Difference between revisions of "2015 AMC 10A Problems/Problem 17"
Mathboy282 (talk | contribs) (→Solution 2) |
(→Solution 1) |
||
Line 7: | Line 7: | ||
==Solution 1== | ==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 <math>\frac{\sqrt{3}}{3}</math> so the third must be <math>-\frac{\sqrt{3}}{3}</math>. Since this third line passes through the origin, its equation is simply <math>y = -\frac{\sqrt{3}}{3}x</math>. To find two vertices of the triangle, plug in <math>x=1</math> to both the other equations. | + | 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 <math>\frac{\sqrt{3}}{3}</math> (can be find out easily by using 60 degree angle of the triangle) so the third must be <math>-\frac{\sqrt{3}}{3}</math>. Since this third line passes through the origin, its equation is simply <math>y = -\frac{\sqrt{3}}{3}x</math>. To find two vertices of the triangle, plug in <math>x=1</math> to both the other equations. |
<math>y = -\frac{\sqrt{3}}{3}</math> | <math>y = -\frac{\sqrt{3}}{3}</math> |
Revision as of 06:36, 4 June 2023
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 (can be find out easily by using 60 degree angle of the triangle) 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 problem 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 .
Solution 3 (Intuitive)
We have
With the first condition, we have that
Then, we have
Dividing both sides by on the second and putting over a common denominator gets us The only answer in the answer choices that satisfies this is (D)
Video Solution
~savannahsolver
See Also
Video Solution:
https://www.youtube.com/watch?v=2kvSRL8KMac
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.
a