Difference between revisions of "1995 AHSME Problems/Problem 19"
(New page: ==Problem== Equilateral triangle <math>DEF</math> is inscribed in equilateral triangle <math>ABC</math> such that <math>\overline{DE} \perp \overline{BC}</math>. The reatio of the area of ...) |
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==Problem== | ==Problem== | ||
| − | Equilateral triangle <math>DEF</math> is inscribed in equilateral triangle <math>ABC</math> such that <math>\overline{DE} \perp \overline{BC}</math>. The | + | Equilateral triangle <math>DEF</math> is inscribed in equilateral triangle <math>ABC</math> such that <math>\overline{DE} \perp \overline{BC}</math>. The ratio of the area of <math>\triangle DEF</math> to the area of <math>\triangle ABC</math> is |
<math> \mathrm{(A) \ \frac {1}{6} } \qquad \mathrm{(B) \ \frac {1}{4} } \qquad \mathrm{(C) \ \frac {1}{3} } \qquad \mathrm{(D) \ \frac {2}{5} } \qquad \mathrm{(E) \ \frac {1}{2} } </math> | <math> \mathrm{(A) \ \frac {1}{6} } \qquad \mathrm{(B) \ \frac {1}{4} } \qquad \mathrm{(C) \ \frac {1}{3} } \qquad \mathrm{(D) \ \frac {2}{5} } \qquad \mathrm{(E) \ \frac {1}{2} } </math> | ||
==Solution== | ==Solution== | ||
| − | {{ | + | <asy>draw((0,0)--(3,0)); |
| + | draw((0,0)--(1.5, 2.59807621)); | ||
| + | draw((3,0)--(1.5, 2.59807621)); | ||
| + | draw((1,0)--(1, 1.73205081)); | ||
| + | draw((1,0)--(2.5, 0.866025404)); | ||
| + | draw((1, 1.73205081)--(2.5, 0.866025404));</asy> | ||
| + | |||
| + | Let's take one of the smaller right triangles. Without loss of generality, let the smaller leg be <math>1</math>. Since the triangle is a 30-60-90 right triangle, then the other leg is <math>\sqrt{3}</math> and the hypotenuse is <math>2</math>. The side length of the bigger triangle is <math>1 + 2 = 3</math> and the side length of the smaller triangle is <math>\sqrt{3}</math>. The ratio of the areas of two similar triangles is the square of the ratio of two corresponding side lengths, so the ratio of the area of triangle DEF to the area of triangle ABC is <math>\left(\frac{\sqrt{3}}{3}\right)^2 = \dfrac{1}{3} \rightarrow (C)</math>. | ||
==See also== | ==See also== | ||
| + | {{AHSME box|year=1995|num-b=18|num-a=20}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:58, 10 February 2019
Problem
Equilateral triangle 
is inscribed in equilateral triangle 
such that 
. The ratio of the area of 
to the area of 
is
Solution
![[asy]draw((0,0)--(3,0)); draw((0,0)--(1.5, 2.59807621)); draw((3,0)--(1.5, 2.59807621)); draw((1,0)--(1, 1.73205081)); draw((1,0)--(2.5, 0.866025404)); draw((1, 1.73205081)--(2.5, 0.866025404));[/asy]](http://latex.artofproblemsolving.com/4/e/a/4ea52b08e9988bf4c02b3ca44e919efbf3c5d93c.png)
Let's take one of the smaller right triangles. Without loss of generality, let the smaller leg be 
. Since the triangle is a 30-60-90 right triangle, then the other leg is 
and the hypotenuse is 
. The side length of the bigger triangle is 
and the side length of the smaller triangle is . The ratio of the areas of two similar triangles is the square of the ratio of two corresponding side lengths, so the ratio of the area of triangle DEF to the area of triangle ABC is 
.
See also
| 1995 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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. 
