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Difference between revisions of "1999 AHSME Problems/Problem 2"

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By definition of equilateral triangles, only <math>\boxed{\text{A}}</math> is false.
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==Problem==
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Which of the following statements is false?
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<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>
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<math>\mathrm{(C) \  All\ equilateral\ triangles\ are\ equianguilar.}</math>
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<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>
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==Solution==
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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.
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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.
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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>.
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==See also==
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{{AHSME box|year=1999|num-b=1|num-a=3}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 13:28, 13 February 2019

Problem

Which of the following statements is false?

$\mathrm{(A) \ All\ equilateral\ triangles\ are\ congruent\ to\ each\ other.}$

$\mathrm{(B) \ All\ equilateral\ triangles\ are\ convex.}$

$\mathrm{(C) \ All\ equilateral\ triangles\ are\ equianguilar.}$

$\mathrm{(D) \ All\ equilateral\ triangles\ are\ regular\ polygons.}$

$\mathrm{(E) \ All\ equilateral\ triangles\ are\ similar\ to\ each\ other.}$

Solution

An equilateral triangle is isosceles, and we find that $\angle A=\angle B=\angle C$ 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 $C$ 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 $D$ is true.

Since all of the angles in an equilateral triangle are congruent, all equilateral triangles are similar by AAA similarity. Thus, $E$ is true.

Since $\angle A=\angle B=\angle C$ and $\angle A+\angle B+\angle C=180$, $\angle A=60^{\circ}$. Since no other angles are above $180^{\circ}$, all equilateral triangles are convex. Thus, $B$ is true.

This just leaves choice $\boxed{\mathrm{(A)}}$. This is clearly false: an equilateral triangle with side $1$ is not congruent to an equilateral triangle with side $2$.

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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