Difference between revisions of "2014 AMC 10A Problems/Problem 16"
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==Solution 2== | ==Solution 2== | ||
− | Let the area of the shaded region be <math>x</math>. Let the other two vertices of the kite be <math>I</math> and <math>J</math> with <math>I</math> closer to <math>AD</math> than <math>J</math>. Note that <math> [ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]</math>. The area of <math>ABF</math> is <math>1</math> and the area of <math>DCH</math> is <math>\dfrac{1}{2}</math>. We will solve for the areas of <math>ADI</math> and <math>BCJ</math> in terms of x by noting that the area of each triangle is the length of the perpendicular from <math>I</math> to <math>AD</math> and <math>J</math> to <math>BC</math> respectively. Because the area of <math>x</math> = <math>\dfrac{1}{2} | + | Let the area of the shaded region be <math>x</math>. Let the other two vertices of the kite be <math>I</math> and <math>J</math> with <math>I</math> closer to <math>AD</math> than <math>J</math>. Note that <math> [ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]</math>. The area of <math>ABF</math> is <math>1</math> and the area of <math>DCH</math> is <math>\dfrac{1}{2}</math>. We will solve for the areas of <math>ADI</math> and <math>BCJ</math> in terms of x by noting that the area of each triangle is the length of the perpendicular from <math>I</math> to <math>AD</math> and <math>J</math> to <math>BC</math> respectively. Because the area of <math>x</math> = <math>\dfrac{1}{2} \cdot IJ</math> based on the area of a kite formula, <math>\dfrac{ab}{2}</math> for diagonals of length <math>a</math> and <math>b</math>, <math>IJ = 2x</math>. So each perpendicular is length <math>\dfrac{1-2x}{2}</math>. So taking our numbers and plugging them into <math> [ABCD] =[ABF] + [DCH] - x + [ADI] + [BCJ]</math> gives us <math>2 = \dfrac{5}{2} - 3x</math> Solving this equation for <math>x</math> gives us <math> x = \boxed{\textbf{(E)} \: \frac{1}{6}}</math> |
==Solution 3== | ==Solution 3== | ||
− | 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 | + | 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 = \frac{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>\frac{\frac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}</math>. |
==Solution 4== | ==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>. | + | 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>. |
+ | |||
+ | Lemma: The distance from line segment <math>JI</math> to <math>H</math> is half the distance from <math>JI</math> to <math>F</math> | ||
+ | |||
+ | Proof: Drop perpendiculars of triangles <math>HJI</math> and <math>JIF</math> to line <math>JI</math>, and let the point of intersection be <math>Q</math>. Note that <math>HJI</math> and <math>JIF</math> are similar to <math>HDC</math> and <math>ABF</math>, respectively. Now, the ratio of <math>DC</math> to <math>HF</math> is <math>1:1</math>, which shows that the ratio of <math>JI</math> to <math>HQ</math> is <math>1:1</math>, because of similar triangles as described above. Similarly, the ratio of <math>JI</math> to <math>FQ</math> is <math>1:2</math>. Since these two triangles contain the same base, <math>JI</math>, the ratio of <math>HQ:FQ = 1:2</math>. | ||
+ | |||
+ | Because kite <math>HJFI</math> is orthodiagonal, we multiply <math>\frac{1\cdot\tfrac{1}{3}}{2} = \boxed{\textbf{(E)} \: \frac{1}{6}}</math> | ||
+ | |||
+ | ~Lemma proof by sakshamsethi | ||
+ | |||
+ | ==Solution 5 (Similarity)== | ||
+ | <asy> | ||
+ | import graph; | ||
+ | size(9cm); | ||
+ | pen dps = fontsize(10); defaultpen(dps); | ||
+ | pair D = (0,0); | ||
+ | pair F = (1/2,0); | ||
+ | pair C = (1,0); | ||
+ | pair G = (0,1); | ||
+ | pair E = (1,1); | ||
+ | pair A = (0,2); | ||
+ | pair B = (1,2); | ||
+ | pair H = (1/2,1); | ||
+ | |||
+ | // do not look | ||
+ | pair X = (1/3,2/3); | ||
+ | pair Y = (2/3,2/3); | ||
+ | |||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(G--E); | ||
+ | draw(A--F--B); | ||
+ | draw(D--H--C); | ||
+ | filldraw(H--X--F--Y--cycle,grey); | ||
+ | |||
+ | label("$A$",A,NW); | ||
+ | label("$B$",B,NE); | ||
+ | label("$C$",C,SE); | ||
+ | label("$D$",D,SW); | ||
+ | label("$E$",E,E); | ||
+ | label("$F$",F,S); | ||
+ | label("$G$",G,W); | ||
+ | label("$H$",H,N); | ||
+ | |||
+ | label("$\frac12$",(0.25,0),S); | ||
+ | label("$\frac12$",(0.75,0),S); | ||
+ | label("$1$",(1,0.5),E); | ||
+ | label("$1$",(1,1.5),E); | ||
+ | </asy> | ||
+ | |||
+ | The area of the shaded area is the area of <math>\triangle DHC</math> minus the two triangles on the side. | ||
+ | Extend <math>\overline{DH}</math> so that it hits point <math>B</math>. Call the intersection of <math>\overline{AF}</math> and <math>\overline{DB}</math> point <math>P</math>. <cmath>\triangle APB \sim \triangle FPD</cmath> | ||
+ | Drop altitudes from <math>P</math> down to <math>\overline{DF}</math> and <math>\overline{AB}</math>; call the intersection points <math>L</math> and <math>M</math> respectively. | ||
+ | <cmath>\frac{DF}{AB}=\frac{LP}{PM}=\frac{\frac{1}{2}}{1}=\frac{1}{2}</cmath> | ||
+ | <cmath>LP=\frac{1}{3}\cdot LM=\frac{2}{3}</cmath> | ||
+ | Thus the two triangles on the side have area <math>\frac{1}{2} \cdot \frac{2}{2} \cdot \frac {1}{3} = \frac{1}{6}</math>. Since there are two, their total area is <math>2 \cdot \frac{1}{6} = \frac{1}{3}</math>. The area of <math>\triangle DHC</math> is <math>\frac{1}{2} \cdot 1 \cdot 1=\frac{1}{2}</math>. The shaded region is <math>\frac{1}{2}-\frac{1}{3} = \frac{1}{6}</math> which is <math>\boxed{\textbf{(E)} \frac{1}{6}}</math>. | ||
+ | |||
+ | ~JH. L :) | ||
==See Also== | ==See Also== |
Latest revision as of 19:51, 25 July 2023
Contents
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
.
Lemma: The distance from line segment to
is half the distance from
to
Proof: Drop perpendiculars of triangles and
to line
, and let the point of intersection be
. Note that
and
are similar to
and
, respectively. Now, the ratio of
to
is
, which shows that the ratio of
to
is
, because of similar triangles as described above. Similarly, the ratio of
to
is
. Since these two triangles contain the same base,
, the ratio of
.
Because kite is orthodiagonal, we multiply
~Lemma proof by sakshamsethi
Solution 5 (Similarity)
The area of the shaded area is the area of minus the two triangles on the side.
Extend
so that it hits point
. Call the intersection of
and
point
.
Drop altitudes from
down to
and
; call the intersection points
and
respectively.
Thus the two triangles on the side have area
. Since there are two, their total area is
. The area of
is
. The shaded region is
which is
.
~JH. L :)
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.