Difference between revisions of "2013 AMC 10A Problems/Problem 16"
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− | pair A = (6, 5), B = (8, -3), C = (9, 1), D = (10, 5), E = (7, 1); | + | 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); | draw(A--B--C--cycle^^D--E--B--cycle); | ||
− | dot(A^^B^^C^^D^^E); | + | dot(A^^B^^C^^D^^E^^F); |
label("$A$",A,NW); | label("$A$",A,NW); | ||
label("$B$",B,S); | label("$B$",B,S); |
Revision as of 21:59, 21 March 2023
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.