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AMC 8 DHR
PhoenixMathClub   13
N 5 hours ago by TCIAL_Hi
Hey y'all so I just started preparing for the 2026 AMC 8 and I was wondering if you could give me some suggestions on what I should do so I can get DHR. I barely practice math competition and I just started to get serious this year so my scores might be bad so here it is:

5th Grade: 12
6th Grade: 16
6th Grade Summer/7th Grade: Mocking 20-24 on AMC 8 Tests

I saw many recommendations about buying AoPS books but not many about the number theory so I got the Introduction to Algebra, Introduction to Counting and Probability, and Introduction to Geometry Books. I was wondering what other suggestions that you would have for me to get better at AMC 8.
13 replies
PhoenixMathClub
Jul 10, 2025
TCIAL_Hi
5 hours ago
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Probability
CatsMeow12   10
N Jul 3, 2024 by CatsMeow12
Two pairs of distinct points are chosen from the $3$-by-$3$ grid below. What is the probability that the lines formed by both pairs are perpendicular?

IMAGE

I would greatly appreciate solutions to this so I can check my work. Thanks! :-D
10 replies
CatsMeow12
Jul 3, 2024
CatsMeow12
Jul 3, 2024
Probability
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CatsMeow12
2254 posts
#1
Y by
Two pairs of distinct points are chosen from the $3$-by-$3$ grid below. What is the probability that the lines formed by both pairs are perpendicular?

[asy]unitsize(32);
for (int x=0; x<3; ++x) {
  for (int y=0; y<3; ++y) {
    dot((x,y));
  }
}
[/asy]

I would greatly appreciate solutions to this so I can check my work. Thanks! :-D
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ethanzhang1001
1060 posts
#2
Y by
$\frac{3\cdot 3}{\binom{9}{4}}=\frac{1}{14}$

(oops wrong i spent like 2 seconds before typing this)
This post has been edited 1 time. Last edited by ethanzhang1001, Jul 3, 2024, 1:13 AM
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CatsMeow12
2254 posts
#3
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ethanzhang1001 wrote:
$\frac{3\cdot 3}{\binom{9}{4}}=\frac{1}{14}$

Wait... how did you get this?
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Elephant200
1472 posts
#4
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Are you sure there's only $9$ successful outcomes? I see a lot more.

Also, 4 points don't uniquely determine two pairs of segments.
This post has been edited 1 time. Last edited by Elephant200, Jul 3, 2024, 1:08 AM
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cknk
89 posts
#5
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I, uh... there must be a better way to do this... how would we know that we had all the successful outcomes?
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miyukina
1219 posts
#6 • 1 Y
Y by CatsMeow12
not sure
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CatsMeow12
2254 posts
#7 • 1 Y
Y by miyukina
miyukina wrote:
not sure

I got 4 fewer possibilities... but your solution was almost identical to mine
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miyukina
1219 posts
#8
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CatsMeow12 wrote:
miyukina wrote:
not sure

I got 4 fewer possibilities... but your solution was almost identical to mine

If we both get 8 pairs with slopes of ±1/2 and ±2 , where do you think we differ at?
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CatsMeow12
2254 posts
#9
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miyukina wrote:
CatsMeow12 wrote:
miyukina wrote:
not sure

I got 4 fewer possibilities... but your solution was almost identical to mine

If we both get 8 pairs with slopes of ±1/2 and ±2 , where do you think we differ at?

In the middle term of the numerator, the second term is 1 x 5... I only got one possibility there. You might be correct though, what does the 1 x 5 term represent?
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miyukina
1219 posts
#10
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CatsMeow12 wrote:
miyukina wrote:
CatsMeow12 wrote:

I got 4 fewer possibilities... but your solution was almost identical to mine

If we both get 8 pairs with slopes of ±1/2 and ±2 , where do you think we differ at?

In the middle term of the numerator, the second term is 1 x 5... I only got one possibility there. You might be correct though, what does the 1 x 5 term represent?

The longest diagonal (0, 0) to (2, 2) paired with
1) the other longest diagonal
2) (0, 1) to (1, 0)
3) (0, 2) to (1, 1)
4) (1, 1) to (2, 0)
5) (1, 2) to (2, 1)
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CatsMeow12
2254 posts
#11
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miyukina wrote:
The longest diagonal (0, 0) to (2, 2) paired with
1) the other longest diagonal
2) (0, 1) to (1, 0)
3) (0, 2) to (1, 1)
4) (1, 1) to (2, 0)
5) (1, 2) to (2, 1)

Cases (2) and (5) should be already covered by the 4 x 3 term, but cases (3) and (4) weren't in my solution... these two cases can also be applied by picking the other diagonal, so your answer is actually correct :-D
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