tetrahedron with faces different isosceles triangles (Chile 2003 L1 P4)

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0 replies

jlacosta

Apr 2, 2025

0 replies

Eazy equation clap

giangtruong13 0

Today at 4:03 PM

Find all $x,y,z$ satisfy that: $\frac{x}{y+z}=2x-1; \frac{y}{x+z}=3y-1;\frac{z}{x+y}=5x-1$

0 replies

giangtruong13

Today at 4:03 PM

0 replies

inequality

revol_ufiaw 3

N Today at 2:55 PM by MS_asdfgzxcvb

Prove that that for any real $x \ge 0$ and natural number $n$ ,
$x^n (n+1)^{n+1} \le n^n (x+1)^{n+1}.$

3 replies

revol_ufiaw

Today at 2:05 PM

MS_asdfgzxcvb

Today at 2:55 PM

How to judge a number is prime or not?

mingzhehu 0

Today at 2:45 PM

A=（10X1+1）(10X+1），X1，X∈N+
B=（10 X1+3）（10X+7），X∈N，X1∈N
C=(10 X1+9)(10X+9), X∈N，X1∈N
D=(10 X1+1)(10X+3), X1∈N+，X∈N
E=（10 X1+7）(10X+9），X∈N，X1∈N
F=（10 X1+1）（10X+7），X1∈N+，X∈N
G=（10 X1+3）（10X+9），X∈N，X1∈N
H=（10 X1+1）10X+9），X1∈N+，X∈N
I=（10 X1+3）（10X+3），X1∈N，X∈N
J=( 10X1+7)(10X+7),X∈N，X1∈N

For any natural number P∈{P=10N+1，n∈N}，make P=A or B or C
If P can make the roots of function group(ABC) without any root group completely made up of integer, P will be a prime
For any natural number P∈{P=10N+3，n∈N}，make P=D or E
If P can make the roots of function group(DE) without any root group completely made up
of integer, P will be a prime
For any natural number P∈{P=10N+7，n∈N}，make P=F or G
If P can make the roots of function group(FG) without any root group completely made up
of integer, P will be a prime
For any natural number P∈{P=10N+9，n∈N}，make P=H or I or J
If P can make the roots of function group(GIJ) without any root group completely made up
of integer, P will be a prime

0 replies

mingzhehu

Today at 2:45 PM

0 replies

What is an isogonal conjugate and why is it useful?

EaZ_Shadow 6

N Today at 2:40 PM by maxamc

What is an isogonal conjugate and why is it useful? People use them in Olympiad geometry proofs but I don’t understand why and what is the purpose, as it complicates me because of me not understanding it.

6 replies

EaZ_Shadow

Dec 28, 2024

maxamc

Today at 2:40 PM

Any nice way to do this?

NamelyOrange 3

N Today at 2:00 PM by pooh123

Source: Taichung P.S.1 math program tryouts

How many ordered pairs $(a,b,c)\in\mathbb{N}^3$ are there such that $c=ab$ and $1\le a\le b\le c\le60$ ?

3 replies

NamelyOrange

Apr 2, 2025

pooh123

Today at 2:00 PM

Inequalities

sqing 3

N Today at 2:00 PM by sqing

Let $a,b,c> 0$ and $\frac{a}{a^2+ab+c}+\frac{b}{b^2+bc+a}+\frac{c}{c^2+ca+b} \geq 1$ . Prove that
$a+b+c\leq 3$

3 replies

sqing

Yesterday at 3:52 AM

sqing

Today at 2:00 PM

Inequalities

sqing 0

Today at 1:10 PM

Let $a,b$ be real numbers such that $a^2+b^2+a^3 +b^3=4 .$ Prove that
$a+b \leq 2$ Let $a,b$ be real numbers such that $a+b + a^2+b^2+a^3 +b^3=6 .$ Prove that
$a+b \leq 2$

0 replies

sqing

Today at 1:10 PM

0 replies

that statement is true

pennypc123456789 3

N Today at 12:32 PM by sqing

we have $a^3+b^3 = 2$ and $3(a^4+b^4)+2a^4b^4 \le 8$ , then we can deduce $a^2+b^2$ \le 2 $ ?

3 replies

pennypc123456789

Mar 23, 2025

sqing

Today at 12:32 PM

Distance vs time swimming problem

smalkaram_3549 1

N Today at 11:54 AM by Lankou

How should I approach a problem where we deal with velocities becoming negative and stuff. I know that they both travel 3 Lengths of the pool before meeting a second time.

1 reply

smalkaram_3549

Today at 2:57 AM

Lankou

Today at 11:54 AM

.problem.

Cobedangiu 4

N Today at 11:40 AM by Lankou

Find the integer coefficients after expanding Newton's binomial:
$(\frac{3}{2}-\frac{2}{3}x^2)^n (n \in Z)$

4 replies

Cobedangiu

Yesterday at 6:20 AM

Lankou

Today at 11:40 AM

inequalities - 5/4

pennypc123456789 2

N Today at 11:35 AM by sqing

Given real numbers $x, y$ satisfying $|x| \le 3, |y| \le 3$ . Prove that:
$0 \le (x^2 + 1)(y^2 + 1) + 4(x - 1)(y - 1) \le 164.$

2 replies

pennypc123456789

Today at 8:57 AM

sqing

Today at 11:35 AM

Source of a combinatorics problem

isodynamicappolonius2903 0

Today at 6:47 AM

Does anyone know exactly the source of this problem?? I just remember that it from a combinatorics book of Russia.

0 replies

isodynamicappolonius2903

Today at 6:47 AM

0 replies

KSEA NMSC Mock Contest Group B (Last Problem)

Shiyul 5

N Today at 3:26 AM by Shiyul

Let $a_n$ be a sequence defined by $a_n = a^2 + 1$ . Then the product of four consecutive terms in $a_n$ can be written as the product of two terms in $a_n$ . Find $p + q$ if $(a_(11))(a_(12))(a_(13))(a_(14)) = (a_p)(a_q)$ .

5 replies

Shiyul

Yesterday at 3:09 PM

Shiyul

Today at 3:26 AM

Regarding IMO prepartion

omega2007 1

N Today at 2:49 AM by omega2007

<Hey Everyone'>
I'm 10 grader student and Im starting prepration for maths olympiad..>>> From scratch (not 2+2=4 )

Do you haves compilled resources of Handouts,
PDF,
Links,
List of books topic wise
which are shared on AOPS (and from your prespective) for maths olympiad and any useful thing, which will help me in boosting Maths olympiad prepration.

1 reply

omega2007

Yesterday at 3:14 PM

omega2007

Today at 2:49 AM

tetrahedron with faces different isosceles triangles (Chile 2003 L1 P4)

parmenides51 2

N Sep 3, 2022 by parmenides51

Investigate if there exists a tetrahedron $ABCD$ such that all its faces are different isosceles triangles.

2 replies

parmenides51

May 28, 2019

parmenides51

Sep 3, 2022

tetrahedron with faces different isosceles triangles (Chile 2003 L1 P4)

G H J

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parmenides51

30629 posts

parmenides51

#1 h May 28, 2019, 6:52 PM • 1 Y

Y by Adventure10

Investigate if there exists a tetrahedron $ABCD$ such that all its faces are different isosceles triangles.

This post has been edited 1 time. Last edited by parmenides51, May 29, 2019, 2:13 AM
Reason: name typo

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NikoIsLife

9657 posts

NikoIsLife

#2 h May 29, 2019, 8:39 AM • 1 Y

Y by Adventure10

Solution

(This solution is pretty much just Case Bashing.)

The answer is $\boxed{\mathrm{No}}$ . To see this, suppose that such a tetrahedron $ABCD$ exists.

We have two cases:

Case I. At least one of the faces is equilateral.

Without loss of generality, suppose that $\triangle ABC$ is equilateral. Let $AB=BC=AC=x$ .

Now, consider triangle $ABD$ . This triangle can't be equilateral, otherwise, we get that $\triangle ABD$ is congruent to $\triangle ABC$ , and we're not allowed to do that. We have three cases.

Case I.I. $AB=AD$ .

This means, $AB=AD=x$ . Let $BD=y$ , where $x\ne y$ .

Now, look at triangle $ACD$ . We already know that $AC=AD=x$ . This shows that $CD\ne y$ , otherwise, $\triangle ACD$ would be congruent to $ABD$ . Let $CD=z$ where $y\ne z$ .

And now, we look at the last triangle, $\triangle BCD$ . Notice that $BC=x$ , $BD=y$ and $CD=z$ . This implies that $\triangle BCD$ is a scalene triangle, which is a contradiction.

Case I.II. $AB=BD$

This is just a mirrored version of Case I.I.

Case I.III.[ $AD=BD$

This means, $AB=AD=y$ , where $x\ne y$ . (Otherwise, if $x=y$ , then $\triangle ABD$ would be congruent to $\triangle ABC$ )

However, look at triangles $\triangle BCD$ and $\triangle ACD$ . We know that $BC=AC=x$ and $BD=AD=y$ . Therefore, $\triangle BCD\cong\triangle ACD$ , and we're not allowed to have this.

Case II. None of the faces are equilateral.

Consider triangle $ABC$ . Without loss of generality, suppose that $AB=AC=x$ . Then, since none of the triangles must be equilateral, this means $BC=y$ , where $x\ne y$ .

Now, we look at triangle $BCD$ . We have three cases.

Case II.I. $BC=BD$

This means $BC=BD=y$ .

Next, consider triangle $ABD$ . Since $\triangle ABD$ is isosceles and $AB\ne BD$ , then either $AD=AB$ or $AD=BD$ . We can't have the former, otherwise, $AD=AB=x$ , and we would have $\triangle ABC\cong\triangle ABD$ . Therefore, $AD=BD=y$ .

Now, we look at the last triangle, $\triangle ACD$ . We only have two choices, either $CD=AD$ or $CD=AC$ . If $CD=AD$ , then this would mean that $CD=y$ , so $BC=CD=BD=y$ , which makes $\triangle BCD$ equilateral. We can't have this, since we assumed that none of the triangles must be equilateral.

This leaves us with $CD=AC$ , so $CD=x$ . However, we also cannot have this, since this would imply that $\triangle ACD\cong\triangle BAC$ .

Case II.II. $BC=CD$

This is just a mirrored version of Case II.I.

Case II.III. $BD=CD$

This means, $BD=CD=z$ where $z\ne y$ .

Now, we look at triangles $\triangle ABD$ and $\triangle ACD$ . Notice that $AB=AC=x$ and $BD=CD=z$ . Therefore, $\triangle ABD\cong\triangle ACD$ , and we cannot have this.

Finally, from the casework we did above, we see that it's impossible to have a tetrahedron $ABCD$ such that all its faces are different isosceles triangles.

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parmenides51

30629 posts

parmenides51

#3 h Sep 3, 2022, 5:40 PM

Y by

another solution is here

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