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Cyclic Quads and Parallel Lines
gracemoon124   18
N 10 minutes ago by busy-beaver
Source: 2015 British Mathematical Olympiad?
Let $ABCD$ be a cyclic quadrilateral. Let $F$ be the midpoint of the arc $AB$ of its circumcircle which does not contain $C$ or $D$. Let the lines $DF$ and $AC$ meet at $P$ and the lines $CF$ and $BD$ meet at $Q$. Prove that the lines $PQ$ and $AB$ are parallel.
18 replies
gracemoon124
Aug 16, 2023
busy-beaver
10 minutes ago
Shortest number theory you might've seen in your life
AlperenINAN   2
N 13 minutes ago by zuat.e
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)$ is a perfect square, then $pq + 1$ is also a perfect square.
2 replies
AlperenINAN
an hour ago
zuat.e
13 minutes ago
Is this even algebra or geometry
Sadigly   3
N 14 minutes ago by Moon_settler
Source: Azerbaijan Junior NMO 2019
Alice creates the graphs $y=|x-a|$ and $y=c-|x-b|$ , where $a,b,c\in\mathbb{R^+}$. She observes that these two graphs and $x$ axis divides the positive side of the plane ($x,y>0$) into two triangles and a quadrilateral. Find the ratio of sums of two triangles' areas to the area of quadrilateral.
3 replies
Sadigly
23 minutes ago
Moon_settler
14 minutes ago
Prove or disprove the existence of such a
Sadigly   0
21 minutes ago
Source: Azerbaijan NMO
A positive number $a$ is given, such that $a$ could be expressed as difference of two perfect squares ($a=\frac1{n^2}-\frac1{m^2}$). Is it possible for $2a$ to be expressed as difference of two perfect squares?
0 replies
Sadigly
21 minutes ago
0 replies
Points on the sides of cyclic quadrilateral satisfy the angle conditions
AlperenINAN   1
N 26 minutes ago by ehuseyinyigit
Source: Turkey JBMO TST 2025 P1
Let $ABCD$ be a cyclic quadrilateral and let the intersection point of lines $AB$ and $CD$ be $E$. Let the points $K$ and $L$ be arbitrary points on sides $CD$ and $AB$ respectively, which satisfy the conditions
$$\angle KAD = \angle KBC \quad \text{and} \quad \angle LDA = \angle LCB.$$Prove that $EK = EL$.
1 reply
AlperenINAN
an hour ago
ehuseyinyigit
26 minutes ago
Writing quadratic trinomials inside cells
Sadigly   0
29 minutes ago
Source: Azerbaijan Junior NMO 2019
A $6\times6$ square is given, and a quadratic trinomial with a positive leading coefficient is placed in each of its cells. There are $108$ coefficents in total, and these coefficents are chosen from the set $[-66;47]$, and each coefficient is different from each other. Prove that there exists at least one column such that the sum of the six trinomials in that column has a real root.
0 replies
Sadigly
29 minutes ago
0 replies
Product of consecutive terms divisible by a prime number
BR1F1SZ   1
N 30 minutes ago by IndoMathXdZ
Source: 2025 Francophone MO Seniors P4
Determine all sequences of strictly positive integers $a_1, a_2, a_3, \ldots$ satisfying the following two conditions:
[list]
[*]There exists an integer $M > 0$ such that, for all indices $n \geqslant 1$, $0 < a_n \leqslant M$.
[*]For any prime number $p$ and for any index $n \geqslant 1$, the number
\[
a_n a_{n+1} \cdots a_{n+p-1} - a_{n+p}
\]is a multiple of $p$.
[/list]


1 reply
1 viewing
BR1F1SZ
Today at 12:09 AM
IndoMathXdZ
30 minutes ago
Minimum value of a 3 variable expression
bin_sherlo   3
N 34 minutes ago by ehuseyinyigit
Source: Türkiye 2025 JBMO TST P6
Find the minimum value of
\[\frac{x^3+1}{(y-1)(z+1)}+\frac{y^3+1}{(z-1)(x+1)}+\frac{z^3+1}{(x-1)(y+1)}\]where $x,y,z>1$ are reals.
3 replies
bin_sherlo
an hour ago
ehuseyinyigit
34 minutes ago
Incenter is the foot of altitude
Sadigly   0
an hour ago
Source: Azerbaijan JBMO TST 2023
Let $ABC$ be a triangle and let $\Omega$ denote the circumcircle of $ABC$. The foot of altitude from $A$ to $BC$ is $D$. The foot of altitudes from $D$ to $AB$ and $AC$ are $K;L$ , respectively. Let $KL$ intersect $\Omega$ at $X;Y$, and let $AD$ intersect $\Omega$ at $Z$. Prove that $D$ is the incenter of triangle $XYZ$
0 replies
Sadigly
an hour ago
0 replies
System of equations in juniors' exam
AlperenINAN   1
N an hour ago by AlperenINAN
Source: Turkey JBMO TST 2025 P3
Find all positive real solutions $(a, b, c)$ to the following system:
$$
\begin{aligned}
a^2 + \frac{b}{a} &= 8, \\
ab + c^2 &= 18, \\
3a + b + c &= 9\sqrt{3}.
\end{aligned}
$$
1 reply
AlperenINAN
an hour ago
AlperenINAN
an hour ago
reals associated with 1024 points
bin_sherlo   0
an hour ago
Source: Türkiye 2025 JBMO TST P8
Pairwise distinct points $P_1,\dots,P_{1024}$, which lie on a circle, are marked by distinct reals $a_1,\dots,a_{1024}$. Let $P_i$ be $Q-$good for a $Q$ on the circle different than $P_1,\dots,P_{1024}$, if and only if $a_i$ is the greatest number on at least one of the two arcs $P_iQ$. Let the score of $Q$ be the number of $Q-$good points on the circle. Determine the greatest $k$ such that regardless of the values of $a_1,\dots,a_{1024}$, there exists a point $Q$ with score at least $k$.
0 replies
bin_sherlo
an hour ago
0 replies
geometry
realitywarper_888   2
N Oct 17, 2022 by PROF65
Let $\Delta ABC$ be a triangle. $E, F$ are points such that $EA \perp AB, EB \perp BC, FA \perp AC, FC \perp BC$. $M$ is the midpoint of $BC$, $L$ lies on $BC$ such that $AL$ is the symmedian of $\Delta ABC$. Perpendicular bisector of $LM$ cut $AM$ at $K$. Prove that $(KLM)$ touches two circles with diameters $BE$ and $CF$
2 replies
realitywarper_888
Oct 16, 2022
PROF65
Oct 17, 2022
geometry
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realitywarper_888
31 posts
#1
Y by
Let $\Delta ABC$ be a triangle. $E, F$ are points such that $EA \perp AB, EB \perp BC, FA \perp AC, FC \perp BC$. $M$ is the midpoint of $BC$, $L$ lies on $BC$ such that $AL$ is the symmedian of $\Delta ABC$. Perpendicular bisector of $LM$ cut $AM$ at $K$. Prove that $(KLM)$ touches two circles with diameters $BE$ and $CF$
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on_gale
65 posts
#2 • 2 Y
Y by realitywarper_888, Mango247
Let $X, Y$ be the second points of intersection of $AL, AM$ with $(ABC)$.
Let $Z$ be a point in $AL$ such that $XK \| ZM$.

Using inversion with center $A$ and mapping $B$ to $C$, we conclude that the inverses of $(EAB), (FAC)$ are the lines $CD, BD$ respectively.
Obviously, $(X, M)$ and $(L, Y)$ are pairs of inverses. Also, since $AZ \cdot AK = AX \cdot AM$, we see that $Z$ is the inverse of $K$.

Therefore, it suffices to prove that $(XYZ)$ is tangent to the lines $BD$ and $CD$.
And that's equivalent to prove that $(XYZ)$ and $(ABC)$ are homothetic. But we already know that $Y$ is the reflection of $X$ over $DM$.

So we just have to prove that $\dfrac{DA}{DX} = \dfrac{DX}{DZ}$ which is the same as saying $DX^2 = DA \cdot DZ$.

Also, since $(A, X; B, C) = -1$, then $\angle{XML} = \angle{LMK} = \angle{KLM}$ so $LK \| XM$.

And that's it, the rest is just bashing:
Let $p = AK$, $q = KM$. Then $AL = pr$, $LX = qr$ for some $r$.

Now, we'll use that $(A, X; L, D) = -1$ to say that $XD = \dfrac{qr(p+q)}{p-q}$.
Furthermore, we have $\dfrac{AX}{XZ} = \dfrac{AK}{KM}$ implying that $XZ = \dfrac{qr(p+q)}{p}$.

So $DZ = DX - XZ = \dfrac{q^2r(p+q)}{p(p-q)}$. And now we just check it:

$DA \cdot DZ = (\dfrac{pr(p+q)}{p-q}) \cdot (\dfrac{q^2r(p+q)}{p(p-q)}) = \dfrac{q^2r^2(p+q)^2}{(p-q)^2} = DX^2$
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PROF65
2016 posts
#3 • 1 Y
Y by realitywarper_888
Let $O_1,O_2 , S$ the centers of $(AEB)$, $(AFC)$ , the intersection of $BC$ and $O_1O_2$; $T,T'$ the tangency points of $(AEB),(AFC)$ with their common tangent other than $BC$ :
it s clear that $S$ is the ex-similicenter of $(AEB),(AFC)$ and $M$ is on their radical axis:
then $SA^2=SB.SC$ so $AS$ is tangent to $(ABC)$ hence $(S,L,B,C)=-1$ implies $ ML.MS=MB^2$ hence the inversion with center $M$ and power $MB^2$ swaps $S$ with $ L$ ; $TT'$ with $(I)$, the common tangent circle to $(O_1),(O_2)$ passing through $L,M$ ; $K'$ with the midpoint of $TT'$ say $G$ where $K'=(I)\cap AM$.
Whence $ MK'L\sim GSM$ but $ SG =SM$ hence $K'M=K'L$ which means $K=K'$ and $ (KLM)$ touches the two circles.
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