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Centrally symmetric polyhedron
genius_007   0
4 minutes ago
Source: unknown
Does there exist a convex polyhedron with an odd number of sides, where each side is centrally symmetric?
0 replies
genius_007
4 minutes ago
0 replies
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A game of digits and seventh powers
v_Enhance   28
N 24 minutes ago by blueprimes
Source: Taiwan 2014 TST3 Quiz 1, P2
Alice and Bob play the following game. They alternate selecting distinct nonzero digits (from $1$ to $9$) until they have chosen seven such digits, and then consider the resulting seven-digit number by concatenating the digits in the order selected, with the seventh digit appearing last (i.e. $\overline{A_1B_2A_3B_4A_6B_6A_7}$). Alice wins if and only if the resulting number is the last seven decimal digits of some perfect seventh power. Please determine which player has the winning strategy.
28 replies
v_Enhance
Jul 18, 2014
blueprimes
24 minutes ago
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n containers distribute gas to a car for a single loop along the track
parmenides51   16
N 33 minutes ago by v_Enhance
Source: Spanish Mathematical Olympiad 1997 P6
The exact quantity of gas needed for a car to complete a single loop around a track is distributed among $n$ containers placed along the track. Prove that there exists a position starting at which the car, beginning with an empty tank of gas, can complete a loop around the track without running out of gas. The tank of gas is assumed to be large enough.
16 replies
parmenides51
Jul 31, 2018
v_Enhance
33 minutes ago
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Factorial: n!|a^n+1
Nima Ahmadi Pour   68
N 39 minutes ago by blueprimes
Source: IMO Shortlist 2005 N4, Iran preparation exam
Find all positive integers $ n$ such that there exists a unique integer $ a$ such that $ 0\leq a < n!$ with the following property:
\[ n!\mid a^n + 1 \]

Proposed by Carlos Caicedo, Colombia
68 replies
Nima Ahmadi Pour
Apr 24, 2006
blueprimes
39 minutes ago
J
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Geometry problem
Whatisthepurposeoflife   1
N an hour ago by Royal_mhyasd
Source: Derived from MEMO 2024 I3
Triangle ∆ABC is scalene the circle w that goes through the points A and B intersects AC at E BC at D let the Lines BE and AD intersect at point F. And let the tangents A and B of circle w Intersect at point G.
Prove that C F and G are collinear
1 reply
Whatisthepurposeoflife
an hour ago
Royal_mhyasd
an hour ago
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Inspired by SunnyEvan
sqing   2
N an hour ago by SunnyEvan
Source: Own
Let $ a,b,c $ be reals such that $ a^2+b^2+c^2=2(ab+bc+ca). $ Prove that$$ \frac{1}{12} \leq \frac{a^2b+b^2c+c^2a}{(a+b+c)^3} \leq \frac{5}{36} $$Let $ a,b,c $ be reals such that $ a^2+b^2+c^2=5(ab+bc+ca). $ Prove that$$ -\frac{1}{25} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{197}{675} $$
2 replies
sqing
May 17, 2025
SunnyEvan
an hour ago
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\frac{x^2}{x+y}+\frac{y^2}{1-x}+\frac{(1-x-y)^2}{1-y}\geq\frac{1}{2} if 0<x,y<1
parmenides51   6
N an hour ago by JH_K2IMO
Source: KJMO 2009 p3
For two arbitrary reals $x, y$ which are larger than $0$ and less than $1.$ Prove that$$\frac{x^2}{x+y}+\frac{y^2}{1-x}+\frac{(1-x-y)^2}{1-y}\geq\frac{1}{2}.$$
6 replies
parmenides51
May 2, 2019
JH_K2IMO
an hour ago
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3 var inequality
SunnyEvan   9
N an hour ago by SunnyEvan
Let $ a,b,c \in R $ ,such that $ a^2+b^2+c^2=4(ab+bc+ca)$Prove that :$$ \frac{7-2\sqrt{14}}{48} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{7+2\sqrt{14}}{48} $$
9 replies
SunnyEvan
May 17, 2025
SunnyEvan
an hour ago
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2-var inequality
sqing   6
N an hour ago by sqing
Source: Own
Let $ a,b>0 , a^2+b^2-ab\leq 1 . $ Prove that
$$a^3+b^3 -\frac{a^4}{b+1} -\frac{b^4}{a+1} \leq 1 $$
6 replies
sqing
Yesterday at 2:15 AM
sqing
an hour ago
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A sharp one with 3 var (3)
mihaig   2
N 2 hours ago by mihaig
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$a^2+b^2+c^2+5abc\geq8.$$
2 replies
mihaig
Yesterday at 5:17 PM
mihaig
2 hours ago
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an equation from the a contest
alpha31415   2
N 2 hours ago by User141208
Find all (complex) roots of the equation:
(z^2-z)(1-z+z^2)^2=-1/7
2 replies
alpha31415
May 21, 2025
User141208
2 hours ago
J
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J
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Angle EBA is equal to Angle DCB
WakeUp   6
N Apr 10, 2025 by Nari_Tom
Source: Baltic Way 2011
Let $ABCD$ be a convex quadrilateral such that $\angle ADB=\angle BDC$. Suppose that a point $E$ on the side $AD$ satisfies the equality
\[AE\cdot ED + BE^2=CD\cdot AE.\]
Show that $\angle EBA=\angle DCB$.
6 replies
WakeUp
Nov 6, 2011
Nari_Tom
Apr 10, 2025
J
Angle EBA is equal to Angle DCB
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Reveal topic tags
geometry proposed
geometry
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G H BBookmark kLocked kLocked NReply
Source: Baltic Way 2011
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WakeUp
1347 posts
WakeUp
#1 h  Nov 6, 2011, 1:25 PM • 3 Y
Y by anantmudgal09, Adventure10, Mango247
Let $ABCD$ be a convex quadrilateral such that $\angle ADB=\angle BDC$. Suppose that a point $E$ on the side $AD$ satisfies the equality
\[AE\cdot ED + BE^2=CD\cdot AE.\]
Show that $\angle EBA=\angle DCB$.
Z K Y
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newsun
1418 posts
newsun
#2 h  Nov 6, 2011, 1:58 PM • 3 Y
Y by IstekOlympiadTeam, Adventure10, Mango247
WakeUp wrote:
Let $ABCD$ be a convex quadrilateral such that $\angle ADB=\angle BDC$. Suppose that a point $E$ on the side $AD$ satisfies the equality
\[AE\cdot ED + BE^2=CD\cdot AE.\]
Show that $\angle EBA=\angle DCB$.
Let $F$ on the side $CD$ such that $DE=DF$ then we have $ \triangle{DEB} = \triangle{DFB} $
Therefore, $BE=BF$ and $\angle{BEA} = \angle{BFC} $
It is sufficient to show that $ \triangle{AEB} \sim \triangle{BFC} $
indeed,
$ AE\cdot ED + BE^2=CD\cdot AE \Leftrightarrow AE(CD-DE) = BE.BF \Leftrightarrow AE(CD-DF) = BE.BF \Leftrightarrow AE.FC = BE.BF \Leftrightarrow \frac{AE}{EB} = \frac{BF}{FC} $ , as required.
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Kimchiks926
256 posts
Kimchiks926
#3 h  Oct 4, 2020, 6:50 AM
Y by
Solved with @blastoor
Let $BE$ intersect $\odot(ABD)$ at point $X \neq B$. From PoP we have that $AE \cdot ED = BE \cdot EX$. Thus original equality becomes:
\begin{align*} AE \cdot ED + BE^2 = CD \cdot AE \\ BE \cdot EX + BE^2 = CD \cdot AE \\ BE \cdot BX = CD \cdot AE \qquad (2) \\ \end{align*}
Claim: $\triangle BXA \sim \triangle CDB$
Proof: It is easy to see that $\triangle AEX \sim \triangle BED $ and that $\angle AXB =\angle ABD = \angle BDC$. Consequently we have:
$$ \frac{AE}{AX} = \frac{BE}{BD} \implies BE \cdot AX = AE \cdot BD \qquad (20 $$Dividing relations $(1)$ and $(2)$ we get that:
$$ \frac{BX}{AX} = \frac{CD}{BD} $$This implies that $\triangle BXA \sim \triangle CDB$ and thus the claim is proved.

From claim follows $\angle XBA =\angle EBA =\angle DCB$ as desired.
This post has been edited 1 time. Last edited by Kimchiks926, Oct 13, 2020, 1:15 PM
Reason: typo
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rafaello
1079 posts
rafaello
#4 h  Aug 13, 2021, 9:14 PM
Y by
Construct $L$ on $AD$ such that $DL=DC$. Thus,
$$BE^2=CD\cdot AE-AE\cdot AD=EA\cdot EL,$$thus $BE$ is tangent to $(ABL)$.
As $BD$ is the angle bisector of $\triangle CDL$, we also have $BD$ the perpendicular bisector of $LC$.
Hence, $$\measuredangle EBA=\measuredangle BLD=\measuredangle DCB,$$we are done.
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HamstPan38825
8868 posts
HamstPan38825
#5 h  Sep 13, 2022, 5:18 PM
Y by
We prove the converse, as such a point $E$ is clearly unique. Let $E'$ be the reflection of $E$ over $\overline{BD}$ and define $A'$ similarly. Then $$AE \cdot ED + BE^2 = DE' \cdot E'A' + BE'^2.$$Because $\overline{BE'}$ is tangent to $(A'BC)$ by the angle condition, $$DE' \cdot E'A' + BE'^2 = DE' \cdot E'A' + E'A' \cdot E'C = EA \cdot CD,$$as needed.
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AshAuktober
1009 posts
AshAuktober
#6 h  Dec 18, 2024, 5:51 PM
Y by
Reflect $C$ about $BD$ to get $C' \in \overrightarrow{DA}$. Then the condition becomes $EA \cdot EC' = EB^2$, and what is to be proven is the Alternate Segment Theorem, so we're done.
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Nari_Tom
117 posts
Nari_Tom
#7 h  Apr 10, 2025, 9:38 AM
Y by
I had following simple solution. If we lock the points $B, E, D,C$, then there is will be unique $A$ which requires the given condition. So let's assume that $\angle EBA=\angle DCB$, and prove the given condition. Let $B'$ be the reflection of $B$ over $E$. Let $F=(ABB') \cap AE$. $G=BD \cap B'F$.
Let $H$ be the midpoint of $BG$. So it's clear that $2*EH=BD+DG$. On the other hand we have: $\triangle FDG \sim \triangle FEH \sim \triangle CDB$. Which means that $2*EH=BD+DG$ $\implies$ $2*EF=DF+DC$ $\implies$ $CD-DE=ED+DF=EF$. On the other hand by Thales theorem: $BE*BE'=AE*EF$ $\implies$ $BE^2=AE(CD-DE)$, and conclusion follows.
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