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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
d+2 pts in R^d can partition
EthanWYX2009   0
28 minutes ago
Source: Radon's Theorem
Show that: any set of $d + 2$ points in $\mathbb R^d$ can be partitioned into two sets whose convex hulls intersect.
0 replies
EthanWYX2009
28 minutes ago
0 replies
hard inequality omg
tokitaohma   4
N an hour ago by arqady
1. Given $a, b, c > 0$ and $abc=1$
Prove that: $ \sqrt{a^2+1} + \sqrt{b^2+1} + \sqrt{c^2+1} \leq \sqrt{2}(a+b+c) $

2. Given $a, b, c > 0$ and $a+b+c=1 $
Prove that: $ \dfrac{\sqrt{a^2+2ab}}{\sqrt{b^2+2c^2}} + \dfrac{\sqrt{b^2+2bc}}{\sqrt{c^2+2a^2}} + \dfrac{\sqrt{c^2+2ca}}{\sqrt{a^2+2b^2}} \geq \dfrac{1}{a^2+b^2+c^2} $
4 replies
tokitaohma
Yesterday at 5:24 PM
arqady
an hour ago
ISI UGB 2025 P4
SomeonecoolLovesMaths   6
N an hour ago by Atmadeep
Source: ISI UGB 2025 P4
Let $S^1 = \{ z \in \mathbb{C} \mid |z| =1 \}$ be the unit circle in the complex plane. Let $f \colon S^1 \longrightarrow S^2$ be the map given by $f(z) = z^2$. We define $f^{(1)} \colon = f$ and $f^{(k+1)} \colon = f \circ f^{(k)}$ for $k \geq 1$. The smallest positive integer $n$ such that $f^{(n)}(z) = z$ is called the period of $z$. Determine the total number of points in $S^1$ of period $2025$.
(Hint : $2025 = 3^4 \times 5^2$)
6 replies
SomeonecoolLovesMaths
Yesterday at 11:24 AM
Atmadeep
an hour ago
An innocent-looking inequality
Bryan0224   0
an hour ago
Source: Idk
If $\{a_i\}_{1\le i\le n }$ and $\{b_i\}_{1\le i\le n}$ are two sequences between $1$ and $2$ and they satisfy $\sum_{i=1}^n a_i^2=\sum_{i=1}^n b_i^2$, prove that $\sum_{i=1}^n\frac{a_i^3}{b_i}\leq 1.7\sum_{i=1}^{n} a_i^2$, and determine when does equality hold
Please answer this @sqing :trampoline:
0 replies
+1 w
Bryan0224
an hour ago
0 replies
No more topics!
INMO 2018 -- Problem #1
integrated_JRC   19
N Dec 30, 2022 by dheerstar12
Source: INMO 2018
Let $ABC$ be a non-equilateral triangle with integer sides. Let $D$ and $E$ be respectively the mid-points of $BC$ and $CA$ ; let $G$ be the centroid of $\Delta{ABC}$. Suppose, $D$, $C$, $E$, $G$ are concyclic. Find the least possible perimeter of $\Delta{ABC}$.
19 replies
integrated_JRC
Jan 21, 2018
dheerstar12
Dec 30, 2022
INMO 2018 -- Problem #1
G H J
G H BBookmark kLocked kLocked NReply
Source: INMO 2018
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integrated_JRC
3465 posts
#1 • 3 Y
Y by The-X-squared-factor, Adventure10, Mango247
Let $ABC$ be a non-equilateral triangle with integer sides. Let $D$ and $E$ be respectively the mid-points of $BC$ and $CA$ ; let $G$ be the centroid of $\Delta{ABC}$. Suppose, $D$, $C$, $E$, $G$ are concyclic. Find the least possible perimeter of $\Delta{ABC}$.
This post has been edited 2 times. Last edited by integrated_JRC, Jan 22, 2018, 2:10 AM
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Supercali
1261 posts
#2 • 1 Y
Y by Adventure10
Looks like we posted at the same time......
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Supercali
1261 posts
#3 • 1 Y
Y by Adventure10
I will delete my post. Please make a contest collection.
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62861
3564 posts
#4 • 4 Y
Y by opptoinfinity, Aryan-23, Adventure10, Mango247
jrc1729 wrote:
$\mathbf{1)}$ Let $ABC$ be a non-equilateral triangle with integer sides. Let $D$ and $E$ be respectively the mid-points of $BC$ and $CA$ ; let $G$ be the centroid of $\Delta{ABC}$. Suppose, $D$, $C$, $E$, $G$ are concyclic. Find the least possible perimeter of $\Delta{ABC}$.

This means $\angle BCA + \angle BGA = 180^{\circ}$, so $G$ is the C-HM point of $\triangle ABC$; i.e. $\odot(ACG)$ and $\odot(BCG)$ are tangent to $\overline{AB}$.

It is well-known that this implies $AC^2 + BC^2 = 2AB^2$, but I post the solution here for completeness. Letting $M$ be the midpoint of $\overline{AB}$, we need $\tfrac{1}{3} MA^2 = MG \cdot MA = MB^2 = MC^2$, or $\tfrac{1}{3} \tfrac{2a^2 + 2b^2 - c^2}{4} = \tfrac{c^2}{4}$, so $a^2 + b^2 = 2c^2$.

So $(a + b)^2 + (a - b)^2 = (2c)^2$. It is well-known that a necessary and sufficient condition for this to have integer solutions $(a, b)$ is for $c$ to have a prime factor congruent to 1 mod 4. It is clear that in the minimal perimeter triangle $\gcd(a, b, c) = 1$. The smallest possible values of $c$ are 5 and 13. $c = 5$ leads to $\{a, b\} = \{1, 7\}$, not a triangle. $c = 13$ leads to $\{a, b\} = \{7, 17\}$, which indeed works. The answer is $a + b + c = \boxed{37}$.
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Vrangr
1600 posts
#5 • 5 Y
Y by AnArtist, Wizard_32, Abhyuday, Adventure10, Mango247
By Power of a point,
\[1 = \frac{BG\cdot BE}{BD\cdot BC} = \frac{\frac23 BE^2}{\frac12 BC^2} \implies \frac{BE}{BC}= \frac{\sqrt3}2\]Let $AB = c, BC = a, CA = b$,
By Appolonius' theorem.
\[ a^2 + c^2 = 2(BE^2 + EC^2) = \frac32 a^2 + \frac12 b^2 \iff a^2 + b^2 = 2c^2\]
I couldn't do anything after that in the paper.
I just realized this, now I feel sad :(
\[a^2+b^2=2c^2 \iff (a-b)^2+(a+b)^2 = (2c)^2 \]WLOG $a>b$
Now the smallest pythagorean triple is
\[3,4,5 \implies a-b = 3, a+b = 4, 2c = 5 \implies a = \frac72, b=\frac12, c=\frac52\]\[\text{But } a + c = \frac62 < \frac72 = b \]Therefore, this can not be a solution

Now, the second smallest pythagorean triple is
\[(a-b,a+b, 2c) = (5, 12, 13) \implies a = \frac{17}2, b=\frac72, c=\frac{13}2 \]\[\text{and }b + c = 10 > \frac{17}2 = a\]Therefore $(a:b:c) = (17:7:13)$ Thus, the smallest value of $a,b,c$ is $17,7,13$. Therefore $a+b+c = 37$.
This post has been edited 2 times. Last edited by Vrangr, Jan 21, 2018, 3:15 PM
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sa2001
281 posts
#7 • 2 Y
Y by Adventure10, Mango247
I got $a^2 + b^2 = 2c^2$. Then I wrote -
by putting small values for $c$, we that $(17, 7, 13)$ is a valid solution with the smallest value of $c$.
Then, I used that to get a bound on the value of $c$ (using the triangle inequality) till which we have to check for any other possible solutions.
The bound was not very large. ($c<=18$).
Will they cut any marks for this? :(
(I didn't explicitly check values on the answer sheet. I did it mentally and some of it on rough sheets.)
This post has been edited 2 times. Last edited by sa2001, Jan 21, 2018, 5:29 PM
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hydrohelium
245 posts
#8 • 2 Y
Y by Adventure10, Mango247
CantonMathGuy wrote:
jrc1729 wrote:
$\mathbf{1)}$ Let $ABC$ be a non-equilateral triangle with integer sides. Let $D$ and $E$ be respectively the mid-points of $BC$ and $CA$ ; let $G$ be the centroid of $\Delta{ABC}$. Suppose, $D$, $C$, $E$, $G$ are concyclic. Find the least possible perimeter of $\Delta{ABC}$.

This means $\angle BCA + \angle BGA = 180^{\circ}$, so $G$ is the C-HM point of $\triangle ABC$; i.e. $\odot(ACG)$ and $\odot(BCG)$ are tangent to $\overline{AB}$.

It is well-known that this implies $AC^2 + BC^2 = 2AB^2$, but I post the solution here for completeness. Letting $M$ be the midpoint of $\overline{AB}$, we need $\tfrac{1}{3} MA^2 = MG \cdot MA = MB^2 = MC^2$, or $\tfrac{1}{3} \tfrac{2a^2 + 2b^2 - c^2}{4} = \tfrac{c^2}{4}$, so $a^2 + b^2 = 2c^2$.

So $(a + b)^2 + (a - b)^2 = (2c)^2$. It is well-known that a necessary and sufficient condition for this to have integer solutions $(a, b)$ is for $c$ to have a prime factor congruent to 1 mod 4. It is clear that in the minimal perimeter triangle $\gcd(a, b, c) = 1$. The smallest possible values of $c$ are 5 and 13. $c = 5$ leads to $\{a, b\} = \{1, 7\}$, not a triangle. $c = 13$ leads to $\{a, b\} = \{7, 17\}$, which indeed works. The answer is $a + b + c = \boxed{37}$.

It is not necessary that $\gcd(a, b, c) =1$ is required for least perimeter. The triplet $(6, 8,10)$ was also possible
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ShamimAkhtar212
126 posts
#10 • 6 Y
Y by MathFather1234, Pi-is-3, amar_04, RAMUGAUSS, Adventure10, Mango247
Hydrohelium, for the minimal perimeter, it is enough to notice that the triplet should be a primitive one. :noo:
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ShamimAkhtar212
126 posts
#11 • 3 Y
Y by MathFather1234, Adventure10, Mango247
CantonMathGuy wrote:
jrc1729 wrote:
$\mathbf{1)}$ Let $ABC$ be a non-equilateral triangle with integer sides. Let $D$ and $E$ be respectively the mid-points of $BC$ and $CA$ ; let $G$ be the centroid of $\Delta{ABC}$. Suppose, $D$, $C$, $E$, $G$ are concyclic. Find the least possible perimeter of $\Delta{ABC}$.

This means $\angle BCA + \angle BGA = 180^{\circ}$, so $G$ is the C-HM point of $\triangle ABC$; i.e. $\odot(ACG)$ and $\odot(BCG)$ are tangent to $\overline{AB}$.

It is well-known that this implies $AC^2 + BC^2 = 2AB^2$, but I post the solution here for completeness. Letting $M$ be the midpoint of $\overline{AB}$, we need $\tfrac{1}{3} MA^2 = MG \cdot MA = MB^2 = MC^2$, or $\tfrac{1}{3} \tfrac{2a^2 + 2b^2 - c^2}{4} = \tfrac{c^2}{4}$, so $a^2 + b^2 = 2c^2$.

So $(a + b)^2 + (a - b)^2 = (2c)^2$. It is well-known that a necessary and sufficient condition for this to have integer solutions $(a, b)$ is for $c$ to have a prime factor congruent to 1 mod 4. It is clear that in the minimal perimeter triangle $\gcd(a, b, c) = 1$. The smallest possible values of $c$ are 5 and 13. $c = 5$ leads to $\{a, b\} = \{1, 7\}$, not a triangle. $c = 13$ leads to $\{a, b\} = \{7, 17\}$, which indeed works. The answer is $a + b + c = \boxed{37}$.

Does this also imply that $AC^2+BC^2=2AB^2???$ I read the article about the $HM$ points by anantmudgal09 but i did not see this fact.... can you suggest an article with all the properties of $HM$ points please??
This post has been edited 3 times. Last edited by ShamimAkhtar212, Feb 2, 2019, 3:44 PM
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ShamimAkhtar212
126 posts
#12 • 3 Y
Y by MathFather1234, Adventure10, Mango247
It is easily obtained by power of point, but is it a standard lemma or theorem?
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Pi-is-3
51 posts
#13 • 1 Y
Y by Adventure10
I want to say that instead of achieving this using synthetic results, we can get $a^2+b^2=2c^2$ in 5 to 10 minutes by a barycentric bash.
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MathPassionForever
1663 posts
#14 • 2 Y
Y by Adventure10, Mango247
Pi-is-3 wrote:
I want to say that instead of achieving this using synthetic results, we can get $a^2+b^2=2c^2$ in 5 to 10 minutes by a barycentric bash.
Better: you can guess that beforehand. Read automedian triangles.
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ftheftics
651 posts
#15 • 1 Y
Y by Gerninza
with out losing generility let assume ,
$AB=c,BC=a,CA=b$.
clearly $\angle BGD=\angle DCE =C$.
Again$\angle BDG=\angle BEA=\alpha$(say)

From $\triangle EBG$ we get $\angle EBC =\alpha -c$ and $\angle BAE=180-(\alpha+c)$.


so we get ,

$GD=\frac{\sin (\alpha-c) .a}{2\sin C }$.

$GE=\frac{\sin(\alpha+c) .b}{2\sin C}$.

$ED=\frac{c}{2}$.

$GC =\frac{c\sin \alpha}{2\sin B}$,

since $G,E,C,D$ are concyclic applying ptolemy's theorem get,

$ED.GC=DC.GE+GD.EC $

$\Rightarrow \frac{ab}{4} .(\frac{\sin (\alpha-c) +\sin(\alpha+c)}{\sin C} =\frac{bc\sin\alpha}{4\sin B}$.

$\Rightarrow \frac{2a\sin C}{\cos C}=\frac{c}{\sin B}$.

$\Rightarrow   \frac{2a\cos c}{c}=\frac{c}{b}$

$\Rightarrow 2ab \cos C =c^2$.

using cosine rule get,

$2ab\frac{a^2+b^2-c^2a}{2ab} =c^2$.

$\Rightarrow a^2 +b^2 =2c^2$.

$(\frac{a+b}{2})^2+(\frac{a-b}{2})^2=c^2$.

we know least pythagorian triplet is $3,4,5$.

but in this case we get $a,b=\{1,7\}$.
its not possible since $\angle B$ is obtuse.

next least pythagorian triplet is $\{12,13,5\}$.
we get $a,b=\{17,7\}$.

so least possible perimeter is $13+17+7=37$.
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Math-wiz
6107 posts
#16 • 1 Y
Y by RudraRockstar
Yay another INMO Geo solved! (Or should I say NT :?)
By Apollonius theorem, we have $AD^2=\frac{2(b^2+c^2)-a^2}{4}$
But, $AD=3GD$
So, $GD^2=\frac{2(b^2+c^2)-a^2}{36}$
By PoP, we have $AG\cdot AD=AE\cdot AC\implies 2GD\cdot 3GD=\frac{b}{2}\cdot b$
So, $GD^2=\frac{b^2}{12}$
And hence, $\frac{b^2}{12}=\frac{2(b^2+c^2)-a^2}{36}$
Simplifying gives $a^2+b^2=2c^2$
The NT part
Since $a,b,c$ are positive integers, we take $d=2$ to get $a=17,b=7,c=13$ and hence, the minimal perimeter is $a+b+c=37$ $\blacksquare$.

@below I think that it is obvious
This post has been edited 3 times. Last edited by Math-wiz, Apr 29, 2020, 9:15 AM
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starchan
1609 posts
#17
Y by
Math-wiz wrote:
Yay another INMO Geo solved! (Or should I say NT :?)
By Apollonius theorem, we have $AD^2=\frac{2(b^2+c^2)-a^2}{4}$
But, $AD=3GD$
So, $GD^2=\frac{2(b^2+c^2)-a^2}{36}$
By PoP, we have $AG\cdot AD=AE\cdot AC\implies 2GD\cdot 3GD=\frac{b}{2}\cdot b$
So, $GD^2=\frac{b^2}{12}$
And hence, $\frac{b^2}{12}=\frac{2(b^2+c^2)-a^2}{36}$
Simplifying gives $a^2+b^2=2c^2$
The NT part
Since $a,b,c$ are positive integers, we take $d=2$ to get $a=17,b=7,c=13$ and hence, the minimal perimeter is $a+b+c=37$ $\blacksquare$.

I believe that the NT part of your proof is incomplete
This is because you have not proved that there is no other triple for which the perimeter is less than 37
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Gerninza
2 posts
#18
Y by
ftheftics wrote:
with out losing generility let assume ,
$AB=c,BC=a,CA=b$.
clearly $\angle BGD=\angle DCE =C$.
Again$\angle BDG=\angle BEA=\alpha$(say)

From $\triangle EBG$ we get $\angle EBC =\alpha -c$ and $\angle BAE=180-(\alpha+c)$.


so we get ,

$GD=\frac{\sin (\alpha-c) .a}{2\sin C }$.

$GE=\frac{\sin(\alpha+c) .b}{2\sin C}$.

$ED=\frac{c}{2}$.

$GC =\frac{c\sin \alpha}{2\sin B}$,

since $G,E,C,D$ are concyclic applying ptolemy's theorem get,

$ED.GC=DC.GE+GD.EC $

$\Rightarrow \frac{ab}{4} .(\frac{\sin (\alpha-c) +\sin(\alpha+c)}{\sin C} =\frac{bc\sin\alpha}{4\sin B}$.

$\Rightarrow \frac{2a\sin C}{\cos C}=\frac{c}{\sin B}$.

$\Rightarrow   \frac{2a\cos c}{c}=\frac{c}{b}$

$\Rightarrow 2ab \cos C =c^2$.

using cosine rule get,

$2ab\frac{a^2+b^2-c^2a}{2ab} =c^2$.

$\Rightarrow a^2 +b^2 =2c^2$.

$(\frac{a+b}{2})^2+(\frac{a-b}{2})^2=c^2$.

we know least pythagorian triplet is $3,4,5$.

but in this case we get $a,b=\{1,7\}$.
its not possible since $\angle B$ is obtuse.

next least pythagorian triplet is $\{12,13,5\}$.
we get $a,b=\{17,7\}$.

so least possible perimeter is $13+17+7=37$.

I guess u have made a small mistake whike writing $\alpha$
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stranger_02
337 posts
#19 • 1 Y
Y by Mango247
ftheftics wrote:
with out losing generility let assume ,
$AB=c,BC=a,CA=b$.
clearly $\angle BGD=\angle DCE =C$.
Again$\angle BDG=\angle BEA=\alpha$(say)

From $\triangle EBG$ we get $\angle EBC =\alpha -c$ and $\angle BAE=180-(\alpha+c)$.


so we get ,

$GD=\frac{\sin (\alpha-c) .a}{2\sin C }$.

$GE=\frac{\sin(\alpha+c) .b}{2\sin C}$.

$ED=\frac{c}{2}$.

$GC =\frac{c\sin \alpha}{2\sin B}$,

since $G,E,C,D$ are concyclic applying ptolemy's theorem get,

$ED.GC=DC.GE+GD.EC $

$\Rightarrow \frac{ab}{4} .(\frac{\sin (\alpha-c) +\sin(\alpha+c)}{\sin C} =\frac{bc\sin\alpha}{4\sin B}$.

$\Rightarrow \frac{2a\sin C}{\cos C}=\frac{c}{\sin B}$.

$\Rightarrow   \frac{2a\cos c}{c}=\frac{c}{b}$

$\Rightarrow 2ab \cos C =c^2$.

using cosine rule get,

$2ab\frac{a^2+b^2-c^2a}{2ab} =c^2$.

$\Rightarrow a^2 +b^2 =2c^2$.

$(\frac{a+b}{2})^2+(\frac{a-b}{2})^2=c^2$.

we know least pythagorian triplet is $3,4,5$.

but in this case we get $a,b=\{1,7\}$.
its not possible since $\angle B$ is obtuse.

next least pythagorian triplet is $\{12,13,5\}$.
we get $a,b=\{17,7\}$.

so least possible perimeter is $13+17+7=37$.

$EBG$ is not a triangle but a straight line..
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lazizbek42
548 posts
#20 • 1 Y
Y by MerlinLegion07
$a,b,c$ sides tringle.$\implies 2c^2=a^2+b^2$.
Remaing easy.
This post has been edited 1 time. Last edited by lazizbek42, Feb 6, 2022, 10:16 AM
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Mahdi_Mashayekhi
695 posts
#21
Y by
Nice combination of NT and GEO
Geo part:
Claim1 : $a^2 + b^2 = 2c^2$.
Proof : we have $GD^2=\frac{2(b^2+c^2)-a^2}{36}$ and also from power of point $A$ we have $6GD^2 = b^2/2$ and they simply prove our claim.

NT part:
we have $a^2 + b^2 = 2c^2$ or in fact $(a-b)^2 + (a+b)^2 = (2c)^2$ and we know $a,b,c$ are integer. so triple $(a-b,a+b,2c)$ must be pythagorean.
we know smallest pythagorean triple is $(3x,4x,5x)$. but our triple can't be in this form because then :
$a = \frac{7x}{2}$, $b = \frac{x}{2}$ and $c = \frac{5x}{2}$ and $b+c$ is less than $a$.
next smallest pythagorean triple is $(5x,12x,13x)$. with this we have no inequality contradiction so $(a,b,c) = (17,7,13)$ so least possible perimeter of $ABC$ is $17 + 7 + 13 = 37$.
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dheerstar12
18 posts
#27
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By PoP we get $BE = \frac{\sqrt3 a}{2}$

On using appolonius thm and substituting the above value we get \[a^2 + b^2 = 2c^2\]Now we just have to check for pythagorean triplets and we find the smallest solution to be $17,7,13$
So the ans is 37
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