Difference between revisions of "1995 AHSME Problems/Problem 19"
(→Solution) |
m (→Solution) |
||
Line 12: | Line 12: | ||
draw((1, 1.73205081)--(2.5, 0.866025404));</asy> | 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 | + | 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== |
Latest revision as of 20:58, 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
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.