Difference between revisions of "2013 AMC 10A Problems/Problem 16"
m (→Solution) |
|||
Line 6: | Line 6: | ||
[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
− | ==Solution== | + | ==Solution 1== |
Let <math>A</math> be at <math>(6, 5)</math>, B be at <math>(8, -3)</math>, and <math>C</math> be at <math>(9, 1)</math>. Reflecting over the line <math>x=8</math>, we see that <math>A' = D = (10,5)</math>, <math>B' = B</math> (as the x-coordinate of B is 8), and <math>C' = E = (7, 1)</math>. Line <math>AB</math> can be represented as <math>y=-4x+29</math>, so we see that <math>E</math> is on line <math>AB</math>. | Let <math>A</math> be at <math>(6, 5)</math>, B be at <math>(8, -3)</math>, and <math>C</math> be at <math>(9, 1)</math>. Reflecting over the line <math>x=8</math>, we see that <math>A' = D = (10,5)</math>, <math>B' = B</math> (as the x-coordinate of B is 8), and <math>C' = E = (7, 1)</math>. Line <math>AB</math> can be represented as <math>y=-4x+29</math>, so we see that <math>E</math> is on line <math>AB</math>. | ||
Line 26: | Line 26: | ||
We see that if we connect <math>A</math> to <math>D</math>, we get a line of length <math>4</math> (between <math>(6, 5)</math> and <math>(10,5)</math>). The area of <math>\triangle ABD</math> is equal to <math>\frac{bh}{2} = \frac{4(8)}{2} = 16</math>. | We see that if we connect <math>A</math> to <math>D</math>, we get a line of length <math>4</math> (between <math>(6, 5)</math> and <math>(10,5)</math>). The area of <math>\triangle ABD</math> is equal to <math>\frac{bh}{2} = \frac{4(8)}{2} = 16</math>. | ||
− | Now, let the point of intersection between <math>AC</math> and <math>DE</math> be <math>F</math>. If we can just find the area of <math>\triangle ADF</math> and subtract it from 16, we are done. | + | Now, let the point of intersection between <math>AC</math> and <math>DE</math> be <math>F</math>. If we can just find the area of <math>\triangle ADF</math> and subtract it from <math>16</math>, we are done. |
We realize that because the diagram is symmetric over <math>x = 8</math>, the intersection of lines <math>AC</math> and <math>DE</math> should intersect at an x-coordinate of <math>8</math>. We know that the slope of <math>DE</math> is <math>\frac{5-1}{10-7} = \frac{4}{3}</math>. Thus, we can represent the line going through <math>E</math> and <math>D</math> as <math>y - 1=\frac{4}{3}(x - 7)</math>. Plugging in <math>x = 8</math>, we find that the y-coordinate of F is <math>\frac{7}{3}</math>. Thus, the height of <math>\triangle ADF</math> is <math>5 - \frac{7}{3} = \frac{8}{3}</math>. Using the formula for the area of a triangle, the area of <math>\triangle ADF</math> is <math>\frac{16}{3}</math>. | We realize that because the diagram is symmetric over <math>x = 8</math>, the intersection of lines <math>AC</math> and <math>DE</math> should intersect at an x-coordinate of <math>8</math>. We know that the slope of <math>DE</math> is <math>\frac{5-1}{10-7} = \frac{4}{3}</math>. Thus, we can represent the line going through <math>E</math> and <math>D</math> as <math>y - 1=\frac{4}{3}(x - 7)</math>. Plugging in <math>x = 8</math>, we find that the y-coordinate of F is <math>\frac{7}{3}</math>. Thus, the height of <math>\triangle ADF</math> is <math>5 - \frac{7}{3} = \frac{8}{3}</math>. Using the formula for the area of a triangle, the area of <math>\triangle ADF</math> is <math>\frac{16}{3}</math>. |
Revision as of 20:46, 28 December 2019
Contents
[hide]Problem
A triangle with vertices , , and is reflected about the line to create a second triangle. What is the area of the union of the two triangles?
Solution 1
Let be at , B be at , and be at . Reflecting over the line , we see that , (as the x-coordinate of B is 8), and . Line can be represented as , so we see that is on line .
We see that if we connect to , we get a line of length (between and ). The area of is equal to .
Now, let the point of intersection between and be . If we can just find the area of and subtract it from , we are done.
We realize that because the diagram is symmetric over , the intersection of lines and should intersect at an x-coordinate of . We know that the slope of is . Thus, we can represent the line going through and as . Plugging in , we find that the y-coordinate of F is . Thus, the height of is . Using the formula for the area of a triangle, the area of is .
To get our final answer, we must subtract this from .
Solution 2
First, realize that is the midpoint of and is the midpoint of . Connect to to form . Let the midpoint of be . Connect to . is a median of .
Because is isosceles, is also an altitude of . We know the length of and from the given coordinates. The area of is .
Let the intesection of , and be . is the centroid of . Therefore, it splits into and . The area of quadrilateral
~Zeric Hang
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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.