Q20 in AMC 2012 Junior (1 Viewer)

[TuePhuong]

Please help me to solve Q20 in AMC 2012 Junior!
View attachment Q20-AMC2012-J.doc

 

[si2136]

There can be 2 scenarios in this question.

1: Since all of the distances are less than 30, let m be less than 30.

m < 30

Therefore 10 + 15 + m > 30

Therefore m > 5

2: Since there is no specified shape, let me be more than 30.

m > 30

m < 30 + 15 + 10

m < 55

Combine both scenarios 1 and 2.

5 < x < 55

which is 0 < x < 50

Because distance is positive and there's no such thing as an imaginary distance lol

1 < x < 49

Therefore maximum is 49 AKA B

NOTE: Next time, you can upload it into a picture, instead of a .doc lol

 

[leehuan]

I used the exact same method. However I interpreted the question differently to obtain the extreme cases 5 and 55.

Draw an arbitrary diagram of the quadrilateral PQRS and then consider the extreme case

Lower extreme case: The sides PQ and SR fold into QR

The length of SP will then be 30-15-10 = 5

Upper extreme case: The sides PQ and SR fold out to be collinear with QR

The length of SP will be 30+15+10 = 55

So 55-5+1=51 different cases

Then discard the two cases where they are collinear:
51-2=49

 

[InteGrand]

There can be 2 scenarios in this question.

1: Since all of the distances are less than 30, let m be less than 30.

m < 30

Therefore 10 + 15 + m > 30

Therefore m > 5

2: Since there is no specified shape, let me be more than 30.

m > 30

m < 30 + 15 + 10

m < 55

Combine both scenarios 1 and 2.

5 < x < 55

which is 0 < x < 50

Because distance is positive and there's no such thing as an imaginary distance lol

1 < x < 49

Therefore maximum is 49 AKA B

NOTE: Next time, you can upload it into a picture, instead of a .doc lol

The question isn't asking for a maximum, it's asking for what is the total no. of possible values for m. So I guess you should technically justify why m can indeed attain each of the integer values from 6 to 54 without having any three points in a line, but for the purposes of an MCQ test like this you don't need to as long as you can see that it can take on more than five (option A) of these values, which is decently easy to see.

 

[si2136]

The question isn't asking for a maximum, it's asking for what is the total no. of possible values for m. So I guess you should technically justify why m can indeed attain each of the integer values from 6 to 54 without having any three points in a line, but for the purposes of an MCQ test like this you don't need to as long as you can see that it can take on more than five (option A) of these values, which is decently easy to see.

But isn't the total possible values the maximum anyways?

 

[davidgoes4wce]

What does AMC Stand for?

 

[eyeseeyou]

What does AMC Stand for?

Australian maths competition

 

[InteGrand]

But isn't the total possible values the maximum anyways?

Not a priori, it's possible for something to be bounded between two values without actually attaining all values in between.

 

[davidgoes4wce]

Australian maths competition

I've never heard of it before. Is there any point in doing it?

 

[davidgoes4wce]

My school didn't ever give me the opportunity to participate in that sort of stuff.

 

[eyeseeyou]

I've never heard of it before. Is there any point in doing it?

It's like ICAS (if you know what ICAS is)

 

[leehuan]

My school didn't ever give me the opportunity to participate in that sort of stuff.

-1 like to your school.
______________________

Nah not really. But you do get awards obviously; the questions aren't meant to be just know it, they test if you can actually do maths

 

[TuePhuong]

Thanks a lot!