Difference between revisions of "2014 AMC 10A Problems/Problem 16"
Jskalarickal (talk | contribs) (→Solution 3) |
(Added my own solution.) |
||
Line 98: | Line 98: | ||
From the diagram in Solution 1, let <math>e</math> be the height of <math>XHY</math> and <math>f</math> be the height of <math>XFY</math>. It is clear that their sum is <math>1</math> as they are parallel to <math>GD</math>. Let <math>k</math> be the ratio of the sides of the similar triangles <math>XFY</math> and <math>AFB</math>, which are similar because <math>XY</math> is parallel to <math>AB</math> and the triangles share angle <math>F</math>. Then <math>k = f/2</math>, as 2 is the height of <math>AFB</math>. Since <math>XHY</math> and <math>DHC</math> are similar for the same reasons as <math>XFY</math> and <math>AFB</math>, the height of <math>XHY</math> will be equal to the base, like in <math>DHC</math>, making <math>XY = e</math>. However, <math>XY</math> is also the base of <math>XFY</math>, so <math>k = e / AB</math> where <math>AB = 1</math> so <math>k = e</math>. Subbing into <math>k = f/2</math> gives a system of linear equations, <math>e + f = 1</math> and <math>e = f/2</math>. Solving yields <math>e = XY = 1/3</math> and <math>f = 2/3</math>, and since the area of the kite is simply the product of the two diagonals over <math>2</math> and <math>HF = 1</math>, our answer is <math>\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}</math>. | From the diagram in Solution 1, let <math>e</math> be the height of <math>XHY</math> and <math>f</math> be the height of <math>XFY</math>. It is clear that their sum is <math>1</math> as they are parallel to <math>GD</math>. Let <math>k</math> be the ratio of the sides of the similar triangles <math>XFY</math> and <math>AFB</math>, which are similar because <math>XY</math> is parallel to <math>AB</math> and the triangles share angle <math>F</math>. Then <math>k = f/2</math>, as 2 is the height of <math>AFB</math>. Since <math>XHY</math> and <math>DHC</math> are similar for the same reasons as <math>XFY</math> and <math>AFB</math>, the height of <math>XHY</math> will be equal to the base, like in <math>DHC</math>, making <math>XY = e</math>. However, <math>XY</math> is also the base of <math>XFY</math>, so <math>k = e / AB</math> where <math>AB = 1</math> so <math>k = e</math>. Subbing into <math>k = f/2</math> gives a system of linear equations, <math>e + f = 1</math> and <math>e = f/2</math>. Solving yields <math>e = XY = 1/3</math> and <math>f = 2/3</math>, and since the area of the kite is simply the product of the two diagonals over <math>2</math> and <math>HF = 1</math>, our answer is <math>\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}</math>. | ||
+ | |||
+ | =Solution 4= | ||
+ | Let the unmarked vertices of the shaded area be labeled <math>I</math> and <math>J</math>, with <math>I</math> being closer to <math>GD</math> than <math>J</math>. Noting that kite <math>HJFI</math> can be split into triangles <math>HJI</math> and <math>JIF</math>. Because <math>HJI</math> and <math>JIF</math> are similar to <math>HDC</math> and <math>ABF</math>, we know that the distance from line segment <math>JI</math> to <math>H</math> is half the distance from <math>JI</math> to <math>F</math>. Because kite <math>HJFI</math> is orthodiagonal, we multiply <math>(1*(1/3))/2 = \boxed{\textbf{(E)} \: \frac{1}{6}}</math> | ||
==See Also== | ==See Also== |
Revision as of 16:28, 2 February 2015
Problem
In rectangle ,
,
, and points
,
, and
are midpoints of
,
, and
, respectively. Point
is the midpoint of
. What is the area of the shaded region?
Solution 1
Denote . Then
. Let the intersection of
and
be
, and the intersection of
and
be
. Then we want to find the coordinates of
so we can find
. From our points, the slope of
is
, and its
-intercept is just
. Thus the equation for
is
. We can also quickly find that the equation of
is
. Setting the equations equal, we have
. Because of symmetry, we can see that the distance from
to
is also
, so
. Now the area of the kite is simply the product of the two diagonals over
. Since the length
, our answer is
.
Solution 2
Let the area of the shaded region be . Let the other two vertices of the kite be
and
with
closer to
than
. Note that
. The area of
is
and the area of
is
. We will solve for the areas of
and
in terms of x by noting that the area of each triangle is the length of the perpendicular from
to
and
to
respectively. Because the area of
=
based on the area of a kite formula,
for diagonals of length
and
,
. So each perpendicular is length
. So taking our numbers and plugging them into
gives us
Solving this equation for
gives us
Solution 3
From the diagram in Solution 1, let be the height of
and
be the height of
. It is clear that their sum is
as they are parallel to
. Let
be the ratio of the sides of the similar triangles
and
, which are similar because
is parallel to
and the triangles share angle
. Then
, as 2 is the height of
. Since
and
are similar for the same reasons as
and
, the height of
will be equal to the base, like in
, making
. However,
is also the base of
, so
where
so
. Subbing into
gives a system of linear equations,
and
. Solving yields
and
, and since the area of the kite is simply the product of the two diagonals over
and
, our answer is
.
Solution 4
Let the unmarked vertices of the shaded area be labeled and
, with
being closer to
than
. Noting that kite
can be split into triangles
and
. Because
and
are similar to
and
, we know that the distance from line segment
to
is half the distance from
to
. Because kite
is orthodiagonal, we multiply
See Also
2014 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.