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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
60 posts!(and a question )
kjhgyuio   0
a minute ago
Finally 60 posts :D
0 replies
kjhgyuio
a minute ago
0 replies
find angle
TBazar   5
N 6 minutes ago by TBazar
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
5 replies
TBazar
Yesterday at 6:57 AM
TBazar
6 minutes ago
2 var inquality
Iveela   19
N 28 minutes ago by sqing
Source: Izho 2025 P1
Let $a, b$ be positive reals such that $a^3 + b^3 = ab + 1$. Prove that \[(a-b)^2 + a + b \geq 2\]
19 replies
Iveela
Jan 14, 2025
sqing
28 minutes ago
Set of perfect powers is irreducible
Assassino9931   0
29 minutes ago
Source: Al-Khwarizmi International Junior Olympiad 2025 P4
For two sets of integers $X$ and $Y$ we define $X\cdot Y$ as the set of all products of an element of $X$ and an element of $Y$. For example, if $X=\{1, 2, 4\}$ and $Y=\{3, 4, 6\}$ then $X\cdot Y=\{3, 4, 6, 8, 12, 16, 24\}.$ We call a set $S$ of positive integers good if there do not exist sets $A,B$ of positive integers, each with at least two elements and such that the sets $A\cdot B$ and $S$ are the same. Prove that the set of perfect powers greater than or equal to $2025$ is good.

(In any of the sets $A$, $B$, $A\cdot B$ no two elements are equal, but any two or three of these sets may have common elements. A perfect power is an integer of the form $n^k$, where $n>1$ and $k > 1$ are integers.)

Lajos Hajdu and Andras Sarkozy, Hungary
0 replies
Assassino9931
29 minutes ago
0 replies
Inequalities
sqing   3
N Today at 3:29 AM by sqing
Let $ a,b>0 $ and $\frac{a}{a^2+3}+ \frac{b}{b^2+ 3} \geq \frac{1}{2} . $ Prove that
$$a^2+ab+b^2\geq 3$$$$a^2-ab+b^2 \geq 1 $$Let $ a,b>0 $ and $\frac{a}{a^3+3}+ \frac{b}{b^3+ 3}\geq \frac{1}{2} . $ Prove that
$$a^3+ab+b^3 \geq 3$$$$ a^3-ab+b^3\geq 1 $$
3 replies
sqing
Wednesday at 12:59 PM
sqing
Today at 3:29 AM
exist solutions?
teomihai   6
N Today at 12:05 AM by iwastedmyusername
Find how many perfect squares of five different digits there are, with elements from the set ${0,1,4,6,9}$.
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teomihai
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iwastedmyusername
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A pentagon inscribed in a circle of radius √2
tom-nowy   6
N Yesterday at 11:55 PM by anticodon
Can a pentagon with all rational side lengths be inscribed in a circle of radius $\sqrt{2}$ ?
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tom-nowy
May 6, 2025
anticodon
Yesterday at 11:55 PM
Menelau's theorem
noneofyou34   6
N Yesterday at 11:10 PM by Shan3t
Please can someone help me prove that orthocenter of a triangle exists by using Menelau's Theorem!
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noneofyou34
Yesterday at 5:52 PM
Shan3t
Yesterday at 11:10 PM
a,b,c irrational, f(x)=ax^2+bx+c : [-1,1] to [-1,1] surjective
tom-nowy   0
Yesterday at 11:03 PM
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tom-nowy
Yesterday at 11:03 PM
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Math analytical geometry
JDog22   2
N Yesterday at 8:37 PM by Alex-131
Could you please tell me if this is correct:
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JDog22
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William_Mai   14
N Yesterday at 5:34 PM by hi2024IMOp14
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Find the minimum value of $P = ab + 2bc + 3ca$.

Source: Pham Le Van
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William_Mai
May 3, 2025
hi2024IMOp14
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MTA_2024   7
N Yesterday at 4:56 PM by hi2024IMOp14
Let $ABC$ be a triangle such that $AB=3$,$BC=5$ and $AC=6$.Let $D$ be a point on side $AC$ and $E$ one on side $BC$ so that the line $DE$ is tangent to the incircle of $\triangle ABC$ .
Evaluate the perimeter of triangle $\triangle CDE$.
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MTA_2024
Wednesday at 6:52 PM
hi2024IMOp14
Yesterday at 4:56 PM
four point lie on circle
Kizaruno   0
Yesterday at 3:29 PM
Let triangle ABC be inscribed in a circle with center O. A line d intersects sides AB and AC at points E and D, respectively. Let M, N, and P be the midpoints of segments BD, CE, and DE, respectively. Let Q be the foot of the perpendicular from O to line DE. Prove that the points M, N, P, and Q lie on a circle.
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Kizaruno
Yesterday at 3:29 PM
0 replies
Range if \omega for No Inscribed Right Triangle y = \sin(\omega x)
ThisIsJoe   0
Yesterday at 2:02 PM
For a positive number \omega , determine the range of \omega for which the curve y = \sin(\omega x) has no inscribed right triangle.
Could someone help me figure out how to approach this?
0 replies
ThisIsJoe
Yesterday at 2:02 PM
0 replies
Choose P on (AOB) and Q on (AOC)
MarkBcc168   8
N Apr 30, 2025 by NO_SQUARES
Source: ELMO Shortlist 2024 G5
Let $ABC$ be a triangle with circumcenter $O$ and circumcircle $\omega$. Let $D$ be the foot of the altitude from $A$ to $\overline{BC}$. Let $P$ and $Q$ be points on the circumcircles of triangles $AOB$ and $AOC$, respectively, such that $A$, $P$, and $Q$ are collinear. Prove that if the circumcircle of triangle $OPQ$ is tangent to $\omega$ at $T$, then $\angle BTD=\angle CAP$.

Tiger Zhang
8 replies
MarkBcc168
Jun 22, 2024
NO_SQUARES
Apr 30, 2025
Choose P on (AOB) and Q on (AOC)
G H J
G H BBookmark kLocked kLocked NReply
Source: ELMO Shortlist 2024 G5
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MarkBcc168
1595 posts
#1 • 1 Y
Y by Rounak_iitr
Let $ABC$ be a triangle with circumcenter $O$ and circumcircle $\omega$. Let $D$ be the foot of the altitude from $A$ to $\overline{BC}$. Let $P$ and $Q$ be points on the circumcircles of triangles $AOB$ and $AOC$, respectively, such that $A$, $P$, and $Q$ are collinear. Prove that if the circumcircle of triangle $OPQ$ is tangent to $\omega$ at $T$, then $\angle BTD=\angle CAP$.

Tiger Zhang
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CyclicISLscelesTrapezoid
372 posts
#2 • 7 Y
Y by VicKmath7, Rounak_iitr, CT17, ihatemath123, Funcshun840, ehuseyinyigit, NonoPL
My problem!

[asy]
size(7cm); defaultpen(0.8); defaultpen(fontsize(9pt)); dotfactor*=0.7;
pen bluefill,pinkfill,turquoisedraw,bluedraw,purpledraw,pinkdraw,graydraw;
bluefill = RGB(204,233,255);
pinkfill = RGB(255,204,255);
turquoisedraw = RGB(0,170,153);
bluedraw = RGB(0,102,255);
purpledraw = RGB(170,34,255);
pinkdraw = RGB(255,17,255);
graydraw = RGB(119,119,119);
pair A,B,C,M,N,O,T,P,Q,D;
A = dir(85);
B = dir(195);
C = dir(345);
O = circumcenter(A,B,C);
M = (A+B)/2;
N = (A+C)/2;
path c,d,e,f;
c = circumcircle(A,B,C);
T = intersectionpoints(c,11M-10N--N)[0];
d = circle((O+T)/2,distance(O,T)/2);
e = circumcircle(A,B,O);
f = circumcircle(A,C,O);
P = intersectionpoints(d,e)[0];
Q = intersectionpoints(d,f)[0];
D = foot(A,B,C);
filldraw(c,bluefill,bluedraw);
fill(d,pinkfill);
filldraw(circumcircle(A,M,N),pinkfill,pinkdraw);
draw(d,pinkdraw);
draw(e,purpledraw);
draw(f,purpledraw);
draw(A--B--C--cycle,bluedraw);
draw(A--P,turquoisedraw);
draw(N--T,turquoisedraw);
draw(A--D,graydraw);
dot("$A$",A,1.9*dir(85));
dot("$B$",B,dir(225));
dot("$C$",C,dir(315));
dot("$M$",M,dir(140));
dot("$N$",N,dir(30));
dot("$O$",O,1.9*dir(270));
dot("$T$",T,dir(150));
dot("$P$",P,dir(265));
dot("$Q$",Q,dir(10));
dot("$D$",D,dir(270));
[/asy]

Let $\measuredangle$ denote directed angles modulo $180^\circ$.

Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$. The crux of the problem is the following claim.

Claim: $AMON$ and $TQOP$ are symmetric about the perpendicular bisector of $\overline{AT}$.

Proof: Notice that the circumcircle of $AMN$ has diameter $\overline{AO}$, so it is tangent to $\omega$. Thus, the circumcircles of $AMON$ and $TQOP$ are symmetric about the perpendicular bisector of $\overline{AT}$. Also, we have $\measuredangle OTP=\measuredangle OQP=\measuredangle OCA=\measuredangle NAO$, so $N$ and $P$ are symmetric about the perpendicular bisector of $\overline{AT}$. Similarly, $M$ and $Q$ are symmetric. $\square$

Since $A$, $P$, and $Q$ are collinear, $M$, $N$, and $T$ are collinear. Since $\overline{MN}$ is the perpendicular bisector of $\overline{AD}$, we have $TA=TD$. Let $\overline{AP}$ and $\overline{MN}$ intersect at $X$. By symmetry, $XAT$ is isosceles. Thus, we have
\[\measuredangle BTD=\measuredangle BTA+\measuredangle ATD=\measuredangle BCA+2\measuredangle ATX=\measuredangle MNA+\measuredangle AXN=\measuredangle PAC\]and we are done. $\square$
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MathLuis
1524 posts
#3 • 2 Y
Y by IYxMT, ehuseyinyigit
Nice problem, here is a solution found while watching Portugal vs Turkey Match lol.
Let the tangency point be $T$, let $M,N$ midpoints of $AC, AB$ respectively and let $(OPQ) \cap (ANMO)=G$, note that they are circles with diameters $TO,AO$ respectively and since $TO=AO$ and $\angle TGO=90$ we get $G$ midpoint of $TA$, so these circles are symetric w.t.t. $GO$. Now $\angle APO=\angle ABO=\angle NAO=\angle NMO$ which shows that arcs $QO,NO$ have the same lenght, in the same way arcs $PO,MO$ have the same lenght and thus $(N,Q), (M,P)$ are symetric w.r.t. $GO$, now this means $T,M,N$ colinear. To finish note that $\angle GTP=\angle AQG=\angle TNG=\angle NTB$ (last one is just midbase), therefore $\angle NTD=\angle ATN=\angle BTP$ which means $\angle BTD=\angle PTN=\angle CAP$ as desired thus we are done :cool:.
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DottedCaculator
7351 posts
#4
Y by
Invert at $O$. We have $\overline{p'}=\frac{2t-a-b}{t^2-ab}$, so $p=\frac{t^2-ab}{2t-a-b}$, which means $p-a=\frac{(t-a)^2}{2t-a-b}$. Similarly, $q-a=\frac{(t-a)^2}{2t-a-c}$, so $\frac{2t-a-b}{2t-a-c}$ is real, so $\frac{2t-a-c}{b-c}$ is real, which implies $2t-a-c=-bc\left(\frac2t-\frac1a-\frac1c\right)$, or $$2at^2-(a+c)(a+b)t+2abc=0.$$
We have $d=\frac12\left(a+b+c-\frac{bc}a\right)$. The condition $\angle BTD=\angle CAP$ is equivalent to $\frac{b-t}{d-t}\frac{c-a}{p-a}$ real, or
\begin{align*}
\frac{btca}{(d-t)\frac{(t-a)^2}{2t-a-b}}&=\frac1{\left(\overline d-\frac1t\right)\frac{\overline{(t-a)^2}}{\overline{2t-a-b}}}\\
\frac{abct(2t-a-b)}{d-t}&=\frac{a^2t^2\left(\frac2t-\frac1a-\frac1b\right)}{\overline d-\frac1t}\\
b^2c(2t-a-b)(t\overline d-1)&=t(d-t)(2ab-at-bt)\\
t^3(a+b)-t^2(ad+bd+2ab+2b^2c\overline d)+t(2abd+2b^2c+ab^2c\overline d+b^3c\overline d)-b^2c(a+b)&=0\\
t^3a(a+b)-t^2\frac12((a+b)(a^2+ab+ac-bc)+4a^2b+2b(bc+ca+ab-a^2))+t\frac12(2ab(a^2+ab+ac-bc)+4ab^2c+b(a+b)(ab+bc+ca-a^2))-ab^2c(a+b)&=0\\
t^32a-t^2(a^2+3ab+ac+bc)+tb(a^2+3ac+ab+bc)-2ab^2c&=0\\
(2at^2-(a+c)(a+b)t+2abc)(t-b)&=0.
\end{align*}
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ihatemath123
3446 posts
#5
Y by
Solved with Benjaminxiao, Zhaom.

Let $P'$ be the intersection of lines $OP$ and $AB$, and let $Q'$ be the intersection of lines $OQ$ and $\overline{AC}$; in other words, $P'$ and $Q'$ are the images of $P$ and $Q$ under inversion about $\omega$. Therefore, line $P'Q'$ is tangent to $\omega$ at $T$.

Also, since $A$, $P$ and $Q$ are collinear, it follows that $(AOP'Q')$ is cyclic. By Simson's theorem, the feet from $O$ onto sides $\overline{AB}$, $\overline{AC}$ and $\overline{P'Q'}$ are collinear; in other words, $T$ lies on the $A$-midline of $\triangle ABC$.

Claim: We have $\triangle TPQ \sim \triangle ACB$ (with similarity factor of $\tfrac{1}{2}$.)
Proof: We have \[\angle TPQ = \angle TOQ' = 90^{\circ} - \angle OQ'P' = \angle P'AO - 90^{\circ} = \angle C,\]and likewise for $\angle TQP$ and $\angle B$.

If we define $P_1$ and $Q_1$ as the reflections of $T$ across $P$ and $Q$, respectively, we have that $\triangle T Q_1 P_1$ and $\triangle ABC$ are reflections (and in particular, $\omega$ is the circumcircle of $\triangle TQ_1 P_1$). From this, it follows by symmetry $\angle PAT = \angle (P_1 Q_1, AT) = \angle (AT, BC)$.

Now, we have
\[\angle BTD = \angle ATD - \angle C = \angle PAT + \angle (AT, BC) - \angle C = \angle PAT + \angle (AT, AC) = \angle PAT - \angle CAT = \angle CAP.\]
This post has been edited 1 time. Last edited by ihatemath123, Jul 29, 2024, 5:14 AM
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CyclicISLscelesTrapezoid
372 posts
#6 • 1 Y
Y by ihatemath123
I regret not discovering the following version of the problem (that I found just now), which has a significantly more appealing statement.
ELMO SL 2024 G5 but better wrote:
Let $ABC$ be a triangle with circumcenter $O$ and circumcircle $\omega$. A line through $A$ intersects $\overline{BC}$, the circumcircle of $AOB$, and the circumcircle of $AOC$ at $D$, $E \ne A$, and $F \ne A$, respectively. Suppose the circumcircle of $OEF$ is tangent to $\omega$ at $T$. Prove that $\angle ATD=90^\circ$.
Solution
This post has been edited 1 time. Last edited by CyclicISLscelesTrapezoid, Feb 2, 2025, 11:33 PM
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awesomeming327.
1717 posts
#7
Y by
Let $M$ and $N$ be midpoints of $AB$ and $AC$, respectively.

Claim 1: $ATNP$ and $ATMQ$ are isosceles trapezoids.
Note that $OT$ is the diameter of the circle $(OPQ)$ and $OA$ is the diameter of the circle $(OMN)$ so $\angle TPO=\angle ANO=90^\circ$
Furthermore, we have
\[\measuredangle PTO=\measuredangle PQO=\measuredangle AQO=\measuredangle ACO=\measuredangle OAC=\measuredangle OAN\]Since we also have $OT=OA$, $\triangle OTP\cong\triangle OAN$. Similarly, $\triangle OTQ\cong \triangle OAM$. Thus, our claim is true.

Since $A$, $P$, $Q$ are collinear, this means that $T$, $M$, $N$ are collinear. Let $MN$ intersect $(ABC)$ at $T$ and $T'$. Then $TT'BC$ is an isosceles trapezoid. We have
\[\measuredangle PTA=\measuredangle TAN=\measuredangle TAC=\measuredangle TT'C=\measuredangle BTT'=\measuredangle BTN\]This means that $\measuredangle BTP=\measuredangle NTA$. Since $TN$ is the perpendicular bisector of $AD$,
\[\measuredangle NTA=\measuredangle DTN\]so $\measuredangle BTD=\measuredangle PTN=\measuredangle PAN=\measuredangle PAC$ as desired.
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akasht
84 posts
#8
Y by
skibidi spiral

Let $X$ and $Y$ be the inverses of $P$ and $Q$ upon inverting about $(ABC)$. Then $A$ lies on $(OXY)$ and furthermore $XY$ is tangent to $(ABC)$. This implies that $T$ is the foot of $O$ on $XY$. Additionally, $A,X,B$ and $A,Y,C$ are collinear. Let $A'$ be the other intersection of $(ABC)$ with $(OXY)$.
We now setup some points to do a spiral at $A'$. Define
  • $T'$ as the reflection of $T$ about the midpoint of $XY$.
  • $S, S'$ as the intersection of $AT,A'T$ with $(OXY)$.
  • $O'$ as the antipode of $O$ in $(OXY)$.
Now note that the spiral about $A'$ sending $(ABC)$ to $(OXY)$ sends $A \to O'$, $B \to X$, $C \to Y$, $D \to T'$, $T \to S$. Also, observe that by Reim, $SS' // XY$ so $S,S'$ are symmetric about the perpendicular bisector of $XY$. Thus \[\angle BTD = \angle XST' = \angle YS'T = \angle YS'A' = \angle A'AC = \angle PAC\]as desired
This post has been edited 2 times. Last edited by akasht, Apr 30, 2025, 12:41 PM
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NO_SQUARES
1116 posts
#9
Y by
CyclicISLscelesTrapezoid wrote:
I regret not discovering the following version of the problem (that I found just now), which has a significantly more appealing statement.
ELMO SL 2024 G5 but better wrote:
Let $ABC$ be a triangle with circumcenter $O$ and circumcircle $\omega$. A line through $A$ intersects $\overline{BC}$, the circumcircle of $AOB$, and the circumcircle of $AOC$ at $D$, $E \ne A$, and $F \ne A$, respectively. Suppose the circumcircle of $OEF$ is tangent to $\omega$ at $T$. Prove that $\angle ATD=90^\circ$.
Solution

Also it seems that if $(OEF) \cap \omega =T_1, T_2$, then $\angle AT_1D+\angle AT_2D = 180^\circ$.
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