbers $p$ and $q$, such that:

$\blacksquare$ $4\nmid c$
$\blacksquare$ $p\not\equiv 11\pmod{16}$
$\blacksquare$ $p^aq^b-1=(p+4)^c$
10 replies
1 viewing
p.lazarov06
May 7, 2023
mathbetter
an hour ago
Concurrence in Cyclic Quadrilateral
GrantStar   39
N an hour ago by ItsBesi
Source: IMO Shortlist 2023 G3
Let $ABCD$ be a cyclic quadrilateral with $\angle BAD < \angle ADC$. Let $M$ be the midpoint of the arc $CD$ not containing $A$. Suppose there is a point $P$ inside $ABCD$ such that $\angle ADB = \angle CPD$ and $\angle ADP = \angle PCB$.

Prove that lines $AD, PM$, and $BC$ are concurrent.
39 replies
GrantStar
Jul 17, 2024
ItsBesi
an hour ago
IMO Genre Predictions
ohiorizzler1434   22
N an hour ago by rhydon516
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
22 replies
ohiorizzler1434
Today at 6:51 AM
rhydon516
an hour ago
Inequality
MathsII-enjoy   1
N 2 hours ago by arqady
A interesting problem generalized :-D
1 reply
MathsII-enjoy
3 hours ago
arqady
2 hours ago
Inequality
lgx57   2
N 2 hours ago by mashumaro
Source: Own
$a,b,c>0,ab+bc+ca=1$. Prove that

$$\sum \sqrt{8ab+1} \ge 5$$
(I don't know whether the equality holds)
2 replies
lgx57
2 hours ago
mashumaro
2 hours ago
Find min
lgx57   1
N 2 hours ago by arqady
Source: Own
Find min of $\dfrac{a^2}{ab+1}+\dfrac{b^2+2}{a+b}$
1 reply
lgx57
2 hours ago
arqady
2 hours ago
Product is a perfect square( very easy)
Nuran2010   1
N 2 hours ago by SomeonecoolLovesMaths
Source: Azerbaijan Junior National Olympiad 2021 P1
At least how many numbers must be deleted from the product $1 \times 2 \times \dots \times 22 \times 23$ in order to make it a perfect square?
1 reply
Nuran2010
4 hours ago
SomeonecoolLovesMaths
2 hours ago
smo 2018 open 2nd round q2
dominicleejun   7
N 2 hours ago by Kyj9981
Let O be a point inside triangle ABC such that $\angle BOC$ is $90^\circ$ and $\angle BAO = \angle BCO$. Prove that $\angle OMN$ is $90$ degrees, where $M$ and $N$ are the midpoints of $\overline{AC}$ and $\overline{BC}$, respectively.
7 replies
dominicleejun
Aug 15, 2019
Kyj9981
2 hours ago
trolling geometry problem
iStud   4
N Apr 22, 2025 by GreenTea2593
Source: Monthly Contest KTOM April P3 Essay
Given a cyclic quadrilateral $ABCD$ with $BC<AD$ and $CD<AB$. Lines $BC$ and $AD$ intersect at $X$, and lines $CD$ and $AB$ intersect at $Y$. Let $E,F,G,H$ be the midpoints of sides $AB,BC,CD,DA$, respectively. Let $S$ and $T$ be points on segment $EG$ and $FH$, respectively, so that $XS$ is the angle bisector of $\angle{DXA}$ and $YT$ is the angle bisector of $\angle{DYA}$. Prove that $TS$ is parallel to $BD$ if and only if $AC$ divides $ABCD$ into two triangles with equal area.
4 replies
iStud
Apr 21, 2025
GreenTea2593
Apr 22, 2025
trolling geometry problem
G H J
Source: Monthly Contest KTOM April P3 Essay
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iStud
268 posts
#1
Y by
Given a cyclic quadrilateral $ABCD$ with $BC<AD$ and $CD<AB$. Lines $BC$ and $AD$ intersect at $X$, and lines $CD$ and $AB$ intersect at $Y$. Let $E,F,G,H$ be the midpoints of sides $AB,BC,CD,DA$, respectively. Let $S$ and $T$ be points on segment $EG$ and $FH$, respectively, so that $XS$ is the angle bisector of $\angle{DXA}$ and $YT$ is the angle bisector of $\angle{DYA}$. Prove that $TS$ is parallel to $BD$ if and only if $AC$ divides $ABCD$ into two triangles with equal area.
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iStud
268 posts
#2
Y by
i can prove the $\Longleftarrow$ direction but not pretty sure with the $\Longrightarrow$ direction, can anyone help me?
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mathuz
1520 posts
#3
Y by
Note that $EFGH$ is a parallelogram and $FG\parallel EH \parallel BD$. Therefore, $TS\parallel BD$ iff $ES:SG = HT:TF$.

Assume now $TS\parallel BD$. Then $ES:SG = HT:TF$.
We have $\triangle XCD \sim \triangle XAB$ and $\triangle YBC\sim \triangle YDA$. This implies $XS$ is the angle bisector of $\angle EXG$ and $YT$ is the angle bisector of $\angle HYF$. Therefore, we have \[ \frac{EX}{GX} = \frac{ES}{SG} = \frac{HT}{TF} = \frac{HY}{FY}. \]Again, by the similarity this means \[ \frac{AB}{CD} = \frac{DA}{BC} \quad \Rightarrow \quad AB\cdot BC = CD\cdot DA, \]which essentially means the areas of $ABC$ and $ADC$ are equal.

This gives $\Longrightarrow $ direction.
This post has been edited 1 time. Last edited by mathuz, Apr 22, 2025, 1:35 AM
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iStud
268 posts
#4
Y by
mathuz wrote:
Note that $EFGH$ is a parallelogram and $FG\parallel EH \parallel BD$. Therefore, $TS\parallel BD$ iff $ES:SG = HT:TF$.

Assume now $TS\parallel BD$. Then $ES:SG = HT:TF$.
We have $\triangle XCD \sim \triangle XAB$ and $\triangle YBC\sim \triangle YDA$. This implies $XS$ is the angle bisector of $\angle EXG$ and $YT$ is the angle bisector of $\angle HYF$. Therefore, we have \[ \frac{EX}{GX} = \frac{ES}{SG} = \frac{HT}{TF} = \frac{HY}{FY}. \]Again, by the similarity this means \[ \frac{AB}{CD} = \frac{DA}{BC} \quad \Rightarrow \quad AB\cdot BC = CD\cdot DA, \]which essentially means the areas of $ABC$ and $ADC$ are equal.

This gives $\Longrightarrow $ direction.

demn, all of my Menelause bash worth nothing in front of this dude's solution
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GreenTea2593
236 posts
#5
Y by
Oops, I took this from 2016 Japan MO Finals Problem 2
This post has been edited 1 time. Last edited by GreenTea2593, Apr 22, 2025, 3:12 AM
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