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

(Problem 21)
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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>   
 
<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==
  
 
(1) is the inverse
 
(1) is the inverse

Revision as of 14:32, 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 conditional Therefore, (3) and (4) are correct. $\boxed{\textbf{(E) } (3), (4)}$