Difference between revisions of "2010 AMC 12A Problems/Problem 12"

Latest revision as of 00:40, 28 August 2024

The following problem is from both the 2010 AMC 12A #12 and 2010 AMC 10A #15, so both problems redirect to this page.

Problem

In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.

Brian: "Mike and I are different species."

Chris: "LeRoy is a frog."

LeRoy: "Chris is a frog."

Mike: "Of the four of us, at least two are toads."

How many of these amphibians are frogs?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution 1

Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.

As Mike is a frog, his statement is false, hence there is at most one toad.

As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad.

Hence we must have one toad and $\boxed{\textbf{(D)}\ 3}$ frogs.

Solution 2 (logical reasoning like solution 1, but a different train of thought)

Notice that one of Chris and LeRoy must be a frog: if Chris is a frog, then he lies about LeRoy being a frog. Hence LeRoy is a toad. Alternatively, if Chris is a toad, then he tells the truth about LeRoy being a frog.

Assume Brian is a toad. Then Mike is a frog, and he lies about at least two being toads. This means that none or one of the amphibians is a toad (the opposite of the statement $n\geq2$ is $n<2$ , or $n=0, 1$ ). However, this is absurd because we assumed Brian is a toad, and we know one of Chris and LeRoy is a toad. So our assumption leads to a contradiction.

Hence Brian must be a frog, and he and Mike are the same species. Mike is also a frog. One of Chris and LeRoy is a frog. There are $3$ frogs in total $\Longrightarrow \boxed{\textbf{(D) } 3}$ .

~JH. L

Solution 3: Casework/Proof by contradiction First, we assume that Brian is a toad. Since his statements are truthful, Mike could not be a toad and must be a frog. Since frogs are liars, Brian must be the only toad in the pond, so Chris and Leroy are both frogs. However, they would then be telling the truth about each other, which only toads do. Hence, this case results in a contraction. Hence, Brian is a frog. He is lying, so Mike must also be a frog. Mike's statement is also false, so Chris and Leroy cannot both be toads. As shown in the previous case, them both being frogs is impossible, so one must be a toad and the other a frog. Therefore, there must be a total of 3 frogs. By-Monkey_King

Video Solution by the Beauty of Math

https://youtu.be/kU70k1-ONgM?t=1207

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-== Problem 12 ==
+{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #12]] and [[2010 AMC 10A Problems|2010 AMC 10A #15]]}}
+== Problem ==
 In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
@@ Line 14: / Line 16: @@
 <math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math>
-== Solution ==
+== Solution 1==
+Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.
-=== Solution 1 ===
+As Mike is a frog, his statement is false, hence there is at most one toad.
-We can begin by first looking at Chris and LeRoy.
+As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad.
-Suppose Chris and LeRoy are the same species. If Chris is a toad, then what he says is true, so LeRoy is a frog. However, if LeRoy is a frog, then he is lying, but clearly Chris is not a frog, and we have a contradiction. The same applies if Chris is a frog.
+Hence we must have one toad and <math>\boxed{\textbf{(D)}\ 3}</math> frogs.
-Clearly, Chris and LeRoy are different species, and so we have at least <math>1</math> frog out of the two of them.
-Now suppose Mike is a toad. Then what he says is true because we already have <math>2</math> toads. However, if Brian is a frog, then he is lying, yet his statement is true, a contradiction. If Brian is a toad, then what he says is true, but once again it conflicts with his statement, resulting in contradiction.
+== Solution 2 (logical reasoning like solution 1, but a different train of thought)==
+Notice that one of Chris and LeRoy must be a frog: if Chris is a frog, then he lies about LeRoy being a frog. Hence LeRoy is a toad. Alternatively, if Chris is a toad, then he tells the truth about LeRoy being a frog.
-Therefore, Mike must be a frog. His statement must be false, which means that there is at most <math>1</math> toad. Since either Chris or LeRoy is already a toad, Brain must be a frog. We can also verify that his statement is indeed false.
+Assume Brian is a toad. Then Mike is a frog, and he lies about at least two being toads. This means that none or one of the amphibians is a toad (the opposite of the statement <math>n\geq2</math> is <math>n<2</math>, or <math>n=0, 1</math>). However, this is absurd because we assumed Brian is a toad, and we know one of Chris and LeRoy is a toad. So our assumption leads to a contradiction.
-Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have <math>3</math> frogs total. <math>\boxed{\textbf{(D)}}</math>
+Hence Brian must be a frog, and he and Mike are the same species. Mike is also a frog. One of Chris and LeRoy is a frog. There are <math>3</math> frogs in total <math>\Longrightarrow \boxed{\textbf{(D) } 3}</math>.
-=== Solution 2 ===
+~JH. L
-Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.
+Solution 3: Casework/Proof by contradiction
+First, we assume that Brian is a toad. Since his statements are truthful, Mike could not be a toad and must be a frog. Since frogs are liars, Brian must be the only toad in the pond, so Chris and Leroy are both frogs. However, they would then be telling the truth about each other, which only toads do. Hence, this case results in a contraction.
+Hence, Brian is a frog. He is lying, so Mike must also be a frog. Mike's statement is also false, so Chris and Leroy cannot both be toads. As shown in the previous case, them both being frogs is impossible, so one must be a toad and the other a frog. Therefore, there must be a total of 3 frogs.
+By-Monkey_King
-As Mike is a frog, his statement is false, hence there is at most one toad.
+==Video Solution by the Beauty of Math==
+https://youtu.be/kU70k1-ONgM?t=1207
-As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad.
+== See also ==
+{{AMC12 box|year=2010|num-b=11|num-a=13|ab=A}}
+{{AMC10 box|year=2010|num-b=14|num-a=16|ab=A}}
-Hence we must have one toad and three frogs.
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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