Difference between revisions of "2013 AMC 10A Problems/Problem 16"
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Idk12345678 (talk | contribs) (→Solution 3) |
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label("$D$",D,NE); | label("$D$",D,NE); | ||
label("$E$",E,W); | label("$E$",E,W); | ||
+ | label("$F$", F, N); | ||
Line 43: | Line 44: | ||
~Zeric Hang | ~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== | ==See Also== |
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.