Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Geogbra removed - error in drawing.)
m (→‎Problem: Fixed spacing)
 
(2 intermediate revisions by 2 users not shown)
Line 3: Line 3:
  
 
<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>
  
−
==Solutions==
+
==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, and <math>C</math> is true.
  
Line 23: Line 27:
  
 
[[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?

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

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

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

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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1999_AHSME_Problems/Problem_2&oldid=101996"