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Tangents forms triangle with two times less area
NO_SQUARES   0
2 hours ago
Source: Kvant 2025 no. 2 M2831
Let $DEF$ be triangle, inscribed in parabola. Tangents in points $D,E,F$ forms triangle $ABC$. Prove that $S_{DEF}=2S_{ABC}$. ($S_T$ is area of triangle $T$).
From F.S.Macaulay's book «Geometrical Conics», suggested by M. Panov
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NO_SQUARES
2 hours ago
0 replies
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Woaah a lot of external tangents
egxa   3
N 3 hours ago by NO_SQUARES
Source: All Russian 2025 11.7
A quadrilateral \( ABCD \) with no parallel sides is inscribed in a circle \( \Omega \). Circles \( \omega_a, \omega_b, \omega_c, \omega_d \) are inscribed in triangles \( DAB, ABC, BCD, CDA \), respectively. Common external tangents are drawn between \( \omega_a \) and \( \omega_b \), \( \omega_b \) and \( \omega_c \), \( \omega_c \) and \( \omega_d \), and \( \omega_d \) and \( \omega_a \), not containing any sides of quadrilateral \( ABCD \). A quadrilateral whose consecutive sides lie on these four lines is inscribed in a circle \( \Gamma \). Prove that the lines joining the centers of \( \omega_a \) and \( \omega_c \), \( \omega_b \) and \( \omega_d \), and the centers of \( \Omega \) and \( \Gamma \) all intersect at one point.
3 replies
egxa
Apr 18, 2025
NO_SQUARES
3 hours ago
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2023 Hong Kong TST 3 (CHKMO) Problem 4
PikaNiko   3
N 3 hours ago by lightsynth123
Source: 2023 Hong Kong TST 3 (CHKMO)
Let $ABCD$ be a quadrilateral inscribed in a circle $\Gamma$ such that $AB=BC=CD$. Let $M$ and $N$ be the midpoints of $AD$ and $AB$ respectively. The line $CM$ meets $\Gamma$ again at $E$. Prove that the tangent at $E$ to $\Gamma$, the line $AD$ and the line $CN$ are concurrent.
3 replies
PikaNiko
Dec 3, 2022
lightsynth123
3 hours ago
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Help my diagram has too many points
MarkBcc168   27
N 3 hours ago by Om245
Source: IMO Shortlist 2023 G6
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$. A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $BC$ at $D$. Let $AB$ and $AC$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$. Lines $MP$ and $NQ$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ again at $X\neq K$.

Prove that $\angle BXP = \angle CXQ$.

Kian Moshiri, United Kingdom
27 replies
MarkBcc168
Jul 17, 2024
Om245
3 hours ago
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Geometry, SMO 2016, not easy
Zoom   18
N 4 hours ago by SimplisticFormulas
Source: Serbia National Olympiad 2016, day 1, P3
Let $ABC$ be a triangle and $O$ its circumcentre. A line tangent to the circumcircle of the triangle $BOC$ intersects sides $AB$ at $D$ and $AC$ at $E$. Let $A'$ be the image of $A$ under $DE$. Prove that the circumcircle of the triangle $A'DE$ is tangent to the circumcircle of triangle $ABC$.
18 replies
Zoom
Apr 1, 2016
SimplisticFormulas
4 hours ago
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A touching question on perpendicular lines
Tintarn   2
N 4 hours ago by pi_quadrat_sechstel
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 3
Let $k$ be a semicircle with diameter $AB$ and midpoint $M$. Let $P$ be a point on $k$ different from $A$ and $B$.

The circle $k_A$ touches $k$ in a point $C$, the segment $MA$ in a point $D$, and additionally the segment $MP$. The circle $k_B$ touches $k$ in a point $E$ and additionally the segments $MB$ and $MP$.

Show that the lines $AE$ and $CD$ are perpendicular.
2 replies
Tintarn
Mar 17, 2025
pi_quadrat_sechstel
4 hours ago
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Iran Team Selection Test 2016
MRF2017   9
N Today at 4:22 AM by SimplisticFormulas
Source: TST3,day1,P2
Let $ABC$ be an arbitrary triangle and $O$ is the circumcenter of $\triangle {ABC}$.Points $X,Y$ lie on $AB,AC$,respectively such that the reflection of $BC$ WRT $XY$ is tangent to circumcircle of $\triangle {AXY}$.Prove that the circumcircle of triangle $AXY$ is tangent to circumcircle of triangle $BOC$.
9 replies
MRF2017
Jul 15, 2016
SimplisticFormulas
Today at 4:22 AM
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Circumcircle excircle chaos
CyclicISLscelesTrapezoid   25
N Yesterday at 7:55 PM by bin_sherlo
Source: ISL 2021 G8
Let $ABC$ be a triangle with circumcircle $\omega$ and let $\Omega_A$ be the $A$-excircle. Let $X$ and $Y$ be the intersection points of $\omega$ and $\Omega_A$. Let $P$ and $Q$ be the projections of $A$ onto the tangent lines to $\Omega_A$ at $X$ and $Y$ respectively. The tangent line at $P$ to the circumcircle of the triangle $APX$ intersects the tangent line at $Q$ to the circumcircle of the triangle $AQY$ at a point $R$. Prove that $\overline{AR} \perp \overline{BC}$.
25 replies
CyclicISLscelesTrapezoid
Jul 12, 2022
bin_sherlo
Yesterday at 7:55 PM
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Geometry Problem
Itoz   3
N Yesterday at 3:09 PM by Itoz
Source: Own
Given $\triangle ABC$. Let the perpendicular line from $A$ to $BC$ meets $BC,\odot(ABC)$ at points $S,K$, respectively, and the foot from $B$ to $AC$ is $L$. $\odot (AKL)$ intersects line $AB$ at $T(\neq A)$, $\odot(AST)$ intersects line $AC$ at $M(\neq A)$, and lines $TM,CK$ intersect at $N$.

Prove that $\odot(CNM)$ is tangent to $\odot (BST)$.
3 replies
Itoz
Apr 18, 2025
Itoz
Yesterday at 3:09 PM
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combinatorial geo question
SAAAAAAA_B   2
N Yesterday at 2:35 PM by R8kt
Kuba has two finite families $\mathcal{A}, \mathcal{B}$ of convex polygons (in the plane). It turns out that every point of the plane lies in the same number of elements of $\mathcal{A}$ as elements of $\mathcal{B}$. Prove that $|\mathcal{A}| = |\mathcal{B}|$.

\textit{Note:} We treat segments and points as degenerate convex polygons, and they can be elements of $\mathcal{A}$ or $\mathcal{B}$.
2 replies
SAAAAAAA_B
Apr 14, 2025
R8kt
Yesterday at 2:35 PM
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Equation with powers
a_507_bc   6
N Apr 3, 2025 by EVKV
Source: Serbia JBMO TST 2024 P1
Find all non-negative integers $x, y$ and primes $p$ such that $$3^x+p^2=7 \cdot 2^y.$$
6 replies
a_507_bc
May 25, 2024
EVKV
Apr 3, 2025
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Equation with powers
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Source: Serbia JBMO TST 2024 P1
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a_507_bc
676 posts
a_507_bc
#1 h  May 25, 2024, 5:27 PM
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Find all non-negative integers $x, y$ and primes $p$ such that $$3^x+p^2=7 \cdot 2^y.$$
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NO_SQUARES
1082 posts
NO_SQUARES
#2 h  May 25, 2024, 6:10 PM
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a_507_bc wrote:
Find all non-negative integers $x, y$ and primes $p$ such that $$3^x+p^2=7 \cdot 2^y.$$
If $y=0$ then $x=1, p=2$. If $y=1$ then by mod 3 $x=0$ and there are no solutions. Now let $y>1$, so $4 | RHS$.
Note that since $4|7-3$ we have $2 \not | x$. After this look at mod 8 to get $y<3$ ($p \not = 2$).
This post has been edited 2 times. Last edited by NO_SQUARES, May 25, 2024, 7:30 PM
Reason: was wrong
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RagvaloD
4909 posts
RagvaloD
#3 h  May 25, 2024, 7:21 PM • 2 Y
Y by NO_SQUARES, ehuseyinyigit
$p=2 \to y=0,x=1$
$p$ is odd $\to 3^x+p^2 \equiv 2,4 \pmod {8} \to y<3$
for $y=1$ there are no solutions
For $y=2: x=1,p=5$
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Assassino9931
1248 posts
Assassino9931
#4 h  May 25, 2024, 8:20 PM
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We only need that $p$ is odd if $p \geq 3$. Indeed, mod 8 we have $3^x \equiv 3,1$ and $p^2\equiv 1$ and so $3^x + p^2 \equiv 2, 4$ while $7 \cdot 2^y \equiv 0$ for $y\geq 3$, contradiction. Hence either $p=2$ or $y\leq 2$.

If $p=2$, then parity insists on $y=0$, so $x=1$. If $y=2$, then only $p=5$ and $x=1$ works. If $y=1$, then there are no solutions. If $y=0$, then $p=2$ and $x=1$ works.

Hence all solutions are $(x,y,p) = (1,0,2), (1,2,5)$.
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THE_SOLVER
1 post
THE_SOLVER
#5 h  Jul 31, 2024, 4:16 PM • 1 Y
Y by JelaByteEngineer
Simply by applying mod 8 which restricts or bounds the value of y i.e y<3
By checking manually which gives us 2 solutions i.e (1,2,0);(1,2,5)
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ali123456
51 posts
ali123456
#6 h  Mar 18, 2025, 4:34 PM
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sketch of my solution
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EVKV
52 posts
EVKV
#7 h  Apr 3, 2025, 2:32 AM
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p is odd for $y \neq 0$
and clearly x= 1 , y= 0, p=2 satisfies
$For y \geq 3$
$either 2,4 \equiv 0$ $mod$ $8$ nonsense
now checking remaining
all solutions are $(x,y,p) = (1,0,2), (1,2,5)$.
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