Difference between revisions of "1999 AHSME Problems/Problem 2"
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<math> \mathrm{(A) \ All\ equilateral\ triangles\ are\ congruent\ to\ each\ other.}</math> | <math> \mathrm{(A) \ All\ equilateral\ triangles\ are\ congruent\ to\ each\ other.}</math> | ||
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<math>\mathrm{(B) \ All\ equilateral\ triangles\ are\ convex.}</math> | <math>\mathrm{(B) \ All\ equilateral\ triangles\ are\ convex.}</math> | ||
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<math>\mathrm{(C) \ All\ equilateral\ triangles\ are\ equianguilar.}</math> | <math>\mathrm{(C) \ All\ equilateral\ triangles\ are\ equianguilar.}</math> | ||
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<math>\mathrm{(D) \ All\ equilateral\ triangles\ are\ regular\ polygons.}</math> | <math>\mathrm{(D) \ All\ equilateral\ triangles\ are\ regular\ polygons.}</math> | ||
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<math>\mathrm{(E) \ All\ equilateral\ triangles\ are\ similar\ to\ each\ other.} </math> | <math>\mathrm{(E) \ All\ equilateral\ triangles\ are\ similar\ to\ each\ other.} </math> | ||
− | + | ==Solution== | |
− | + | An equilateral triangle is isosceles, and we find that <math>\angle A=\angle B=\angle C</math> if we use the property of isosceles triangles that if two sides of a triangle are equal then the opposite angles are equal. Thus equilateral triangles are equiangular, and <math>C</math> is true. | |
− | An equilateral triangle is isosceles, and we find that <math>\angle A=\angle B=\angle C</math> if we use the property of isosceles triangles that if two sides of a triangle are equal then the opposite angles are equal. Thus equilateral triangles are equiangular. Regular | + | |
− | + | Regular polygons are both equilateral and equiangular, and so are equilateral triangles are both equilateral (by definition) and equiangular (by the above argument). Thus equilateral triangles are regular polygons and <math>D</math> is true. | |
− | + | ||
+ | Since all of the angles in an equilateral triangle are congruent, all equilateral triangles are similar by AAA similarity. Thus, <math>E</math> is true. | ||
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+ | Since <math>\angle A=\angle B=\angle C</math> and <math>\angle A+\angle B+\angle C=180</math>, <math>\angle A=60^{\circ}</math>. Since no other angles are above <math>180^{\circ}</math>, all equilateral triangles are convex. Thus, <math>B</math> is true. | ||
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+ | This just leaves choice <math>\boxed{\mathrm{(A)}}</math>. This is clearly false: an equilateral triangle with side <math>1</math> is not congruent to an equilateral triangle with side <math>2</math>. | ||
==See also== | ==See also== | ||
+ | {{AHSME box|year=1999|num-b=1|num-a=3}} | ||
+ | |||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:28, 13 February 2019
Problem
Which of the following statements is false?
Solution
An equilateral triangle is isosceles, and we find that if we use the property of isosceles triangles that if two sides of a triangle are equal then the opposite angles are equal. Thus equilateral triangles are equiangular, and is true.
Regular polygons are both equilateral and equiangular, and so are equilateral triangles are both equilateral (by definition) and equiangular (by the above argument). Thus equilateral triangles are regular polygons and is true.
Since all of the angles in an equilateral triangle are congruent, all equilateral triangles are similar by AAA similarity. Thus, is true.
Since and , . Since no other angles are above , all equilateral triangles are convex. Thus, is true.
This just leaves choice . This is clearly false: an equilateral triangle with side is not congruent to an equilateral triangle with side .
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 | ||
All AHSME Problems and Solutions |
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