Difference between revisions of "2015 AMC 10A Problems/Problem 17"
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==Solution 2== | ==Solution 2== | ||
| − | Draw a line from the y-intercept of the equation <math>y=1+ \frac{\sqrt{3}}{3} x</math> perpendicular to the line <math>x=1</math>. There is a square of side length 1 inscribed in the equilateral triangle. The problems becomes reduced to finding the perimeter of an equilateral triangle with a square of side length 1 inscribed in it. The side length is | + | Draw a line from the y-intercept of the equation <math>y=1+ \frac{\sqrt{3}}{3} x</math> perpendicular to the line <math>x=1</math>. There is a square of side length 1 inscribed in the equilateral triangle. The problems becomes reduced to finding the perimeter of an equilateral triangle with a square of side length 1 inscribed in it. The side length is <math>2\left(\frac{1}{\sqrt{3}}\right) + 1</math>. After multiplying the side length by 3 and rationalizing, you get <math>\boxed{\textbf{(D) }3 + 2\sqrt{3}}</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=A|num-b=16|num-a=18}} | {{AMC10 box|year=2015|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:54, 21 January 2019
Contents
Problem
A line that passes through the origin intersects both the line 
and the line 
. The three lines create an equilateral triangle. What is the perimeter of the triangle?
Solution 1
Since the triangle is equilateral and one of the sides is a vertical line, the triangle must have a horizontal line of symmetry, and therefore the other two sides will have opposite slopes. The slope of the other given line is 
so the third must be 
. Since this third line passes through the origin, its equation is simply 
. To find two vertices of the triangle, plug in 
to both the other equations.
We now have the coordinates of two vertices, 
and 
. The length of one side is the distance between the y-coordinates, or 
.
The perimeter of the triangle is thus 
, so the answer is 
Solution 2
Draw a line from the y-intercept of the equation perpendicular to the line . There is a square of side length 1 inscribed in the equilateral triangle. The problems becomes reduced to finding the perimeter of an equilateral triangle with a square of side length 1 inscribed in it. The side length is 
. After multiplying the side length by 3 and rationalizing, you get .
See Also
| 2015 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
