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dchenmathcounts   45
N 7 minutes ago by EpicBird08
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
45 replies
dchenmathcounts
May 23, 2020
EpicBird08
7 minutes ago
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April Fools Geometry
awesomeming327.   6
N an hour ago by GreekIdiot
Let $ABC$ be an acute triangle with $AB<AC$, and let $D$ be the projection from $A$ onto $BC$. Let $E$ be a point on the extension of $AD$ past $D$ such that $\angle BAC+\angle BEC=90^\circ$. Let $L$ be on the perpendicular bisector of $AE$ such that $L$ and $C$ are on the same side of $AE$ and
\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]Let the reflection of $D$ across $AB$ and $AC$ be $W$ and $Y$, respectively. Let $X\in AW$ and $Z\in AY$ such that $\angle XBE=\angle ZCE=90^\circ$. Let $EX$ and $EZ$ intersect the circumcircles of $EBD$ and $ECD$ at $J$ and $K$, respectively. Let $LB$ and $LC$ intersect $WJ$ and $YK$ at $P$ and $Q$. Let $PQ$ intersect $BC$ at $F$. Prove that $FB/FC=DB/DC$.
6 replies
awesomeming327.
Apr 1, 2025
GreekIdiot
an hour ago
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Functional equations
hanzo.ei   14
N an hour ago by jasperE3
Source: Greekldiot
Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$
14 replies
hanzo.ei
Mar 29, 2025
jasperE3
an hour ago
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Problem 1
SlovEcience   2
N an hour ago by Raven_of_the_old
Prove that
\[ C(p-1, k-1) \equiv (-1)^{k-1} \pmod{p} \]for \( 1 \leq k \leq p-1 \), where \( C(n, m) \) is the binomial coefficient \( n \) choose \( m \).
2 replies
SlovEcience
2 hours ago
Raven_of_the_old
an hour ago
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Conditional maximum
giangtruong13   1
N an hour ago by giangtruong13
Source: Specialized Math
Let $a,b$ satisfy that: $1 \leq a \leq2$ and $1 \leq b \leq 2$. Find the maximum: $$A=(a+b^2+\frac{4}{a^2}+\frac{2}{b})(b+a^2+\frac{4}{b^2}+\frac{2}{a})$$
1 reply
1 viewing
giangtruong13
Mar 22, 2025
giangtruong13
an hour ago
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four variables inequality
JK1603JK   0
an hour ago
Source: unknown?
Prove that $$27(a^4+b^4+c^4+d^4)+148abcd\ge (a+b+c+d)^4,\ \ \forall a,b,c,d\ge 0.$$
0 replies
JK1603JK
an hour ago
0 replies
J
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a hard geometry problen
Tuguldur   0
2 hours ago
Let $ABCD$ be a convex quadrilateral. Suppose that the circles with diameters $AB$ and $CD$ intersect at points $X$ and $Y$. Let $P=AC\cap BD$ and $Q=AD\cap BC$. Prove that the points $P$, $Q$, $X$ and $Y$ are concyclic.
( $AB$ and $CD$ are not the diagnols)
0 replies
Tuguldur
2 hours ago
0 replies
J
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Problem 2
SlovEcience   0
2 hours ago
Let \( a, n \) be positive integers and \( p \) be an odd prime such that:
\[ a^p \equiv 1 \pmod{p^n}. \]Prove that:
\[ a \equiv 1 \pmod{p^{n-1}}. \]
0 replies
SlovEcience
2 hours ago
0 replies
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Regarding Maaths olympiad prepration
omega2007   1
N 2 hours ago by GreekIdiot
<Hey Everyone'>
I'm 10 grader student and Im starting prepration for maths olympiad..>>> From scratch (not 2+2=4 )

Do you haves compilled resources of Handouts,
PDF,
Links,
List of books topic wise
which are shared on AOPS (and from your prespective) for maths olympiad and any useful thing, which will help me in boosting Maths olympiad prepration.
1 reply
omega2007
2 hours ago
GreekIdiot
2 hours ago
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Induction
Mathlover_1   2
N 2 hours ago by GreekIdiot
Hello, can you share links of same interesting induction problems in algebra
2 replies
Mathlover_1
Mar 24, 2025
GreekIdiot
2 hours ago
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n-gon function
ehsan2004   10
N 2 hours ago by Zany9998
Source: Romanian IMO Team Selection Test TST 1996, problem 1
Let $ f: \mathbb{R}^2 \rightarrow \mathbb{R} $ be a function such that for every regular $ n $-gon $ A_1A_2 \ldots A_n $ we have $ f(A_1)+f(A_2)+\cdots +f(A_n)=0 $. Prove that $ f(x)=0 $ for all reals $ x $.
10 replies
ehsan2004
Sep 13, 2005
Zany9998
2 hours ago
J
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J
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nice problem
hanzo.ei   3
N Yesterday at 10:47 AM by X.Luser
Source: I forgot
Let triangle $ABC$ be inscribed in the circumcircle $(O)$ and circumscribed about the incircle $(I)$, with $AB < AC$. The incircle $(I)$ touches the sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. A line through $I$, perpendicular to $AI$, intersects $BC$, $CA$, and $AB$ at $X$, $Y$, and $Z$, respectively. The line $AI$ meets $(O)$ at $M$ (distinct from $A$). The circumcircle of triangle $AYZ$ intersects $(O)$ at $N$ (distinct from $A$). Let $P$ be the midpoint of the arc $BAC$ of $(O)$. The line $AI$ cuts segments $DF$ and $DE$ at $K$ and $L$, respectively, and the tangents to the circle $(DKL)$ at $K$ and $L$ intersect at $T$. Prove that $AT \perp BC$.
3 replies
hanzo.ei
Mar 29, 2025
X.Luser
Yesterday at 10:47 AM
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nice problem
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Reveal topic tags
geometry unsolved
geometry
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G H BBookmark kLocked kLocked NReply
Source: I forgot
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hanzo.ei
15 posts
hanzo.ei
#1 h  Mar 29, 2025, 5:58 PM • 1 Y
Y by cubres
Let triangle $ABC$ be inscribed in the circumcircle $(O)$ and circumscribed about the incircle $(I)$, with $AB < AC$. The incircle $(I)$ touches the sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. A line through $I$, perpendicular to $AI$, intersects $BC$, $CA$, and $AB$ at $X$, $Y$, and $Z$, respectively. The line $AI$ meets $(O)$ at $M$ (distinct from $A$). The circumcircle of triangle $AYZ$ intersects $(O)$ at $N$ (distinct from $A$). Let $P$ be the midpoint of the arc $BAC$ of $(O)$. The line $AI$ cuts segments $DF$ and $DE$ at $K$ and $L$, respectively, and the tangents to the circle $(DKL)$ at $K$ and $L$ intersect at $T$. Prove that $AT \perp BC$.
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hanzo.ei
15 posts
hanzo.ei
#2 h  Mar 30, 2025, 2:21 PM
Y by
bump!!!!
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Lil_flip38
48 posts
Lil_flip38
#3 h  Mar 30, 2025, 7:48 PM • 2 Y
Y by hanzo.ei, X.Luser
I dont understand why there are so many unnecessary points defined, but oh well
Let \(S\) be the midpoint of \(BC\), \(R\) be the midpoint of \(AH\) where \(H\) is the foot of the altitude, \(AI\cap BC=P,\), let \(V\) be the point on \(I\) such that \(SV=SD\). Let \(AV\cap BC=Q\). It is well known that \(\angle AVD=90^\circ\) Let \(U=AI\cap DV\), and \(AT\cap BC = H'\)
We first claim that \(V\) lies on \((DKL)\), and that the center is the midpoint of \(BC\). First, consider inversion about \(I\). \(K,L\) are inverses, because \(L\) is mapped to \(AI\cap (CDEI)\) which is \(K\) by Iran lemma. This implies that \((DKL)\) is orthogonal to \(I\), so the center of \(DKL\) lies on \(BC\) as \(ID\perp BC\). Also by Iran lemma, \(BL\perp AI, CK\perp AI\implies BL\parallel CK\). So if we now consider the perpendicular bisector of \(KL\) it passes through \(S\). Thus, \(S\) is the center of \((DKL)\). Now, by definition, \(V\) lies on this circle aswell. Again, it is well known that \(Q\) is the \(A\)-extouch point, so it also lies on \((DKL)\). Note that \((L,K;D,V)=-1\), so \(T\) lies on \(DV\), and \((T, O;D,V)=-1\). Now, as
\[-1=(T,O;D,V) \overset{A}{=}(H',P;D,Q)\]and as its well known that \(R,I,Q\) lie on a line, we also have:
\[-1=(H,A;\infty_{\perp BC},R)\overset{I}{=}(H,P;D,Q)\]It follows that \(H=H'\) as desired.
I believe that most of the well known lemmas i stated throughout the solution are in EGMO chapter 4
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X.Luser
4 posts
X.Luser
#5 h  Yesterday at 10:47 AM
Y by
Lil_flip38 wrote:
I dont understand why there are so many unnecessary points defined, but oh well
Let \(S\) be the midpoint of \(BC\), \(R\) be the midpoint of \(AH\) where \(H\) is the foot of the altitude, \(AI\cap BC=P,\), let \(V\) be the point on \(I\) such that \(SV=SD\). Let \(AV\cap BC=Q\). It is well known that \(\angle AVD=90^\circ\) Let \(U=AI\cap DV\), and \(AT\cap BC = H'\)
We first claim that \(V\) lies on \((DKL)\), and that the center is the midpoint of \(BC\). First, consider inversion about \(I\). \(K,L\) are inverses, because \(L\) is mapped to \(AI\cap (CDEI)\) which is \(K\) by Iran lemma. This implies that \((DKL)\) is orthogonal to \(I\), so the center of \(DKL\) lies on \(BC\) as \(ID\perp BC\). Also by Iran lemma, \(BL\perp AI, CK\perp AI\implies BL\parallel CK\). So if we now consider the perpendicular bisector of \(KL\) it passes through \(S\). Thus, \(S\) is the center of \((DKL)\). Now, by definition, \(V\) lies on this circle aswell. Again, it is well known that \(Q\) is the \(A\)-extouch point, so it also lies on \((DKL)\). Note that \((L,K;D,V)=-1\), so \(T\) lies on \(DV\), and \((T, O;D,V)=-1\). Now, as
\[-1=(T,O;D,V) \overset{A}{=}(H',P;D,Q)\]and as its well known that \(R,I,Q\) lie on a line, we also have:
\[-1=(H,A;\infty_{\perp BC},R)\overset{I}{=}(H,P;D,Q)\]It follows that \(H=H'\) as desired.
I believe that most of the well known lemmas i stated throughout the solution are in EGMO chapter 4
Very beautiful solution, thanks!
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