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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>
  
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==Solutions==
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==Solution==
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===Solution 1===
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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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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 pentagons are both equilateral and equiangular, and so are equilateral triangles. Thus equilateral triangles are regular polygons. Since all of the angles are the same, all equilateral triangles are similar. 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. This just leaves choice <math>\boxed{\mathrm{(A)}}</math>.
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===Solution 2===
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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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Congruent triangles have the same side length. <geogebra>f1a4134248dedfd1e182804e035fd400ef763f04</geogebra> The image above disproves <math>\boxed{\mathrm{(A)}}</math>.
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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>.
  
 
==See also==
 
==See also==
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[[Category:Introductory Geometry Problems]]
 
[[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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