INMO 2018 -- Problem #1

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integrated_JRC

3465 posts

integrated_JRC

#1 h Jan 21, 2018, 11:13 AM • 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

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Supercali

#2 h Jan 21, 2018, 11:15 AM • 1 Y

Y by Adventure10

Looks like we posted at the same time......

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Supercali

1261 posts

Supercali

#3 h Jan 21, 2018, 11:16 AM • 1 Y

Y by Adventure10

I will delete my post. Please make a contest collection.

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62861

3564 posts

62861

#4 h Jan 21, 2018, 11:28 AM • 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

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Vrangr

#5 h Jan 21, 2018, 1:01 PM • 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

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sa2001

#7 h Jan 21, 2018, 5:23 PM • 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

hydrohelium

#8 h Dec 2, 2018, 4:32 AM • 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

ShamimAkhtar212

#10 h Jan 26, 2019, 12:16 PM • 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

ShamimAkhtar212

#11 h Feb 2, 2019, 3:09 PM • 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

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ShamimAkhtar212

#12 h Feb 2, 2019, 3:45 PM • 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

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Pi-is-3

#13 h May 25, 2019, 2:10 AM • 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

MathPassionForever

#14 h May 25, 2019, 3:06 AM • 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

ftheftics

#15 h Apr 1, 2020, 5:32 AM • 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

Math-wiz

#16 h Apr 29, 2020, 7:48 AM • 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

1603 posts

starchan

#17 h Apr 29, 2020, 8:46 AM

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

Gerninza

#18 h Jul 4, 2020, 2:17 PM

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

stranger_02

#19 h Jul 17, 2020, 4:09 PM • 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

lazizbek42

#20 h Feb 6, 2022, 10:16 AM • 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

690 posts

Mahdi_Mashayekhi

#21 h Mar 5, 2022, 6:19 PM

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

dheerstar12

#27 h Dec 30, 2022, 8:21 PM

Y by

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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