Difference between revisions of "2013 AMC 10A Problems/Problem 16"
(→Solution 1) |
Idk12345678 (talk | contribs) |
||
Line 44: | 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> | ||
==See Also== | ==See Also== |
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:
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.