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1999 AHSME Problems/Problem 2: Difference between revisions

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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>
<math>\mathrm{(B) \  All\ equilateral\ triangles\ are\ convex.}</math>
<math>\mathrm{(B) \  All\ equilateral\ triangles\ are\ convex.}</math>
<math>\mathrm{(C) \  All\ equilateral\ triangles\ are\ equianguilar.}</math>
<math>\mathrm{(C) \  All\ equilateral\ triangles\ are\ equianguilar.}</math>
<math>\mathrm{(D) \  All\ equilateral\ triangles\ are\ regular\ polygons.}</math>
<math>\mathrm{(D) \  All\ equilateral\ triangles\ are\ regular\ polygons.}</math>
<math>\mathrm{(E) \  All\ equilateral\ triangles\ are\ similar\ to\ each\ other.}  </math>
<math>\mathrm{(E) \  All\ equilateral\ triangles\ are\ similar\ to\ each\ other.}  </math>


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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

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