Difference between revisions of "2013 AMC 10A Problems/Problem 16"
Countingkg (talk | contribs) (Created page with "==Problem== A triangle with vertices <math>(6, 5)</math>, <math>(8, -3)</math>, and <math>(9, 1) is reflected about the line </math>x=8<math> to create a second triangle. What ...") |
Idk12345678 (talk | contribs) (→Solution 3) |
||
(16 intermediate revisions by 9 users not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | A triangle with vertices <math>(6, 5)</math>, <math>(8, -3)</math>, and <math>(9, 1) is reflected about the line < | + | A triangle with vertices <math>(6, 5)</math>, <math>(8, -3)</math>, and <math>(9, 1)</math> is reflected about the line <math>x=8</math> to create a second triangle. What is the area of the union of the two triangles? |
− | < | + | <math> \textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3} </math> |
+ | [[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>. | ||
+ | |||
+ | <asy> | ||
+ | pair A = (6, 5), B = (8, -3), C = (9, 1), D = (10, 5), E = (7, 1), F = (8, 7/3); | ||
+ | draw(A--B--C--cycle^^D--E--B--cycle); | ||
+ | dot(A^^B^^C^^D^^E^^F); | ||
+ | label("$A$",A,NW); | ||
+ | label("$B$",B,S); | ||
+ | label("$C$",C,SE); | ||
+ | label("$D$",D,NE); | ||
+ | label("$E$",E,W); | ||
+ | label("$F$", F, N); | ||
+ | |||
+ | |||
+ | </asy> | ||
+ | |||
+ | |||
+ | 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 <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>. | ||
+ | |||
+ | To get our final answer, we must subtract this from <math>16</math>. <math>[ABD] - [ADF] = 16 - \frac{16}{3} = \boxed{\textbf{(E) }\frac{32}{3}}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | First, realize that <math>E</math> is the midpoint of <math>AB</math> and <math>C</math> is the midpoint of <math>BD</math>. Connect <math>A</math> to <math>D</math> to form <math>\triangle ABD</math>. Let the midpoint of <math>AD</math> be <math>G</math>. Connect <math>B</math> to <math>G</math>. <math>BG</math> is a median of <math>\triangle ABD</math>. | ||
+ | |||
+ | |||
+ | Because <math>\triangle ABD</math> is isosceles, <math>BG</math> is also an altitude of <math>\triangle ABD</math>. We know the length of <math>AD</math> and <math>BG</math> from the given coordinates. The area of <math>\triangle ABD</math> is <math>\frac{bh}{2} = \frac{4(8)}{2} = 16</math>. | ||
+ | |||
+ | |||
+ | Let the intesection of <math>AC</math>, <math>DE</math> and <math>BG</math> be <math>F</math>. <math>F</math> is the centroid of <math>\triangle ABD</math>. Therefore, it splits <math>BG</math> into <math>BF={2 \over 3}(BG)</math> and <math>FG={1\over 3}(BG)</math>. The area of quadrilateral <math>ABDF = 16\cdot {2 \over 3} = \boxed{\textbf{(E) }\frac{32}{3}}</math> | ||
+ | |||
+ | ~Zeric Hang | ||
+ | |||
+ | ==Solution 3== | ||
+ | Since we have to find the area on a coordinate plane, we can use the [[Shoelace_Theorem|Shoelace Theorem]] to find the area of the intersection. When you reflect it, it makes a quadrilateral. | ||
+ | |||
+ | <asy> | ||
+ | pair A = (6, 5), B = (8, -3), C = (9, 1), D = (10, 5), E = (7, 1), F = (8, 7/3); | ||
+ | draw(A--B--C--cycle^^D--E--B--cycle); | ||
+ | dot(A^^B^^C^^D^^E^^F); | ||
+ | label("$A$",A,NW); | ||
+ | label("$B$",B,S); | ||
+ | label("$C$",C,SE); | ||
+ | label("$D$",D,NE); | ||
+ | label("$E$",E,W); | ||
+ | label("$F$", F, N); | ||
+ | |||
+ | |||
+ | </asy> | ||
+ | |||
+ | Since it is reflected around <math>x=8</math>, the point <math>(8,-3)</math> remains the same on both. The top right corners are <math>(6,5)</math>, and its reflection <math>(10,5)</math>. Now to find the 4th point, point F, we can use the equation of the line DE(<math>\frac{4}{3}x - \frac{25}{3}</math>, and substitute <math>x=8</math>, to get <math>\frac{7}{3}</math>. Now we can use the theorem: <math>\frac{1}{2}(((6*-3)+(10*5)+(8*5)+(\frac{7}{3}*8))-((8*5)+(6*5)+(10*\frac{7}{3}) + (8* -3))) = \boxed{\textbf{(E) }\frac{32}{3}}</math> | ||
+ | |||
+ | ~idk12345678 | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2013|ab=A|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} |
Latest revision as of 15:47, 3 April 2024
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
Solution 3
Since we have to find the area on a coordinate plane, we can use the Shoelace Theorem to find the area of the intersection. When you reflect it, it makes a quadrilateral.
Since it is reflected around , the point remains the same on both. The top right corners are , and its reflection . Now to find the 4th point, point F, we can use the equation of the line DE(, and substitute , to get . Now we can use the theorem:
~idk12345678
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.