Difference between revisions of "1957 AHSME Problems/Problem 21"

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== Problem 21 ==
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Start with the theorem "If two angles of a triangle are equal, the triangle is isosceles," and the following four statements:
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1. If two angles of a triangle are not equal, the triangle is not isosceles.
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2. The base angles of an isosceles triangle are equal.
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3. If a triangle is not isosceles, then two of its angles are not equal.
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4. A necessary condition that two angles of a triangle be equal is that the triangle be isosceles.
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Which combination of statements contains only those which are logically equivalent to the given theorem?
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<math>\textbf{(A)}\ 1,\,2,\,3,\,4 \qquad \textbf{(B)}\ 1,\,2,\,3\qquad \textbf{(C)}\ 2,\,3,\,4\qquad \textbf{(D)}\ 1,\,2\qquad\textbf{(E)}\ 3,\,4  </math>
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==Solution==
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(1) is the inverse
 
(1) is the inverse
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(2) is the converse
 
(2) is the converse
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(3) is the contrapositive
 
(3) is the contrapositive
(4) is a restatement of the original conditional
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Therefore, (3) and (4) are correct: option (E).
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(4) is a restatement of the original theorem.
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Therefore, (3) and (4) are correct.
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<math>\boxed{\textbf{(E) } (3), (4)}</math>
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==See Also==
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{{AHSME 50p box|year=1957|num-b=20|num-a=22}}
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{{MAA Notice}}
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[[Category:AHSME]][[Category:AHSME Problems]]

Latest revision as of 19:55, 10 June 2024

Problem 21

Start with the theorem "If two angles of a triangle are equal, the triangle is isosceles," and the following four statements:

1. If two angles of a triangle are not equal, the triangle is not isosceles.

2. The base angles of an isosceles triangle are equal.

3. If a triangle is not isosceles, then two of its angles are not equal.

4. A necessary condition that two angles of a triangle be equal is that the triangle be isosceles.

Which combination of statements contains only those which are logically equivalent to the given theorem?

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

Solution

(1) is the inverse

(2) is the converse

(3) is the contrapositive

(4) is a restatement of the original theorem.

Therefore, (3) and (4) are correct. $\boxed{\textbf{(E) } (3), (4)}$

See Also

1957 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AHSME Problems and Solutions

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