Difference between revisions of "1957 AHSME Problems/Problem 21"
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(1) is the inverse | (1) is the inverse | ||
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(2) is the converse | (2) is the converse | ||
| − | (3) is the contrapositive | + | |
| + | (3) is the [[contrapositive]] | ||
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(4) is a restatement of the original theorem. | (4) is a restatement of the original theorem. | ||
Therefore, (3) and (4) are correct. | Therefore, (3) and (4) are correct. | ||
| − | <math>\boxed{\textbf{(E) } | + | <math>\boxed{\textbf{(E) } 3, 4}</math> |
==See Also== | ==See Also== | ||
| − | {{AHSME box|year=1957|num-b=20|num-a=22}} | + | {{AHSME 50p box|year=1957|num-b=20|num-a=22}} |
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:AHSME]][[Category:AHSME Problems]] | [[Category:AHSME]][[Category:AHSME Problems]] | ||
Latest revision as of 08:58, 25 July 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?
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.
See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
