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Collinearity with orthocenter
liberator   180
N 43 minutes ago by joshualiu315
Source: IMO 2013 Problem 4
Let $ABC$ be an acute triangle with orthocenter $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_1$ is the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$. Prove that $X,Y$ and $H$ are collinear.

Proposed by Warut Suksompong and Potcharapol Suteparuk, Thailand
180 replies
liberator
Jan 4, 2016
joshualiu315
43 minutes ago
IMO ShortList 2001, combinatorics problem 5
orl   12
N an hour ago by Maximilian113
Source: IMO ShortList 2001, combinatorics problem 5
Find all finite sequences $(x_0, x_1, \ldots,x_n)$ such that for every $j$, $0 \leq j \leq n$, $x_j$ equals the number of times $j$ appears in the sequence.
12 replies
orl
Sep 30, 2004
Maximilian113
an hour ago
Aperiodicity of Divisibility
somebodyyouusedtoknow   0
an hour ago
Source: San Diego Honors Math Contest 2025 Part II, Problem 4
Let $a_1,a_2,a_3,\ldots$ be the sequence $2,1,1,2,\ldots,$ so that $a_i \in \{1,2\}$ for each $i$ and so that the decimal number $\overline{a_n a_{n-1} \cdots a_1}$ is divisible by $2^n$ for each $n \geq 1$. Show that the decimal $0.a_1a_2a_3...$ is irrational.
0 replies
somebodyyouusedtoknow
an hour ago
0 replies
Number of Perfect Matchings in a Graph
somebodyyouusedtoknow   0
an hour ago
Source: San Diego Honors Math Contest 2025 Part II, Problem 3
Consider a collection of $2n$ points on the plane, no three of which are collinear, and some set of line segments between them. We say that a subset of these line segments is called a "pairing" if every one of these ($2n$) points is the endpoint of exactly one of the chosen line segments (in other words, a pairing is a perfect matching).

Show that for every $k \geq 1$, there exists such an arrangement of points and line segments (for some value of $n \geq 1$) such that there are exactly $k$ distinct (but not necessarily disjoint) pairings.
0 replies
somebodyyouusedtoknow
an hour ago
0 replies
Polynomial Factors
somebodyyouusedtoknow   0
an hour ago
Source: San Diego Honors Math Contest 2025 Part II, Problem 2
Let $P(x)$ be a polynomial with real coefficients such that $P(x^n) \mid P(x^{n+1})$ for all $n \in \mathbb{N}$. Prove that $P(x) = cx^k$ for some real constant $c$ and $k \in \mathbb{N}$.
0 replies
somebodyyouusedtoknow
an hour ago
0 replies
weirdest expo ever
GreekIdiot   5
N an hour ago by maromex
Source: own
Solve $5^x-2^y=z^3$ where $x,y,z \in \mathbb {Z}$.
5 replies
GreekIdiot
Mar 6, 2025
maromex
an hour ago
Polygons Which Don't Fit
somebodyyouusedtoknow   0
an hour ago
Source: San Diego Honors Math Contest 2025 Part II, Problem 1
Let $P_1,P_2,\ldots,P_n$ be polygons, no two of which are similar. Show that there are polygons $Q_1,Q_2,\ldots,Q_n$ where $Q_i$ is similar to $P_i$ so that for no $i \neq j$ does $Q_i$ contain a polygon that's congruent to $Q_j$.

Note. Here, the word "contain" means for the construction we have, we cannot select a size for $Q_j$ so that $Q_j$ is wholly contained in $Q_i$, and so it does not intersect the edges of $Q_i$ at all.
0 replies
somebodyyouusedtoknow
an hour ago
0 replies
GCD of 2^n-2, 3^n-3, 4^n-4, 5^n-5, ......
tom-nowy   1
N 2 hours ago by tom-nowy
Source: Own
Find all positive integers n such that the greatest common divisor of the sequence
\[ 2^n -2, \;\; 3^n -3, \;\; 4^n -4, \;\; 5^n-5, \, \ldots \ldots \]is $66$. Also, are there infinitely many n for which the greatest common divisor is $6$?
1 reply
tom-nowy
Aug 29, 2023
tom-nowy
2 hours ago
Similar to iran 1996
GreekIdiot   0
2 hours ago
Let $f: \mathbb R \to \mathbb R$ be a function such that $f(f(x)+y)=f(f(x)-y)+4f(x)y \: \forall x,y \: \in \: \mathbb R$. Find all such $f$.
0 replies
GreekIdiot
2 hours ago
0 replies
Weird ninja points collinearity
americancheeseburger4281   0
2 hours ago
Source: Someone I know
For some triangle, define its Ninja Point as the point on its circumcircle such that its Steiner line coincides with the Euler line of the triangle. For an triangle $ABC$, define:
[list]
[*]$O$ as its circumcentre, $H$ as its orthocentre and $N_9$ as its nine-point centre.
[*]$M_a$, $M_b$ and $M_c$ to be the midpoint of the smaller arcs.
[*]$G$ as the isogonal conjugate of the Nagel point (i.e. the exsimillicenter of the incircle and circumcircle)
[*]$S$ as the ninja point of $\Delta M_aM_bM_c$
[*]$K$ as the ninja point of the contact triangle
[/list]
Prove that:
$(a)$ Points $K$, $N_9$ and $I$ are collinear, that is $K$ is the Feuerbach point.
$(b)$ Points $H$, $G$ and $S$ are collinear
0 replies
americancheeseburger4281
2 hours ago
0 replies
Minimum where the sum of squares is greater than 3
m0nk   1
N 3 hours ago by oolite
Source: My friend
If $a,b,c \in R^+$ and $a^2+b^2+c^2 \ge 3$.Find the minimum of $S=\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+\frac{a+b+c}{9}$
1 reply
m0nk
Yesterday at 4:57 PM
oolite
3 hours ago
geometric inequality with altitudes! CMO 2015 P2
aditya21   17
N Jun 9, 2024 by MagicalToaster53
Source: Canadian mathematical olympiad 2015
Let $ABC$ be an acute-angled triangle with altitudes $AD,BE,$ and $CF$. Let $H$ be the orthocentre, that is, the point where the altitudes meet. Prove that \[\frac{AB\cdot AC+BC\cdot CA+CA\cdot CB}{AH\cdot AD+BH\cdot BE+CH\cdot CF}\leq 2.\]
17 replies
aditya21
Apr 24, 2015
MagicalToaster53
Jun 9, 2024
geometric inequality with altitudes! CMO 2015 P2
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G H BBookmark kLocked kLocked NReply
Source: Canadian mathematical olympiad 2015
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aditya21
717 posts
#1 • 1 Y
Y by Adventure10
Let $ABC$ be an acute-angled triangle with altitudes $AD,BE,$ and $CF$. Let $H$ be the orthocentre, that is, the point where the altitudes meet. Prove that \[\frac{AB\cdot AC+BC\cdot CA+CA\cdot CB}{AH\cdot AD+BH\cdot BE+CH\cdot CF}\leq 2.\]
This post has been edited 7 times. Last edited by djmathman, Jun 16, 2015, 9:26 PM
Reason: rearranged numerator
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ABCDE
1963 posts
#2 • 3 Y
Y by A64298347, Adventure10, Mango247
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aditya21
717 posts
#3 • 4 Y
Y by IMO2019, Jason99, Adventure10, Mango247
here is my solution =
let us denote $a,b,c$ as sides of triangle and $R$ as its circumradius
than by easy trigonometry we get $AH=2RcosA , AD=2RsinBsinC$ etc.
so $AH.AD=4R^2cosAsinBsinC=cbcosA$ etc.

than our inequality converts to proving that
$ab+bc+ca\leq 2(cbcosA+abcosC+cacosB)$

now by cosine law ,we have $2cbcosA=b^2+c^2-a^2$ etc.
substituting,our inequality becomes

$ab+bc+ca\leq a^2+b^2+c^2$
which is obvious as it is equivalent to $\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\ge 0$

so we are done :D
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JuanOrtiz
366 posts
#4 • 3 Y
Y by MillenniumFalcon, Adventure10, Mango247
Notice $ 2AH*AD=2AE*AC=2bc*cosA=c^2+b^2-a^2$ and so the problem translates to $a^2+b^2+c^2 \ge ab+bc+ca$, done.
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MexicOMM
292 posts
#5 • 2 Y
Y by Adventure10, Mango247
You three had the exact same solution, and still posted it. A new approach, instead of using $cos$, we see that $2AH \cdot AD= AF \cdot AB + AE \cdot AC$, analogously $2BH \cdot BE= BF \cdot AB + BD \cdot BC$, and $2CH \cdot CF= CE \cdot AC + CD \cdot BC$, but $AF+BF=AB$, $AE+CE=AC$, and $CD+BD=BC$, so we need to prove $ab+bc+ca \le a^2+b^2+c^2$, which is well known.
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AMN300
563 posts
#6 • 2 Y
Y by Adventure10, Mango247
not really contributing anything here, but might as well post my solution
also, this seems too easy for an oly problem?
solution
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Virgil Nicula
7054 posts
#7 • 2 Y
Y by Adventure10, Mango247
See here the proposed problem P2.
This post has been edited 3 times. Last edited by Virgil Nicula, May 12, 2016, 10:58 PM
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programjames1
3046 posts
#8 • 2 Y
Y by Adventure10, Mango247
ABCDE wrote:
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How does Power of a Point give you $AH\cdot AD=AF\cdot AB?$
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Kaskade
469 posts
#9 • 2 Y
Y by Adventure10, Mango247
programjames1 wrote:
ABCDE wrote:
Click to reveal hidden text

How does Power of a Point give you $AH\cdot AD=AF\cdot AB?$

Quadrilateral $HDBF$ is cyclic (from $\angle BFH = \angle HDB = 90$), and the sides $HD$ and $BF$ extended meet at $A$.
This post has been edited 1 time. Last edited by Kaskade, Jun 6, 2018, 5:54 PM
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Jzhang21
308 posts
#10 • 2 Y
Y by Adventure10, Mango247
Note that it suffices to prove $2(AH\cdot AD+BH\cdot BE+CH\cdot CF)\geq ab+bc+ca.$

Claim 1: $AD\cdot AH=\frac{-a^2+b^2+c^2}{2}, BH\cdot BE=\frac{a^2-b^2+c^2}{2},$ and $CH\cdot CE=\frac{a^2+b^2-c^2}{2}$.
Proof: Note that $AD=b\sin C$ and $CDHE$ is a cyclic quadrilateral. Then, $\angle AHE=\angle C$ so $AH=\frac{AE}{\sin C}=\frac{c\cos A}{\sin C}.$ Hence, $$AD\cdot AH=b\sin C\cdot \frac{c\cos A}{\sin C}=bc\cos A=bc(\frac{-a^2+b^2+c^2}{2bc})=\frac{-a^2+b^2+c^2}{2},$$as desired. The others follow similarly. $\Box$

By Claim 1, we simplify the equation to $2(\frac{-a^2+b^2+c^2}{2}+\frac{a^2-b^2+c^2}{2}+\frac{a^2+b^2-c^2}{2})=a^2+b^2+c^2\geq ab+bc+ca.$ This follows by the Trivial Inequality $$\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\geq 0.$$$\blacksquare$
This post has been edited 1 time. Last edited by Jzhang21, Dec 27, 2018, 5:00 PM
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Pluto1708
1107 posts
#11 • 2 Y
Y by Adventure10, Mango247
Easy problem for Canado MO :(
Just note that the $DHS$Joke is equivalent to $\sum{AH \cdot AD}$ = $\sum{bc \cdot \cos{A}}$ = $\frac{a^2 + b^2 + c^2}{2}$
Therefore the expression becomes $a^2 + b^2 + c^2 \ge ab + bc + ca$ which is trivial .
This post has been edited 1 time. Last edited by Pluto1708, Dec 27, 2018, 5:57 PM
Reason: \delta
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Maffeseater
11 posts
#12 • 2 Y
Y by Adventure10, Mango247
ABCDE wrote:
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What is PoP in this context? Might be obvious but I simply can't make any connexions to any known results :maybe:
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Pluto1708
1107 posts
#13 • 2 Y
Y by Adventure10, Mango247
That's pop applied to cylic quad $BFEC$
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khanhnx
1618 posts
#14 • 1 Y
Y by Adventure10
We have: $AB$ . $CA$ + $BC$ . $CA$ + $CA$ . $BC$ $\le$ $AB^2$ + $BC^2$ + $CA^2$ = $AF$ . $AB$ + $BF$ . $AB$ + $BD$ . $BC$ + $CD$ . $BC$ + $CE$ . $CA$ + $AE$ . $CA$ = 2 ($AH$ . $AD$ + $BH$ . $BE$ + $CH$ . $CF$)
So: $\dfrac{AB . CA + BC . CA + CA . BC}{AH . AD + BH . BE + CH . CF}$ $\le$ 2
Equality holds when: $AB$ = $BC$ = $CA$ or $\triangle$ $ABC$ is equilateral
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heheXD1
429 posts
#15
Y by
aditya21 wrote:
here is my solution =
let us denote $a,b,c$ as sides of triangle and $R$ as its circumradius
than by easy trigonometry we get $AH=2RcosA , AD=2RsinBsinC$ etc.
so $AH.AD=4R^2cosAsinBsinC=cbcosA$ etc.

than our inequality converts to proving that
$ab+bc+ca\leq 2(cbcosA+abcosC+cacosB)$

now by cosine law ,we have $2cbcosA=b^2+c^2-a^2$ etc.
substituting,our inequality becomes

$ab+bc+ca\leq a^2+b^2+c^2$
which is obvious as it is equivalent to $\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\ge 0$

so we are done :D

"by easy trigonometry we get $AH=2RcosA , AD=2RsinBsinC$." I've been looking at this for like 2 days and I don't see how this is true... Could someone please provide some insight or hints.
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NoctNight
108 posts
#16
Y by
By power of a point on cyclic quads $BDHF$ and $CDHE$, $AH\cdot AD=AF\cdot AB=AE\cdot AC$. Multiplying the denominator on the LHS to the RHS gives
$$
\sum_{cyc} AB\cdot BC\leq \sum_{cyc} 2(AH\cdot AD)=\sum_{cyc} AF\cdot AB+AE\cdot AC=\sum_{cyc} AB^2
$$AM-GM completes the proof.
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Mogmog8
1080 posts
#17 • 2 Y
Y by centslordm, megarnie
By PoP and AM-GM,\begin{align*}2(AH\cdot AC+BH\cdot BE+CH\cdot CF)&=AF\cdot AB+AE\cdot AC+BF\cdot BA+BD\cdot BC+CE\cdot CA+CD\cdot CB\\&=BC(CD+DB)+AC(DE+AE)+AB(AF+BF)\\&=BC^2+AC^2+AB^2\\&\ge AB\cdot AC+BC\cdot CA+CA\cdot CB\end{align*}$\square$
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MagicalToaster53
159 posts
#18
Y by
Let $a = \overline{BC}, b = \overline{CA}, c = \overline{AB}$. Without loss of generality, assume $a \leq b \leq c$. Observe that $AF = AC \cos \angle A$, and also $\triangle AFH \sim \triangle ADB$ so that \[\frac{AF}{AH} = \frac{AD}{AB} \implies \frac{AB}{AH} = \frac{AD}{AF} = \frac{AD}{AC \cos \angle A} \implies AH \cdot AD = \frac{b^2 + c^2 - a^2}{2}.\]Similarly, \[BH \cdot BE = \frac{c^2 + a^2 - b^2}{2}, \text{ and } \frac{a^2 + b^2 - c^2}{2}.\]Therefore our initial inequality is equivalent to \[\sum_{cyc}^{3} ab \leq \sum_{cyc}^{3} a^2, \]which is trivial by Chebyshev's rearrangement inequality. $\blacksquare$
Remark
This post has been edited 1 time. Last edited by MagicalToaster53, Jun 9, 2024, 10:16 PM
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