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Prove XBY equal to angle C
nataliaonline75   2
N 24 minutes ago by starchan
Let $M$ be the midpoint of $BC$ on triangle $ABC$. Point $X$ lies on segment $AC$ such that $AX=BX$ and $Y$ on line $AM$ such that $XY//AB$. Prove that $\angle XBY = \angle ACB$.
2 replies
nataliaonline75
Yesterday at 2:47 PM
starchan
24 minutes ago
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old one but good one
Sunjee   2
N 43 minutes ago by ehuseyinyigit
If $x_1,x_2,...,x_n $ are positive numbers, then prove that
$$\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\cdots+ \frac{x_n}{1+x_1^2+\cdots+x_n^2}\geq \sqrt{n}$$
2 replies
Sunjee
3 hours ago
ehuseyinyigit
43 minutes ago
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Inspired by giangtruong13
sqing   0
an hour ago
Source: Own
Let $ a,b>0 .$ Prove that$$ \frac{a}{b}+\frac{b}{a}+\frac{a^3}{2b^3+kab^2}+\frac{2b^3}{a^3+b^3+kab^2} \geq \frac{2k+7}{k+2}$$Where $ k\geq 0. $
0 replies
sqing
an hour ago
0 replies
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Functional equation
Math-wiz   24
N an hour ago by nmoon_nya
Source: IMOC SL A1
Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all $x,y\in\mathbb{R}$,
$$f(xy+f(x))=f(xf(y))+x$$
24 replies
Math-wiz
Dec 15, 2019
nmoon_nya
an hour ago
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we can find one pair of a boy and a girl
orl   18
N an hour ago by bin_sherlo
Source: Vietnam TST 2001 for the 42th IMO, problem 3
Some club has 42 members. It’s known that among 31 arbitrary club members, we can find one pair of a boy and a girl that they know each other. Show that from club members we can choose 12 pairs of knowing each other boys and girls.
18 replies
orl
Jun 26, 2005
bin_sherlo
an hour ago
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geometry
blug   0
an hour ago
In trapezius $ABCD$, segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle B$ and $\angle D$ intersect at $P$. Circumcircles of $ABP$ and $CDP$ meet again at $Q$. Angle bisector of $\angle D$ cuts $AB$ at $S$. Prove that
$$1. QD=QS,$$$$2. \angle DCQ=\angle BCQ,$$$$3. \angle BAQ=\angle QAD.$$
0 replies
blug
an hour ago
0 replies
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Eulerline problem
Retemoeg   0
an hour ago
Source: Extension from a problem I read in a book
Show that the isogonal conjugate of the isotomic conjugate of the orthocenter lies on the Euler line.
0 replies
Retemoeg
an hour ago
0 replies
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Nice numer theory
GeoArt   4
N 2 hours ago by Blackbeam999
$p$ is a prime number, $m, x, y$ are natural numbers ($m, x, y > 1$). It is known that $\frac{x^p + y^p}{2}$ $=$ $(\frac{x+y}{2} )^m$. Prove that $p = m$.
4 replies
GeoArt
Jan 7, 2021
Blackbeam999
2 hours ago
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Inequality
MathsII-enjoy   6
N 2 hours ago by MathsII-enjoy
A interesting problem generalized :-D
6 replies
MathsII-enjoy
Saturday at 1:59 PM
MathsII-enjoy
2 hours ago
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Number Theory
fasttrust_12-mn   8
N 2 hours ago by ErTeeEs06
Source: Pan African Mathematics Olympiad P1
Find all positive intgers $a,b$ and $c$ such that $\frac{a+b}{a+c}=\frac{b+c}{b+a}$ and $ab+bc+ca$ is a prime number
8 replies
fasttrust_12-mn
Aug 15, 2024
ErTeeEs06
2 hours ago
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My Unsolved FE on R+
ZeltaQN2008   4
N 2 hours ago by mashumaro
Source: IDK
Give $a>0$. Find all funcitions $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that for all any $x,y\in (0,\infty):$
$$f(xf(y)+a)=yf(x+y+a)$$
4 replies
ZeltaQN2008
4 hours ago
mashumaro
2 hours ago
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J
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two 3D problems in one day
egxa   1
N Apr 18, 2025 by sami1618
Source: All Russian 2025 11.2
A right prism \(ABCA_1B_1C_1\) is given. It is known that triangles \(A_1BC\), \(AB_1C\), \(ABC_1\), and \(ABC\) are all acute. Prove that the orthocenters of these triangles, together with the centroid of triangle \(ABC\), lie on the same sphere.
1 reply
egxa
Apr 18, 2025
sami1618
Apr 18, 2025
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two 3D problems in one day
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Reveal topic tags
geometry
3D geometry
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G H BBookmark kLocked kLocked NReply
Source: All Russian 2025 11.2
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egxa
210 posts
egxa
#1 h  Apr 18, 2025, 9:44 AM • 2 Y
Y by buratinogigle, Miquel-point
A right prism \(ABCA_1B_1C_1\) is given. It is known that triangles \(A_1BC\), \(AB_1C\), \(ABC_1\), and \(ABC\) are all acute. Prove that the orthocenters of these triangles, together with the centroid of triangle \(ABC\), lie on the same sphere.
Z K Y
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sami1618
902 posts
sami1618
#2 h  Apr 18, 2025, 2:41 PM
Y by
Let $H_A$, $H_B$, $H_C$, and $H$ be the orthocenters of triangles $A_1BC$, $AB_1C$, $ABC_1$, and $ABC$, respectively. Let $G$ be the centroid of triangle $ABC$.

Claim. The point $H_A$ is the foot of $H$ onto the plane $A_1BC$.
Proof. Let $H'$ be the foot of $H$ onto the plane $A_1BC$. Notice that $BH$ is perpendicular to plane $ACC_1A_1$. Thus $BH\perp A_1C$, which implies $BH'\perp A_1C$. Similarly, $CH'\perp A_1B$. Thus $H'=H_A$, as required.

An identical result holds for $H_B$ and $H_C$. Notice the centroids of triangles $A_1BC$, $AB_1C$, and $ABC_1$ all coincide at some point $X$. Since $G$ is the foot of $X$ onto the plane $ABC$, it follows that $\angle HGX=90^{\circ}$. By the claim, we also get that $\angle HH_AX=HH_BX=HH_CX=90^{\circ}$. Thus all five points lie on the sphere with diameter $HX$, concluding the proof.
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