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Geometry
Lukariman   6
N an hour ago by Curious_Droid
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
6 replies
Lukariman
Yesterday at 12:43 PM
Curious_Droid
an hour ago
question4
sahadian   5
N 2 hours ago by Mamadi
Source: iran tst 2014 first exam
Find the maximum number of Permutation of set {$1,2,3,...,2014$} such that for every 2 different number $a$ and $b$ in this set at last in one of the permutation
$b$ comes exactly after $a$
5 replies
sahadian
Apr 14, 2014
Mamadi
2 hours ago
Find all functions $f$: \(\mathbb{R^+}\) \(\rightarrow\) \(\mathbb{R^+}\) such
guramuta   5
N 2 hours ago by jasperE3
Source: Balkan MO SL 2021
A5: Find all functions $f$: \(\mathbb{R^+}\) \(\rightarrow\) \(\mathbb{R^+}\) such that:
$$f(xf(x+y)) = xf(y) + 1 $$
5 replies
guramuta
3 hours ago
jasperE3
2 hours ago
number theory
frost23   3
N 2 hours ago by frost23
given any positive integer n show that there are two positive rational numbers a and b not equal to b which are such that a-b, a^2- b^2....................a^n-b^n are all integers
3 replies
frost23
3 hours ago
frost23
2 hours ago
partitioned square
moldovan   8
N 2 hours ago by cursed_tangent1434
Source: Ireland 1994
If a square is partitioned into $ n$ convex polygons, determine the maximum possible number of edges in the obtained figure.

(You may wish to use the following theorem of Euler: If a polygon is partitioned into $ n$ polygons with $ v$ vertices and $ e$ edges in the resulting figure, then $ v-e+n=1$.)
8 replies
moldovan
Jun 29, 2009
cursed_tangent1434
2 hours ago
Geometry
Lukariman   0
3 hours ago
Given acute triangle ABC ,AB=b,AC=c . M is a variable point on side AB. The circle circumscribing triangle BCM intersects AC at N.

a)Let I be the center of the circle circumscribing triangle AMN. Prove that I always lies on a fixed line.

b)Let J be the center of the circle circumscribing triangle MBC. Prove that line segment IJ has a constant length.
0 replies
Lukariman
3 hours ago
0 replies
Finding positive integers with good divisors
nAalniaOMliO   3
N 3 hours ago by nAalniaOMliO
Source: Belarusian National Olympiad 2025
For every positive integer $n$ write all its divisors in increasing order: $1=d_1<d_2<\ldots<d_k=n$.
Find all $n$ such that $2025 \cdot n=d_{20} \cdot d_{25}$.
3 replies
nAalniaOMliO
Mar 28, 2025
nAalniaOMliO
3 hours ago
Concurrent lines
MathChallenger101   4
N 3 hours ago by oVlad
Let $A B C D$ be an inscribed quadrilateral. Circles of diameters $A B$ and $C D$ intersect at points $X_1$ and $Y_1$, and circles of diameters $B C$ and $A D$ intersect at points $X_2$ and $Y_2$. The circles of diameters $A C$ and $B D$ intersect in two points $X_3$ and $Y_3$. Prove that the lines $X_1 Y_1, X_2 Y_2$ and $X_3 Y_3$ are concurrent.
4 replies
MathChallenger101
Feb 8, 2025
oVlad
3 hours ago
Find the value
sqing   7
N 4 hours ago by giangtruong13
Source: Own
Let $a,b,c$ be distinct real numbers such that $ \frac{a^2}{(a-b)^2}+ \frac{b^2}{(b-c)^2}+ \frac{c^2}{(c-a)^2} =1. $ Find the value of $\frac{a}{a-b}+ \frac{b}{b-c}+ \frac{c}{c-a}.$
Let $a,b,c$ be distinct real numbers such that $\frac{a^2}{(b-c)^2}+ \frac{b^2}{(c-a)^2}+ \frac{c^2}{(a-b)^2}=2. $ Find the value of $\frac{a}{b-c}+ \frac{b}{c-a}+ \frac{c}{a-b}.$
Let $a,b,c$ be distinct real numbers such that $\frac{(a+b)^2}{(a-b)^2}+ \frac{(b+c)^2}{(b-c)^2}+ \frac{(c+a)^2}{(c-a)^2}=2. $ Find the value of $\frac{a+b}{a-b}+\frac{b+c}{b-c}+ \frac{c+a}{c-a}.$
7 replies
sqing
Mar 17, 2025
giangtruong13
4 hours ago
2025 HMIC-5
EthanWYX2009   0
4 hours ago
Source: 2025 HMIC-5
Compute the smallest positive integer $k > 45$ for which there exists a sequence $a_1, a_2, a_3, \ldots ,a_{k-1}$ of positive integers satisfying the following conditions:[list]
[*]$a_i = i$ for all integers $1 \le i \le 45;$
[*] $a_{k-i} = i$ for all integers $1 \le i \le 45;$
[*] for any odd integer $1 \le n \le k -45,$ the sequence $a_n, a_{n+1}, \ldots  , a_{n+44}$ is a permutation of
$\{1, 2, \ldots  , 45\}.$[/list]
Proposed by: Derek Liu
0 replies
EthanWYX2009
4 hours ago
0 replies
I need the technique
DievilOnlyM   14
N 4 hours ago by Entei
Let a,b,c be real numbers such that: $ab+7bc+ca=188$.
FInd the minimum value of: $5a^2+11b^2+5c^2$
14 replies
DievilOnlyM
May 23, 2019
Entei
4 hours ago
Lines AD, BE, and CF are concurrent
orl   48
N Apr 7, 2025 by Avron
Source: IMO Shortlist 2000, G3
Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that \[ OD + DH = OE + EH = OF + FH\] and the lines $AD$, $BE$, and $CF$ are concurrent.
48 replies
orl
Aug 10, 2008
Avron
Apr 7, 2025
Lines AD, BE, and CF are concurrent
G H J
Source: IMO Shortlist 2000, G3
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orl
3647 posts
#1 • 8 Y
Y by Davi-8191, mathematicsy, Adventure10, chessgocube, HWenslawski, ImSh95, Mango247, Amir Hossein
Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that \[ OD + DH = OE + EH = OF + FH\] and the lines $AD$, $BE$, and $CF$ are concurrent.
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mr.danh
635 posts
#2 • 6 Y
Y by Adventure10, chessgocube, HWenslawski, ImSh95, Mango247, ehuseyinyigit
Solution
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Valentin Vornicu
7301 posts
#3 • 8 Y
Y by SherlockBond, raknum007, megarnie, chessgocube, ImSh95, Adventure10, Mango247, and 1 other user
We know that the orthocenter reflects over the sides of the triangle on the circumcircle. Therefore the minimal distance $ OD+HD$ equals $ R$. Obviously we can achieve this on all sides, so we assume that $ D,E,F$ are the intersection points between $ A',B',C'$ the reflections of $ H$ across $ BC,CA,AB$ respectively. All we have to prove is that $ AD$, $ BE$ and $ CF$ are concurrent.

In order to do that we need the ratios $ \dfrac {BD}{DC}$, $ \dfrac {CE}{EA}$ and $ \dfrac {AF}{FB}$, and then we can apply Ceva's theorem.

We know that the triangle $ ABC$ is acute, so $ \angle BAH = 90^\circ- \angle B = \angle OAC$, therefore $ \angle HAO = |\angle A - 2(90^\circ -\angle B)| = |\angle B- \angle C|$. In particular this means that $ \angle OA'H = |\angle B-\angle C|$. Since $ \angle BA'A = \angle C$ and $ \angle AA'C = \angle B$, we have that $ \angle BA'D = \angle B$ and $ \angle DA'C = \angle C$.

By the Sine theorem in the triangles $ BA'D$ and $ DA'C$, we get
\[ \dfrac {BD}{DC} = \dfrac { \sin B }{\sin C }.\]

Using the similar relationships for $ \dfrac {CE}{EA}$ and $ \dfrac {AF}{FB}$ we get that those three fractions multiply up to 1, and thus by Ceva's, the lines $ AD, BE$ and $ CF$ are concurrent.
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Zhero
2043 posts
#4 • 7 Y
Y by p1a, chessgocube, ImSh95, Adventure10, Mango247, ehuseyinyigit, and 1 other user
Lemma 1: If an ellipse is inscribed triangle $ABC$, tangent to $BC$, $AC$, and $AB$ at $D$, $E$, and $F$, respectively, then $AD$, $BE$, and $CF$ are concurrent.
Proof: Scale the ellipse about its major axis so that it becomes a circle; here, $A'D'$, $B'E'$, and $C'F'$ concur at the Gergonne point of $A' B' C'$. Scaling preserves incidence, so $AD$, $BE$, and $CF$ concur.

Lemma 2: Let $\ell$ be any line, and let $X$ and $Y$ be any points on the same side of $\ell$. Let $Z$ be the reflection of $X$ across $\ell$, and let $W$ be the intersection of $YZ$ and $\ell$. Then the ellipse with foci $X$ and $Y$ that passes through $W$ is tangent to $\ell$.
Proof: Suppose that the ellipse intersected line $\ell$ at another point, $W'$. Then $XW' + YW' = YW' + ZW' > YZ$ by the triangle inequality, since $W'$, $Y$, and $Z$ are not collinear. On the other hand, by definition of ellipse, $XW' + YW' = XW + YW = ZW + YW = YZ$, so $YZ > YZ$, which is a contradiction.

By lemma 1, it is sufficient to show that there exists an ellipse with foci $O$ and $H$ that are tangent to the sides $BC$, $AC$, and $AB$ at $D$, $E$, and $F$, respectively.

Reflect $H$ across $BC$, $CA$, and $AC$ to get $A'$, $B'$, and $C'$, respectively, and let $OA'$ hit $BC$ at $D$, $OB'$ hit $CA$ at $E$, and $OC'$ hit $AB$ at $F$. . $A',B',C'$ lie on the circumcircle of $ABC$, so $R = OA' = OB' = OC' = HD + DO = HE + EO = HF + FO$, where $R$ is the circumradius of $\triangle ABC$ (since $A', B', C'$ are the reflections of $H$ across the sides of the triangle.)

Consider the ellipse with foci $O$ and $H$ with a major axis of length $R$. By definition, the ellipse passes through $D$, $E$, and $F$. However, by lemma 2, it is tangent to sides $BC$, $CA$, and $AB$, so by lemma 1, we are done.

Some motivation
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Rofler
802 posts
#5 • 2 Y
Y by ImSh95, Adventure10
@Zhero: Instead of scaling, it is more elegant to just use Brainchon's theorem.

Cheers,

Rofler
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Zhero
2043 posts
#6 • 2 Y
Y by ImSh95, Adventure10
Probably. I believe we could also just centrally project it to a circle as well (my first approach). :P I chose to scale because I felt it was the most elementary solution; anyone who knows anything about ellipses should be able to understand the solution I gave.
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goodar2006
1347 posts
#7 • 5 Y
Y by mrdriller, ImSh95, Adventure10, and 2 other users
since $H$ and $O$ are two isogonal conjugate points, there exists an ellipse which these two points are its foci and its tangent to sides of the triangle.
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StefanS
149 posts
#8 • 3 Y
Y by ImSh95, Adventure10, Mango247
Valentin Vornicu wrote:
By the Sine theorem in the triangles $ BA'D$ and $ DA'C$, we get
\[ \dfrac {BD}{DC} = \dfrac { \sin B }{\sin C }.\]

By the law of sines for triangles $~$ $ \triangle{BA'D} $ $~$ and $~$ $ \triangle{DA'C} $ $~$ we get:
\[ \frac { BD } { \sin B } = \frac { A'D } { \sin \angle{DBA'} } \quad \wedge \quad \frac { DC } { \sin C } = \frac { A'D } { \sin \angle{DCA'} } \]
So for:
\[ \dfrac {BD}{DC} = \dfrac { \sin B }{\sin C }.\]
to be true, the following:
\[ \sin \angle{DBA'} = \sin \angle{DCA'} \]
should be true as well. I don't think that's correct. :maybe:
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Virgil Nicula
7054 posts
#9 • 3 Y
Y by ImSh95, Adventure10, Mango247
You are right. See here PP8.
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StefanS
149 posts
#10 • 3 Y
Y by ImSh95, Adventure10, Mango247
Virgil Nicula wrote:
You are right. See here PP8.

Wow you had a math blog!!! :w00t: I unfortunately can't read the entire problems. Only half of the picture is displayed. Do you know why? :(
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Virgil Nicula
7054 posts
#11 • 3 Y
Y by ImSh95, Adventure10, Mango247
See click on the title of message.
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exmath89
2572 posts
#12 • 2 Y
Y by ImSh95, Adventure10
Solution
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IDMasterz
1412 posts
#13 • 5 Y
Y by DPS, ImSh95, Adventure10, Mango247, and 1 other user
Sol
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sayantanchakraborty
505 posts
#14 • 2 Y
Y by ImSh95, Adventure10
orl wrote:
Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that \[ OD + DH = OE + EH = OF + FH\] and the lines $AD$, $BE$, and $CF$ are concurrent.

This problem also appeared in one of the Indian TSTs!!!
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bonciocatciprian
41 posts
#15 • 3 Y
Y by ImSh95, Adventure10, silouan
Take $H_a, H_b, H_c$ be the mirror images of $H$ about $BC, AC, $ and $AB$ respectively. They lie on the circumcircle of $\Delta ABC$. Now, take $\{D\} = OH_a \cap BC$, $\{E\} = OH_b \cap AC$ and $\{F\} = OH_c \cap AB$. Now, using Pool's Theorem, we get that $OD + DH = OE + EH = OF + FH$. Moreover, in this case the sum $OD+DH$ is minimal, so the points $D, E, F$ are situated on an ellipse of foci $O$ and $H$, having the constant sum equal to $R$ (the ray of the circumcircle of $\Delta ABC$), and which is tangent to the edges of the triangle. Now, this is just Gergonne's theorem, under a projective transformation.
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