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Prove excircle is tangent to circumcircle
sarjinius   7
N 29 minutes ago by markam
Source: Philippine Mathematical Olympiad 2025 P4
Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on side $BC$. Points $X$ and $Y$ are chosen on lines $BI$ and $CI$ respectively such that $DXIY$ is a parallelogram. Points $E$ and $F$ are chosen on side $BC$ such that $AX$ and $AY$ are the angle bisectors of angles $\angle BAE$ and $\angle CAF$ respectively. Let $\omega$ be the circle tangent to segment $EF$, the extension of $AE$ past $E$, and the extension of $AF$ past $F$. Prove that $\omega$ is tangent to the circumcircle of triangle $ABC$.
7 replies
sarjinius
Mar 9, 2025
markam
29 minutes ago
Distinct Integers with Divisibility Condition
tastymath75025   15
N 30 minutes ago by cursed_tangent1434
Source: 2017 ELMO Shortlist N3
For each integer $C>1$ decide whether there exist pairwise distinct positive integers $a_1,a_2,a_3,...$ such that for every $k\ge 1$, $a_{k+1}^k$ divides $C^ka_1a_2...a_k$.

Proposed by Daniel Liu
15 replies
+1 w
tastymath75025
Jul 3, 2017
cursed_tangent1434
30 minutes ago
hard problem
Cobedangiu   4
N an hour ago by Cobedangiu
Let $a,b,c>0$ and $a+b+c=3$. Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$
4 replies
Cobedangiu
3 hours ago
Cobedangiu
an hour ago
An easy FE
oVlad   1
N an hour ago by pco
Source: Romania EGMO TST 2017 Day 1 P3
Determine all functions $f:\mathbb R\to\mathbb R$ such that \[f(xy-1)+f(x)f(y)=2xy-1,\]for any real numbers $x{}$ and $y{}.$
1 reply
oVlad
3 hours ago
pco
an hour ago
Fractions and reciprocals
adihaya   34
N an hour ago by de-Kirschbaum
Source: 2013 BAMO-8 #4
For a positive integer $n>2$, consider the $n-1$ fractions $$\dfrac21, \dfrac32, \cdots, \dfrac{n}{n-1}$$The product of these fractions equals $n$, but if you reciprocate (i.e. turn upside down) some of the fractions, the product will change. Can you make the product equal 1? Find all values of $n$ for which this is possible and prove that you have found them all.
34 replies
adihaya
Feb 27, 2016
de-Kirschbaum
an hour ago
GCD Functional Equation
pinetree1   60
N an hour ago by cursed_tangent1434
Source: USA TSTST 2019 Problem 7
Let $f: \mathbb Z\to \{1, 2, \dots, 10^{100}\}$ be a function satisfying
$$\gcd(f(x), f(y)) = \gcd(f(x), x-y)$$for all integers $x$ and $y$. Show that there exist positive integers $m$ and $n$ such that $f(x) = \gcd(m+x, n)$ for all integers $x$.

Ankan Bhattacharya
60 replies
1 viewing
pinetree1
Jun 25, 2019
cursed_tangent1434
an hour ago
Inequality
giangtruong13   3
N an hour ago by KhuongTrang
Let $a,b,c >0$ such that: $a^2+b^2+c^2=3$. Prove that: $$\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}+abc \geq 4$$
3 replies
giangtruong13
Today at 8:01 AM
KhuongTrang
an hour ago
Easy geo
oVlad   3
N an hour ago by Primeniyazidayi
Source: Romania EGMO TST 2019 Day 1 P1
A line through the vertex $A{}$ of the triangle $ABC{}$ which doesn't coincide with $AB{}$ or $AC{}$ intersectes the altitudes from $B{}$ and $C{}$ at $D{}$ and $E{}$ respectively. Let $F{}$ be the reflection of $D{}$ in $AB{}$ and $G{}$ be the reflection of $E{}$ in $AC{}.$ Prove that the circles $ABF{}$ and $ACG{}$ are tangent.
3 replies
oVlad
3 hours ago
Primeniyazidayi
an hour ago
abc(a+b+c)=3, show that prod(a+b)>=8 [Indian RMO 2012(b) Q4]
Potla   28
N an hour ago by mihaig
Let $a,b,c$ be positive real numbers such that $abc(a+b+c)=3.$ Prove that we have
\[(a+b)(b+c)(c+a)\geq 8.\]
Also determine the case of equality.
28 replies
Potla
Dec 2, 2012
mihaig
an hour ago
NT with repeating decimal digits
oVlad   1
N an hour ago by kokcio
Source: Romania EGMO TST 2019 Day 1 P2
Determine the digits $0\leqslant c\leqslant 9$ such that for any positive integer $k{}$ there exists a positive integer $n$ such that the last $k{}$ digits of $n^9$ are equal to $c{}.$
1 reply
oVlad
3 hours ago
kokcio
an hour ago
Inequalities make a comeback
MS_Kekas   2
N an hour ago by ZeroHero
Source: Kyiv City MO 2025 Round 1, Problem 11.5
Determine the largest possible constant \( C \) such that for any positive real numbers \( x, y, z \), which are the sides of a triangle, the following inequality holds:
\[
\frac{xy}{x^2 + y^2 + xz} + \frac{yz}{y^2 + z^2 + yx} + \frac{zx}{z^2 + x^2 + zy} \geq C.
\]
Proposed by Vadym Solomka
2 replies
MS_Kekas
Jan 20, 2025
ZeroHero
an hour ago
Incenter and concurrency
jenishmalla   7
N Yesterday at 4:06 PM by brute12
Source: 2025 Nepal ptst p3 of 4
Let the incircle of $\triangle ABC$ touch sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Let $D'$ be the diametrically opposite point of $D$ with respect to the incircle. Let lines $AD'$ and $AD$ intersect the incircle again at $X$ and $Y$, respectively. Prove that the lines $DX$, $D'Y$, and $EF$ are concurrent, i.e., the lines intersect at the same point.

(Kritesh Dhakal, Nepal)
7 replies
jenishmalla
Mar 15, 2025
brute12
Yesterday at 4:06 PM
Incenter and concurrency
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Source: 2025 Nepal ptst p3 of 4
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jenishmalla
5 posts
#1
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Let the incircle of $\triangle ABC$ touch sides $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Let $D'$ be the diametrically opposite point of $D$ with respect to the incircle. Let lines $AD'$ and $AD$ intersect the incircle again at $X$ and $Y$, respectively. Prove that the lines $DX$, $D'Y$, and $EF$ are concurrent, i.e., the lines intersect at the same point.

(Kritesh Dhakal, Nepal)
This post has been edited 2 times. Last edited by jenishmalla, Mar 15, 2025, 3:00 PM
Reason: formatting
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AshAuktober
993 posts
#2
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Main idea for nuke
This post has been edited 2 times. Last edited by AshAuktober, Mar 15, 2025, 3:00 PM
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Thapakazi
58 posts
#3
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My problem. We present two solutions :love:

Solution 1: Let lines $D'Y$ and $DX$ intersect at $T$. Define $I$ as the incenter of $\triangle{ABC}$ and let $Z = DD' \cap AT$. First, observe that $D'$ is the orthocenter of $\triangle{TAD}.$ Since $IZ \perp AT$, it follows that $Z$ lies in a circle with diameter $AI$. Consequently, $AZFIE$ is a cyclic pentagon.

Now consider the radical axes of the circles $(AZFE), (EFXD)$, and $(AZXD)$. This implies that $AZ$, $EF$, and $DX$ are concurrent at $T$, meaning $EF$ and $DX$ intersect at $T$. However, since $YD'$ and $DX$ also intersect at $T$, it follows that $DX$, $D'Y$, and $EF$ are concurrent at $T$, as desired.

Solution 2: Observe that $FYED$ is a harmonic quadrilateral. Brianchon's Theorem on hexagon $YD'FXDE$, we see that $YX$ and $D'D$ concur on line $FE$. Lastly, by Pascal’s Theorem on hexagon $D'YYXDD$, the desired result follows.
This post has been edited 1 time. Last edited by Thapakazi, Mar 15, 2025, 5:02 PM
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Captainscrubz
57 posts
#4
Y by
Let $G=D'Y\cap XD$
For the complete quadrilateral $D'XDY$ Let $J$ be the center of spiral similarity that sends $D'Y\rightarrow XD$
$\implies$ that $AG\parallel BC$ as $D'XDY$ is cylcic
Let $GE$ meet the incircle again at $F'$
$\implies GE\cdot GF'=GX\cdot GD=GJ\cdot GA$
$\therefore F\equiv F'$
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Double07
76 posts
#5
Y by
This can also be easily bashed with complex numbers:

Take the incircle of $\Delta ABC$ to be the unit circle.

Then $a=\frac{2ef}{e+f}, b=\frac{2df}{d+f}, c=\frac{2de}{d+e}, d'=-d$.

$A, D', X$ are collinear $\implies -d'x=\frac{a-d'}{\overline{a}-\overline{d'}}\implies dx=\frac{\frac{2ef}{e+f}+d}{\frac{2}{e+f}+\frac{1}{d}}=d\cdot\frac{2ef+de+df}{2d+e+f}\implies x=\frac{2ef+de+df}{2d+e+f}$.

$A, D, Y$ are collinear $\implies -dy=\frac{a-d}{\overline{a}-\overline{d}}\implies -dy=d\cdot\frac{2ef-de-df}{2d-e-f}\implies y=\frac{2ef-de-df}{e+f-2d}$.

Let $P=EF\cap DX\implies p=\frac{ef\cdot\frac{2d^2+2de+2df+2ef}{2d+e+f}-\frac{d(2ef+de+df)}{2d+e+f}(e+f)}{ef-\frac{d(2ef+de+df)}{2d+e+f}}=\frac{2e^2f^2-d^2e^2-d^2f^2}{(e+f)(ef-d^2)}$.

Let $Q=EF\cap D'Y\implies q=\frac{ef\cdot\frac{2d^2-2de-2df+2ef}{e+f-2d}+\frac{d(2ef-de-df)}{e+f-2d}(e+f)}{ef+\frac{d(2ef-de-df)}{e+f-2d}}=\frac{2e^2f^2-d^2e^2-d^2f^2}{(e+f)(ef-d^2)}=p$, so $P=Q$ and we're done.
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aidenkim119
32 posts
#6
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Trivial by Six-point-line!
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ihategeo_1969
199 posts
#7
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All pole-polars are w.r.t incircle. It is well known midpoint of $\overline{BC}$ say $M$ is pole of $\overline{DX}$ (and $A$ is pole of $\overline{EF}$); hence we just need to prove pole of $\overline{YD'}$ lies on $\overline{AM}$ by La Hire. Now $\overline{AM}$, $\overline{DD'}$, $\overline{EF}$ concurrence is well known and since $(YD;EF)=(D'X;EF)=-1$, we have the concurrency point also lies on $\overline{YX}$. Now Pascal at $YYDD'D'X$ finishes.
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brute12
1 post
#9
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So, like one little solution which i noticed during exam time :blush:
Define:
\[
P = EF \cap AD, \quad Q = EF \cap AD'.
\]
We claim that lines $DX$, $D'Y$, and $EF$ are concurrent.

Note that the cross-ratios satisfy:
\[
(A, Q; D', X) = (A, P; Y, D) = -1.
\]
This follows because $P$ and $Q$ are defined by the intersections of $AD$ and $AD'$ with $EF$, and $(A, P; Y, D)$ and $(A, Q; D', X)$ are harmonic divisions.

Moreover, since $D'$ is the reflection of $D$ across the incenter $I$ of triangle $ABC$, it follows that $D'$ and $Y$ lie on the same side of $EF$, and likewise for $D$ and $X$.

Now, by the **Prism Lemma**, which states that if $(A, P; Y, D) = -1$ and $(A, Q; D', X) = -1$ and if $D'$ and $X$ lie on the same side of $EF$, then the lines $DX$, $D'Y$, and $EF$ concur, the result follows.


REMARK: : we haven't used the property that D is the intouch point of the circle with side BC, and D' is the antipode of D, It holds for general point on the incircle, this property of D opens the path for the nice synthetic solution by radical axis, only required property was that EDFD' harmonic!
This post has been edited 1 time. Last edited by brute12, Yesterday at 4:13 PM
Reason: just for adding nice remark
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